Chapter 1: Basic Definitions, Terminologies and Concepts --------------------------------------1.1 Thermodynamics: It is a basic science that deals with: 1. Energy transformation from one form to another. 2. Various properties of substances. U U Thermodynamics can be classified into: U 1. 2. Macroscopic or Classical Thermodynamics: It does not require the knowledge of the molecules or atoms behavior. Microscopic or Statistical Thermodynamics: It is based on the behavior of the molecules and atoms. Applications: Thermodynamic laws and principles are very important tools to get the energy transfer or fluid properties involved in many engineering systems such as: aero planes, power plants, cars, rockets, heaters …… etc U U Units: the units that will be used in this course are the SI units: SI: kg, N, m, s (primary), J, W…(Secondary) Joule, J = N.m = kg.m2/s2 Watt, w = J/s = kg.m2/s3 U U Dimensional homogeneity: This rule is very simple, yet very powerful in any engineering analysis. It states that every term in an equation must have the same units of the other terms. For example: E= (This equation is wrong because different units are added together) 1/2 mv2 + gz kJ + kJ/kg In the above equation “E” stands for the total energy (kJ), the term (1/2mv2 ) is the kinetic energy (kJ), but the term (gz) is the potential energy per unit mass (kJ/kg). As such, adding the first term to the second term is wrong and causes unrealistic engineering results. In summary: you have to be sure that all the terms in any equation must have the same units. 1.2 System: In thermodynamic studies, a quantity of matter or a region in space chosen for studying energy transformation is selected. This quantity or region is called system. The region outside the system is called the surroundings. The real or imaginary surface that separates the system from the surroundings is called the boundary (Fig.1.1). Remember that the boundary can be fixed (steam in a rigid tank) or moving (a gas in an elastic balloon). In the moving boundary, the balloon surface (boundary) changes in size or shape. Fig. 1 1.2.1 Closed system: The mass of the fluid such as air or steam doesn’t cross (enter or leave) the system boundaries, but energy (heat, work…) can cross the boundary (to or from the system). The boundary of the closed system can be fixed (Fig.1.2) or moving (Fig. 1.3) Examples of closed system: heated or cooled fluid in a piston-cylinder device or in a closed vessel (tank or container). Fig 1.2: Mass cannot cross the boundaries of a closed system, but energy can. Fig 1.3 A closed system with a moving boundary 1 1.2.2 Open system (control volume): It is a region in the space that involves mass and energy flow. This means that mass or fluid can cross the system boundaries (Fig. 1.4) Examples: Energy and mass cross the boundaries of the following open systems: Turbine, compressor, pump, fan, blower, boiler, condenser, heat exchanger, water heater, flow in pipes (ducts or tubes), nozzle, diffuser ..etc. 1.2.3 Isolated system: It is a system with no heat or work, or mass transfer across the boundaries. Fig 1.4: An open system (a control volume) with one inlet and one exit. 1.3 Property of a system: It is any macroscopic characteristic (quantity) of a system such as: - Pressure (P); Pa or kPa, MPa, bar, atm - Temperature (T); K = t (oC ) + 273 - Specific volume (v): volume per unit mass. - v = total volume/mass = V / m = 1/ ρ m3/kg - Density (ρ) : mass per unit volume. ρ = mass / total volume = m / V kg/m3 13.1 Intensive properties: are those, which are independent of the size of a system (T, P, ρ). 1.3.2 Extensive properties: vary directly with the size of the system (mass, volume) (Fig. 1.5). Intensive (specific) property = Extensive property / mass e.g. v = V/m Fig 1.5: Criteria to differentiate intensive and extensive properties. 1.4 State: It is the condition of a system as described by any two independent properties (P, T, v, u, ….. ) (Fig. 1.6). Fig 1.6: A system at two different states NOTE: Temperature & pressure are dependent properties during phase change of boiling or condensation. i.e. T = ƒ(p). This means that there is a known temperature at a certain pressure during boiling or condensation. Equation of State: Any relationship among thermodynamic properties is called an equation of state. For example, the equation of state for ideal gas is Pv = RT . 1.5 Equilibrium: a state of balance, i.e. no driving force within the system. It happens when the properties of the system don’t change with time and the properties are the same at any point within the system. Equilibrium state: properties are uniform throughout the entire system. 1.5.1 Thermal equilibrium: temperature is uniform throughout the entire system. 1.5.2 Mechanical equilibrium: pressure is uniform throughout the entire system. 1.5.3 Phase equilibrium: phase (liquid, gas, solid) is uniform throughout the entire system. 1.5.4 Chemical equilibrium: chemical composition is uniform throughout the entire system. 1.5.5 Thermodynamic equilibrium: system is isolated and properties do not change with time. 2 1.6 Process: It is the transformation of a system from one state to another (Fig. 1.7). The process (1-2) in that figure is a compression process since the volume decreases and the pressure increases. State 1 is the initial state, state 2 is the final state and the process goes from 1 to 2. Fig 1.7 1.6.1 Types of process path: - Constant pressure process: (Isobaric; P = P1 = P2 = C) - Constant temperature process: (Isothermal; T = T1 = T2 = C) - Constant volume process: (Isochoric; v1 = v2 = C) - Constant entropy process: (Isentropic; s = s1 = s2 =C) γ = cp / cv ( for a perfect gas) Pvγ = c ; n = polytropic index, n takes any value. n = 0, 1, 1.4 = γ , 2 …. ∞ Pvn = c ; - Adiabatic process: No heat transfer process = insulated surface (Q = 0) 1.7 Cycle: a sequence of processes that returns the system to its initial state. Final state = initial state (Fig. 1.8) 1.8 Symbol convention Fig 1.8 Property: Capital Extensive, e.g., V, KE, PE Lower Intensive, e.g., v, ke, pe NOTE: temperature (T) and pressure (P) are always intensive and capital Path: Exact differential change Exact finite change Inexact differential change Inexact finite change Cyclic process d (ds, du, dh) (point function), change of a property Δ (Δs, Δh, ΔU) δ(δw, δq) change of a quantity which is not a property, The change depends on the path (process) subscript 12 (change from state 1 to state 2) , e.g., W12 § 1.9 Temperature & the Zeroth Law Although we are familiar with temperature as a measure of “hotness” or “coldness,” it is not easy to give an exact definition of it. However, temperature is considered as a thermodynamic property that is the measure of the energy content of a mass. When heat energy is transferred to a body, the body's energy content increases and so does its temperature. In fact it is the difference in temperature that causes energy transfer, called heat transfer, to flow from a hot body to a cold body. Two bodies are in thermal equilibrium when they have reached the same temperature. If two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. This simple fact is known as the zeroth law of thermodynamics. body A body B The zeroth law states that: If TA = TC and TB = TC in Fig. 1.9 Then TA = TB A, B, and C are in thermal equilibrium Fig. 1.9 body C Absolute Temp. T(K) = T (◌C) ْ + 273 3 1.10 Pressure: is the force exerted by a fluid per unit normal area 1 Pa = 1 N/m2 ( pascal) 1 bar = 105 Pa = 0.1 MPa = 100 kPa 1 atm = 101.325 Pa The pressure used in all calculations of state is the absolute pressure measured relative to absolute zero pressure. However, pressures are often measured relative to atmospheric pressure, called gage or vacuum pressures. 1.10.1 Absolute pressure Pabs (Fig. 1.10) : calculated relative to absolute vacuum. 1.10.2 Gage pressure Pgage:difference between absolute pressure and atmospheric pressure (Pabs > Patm). 1.10.3 Vacuum pressure Pvac: difference between atmospheric and absolute pressure. This means Pgage = Pabs - Patm Pvac = Patm - Pabs Fig 1.10: Absolute, gage, and vacuum pressures. 1.11 U- tube Manometer (Fig.1.11): Small to moderate pressure differences are measured by a manometer and a differential fluid column of height h. This height corresponds to a pressure difference between the system and the surroundings of the manometer. (N/m2) ∆P = P1 – Patm = ρgh = γh where: γ = ρg is called (the specific weight) Specific gravity (sp. gr.) : sp.gr. = γ γH = 2O ρ ρH 2O Fig 1.11: U-tube manometer. 4 PROBLEMS 1. Define enthalpy, flow work, extensive property, and thermal equilibrium. 2. Define the isothermal, isobaric, adiabatic, isochoric processes, closed system, open system, heat, work, intensive property, process, and cycle. 3. What is the zeroth law of thermodynamics ? 4. A manometer is used to measure the pressure in a tank. The manometer fluid has a specific gravity of 0.85, and the manometer column height is 55 cm. If the local atmospheric pressure is 96 kPa, determine the absolute pressure within the tank. (Answer:100 kPa) 5. A gas is contained in a vertical, frictionless piston-cylinder device. The piston has a mass of 5 kg and a cross-sectional area of 25 cm2. A compressed spring above the piston exerts a force of 75 N on the piston. If the atmospheric pressure is 98 kPa, determine the pressure inside the cylinder. (Answer:147.6 kPa) 6. The piston of a piston-cylinder device containing a gas has a mass of 60 kg and a cross-sectional area of 0.04 m2. The local atmospheric pressure is 0.97 bar, and the gravitational acceleration is 9.8 m/s2: a) Determine the pressure inside the cylinder. b) If some heat is transferred to the gas until its volume doubles, do you expect the pressure inside the cylinder will change? (Answer:1.117 bars) 5 Chapter 2: Properties of pure substances 2.1 Pure substance: It is a substance that has a homogeneous and fixed chemical composition through out its material (H2O, N2, air, CO2). Pure substance may exist at the same time in more than one phase in equilibrium (ice in water or liquid droplets in vapor). This means that a mixture of ice and liquid water is a pure substance and a mixture of water vapor and liquid water is a pure substance. - Properties of ideal gas are related by: Pv = RT or P = ρRT - Incompressible substance: is the substance that its volume does not change much with increasing pressure. This means that its density (or specific volume) is nearly constant. i.e., v = constant or ρ = constant. 2.2 Phase change: Substances can exist at solid, liquid or gaseous state. In other words substances can exist at three phases or as a mixture of two phases. Substances can also be changed from one phase to another by heating or cooling. For example, ice (solid phase) can change into water (liquid phase) by heating and the water can change into water vapor (gaseous phase) by more heating. In steam power plants, phase changes occur in the boiler and condenser. Examples of pure substances: 1. Water (solid, liquid, and vapor phases) 2. Mixture of liquid water and water vapor 3. Carbon dioxide, CO2 4. Nitrogen, N2 5. Mixtures of gases, such as air, as long as there is no change of phase. - Condenser: It is a device used in power plant stations to condense steam (water vapor) and hence it changes gas phase (vapor) into liquid phase (water). - Boiler: It is a device used in power plant stations to evaporate water and hence it changes liquid phase (water) into gas phase (vapor). Real substances that readily change phase from solid to liquid to gas such as water, refrigerant-12, and ammonia cannot be treated as ideal gases in general. The pressure, volume, temperature relation, or equation of state for these substances is generally very complicated, and the thermodynamic properties are given in table form. The properties of these substances may be illustrated by the functional relation F(P,v,T)=0, called an equation of state. Figure 2.1 shows three regions where a substance like water may exist as a solid, liquid or gas (or vapor). Also this figure shows that a substance may exist as a mixture of two phases during phase change, solid -vapor, solid-liquid, and liquidvapor. Fig 2.1 Now consider cold water inside a cylinder with a piston of negligible weight (to have H2O at atmospheric pressure) (Figure 2.2). As thus the heating process occurs at constant pressure (atmospheric). Boiling at this pressure occurs at 100 o C which is called saturation temperature at this pressure. This constant pressure heating process is illustrated in Figure 2.3. Fig 2.2: Phase change at constant pressure 6 Process 1-2: The temperature and specific volume will increase from the compressed liquid, or subcooled liquid, state 1, to the saturated liquid state 2. In the compressed liquid region, the properties of the liquid are approximately equal to the properties of the saturated liquid state at the saturation temperature. Process 2-3: At state 2 the liquid has reached the temperature at which it begins to boil, called the saturation temperature, and is said to exist as a saturated liquid. Properties at the saturated liquid state are noted by the subscript f and v2 = vf. During the phase change both the temperature and pressure remain constant (water it boils at 100°C when the pressure is 1 atm or 101.325 kPa). At state 3 the liquid and vapor phase are in equilibrium and any point on the line between states 2 and 3 has the same temperature and pressure. Process 3-4: At state 4, a saturated vapor exists and vaporization is complete. The subscript g will always denote a saturated vapor state. Note v4 = vg. Thermodynamic properties at the saturated liquid state and saturated vapor state are given in the saturated temperature table (table 1) and the saturated pressure table (table 2). These two tables contain the same information. In the first, the saturation temperature is the independent property, and in the second, the saturation pressure is the independent property. The saturation pressure is the pressure at which phase change will occur at a given temperature. In the saturation region the temperature and pressure are dependent properties which mean that if one property is known, then the other is automatically known. Process 4-5: If the constant pressure heating is continued, the temperature increases above the saturation temperature, 100 °C in this example, and the volume also increases. State 5 is called a superheated state because T5 is greater than the saturation temperature for the given pressure and the vapor is not about to condense. Thermodynamic properties for water in the superheated region are given in the superheated steam tables (table 3) Fig 2.3: Heating at constant pressure process Fig 2.4: Heating at different constant pressures Fig 2.5: T-V property diagram Consider repeating this process for other constant pressure lines as shown in Figure 2.4. If all of the saturated liquid states are connected, the saturated liquid line is established as can be seen in figures 2.5 and 2.6. If all of the saturated vapor states are connected, the saturated vapor line is established. These two lines intersect at the critical point and form what is often called the “steam dome.” The region between the saturated liquid line and the saturated vapor line is called by these terms: saturated liquid-vapor mixture region, wet region (i.e., a mixture of saturated liquid and saturated 7 Fig 2.6: P-V property diagram vapor), two-phase region, and just the saturation region. Notice that the trend of the temperature following a constant pressure line is to increase with increasing volume and the trend of the pressure following a constant temperature line is to decrease with increasing volume. The region to the left of the saturated liquid line and below the critical temperature is called the compressed liquid region. The region to the right of the saturated vapor line and above the critical temperature is called the superheated region. Property Tables In addition to the temperature, pressure, and volume data, property tables contain the data for the specific internal energy u, the specific enthalpy h, and the specific entropy s. The enthalpy is a convenient grouping of the internal energy, pressure, and volume and is given by: H = U + PV and the enthalpy per unit mass is h = u + Pv The enthalpy h is quite useful in calculating the energy of mass streams flowing into and out of control volumes. The enthalpy has units of energy per unit mass, kJ/kg. Entropy (s) is an important property that measures randomness. For instance organized people lead to low-entropy lives while mechanical friction of a moving piston through a cylinder with high friction generates entropy. The efficiency of any process decreases with entropy generation and the amount of entropy generation measures how much bad is the process. Saturated Water Tables Since temperature and pressure are dependent properties using the phase change, two tables are given for the saturation region. Temperature table (table 1) has temperature as the independent property (first column); Pressure table (table 2) has pressure as the independent property (first column). These two tables contain the same information and often only one table is given. Table (1): Table (2): For the complete Tables, the last entry is the critical point at 22.09 MPa. and 374.14 oC. The subscript fg used in Tables refers to the difference between the saturated vapor value and the saturated liquid value region. That is, ufg = ug - uf = increase in internal energy due to vaporization hfg = hg - hf = increase in enthalpy due to vaporization sfg = sg - sf = increase in entropy due to vaporization The quantity hfg is called the enthalpy of vaporization (or latent heat of vaporization). It represents the amount of energy needed to vaporize a unit of mass of saturated liquid at a given temperature or pressure. It decreases as the temperature or pressure increases, and becomes zero at the critical point. 8 2.3 Some terminologies and definitions: Saturation Temperature or Pressure: It is the T at which a pure substance starts boiling at a given P, and this T is called Saturation T. Likewise, at a given temperature, the pressure at boiling is called the saturation pressure Psat e.g., H2O at 1 atm, Tsat =100 0C (H2O at 100 0C, Psat = 1 atm) Remember : Saturation temperature depends on pressure. For example. If P = 0.5 bar, water boils at 81.35°C and if P = 10 bar, Tsat = 179.88°C. Saturated Liquid (Subcooled/ Compressed Liquid): If a substance exists as liquid at the saturation T and P, it is called saturated liquid. If the T of the liquid is lower than Tsat for the existing P, it is either called a subcooled liquid (implying T < Tsat for the given P) or a compressed liquid (implying P > Psat for the given T). The saturated liquid is the liquid which is about to vaporize while compressed liquid is the liquid that is not to vaporize. Quality and Saturated Liquid-Vapor Mixture Now, let’s review the constant pressure heat addition process for water shown in Figure 2-3. Since state 3 is a mixture of saturated liquid and saturated vapor, how do we locate it on the T-v diagram? To establish the location of state 3 a new parameter called the steam quality or dryness fraction x which is defined as m mg mass of vapor x = vapor = = mtotal mg + m f mass of vapor + mass of liquid The quality ( x) is zero for the saturated liquid and one for the saturated vapor (0 ≤ x ≤1). The average specific volume ( or any property) at any state in the wet region as state 3 is given in terms of the quality ( x) as follows: Consider a mixture of saturated liquid and saturated vapor. The liquid has a mass mf and occupies a volume Vf. The vapor has a mass mg and occupies a volume Vg. mtotal = mliquid + mvapor mt = m1 = mf + mg Also V =Vf + Vg m1v = mfvf + mgvg v = (m1 – mg) vf + mg vg m1 m1 v = (1 – x) vf + x vg v = vf + x (vg – vf) Similarly h = hf + x (hg – hf) u = uf + x (ug – uf) s = sf + x (sg – sf) Superheated Water Table (Table 3): A substance is said to be superheated if its temperature is greater than the saturation temperature for the given pressure. State 5 in Figure 2-3 is a superheated state. In the superheated water table, T and P are the independent properties. The value of temperature to the right of the pressure is the saturation temperature for the pressure. The first entry in the table is the saturated vapor state at the pressure. 9 Compressed Liquid Water Table A substance is said to be a compressed liquid when its pressure is greater than the saturation pressure for the temperature. It is now noted that state 1 in Figure 2-3 is called a compressed liquid state because the saturation pressure for the temperature T1 is less than P1. At pressures below 5 MPa for water, the data are approximately equal to the saturated liquid data at the given temperature. We approximate intensive parameter y, that is v, u, h, and s data as: y ≈ yf @ T How to Choose the Right Table The correct table to use to find the thermodynamic properties of a real substance can always be determined by comparing the known state properties to the properties in the saturation region. Given the temperature or pressure and one other property from the group v, u, h, and s, the following procedure is used. For example if the pressure and specific volume are specified, three questions are asked: For the given pressure, You go to table 1 or 2 and check the following: - If v < vf the state is subcooled or compressed liquid, then you use table 3. - If v = vf the state is saturated liquid, x = 0 and you use table 1 or 2. - If v = vg the state is saturated vapor (dry saturated), x = 1 and you use table 1 or 2. - If vf < v < vg the state is wet steam and 0 < x < 1 and you have to get the exact value by using the data from table 1 and 2 to get vf and vg then v = vf + x (vg - vf ). After that you use x to get h, u, s using the relation h = hf + x hfg , u = uf + xufg , s = sf + x sfg. - If v > vg the state is superheated and you use table table 3. Example 2-1 Find the internal energy of water at the given states for 7 MPa and plot the states on T-v, and P-v diagrams. a. P = 7 MPa, dry saturated or saturated vapor: Using Table 1 or 2: u = ug = 2580.5 kJ/kg b. P = 7 MPa, wet saturated or saturated liquid: Using Table 1 or 2: u = uf = 1257.6 kJ/kg c. Moisture = 5%, P = 7 MPa: let moisture be y, defined as y = mf/m = 0.05, then x = 1- y = 1.0 – 0.05 = 0.95, using table 1 to get uf and ug , then: u = uf + x(ug – uf) = 1257.6 + 0.95(2580.5 – 1257.6) = 2514.4 kJ/kg. d. P = 7 MPa, T = 600°C, For P = 7 MPa: Table 1 gives Tsat = 285.9°C. Since 600°C > Tsat for this pressure, the state is superheated. Use Table 3, u = 3260.7 kJ/kg. e. P = 7 MPa, T = 100 oC: Using Table 1, At T = 100 oC, Psat = 0.10132 MPa. Since P > Psat, he state is compressed liquid. approximate solution: u = uf @ T = 100 = 418.94 kJ/kg. We do linear interpolation to get the value at 100 °C. (We will demonstrate how to do linear interpolation with this problem even though one could accurately estimate the answer.) u = 418.96 kJ/kg. 7. f. P = 7 MPa, T = 460°C: since 460°C > Tsat at P = 7 MPa: the state is superheated. Using Table 3, we do linear interpolation to get u = 2997.1 kJ/kg Using the above table, form the following ratios. 10 Example 2-2 Determine the enthalpy of 1.5 kg of water contained in a volume of 1.2 m3 at 200 kPa. Recall we need two independent, intensive properties to specify the state of a simple substance. Pressure P is one intensive property and specific volume is another. Therefore, we calculate the specific volume. v = volume/mass = 1.2/1.5 = 0.8 m3/kg Using Table 1 at P = 200 kPa, vf = 0.001061 m3/kg , vg = 0.8857 m3/kg Now : Is v < vf ? no. Is vf < v < vg ? yes Is v > vg ? no. This means that the state is a wet steam and we must get the quality x. v = vf + x (vg – vf) 0.8 = 0.001061 + x ( 0.8857 - 0.001061) , x = 0903 Similarly: h = hf + x (hg – hf) = = 504 .7 + 0 903 (2201.9) = 2493.3 kJ/kg Example 2-3 Consider the closed, rigid container of water shown below. The pressure is 700 kPa, the mass of the saturated liquid is 1.78 kg, and the mass of the saturated vapor is 0.22 kg. Heat is added to the water until the pressure increases to 8 MPa. Find the final temperature, enthalpy, and internal energy of the water. Does the liquid level rise or fall? Plot this process on a P-v diagram with respect to the saturation lines and the critical point. Let’s introduce a solution procedure that we will follow throughout the course. System: A closed system composed of the water enclosed in the tank Property Relation: Steam Tables Process: Volume is constant (rigid container) For the closed system the total mass is constant and since the process is one in which the volume is constant, the average specific volume of the saturated mixture during the process is given by v = V/m = constant or v1 = v2 x = mg /(mg + mf) = 0.22/(1.78 + 0.22) = 0.1 Then, at P = 700 kPa v = vf + x (vg – vf) = 0.001108 + 0.11( 0. 2729 - 0.001108) = 0.031 m3/kg State 2 is specified by: P2 = 8 MPa, v2 = 0.031 m3/kg At 8 MPa, vf = 0.001384 m3/kg, vg = 0.002352 m3/kg, but v2 = 0.031 m3/kg; therefore, v2 > vg and hence the state is superheated. Interpolating in the superheated tables at 8 MPa gives, T2 = 362 °C, h2 = 3024 kJ/kg, u2 = 2776 kJ/kg Since state 2 is superheated, the liquid level falls. Extra Assignment Complete the following table for properties of water. Sketch a T-v or P-v diagram for each state. Describe the phase as compressed liquid, saturated mixture, or superheated vapor. If the state is saturated mixture, give the quality. 11 The enthalpy – entropy (hs) chart: The coordinate of an h-s or Mollier chart represent the two major properties of interest in thermodynamic analysis of open systems (boilers, turbines, condensers, pumps). The vertical distance between two states on this diagram is a measure of Δ h and the horizontal distance measures Δ s (the degree of irreversibility for an adiabatic process). It can also be used to present data and getting properties with reasonable accuracy especially in the superheated vapor region. By tracing the lines of different coloure, you can trace the lines (curves) of steam quality (x) pressure (p), temperature (t,°C), specific volume (dm/kg). Note 1. You need to multiply the v data you get from the chart by 10-3 to be in m3/kg. 2. saturated liquid line is not shown, because the lines are very crowded. Example 2.4: steam at 10 bar & 250 C expands isentropically to 0.16 bars. Find the change in enthalpy and the final dryness fraction. Solution: by plotting state (1) on Mollier chart and going vertically to P = 0.16 bar. From the chart:Δ h = h2 – h1 = 2861 – 2242 = 619 kJ/kg x2 = 0.85 & t2 = 55 C. v2 = 8000 dm3/kg = 8 m3/kg h-s or Mollier diagram 12 Equations of State The relationship among the state variables, temperature, pressure, and specific volume is called the equation of state. We now consider the equation of state for the vapor or gaseous phase of simple compressible substances. Ideal Gas Based on our experience in chemistry and physics we recall that the combination of Boyle’s and Charles’ laws for gases at low pressure result in the equation of state for the ideal gas as where R is the constant of proportionality and is called the gas constant and takes on a different value for each gas. If a gas obeys this relation, it is called an ideal gas. We often write this equation as or Here or P = absolute pressure in MPa, or kPa v = specific volume in m3/kg T = absolute temperature in K Ru = universal gas constant = 8.314 kJ/(kmol K) The gas constant for ideal gases (R) is related to the universal gas constant Ru through: Where M is the molecular weight, The mass, m, is related to the number of moles (N) by m = NM. Mair = 29, Mo2 = 32, MN2 = 28, Mco = 28, Mco2 = 44, MH2 = 2 Example 2-5 Determine the particular gas constant for air and hydrogen. The ideal gas equation of state is used when (1) the pressure is small compared to the critical pressure or (2) when the temperature is twice the critical temperature and the pressure is less than 10 times the critical pressure. The critical point is that state where there is an instantaneous change from the liquid phase to the vapor phase for a substance. Critical point data are given in Tables. Compressibility Factor To understand the above criteria and to determine how much the ideal gas equation of state deviates from the actual gas behavior, we introduce the compressibility factor Z as follows. Pv = Z Ru T or For an ideal gas Z = 1, and the deviation of Z from unity measures the deviation of the actual P-V-T relation from the ideal gas equation of state. The compressibility factor is expressed as a function of the reduced pressure and the reduced temperature. The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure, which are defined Pr = P/Pr as Tr = T/Tcr and where Pcr and Tcr are the critical pressure and temperature, respectively. The critical constant data for various substances are given in Tables. 13 Z =Pv/RuT The figure below shows the percentage of error for the – volume ((vtable videal/vtable)x100) for assuming water (superheated steam) to an ideal gas. be We see that the region for which water behaves as an ideal gas is in the superheated region and depends on both T and P. We must be cautioned that in this course, when water the working fluid, the ideal gas assumption may not be used to solve problems. We must use the real gas relations, i.e., the property tables. is Useful Ideal Gas Relation: The Combined Gas Law By writing the ideal gas equation twice for a fixed mass and simplifying, the properties of an ideal gas at two different states are related by m1 = m2 or 14 PROBLEMS 1. Define subcooled liquid, wet steam, saturated liquid, saturated vapor, and super heated vapor. 2. .A househusband is cooking a meet for his family in a pan which is (a) uncovered, (b) covered with a light lid, and (c) covered with a heavy lid. For which case will the cooking time be the shortest ? Why? 3. Complete the following table for H2O: T, oC P, kPa v, m3/kg u, kJ/kg h, kJ/kg X Phase descriptio 60 ….. 200 110 110 ….. 250 90 ….. 140 ….. 80 ….. ….. 150 300 500 ….. 300 200 800 200 ….. 1000 400 600 3.25 ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. 2300 ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. 1800 ….. ….. ….. 3165.7 ….. ….. ….. Saturated liquid ….. ….. ….. ….. ….. ….. Saturated vapor ….. ….. ….. ….. ….. ….. ….. 0.7 0.0 ….. ….. ….. 4. A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 0.5 Kg lid. If the local atmospheric pressure is 101 kPa, determine the temperature at which the water will start boiling when it is heated. (Answer: 99.9 oC) 3 3 5. A piston-cylinder device contains 0.1 m of liquid water and 0.9 m of water vapor in equilibrium at 500 kPa. Heat is added at constant pressure until the temperature reaches 200 oC. - What is the initial temperature of the water? - Determine the total mass of the water. - Calculate the final volume. - Show the process on a P-v diagram with respect to saturation lines. 6. Superheated water vapor at 1 MPa and 300oC is allowed to cool at constant volume until the temperature drops to 150 oC. At the final state, determine (a) the pressure, (b) the quality, and (c) the enthalpy. Also show the process on a T-v diagram with respect to saturation lines. (Answers: (a) 475.8 kPa, (b) 0.656, (c) 2030.5 KJ/kg) 7. A rigid enclosure, 50 cm on each side, contains a wet mixture of water vapor at 90oC and 20 percent quality. Heat is added until the pressure is raised to 500 kPa. Determine the final state and the quantity of heat added. (Answer: 507.7 kJ) 8. A rigid container is filled with steam at 700 kPa and 200oC. At what temperature will the steam start to condense when the container is cooled? To what temperature must the container be cooled to condense 50 percent of the steam mass? (Answer: T=134.3oC) 3 9. 0.05 kg of steam at 15 bar is contained in a rigid vessel of volume 0.0076 m What is the temperature of the steam? If the vessel is cooled at what temperature will the steam be just dry saturated? Cooling is continued until the pressure in the vessel is 11 bar, calculate the final dryness fraction of the steam. 10. The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25 oC, the pressure gage reads 210 kPa. If the volume of the tire is 0.65 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 50 oC (assame the valume will not change) Also determine the amount of air that must go out to get the original pressure at the new temperature (50oC). Assume the atmospheric pressure to be 100 kPa. 11. A 20-m3 tank contains nitrogen at 25oC and 800 kPa. Some nitrogen is allowed to escape until the pressure in the tank drops to 600 kPa. If the temperature at this point is 20oC, 12. A 1-m3 tank containing air at 25oC and 500 kPa is connected through a valve to another tank containing 5 kg of air at 35oC and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air. (Answer: 284.1 kPa) 15 Chapter 3 The First Law of Thermodynamics 3.1 Energy conservation principle (first law of thermodynamics): Energy can neither be created nor destroyed; it can only change forms 3.2 Energy conservation on a system undergoing a cycle: during any cycle a system undergoes, the cyclic integral of the heat is proportional to the cyclic integral of the work. Consider the Following cycle: For any cycle, the system returns back to its initial state and hence there is no energy stored for the cycle. ∫ Q = ∫W z Thus the first law for a cycle states that the net heat transfer to the system equals the net work done by the system similar to Carnot cycle. Process Description of Carnot Cycle 1-2 Isothermal heat addition, Qin 2-3 Isentropic expansion, Wout 3-4 Isothermal heat rejection, Qout 4-1 Isentropic compression, Win Applying the first law on Carnot cycle: δ Qin – Qout = Wo ut - Win Or Energy input = Energy output (no energy stored in the system) Qin + Win = Qout + Wout 3.3 Energy conservation on a system undergoing process: for a system undergoing a process (change from one state to another) and does not make a cycle the1st law states: Net energy transfer to the system as heat and work = Net increase in the total energy of the system Q – ∆W = ∆E 3.4 Closed system: such as fluids (gases or liquids) contained in piston – cylinder device, rigid vessel, and elastic vessel with no mass transfer. ∑ Q − ∑W = ∆E = change in int errnal energy + change in kinetic energy + change in potential energy (Q1 − Q2 ) − (W2 − W1 ) = ∆U + ∆kE + ∆PE 1 2 2 = (U 2 − U 1 ) + m(V2 − V1 ) + mg ( Z 2 − Z1 ) kJ 2 Net heat transfer across the boundary Net work done in all forms 1 2 W2 W1 Q1 Q2 Note: V1 & V2 are the velocities of the system at states 1 & 2 respectively in m/s. Z1 & Z2 are the altitudes of the system at states 1 & 2 respectively in m. 16 3.5 Energy equation in the rate form: δQ − δW = dE − W = dE ⇒Q dt dt − W = dU + d ( KE ) + d ( PE ) Q dt dt dt δW = δQ , where : Q W = dt dt For a closed system KE & PE can be neglected ∴ ∑ Q − ∑ W = m(u 2 − u1 ) or ∑ q − ∑ w = (u 2 − u1 ) or net heat transfer into the system – net work done by the system = net change in the system internal energy. 3.6 Types of work: There are five types of work: Moving boundary work, mechanical work, shaft work, spring work, and electrical work. (1) Moving Boundary work (PdVwork ): Similar to the moving boundary of a fluid enclosed in a cylinder-piston arrangement. When the piston moves, it causes a moving boundary work. Some Typical Processes Constant volume: If the volume is held constant, dV = 0, and boundary work equation becomes the Constant pressure: If the pressure is held constant, the boundary work equation becomes 17 Constant temperature, ideal gas If the temperature of an ideal gas system is held constant, then the equation of state provides the pressurevolume relation, then, the boundary work is: The polytropic process The polytropic process is one in which the pressure-volume relation is given as PV n = constant The exponent n (polytropic index) may have any value from minus infinity to plus infinity depending on the process. Some of the more common values are given below. Process Exponent n For constant pressure, n = 0 For constant volume, n = ∞ For isothermal & ideal gas, n = 1 For adiabatic & ideal gas, n = k = CP/CV Here, k is the ratio of the specific heat at constant pressure CP to specific heat at constant volume CV. The specific heats will be discussed later. The boundary work done during the polytropic process is found by substituting the pressure-volume relation into the boundary work equation. The result is: For an ideal gas undergoing a polytropic process, the boundary work is Notice that the results we obtained for an ideal gas undergoing a polytropic process when n = 1 are identical to those for an ideal gas undergoing the isothermal process. (2) Mechanical work: This work is similar to the work required to move a block of mass m kg by pulling this block by excreting a force F on it that moves it a distance. δW = F . ds 2 W12 = ∫ F . ds KJ 1 δW F . ds δW = = m m 2 w12 = ∫ 1 F . ds m KJ / kg (3) Shaft work: This is similar to the work done by the crank-shaft of an automobile. 18 A force acting through arm r generates a torque τ (Fig. 3.4). The work done during n revolution is given by Wsh = F .S = si ce τ =Fr τ Fig 3.4: Shaft work is proportional Wsh = FS = 2πnr to the torque applied and the r number of revolutions of the shaft. = 2πnτ kJ and the power transmitted through the shaft is given by W = wt kW then sh where w = 2πn / 60 is the number of revolution per unit time ortheanglervelocityamdnisthenumberofrevolutionper min ate( rpm). Example 1: Determine the power transmitted through the shaft of a car when the torque applied is 200 N.m and the shaft rotates at a rate of 400 revolutions per minute (rpm). Solution: The angular velocity (ω) = Power = τ ω = 418.5* 200 = 83.7 kw (or 112.2 hp) 2πn = (2π)(400/60) = 418.5 1/s 60 (4) Spring work: When a force is applied on a spring, the length of the spring changes and the work done is given by δwspring = F dx For a linear elastic spring, F = kx, where k is called spring constant and has the unit N/m 2 Wspring = ∫ Kx dx 1 = 1 / 2 K ( x2 − x1 ) 2 2 kJ (5)Electric work: This is similar to the work used to heat air or water by an electrical resistance produced by passing an electric current (I) over a time period because of an electric potential or volt (ε) δW = εIdt W12 = t ≡ time t2 ∫ εIdt = εI (t t1 19 2 − t1 ) Example 2: A well insulated piston cylinder device contains 2 kg of air at 100 kpa. An electric heater placed in the tank is connected to a 220-volt source, and an electric current of 5 A flows through the resistant heater for 20 min. Find the work done on the system. W12= εI (Δt) = 220×5×20×60 =264 kJ 3.7 Heat: is the energy transferred (crossed) the boundaries due to temperature difference (not associated with mass and not stored in the system) Sign convention: If heat is transferred to the system :then Q is +ve as in the case of boilers, solar heaters, combustion chambers, tea pots & cooking pans. W + ve Q + ve W - ve Q - ve If heat is transferred to the system (heat is last form the system): then Q is -ve such as condensers, human body & hot Pizza. Example 3: Air is compressed reversibly in a cylinder according to the process equation PV1.3 = constant. The air is initially at a pressure and temperature of 1 bar, 300 K and the final pressure is 6 bar. (i) Calculate the heat transfer between the air and its surroundings per unit mass during the process and state its sense (direction). (ii) What would have been the final pressure had the air been compressed through the same volume ratio adiabatically? Solution: (i) The process is polytropic with n = 1.3 and ( n −1) / n p = 300 × 6 0.3 / 1.3 = 43.62 K T2 = T1 2 p1 R(T1 − T2 ) 0.287(300 − 453.62) kJ = = −146.97 1 w2 = 0.3 n −1 kg Applying the first law on a closed system: 1 q2 −1 w2 = u 2 − u1 = cv (T2 − T1 ) q = −146.97 + 0.718(453.62 − 300) = −146.97 + 110.3 1 2 q = −36.67 1 2 kJ kg (rejected by the gas ) for a reversible adiabatic = isentropic process, PV γ = const. Now v1/v2 = (p2/p1)1/n = 6 1/1.3 = 3.9681 20 Thus p2/p1 = (v1/v2)γ = 3.96811.4 = 6.887 Which means p2 = 6.887 bar compared to P2 = 6 bar. This means that we get higher pressure in the case of isentropic compression because in the case of non-adiabatic process, the pressure is reduced due to the associated heat loss. Example 4: ْ A mass of 2 kg of steam (H2O) at 10 bar, 160 ◌C, is heated at constant pressure to 250 ْ◌C and then cooled at constant volume to 4 bar. Calculate the magnitudes and senses of the heat and work transfers in each of these processes. Solution: From steam tables at 10 bar, Tsat = 179.9 ◌ْ C But T1 = 160◌ْ C thus the H2O at the given state is subcooled liquid at state 1, and from steam tables u1 = 674.2 kJ/kg, v1 = 0.00102 m3/kg At 10 bar, T2 = 250 ◌ْ C thus the H2O is superheated vapor at state 2. And from steam tables v2 = v3 = 0.2328 m3/kg & u2 = 2711 kJ/kg. A fixed mass (2 kg) is heated at constant pressure means that it is enclosed in a pistoncylinder device and the piston should move during the heating process to keep the pressure constant and hence there is a moving boundary work. Process 1-2: W12 = m P(v2 – v1) Or W12 = 2 ×10× 105 ×(0.2328 – 0.001102) J = +463.4 kJ Thus Q12 = W12 + U2 – U1 = W12 + m (u2 – u1) = + 463.4 + 2 (2711 – 674.2) = + 453.6 kJ Process 2-3: W23 = 0 (since volume is held constant). From steam Tables at P3 = 4 bar, vg = 0.4623 m3/kg, vf = 0.001 m3/kg, uf = 605 kJ/kg, ug = 2554 kJ/kg. Since v3 = 0.2328 < vg = 0.4623, then state 3 is wet steam and we have to calculate the steam quality (x) to get u3 u3 = uf + x3 (ug – uf ) note uf is very small compared with ug and hence can be neglected and hence x3 = v3/vg,3 = v2/vg,3 = 0.2328/0.4623 =0.502 Now u3 = uf,3 + x3(ug,3 – uf,3) = 605 + 0.502(2554 – 605) kJ/kg u3 = 1583.4 kJ/kg Thus 2Q3 = 2W3 + U3 – U2 = 2W3 + m (u3 – u2) 2Q3 = 0 + 2(1583.4 – 2711) = - 2255.2 kJ 21 Control Volume (Open System) Energy Analysis 4.1 Steady state open system: This section covers the application of the first low of thermodynamics on open systems or control volumes such as the flow in pipes, ducts, compressors, pumps, nozzles diffusers; heat exchangers, condensers, combustion chambers, heaters, coolers, pumps, ………etc. - Mass crosses the boundary of these systems and heat and work may be exchanged across the control volume of these systems. The first law of thermodynamics controls the transformation of the different energies over these systems. - Steady state means: temperature and all properties are uniform through out the system and do not change over the time at state (1) and state (2). However the change occurs during process 1 2 only i.e d /dt = 0 - Also masses may get into or out of the system but there is no mass change of the system with time i.e there is a mass conservation. Which means that the mass of the system does not increase or decrease during the process. 4.2 Mass conservation: ∑m − ∑ m = For steady state condition, dm = 0 i ∑ dm For one inlet & one exit to the system i = m = m m A: cross – sectional area, m2 V: velocity, m/s ρ: density, kg/m3 v: specific volume = 1/ ρ ,m3/kg ρ1Α1V1 = ρ 2 Α 2V2 Α1V1 Α 2V2 = v1 v2 4.3 Energy conservation: 1st law of thermodynamics Notes: b ∑ Q − ∑ W = m[∆h + ∆k .e + ∆p.e] The sign ∑ is removed for simplicity w = power note W = m ( ) 1 Q − W = m (h2 − h1 ) + V22 − V12 + g ( z 2 − z1 ) 2 or 1 q − w = (h2 − h1 ) + V22 − V12 + g ( z 2 − z1 ) 2 ( ) Any term of the above equation can be neglected at certain conditions or applications. Units are very important (dimensional homogeneity) 1st law & mass conservation are coupled through the specific volume (v) & velocity (V). This means that these two properties appear in both equations. Adiabatic process or insulated system means Q = q = 0 22 For ideal gas du = cv dT volume dh = cp dt or (u2 – u1) = Cv (T2 – T1) where Cv is the specific heat at constant or (h2 – h1) = CP (T2 – T1) where CP is the specific heat at constant pressure. Cp – Cv = R and cp/cv = γ = Specific heat ratio 4.4 Engineering Applications Involving Steady-State Open Systems: Turbines: superheated steam is used in steam turbines to produce work. On the other hand air at high pressure and temperature (high enthalpy) is used in gas turbine to produce work. In general, heat losses can be neglected (Q = 0 ) in both steam or as gas turbines. Moreover, expansion in turbines may be assumed isentropic. Compressors: They are used to increase the pressure of gases such as air. Because of compression, gas temperature increases and the compressor loose heat (Q is – ve). Compressors are driven by electric motors and hence the compressor work is negative. Reversible Steady-flow Work: When neglecting the changes in kinetic and potential energies, the work done for steadyflow devices undergoing an internally reversible process can be written as: wrev = - ∫ vdp kJ/kg The difference between v dp and p dv should be clear. Pdv is associated with reversible boundary work in closed systems while vdp is the reversible work for steady flow open systems when neglecting changes in kinetic and potential energles. The reversible work done for a polytropic process (PVn = c) in steady flow open systems can be calculated as follow: W12 = - 21 ∫ V dP and PVn =C or VP1/n = C = - 21 ∫ (C/P1/n) dP = -C 21 ∫ (p-1/n) dP W12 = - C [p 12−(1 / n ) - p 11−(1 / n ) ]/[1 – (1/n)] W12 = -C[p 12−(1 / n ) - p 11−(1 / n ) ]/[(n-1)/n] From P1Vn1 = P2Vn2 or V1P 11 / n = V2P2 Then, C = V1P 11 / n = V2P 12/ n W12 = -[V2P 12/ n P 12−(1 / n ) - V1P 11 / n P 11−(1 / n ) ]/[(n/1)/n] W12 = - n[ P2V2 – P1V1]/ (n – 1) and for ideal gases (PV = m R T), W12 = - m n R [T2 – T1]/(n-1) kJ ο Power = m n R [T2 – T1]/(n – 1) kw w = - Rn [T2 – T1]/(n-1) kJ / kg 23 kJ Some Steady-Flow Engineering Devices Below are some engineering devices that operate essentially as steady-state, steady-flow control volumes. Pumps: It is similar to compressors, except it compresses liquids such as water or oil. Nozzles: The area of a nozzle decreases in the flow direction to increase the fluid velocity (kinetic energy) and as thus the pressure or enthalpy decreases. Diffusers: It is the opposite of a nozzle. The area increases in the flow direction and as thus, the pressure increases and the velocity decreases. Boilers: Fuels are burned in boilers to evaporate water and produce steam or superheated steam (Q is negative). Condensers: Condensers are used to condense steam. Throttling valve: It is a device used to make a sudden drop in pressure. The enthalpy before and after the throttling valve is the same. Example 5: Air compressor is used to compress 0.03 kg/s atmospheric air from 1 bar and 25 ºC to 10 bar. If the compression process is polytrophic and follows the relation PV1.3 = constant, calculate the power required to drive the compressor and the heat losses. Consider the pipe diameter at the inlet is 5 cm and that at the exit is 2 cm. Solution: To find the exit temperature T, use the relation: (T2/T1) = (P2/P1)(n-1)/n , with T1 =25 + 273 = 298 k T2 = T1 (P2/P1) (n-1)/n = 298 (10/1)(1.3-1)/1.3 = 506.97 K Then, w = - nR(T2 – T1)/(n-1) w = - 0.287 × 1.3 [506.97 – 298]/(1.3-1) w = -259.89 kJ / kg power = m w = 0.03 × (-259.89) Wo ~ 7.797 kW To calculate the heat losses, apply the first law of thermodynamics for steady flow open system: Qo – Wo = m [(h2 – h1) + (V22 – V12) / 2 + g (z2 – z1)] In this equation the potential energy is neglected but the law of kinetic energy should be calculated because the inlet and exit diameters are given. So, m = p1V1A1 = p2 V2 A2 24 Then, ρ1 = 1/v1 = P1 / (RT1) and ρ 2 = 1/v2 = P2 /(RT2) m = [P1 / (RT1)] V1 A1 = [P2 / (RT2)] V2 A2 m = ρ 1 A1 V1 = A1 V1 / v1 ( P1 v1 = RT1) V1 = m RT1 / [P1 A1 ] = 0.03 × 0.287 × 289 /[1000 × 3.14(0.05)2/4] = 13.07 m/s In the same way: V2 = m RT2 / [P2 A2] = 0.03 × 0.287 × 506.97 / [1000 / 3.14(0.02)2/4] = 13.89 m /s The difference between V1 and V2 is small and thus the change in kinetic energy can be neglected. This can be calculated as: Kinetic energy = [(V2)2 – (V1)2] / 2 (m2 / S2) = 11.148 J/kg = 0.0112 kJ / kg This is too small relative to the work ( w = 259.89 kJ/kg) and can be neglected as mentioned above. Thus, the first law on this steady state open system can written as: Q° - W° = m [(h2 – h1 ) + (V22 – V21) / 2 + g (z2 – z1)] And since it is an air which can be assumed as an ideal gas and in this case: h2 – h1 = Cp ( T2 – T 1) where Cp = 1.0035 kJ/kg K Q° - (-7.797) = 0.03 [CP (T1 - T2) + 0 +0] = 0.03 × 1.0035 (506.97 – 298) = 6.29 Q° = 6.29 – 7.797 = 1.51 kW The negative sign means that the heat will be transferred (lost) from the system. Example 6: Air at 10 °C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine (a) the mass flow rate of the air and (b) the temperature of the air leaving the diffuser. Solution: (a) To determine the mass flow rate, we need to find the specific volume of the air first which is determined from the ideal-gas equation at the inlet condition: RT1 [0.287 x283] v1 = = = 1.015 m3 /kg P1 80 1 1 V1 A1 = (200 )(0.4 ) = 78.8 kg/s m · = v1 1.015 (b) diffuser normally involves no shaft or electrical work (w = 0), negligible heat transfer (q = 0), and a small (if any) elevation change between the inlet and the exit (Δpe = 0 ), then the first law of thermodynamics or the conservation of energy relation on a unit-mass basis for this single-stream steady-flow open system reduces to q - w = Δh + Δke + Δpe (V22 − V12 ) 0 = h2 - h1 + 2 25 The exit velocity of a diffuser is usually small compared to the inlet velocity (V2<<V1); thus the kinetic energy at the exit can be neglected. 0 = Cp (T2 – T1) + V22 − V12 = 1003.5 (T2 – 283) - 0 − (200) 2 2 2 So, T2 = 302.93 K Example 7: Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressor. Solution: It is stated that Δke ≡ 0 and Δ pe ≡ 0. then the conservation of energy equation for this single-stream steady-flow open system (first law of thermodynamics) reduces to q - w = Δh + Δke + Δpe = h2 - h1 The enthalpy of an ideal gas can be calculated from Δh =Cp(T2 T1) Substituting yields - 16 – w = 1.0035 (400 – 280) w = - 136.42 kJ/kg This is the work done on the air per unit mass. The power input to the compressor is determined by multiplying this value by the mass flow rate: W = m w = (0.02) (- 136.42) = -2.73 kw Example 8: The electric heating systems used in many houses consist of a simple duct with resistance wires. Air flows over resistance wires. Consider a 15-kw electric heating system. Air enters the heating section at 100 kPa and 17 ºC with a volume flow rate of 150 m3/min. If heat is lost from the air in the duct to the surroundings at a rate if 200 W, determine the exit temperature of air. Solution: The conservation of energy equation for this single-stream steady-flow open system (noticing that there are no change in kinetic and potential energies) simplifies to Qo – Wo = m [(h2 – h1) + (V22 – V12) / 2 + g (z2 – z1)] = m (h2 – h1) Then the specific volume of the air at the inlet becomes. RT1 [0.287 x290] v1 = = = 0.832 m3 /kg P1 100 ⋅ V 1 150 1 = m = v1 0.832 60 The first law of thermodynamic on the air (Δh = Cp ΔT) reads: Q· - W = m Cp (T2 – T1) -0.2 – (-15) = (3) x 1.005 (T2 – 17), T2 = 21.9 ºC 26 Problems on first law of thermodynamics 1. Show that for a closed system with a moving boundary and during a constant pressure process, the heat transfer is equal to the change in enthalpy. (Answer: -55.45 kJ) 2. A rigid tank contains a hot fluid which is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the liquid. Neglect the energy stored in the paddle wheel. 3. A piston-cylinder device contains 25 g of saturated water vapor which is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. Determine the final temperature of the steam. 4. A piston-cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27 oC. An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of the nitrogen. 5. 0.05 kg of a certain fluid is heated at a constant pressure of 2 bar until the volume occupied is 0.0658 m3. Calculate the heat supplied and the work done, a) When the fluid is steam, initially dry saturated. b) When the fluid is air, initially at 130oC. (Answer: a) Q=18.25 kJ, W=4.304 kJ, b) Q=25.83 kJ, W=7.38 kJ) 6. Steam at 7 bar and dryness fraction 0.9 expands in a cylinder behind a piston isothermally and reversibly to a pressure of 1.5 bar. Calculate the change of internal energy and the change of enthalpy per kg of steam. The heat supplied during the process is found to be 400 kJ/kg. Calculate the work done per kg of steam. (Answer: 182.5 kJ/kg) o 7. 1 kg of steam at 100 bar and 375 C expands reversibly in a perfectly thermally insulated cylinder behind a piston until the pressure is 38 bar and the steam is then dry saturated. Calculate the work done by the steam. (Answer: 169.7 kJ/kg) 8. Air at 1.02 bar, 22 oC, initially occupying a cylinder volume of 0.015 m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate the final temperature, the final volume, and the work done on the mass of air in the cylinder. (Answer: T=234.5oC , V= 0.00388 m3 , W=2.76 kg) 9. In a steam engine, the steam at the beginning of the expansion process is at 7 bar, dryness fraction 0.95, and the expansion follows the law pv1.1=constant, down to a pressure of 0.34 bar. Calculate the work done per kg of steam during the expansion, and the heat flow per kg of steam to or from the cylinder walls during the expansion. (Answer: W= 436 kJ/kg, H=155.6 kJ/kg) 10. 0.468 kg of air at 200 kPa and 25 ºC is contained in a cylinder at one side of a frictionless piston while a 0.181 kg of saturated vapor is contained in the cylinder at the other side. Neglecting piston weight, calculate the volumes of both air and vapor. If heat is transferred to the vapour until the air volume is reduced to half its initial value. Calculate the final air temperature is 25 ºC. 11. Consider an ordinary shower where hot water at 60oC is mixed with cold water at 10oC. If it is desired that a steady stream of warm water at 43oC be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the heat losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 137.9 kPa. 12. An air compressor takes in air at 1 atm and 20oC and discharges into a line having an inside diameter of 1 cm. The average air velocity in the line at a point close to the discharge is 7 m/s, and the discharge pressure is 3.5 atm. Assuming that the compression occurs quasistatically and adiabatically, calculate the work input to the compressor. Assume that the inlet air velocity is very small. (Answer: 205.1 W) 27 13. Steam undergoes an adiabatic and steady-flow process in a turbine from 6.0 Mpa, 500oC, to a pressure of 10 kPa and a quality of 90 percent. Calculate the work output for a flow of 1.0 kg/s. What flow rate would be required to produce a work output of 400 kW ? 14. A centrifugal blower receives air at 1 atm and 20oC in a volume flow rate of 0.7 m3/s. The air enters at 1.0 m/s and discharges at 10.0 m/s, and temperature is essentially constant. Calculate the input power requirements. State assumptions. 15. Air at 6.9 bar, 260oC is throttled to 5.5 bar before expanding through a nozzle to a pressure of 1.1 bar. Assuming that the air flows reversibly in steady flow through the nozzle, and that no heat is rejected, calculate the velocity of the air at exit from the nozzle when the inlet velocity is 100 m/s. (Answer: 637 m/s) 16. 225 kg/h of air at 40oC enter a mixing chamber where it mixes with 540 kg/h of air at 15oC. Calculate the temperature of the air leaving the chamber, assuming steady flow conditions. Assume that the heat loss is negligible. (Answer: 22.4oC) 17. Steam from a superheater at 7 bar, 300oC is mixed in steady adiabatic flow with wet steam at 7 bar, dryness fraction 0.9. Calculate the mass of wet steam required per kg of superheated steam to produce steam at 7 bar, dry saturated. (Answer: 1.43 kg) 18. A fan is designed to take air from a large room at a volumetric flow rate (based on the inlet state to the fan) of 200 m3 / min. The fan must raise the inlet pressure by 1.0 kPa. The air enters the fan at 25 ºC and increases in temperature by a negligible amount in passing through the fan. The air leaves the fan in a duct of 1.00-m2 cross sectional area. What power (kW) is required to drive the fan. 19. A large pump is used to take water at 25 ºC from a nearby lake at a rate of 1 m3/s and raise its pressure from 120 to 700 kPa so that it can be fed into a fire safety main. If the pump is adiabatic and frictionless, how much power is necessary to drive the pump? 28 Chapter 4: The second law of thermodynamics Comparison between 1st and 2nd law: 1st law Energy conservation says that Burning papers can change completely to work (Fig. 4.1) 2nd law • Energy has a quality and a quantity. • A process can change in a certain direction. • The processes in Figs. (4.1) can not be done although the first law is satisfied. Fig 4.1 Transferring heat to a wire will not generate electricity. The above two processes satisfy the 1st law but do not satisfy 2nd law and therefore a process can’t take place unless it satisfies both laws. 4.1 Definitions: Thermal energy reservoirs = heat reservoirs = reservoirs (Fig. 4.2) A hypothetical system with a relatively large thermal energy capacity (mass × specific heat × temp) that can supply or absorb finite amounts of energy or heat without any change in system temperature. Reservoirs are classified as sources or sinks. Examples: (Atmosphere, river, ocean……etc ) Fig. 4.2 Bodies with relatively large thermal capacity can be modeled as thermal energy reservoirs. Source = a reservoir that supplies energy in the form of heat. Sink = a reservoir that absorbs energy in the form of heat. Heat engines (Fig. 4.3) are characterized by the following:1- They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor, etc.) 2- They convert part of this heat to work (usually to a rotating shaft). 3- They reject the remaining waste heat to a low – temperatures sink (the atmosphere, rivers, etc.). 4- They operate on a cycle. Fig 4.3 Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink. 29 4.2 Performance Desired output Required input Performance parameter = Net work output Total heat input Thermal efficiency = ηth = Wnet,out Qin =1- Qout Qin Cyclic devices (heat engines, refrigeration, heat pumps…etc.) operate between a high temperature reservoir (TH) and a low temperature reservoir (TL) QH = magnitude of heat transfer between cyclic device and high – temperature medium at temperature TH QL = magnitude of heat transfer between cyclic device and low – temperature medium at temperature TL Note QL and QH are magnitudes (positive quantities) ηth =1- QL QH 4.3 Statements of the second law of thermodynamics: 4.3.1 Kelvin – Plank statement (power cycles): It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce an equivalent amount of work. Or No heat engine can have a thermal efficiency of 100 percent Kelvin – Plank statement (power cycles): 4.3.2 Clausius statement (Ref, A/C, Heat Pump): It is impossible to construct a device that operate in a cycle and produce no effect other than transfer of heat from a lower temperature body to a higher temperature body. 4.3.2.1 Refrigerators (Fig.4.4): 4.3.2.2 Heat pump: β H .P = QH QH = W QH − QL 30 Refrigerators and heat pumps operate on the same cycle but differ in their objectives. The objective of Refrigerator is to maintain the refrigerated space (at TL) at low temperatures by remove heat from it. The objective of a heat pump is to maintain a heated space at a high temperature by adding (rejecting) heat to it. 4.3.2.3 Air conditioners: Are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment. 4.4 Coefficient of performance (cop): β R = COPR = desired output QL = required input Wnet ,in Wnet ,in = QH − QL β R = COPR = QL 1 = QH QH − QL −1 QL 4.5 Equivalence of the two statements 1-The Kelvin – Planck and the Clausius statements are equivalent in their consequences, and either statement can be used as the expression of the second law of thermodynamics. Both statements are negative because they use it is impossible. 2-Any device that violates the Kelvin – Planck statement also violates the Clausius statement, and vice versa. 4.6 Prove of the second observation: In order to prove that both statements are equivalent: If a device violates one statement, it must violate the other statement (if one goes false the other automatically goes false). Prove (Fig. 4.5) - Assume a device that violates Clausius statement (it has QL= QH). Add a heat engine device that has the same QL. The resultant device produced by adding the tow devces together violates Kelvin-Plank statement. Fig 4.5 From the above analysis, violation of Clausius statement implies (leads to) the violation of KelvinPlanck statement and therefore both statements are equivalent. 31 What is the maximum (η) or (β) for any cycle? The maximum (η) or (β)any cycle that works on two temperature limits is the Carnot (ideal) cycle that works on the same temperature limits. For example for a steam power cycle that works between 1400°C and 25°C, 2maximum = ncarnot = 1- 25 + 273 = 82.4% 1400 + 273 For heat pump: β H .P = QH QH = W QH − QL For heat engine η = W QH − QL = QH QH Note: When we substitute in the efficiency relation, the sign of QH& QL is neglected i.e : η = QH − QL QH Also take absolute values when substituting in (β) • i.e : β RF = QL β H .P = QH W QL = QH − QL QH − QL it can be proved that : βH.P = βRef + 1 4.12 The Carnot principle and Carnot cycle 4.12.1 Carnot principles: 1- The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs or same temp. limits of TL and TH. 2- The efficiencies of all reversible heat engines operating between the same two reservoirs same temp. limits of TL and TH are the same. 4.12.2 The Carnot cycle: Heat is added isothermally and also heat is rejected isothermally. Therefore, heat transferred is through constant temperature i.e. TH = C1 & TL = C2 From (1) to (2): (reversible adiabatic) expansion (turbine). From (2) to (3): isothermal heat rejected at TL (condenser or cooler). From (3) to (4): Isentropic (reversible and adiabatic) compression (pump). From (4) to (1): isothermal heat addition at TH (boiler). QH = TH ∆S QL = TL ∆S Fig 4.6 32 4.12.3 The efficiency of Carnot cycle: η =1− W QH − QL η= = QH QH QL QH where QH ==> from hot reservoir at TH where QC ==> to cold reservoir at TC since heat reservoir is a function of T only, this fact provides the basis for such an absolute temperature scale. Accordingly, The Carnot Efficiency: This is the Max. ηth a power cycle can get. NOTE: It is QC/QH = TC/TH, not QC = TC and QH = TH Example 1(Fig. 4.7): A heat pump with a COP of 1.50 is used to supply 270.000 kJ/h of energy to a small industrial process operating at a temperature of a few hundred degrees above the atmospheric air temperature of 2 ◌C. ْ Determine (a) the power required in kilowatts to drive the heat pump, (b) the rate at which energy is removed from the atmosphere, in kJ/h, and (c) the cost of continuous operation for 1 h if electricity costs 15 piasters per kilowatt-hour. Solution: (a) The heat-pump COP on a rate basis leads to Q out 270.000 kJ / h W net ,in = = COPHP 1.5 = 180.000 kJ / h × 1 kW = 50 kW 3600 kJ / h (b) the heat-transfer rate Qin,HP is found from an energy balance on the heat pump. dE =Q net ,in + Wnet ,in = Qin − Qout + Wnet ,in dt (C) the cost of operation for 1 h is Q in = Q out − W net ,in = ( 270.000 − 180.000) kJ / h = 90.000 kJ / h Cost = 50 kW ×1 h × L.E 0.15/kW.h = L.E. 7.5 33 Fig 4.7 Example 2(Fig. 4.8): A simple steam power cycle receives 100.000 kJ/min as heat transfer to the working fluid at 800 K, and rejects energy as heat transfer from the working fluid at 320 K. If the pump power required is 1400 kJ/min, determine (a) the thermal efficiency of an internally-reversible cycle and (b) the turbine power output in kilowatts. Equipment schematic and data for. Example 2 Fig 4.8 Solution: (a) Internally reversible cycle means carnot cycle and its efficiency is η th , rev = η th ,Carnot = 1 − TL 320 =1− = 0.625 TH 800 (b) ηth = W net ,out / Q in ∴W net ,out = ηth Q boil ,in = 0.625 (100.000 kJ / min ) = 62.500 kJ / min = W − W For this system, the net power output is W net ,out T ,out P ,in ∴WT ,out = W net ,out + W P ,in = (62.500 + 1.400) kJ / min 1 kW . min = 1065 kW = (63.900 kJ / min) 60 kJ Example 3 : An internally reversible refrigerator is used to maintain food in a refrigerated area. Heat transfer into the refrigerator fluid occurs at 2 ◌C. ْ It is also desired to maintain some frozen foods in a freezer. In this case heat transfer into the cyclic device occurs at - 17 ◌C, ْ and heat transfer out occurs again at 27 ◌C. ْ what percentage increase in work input will be required for the frozen-food unit over the refrigerated unit for the same quantity of heat QL removed from the cold regions? Solution: For a reversible refrigerator QH QL = TH TL & W = Q H - QL ∴ W = Q L TH - 1 and hence: TL W = QL,in (300/275 – 1) = 0.0909 QL W = QL,in (300/256 – 1) = 0.1719 QL The percentage increase of W’ over W is given by 0.1719 QL − 0.0909QL W '−W = 0.0909QL W = 0.891 This means that the increase is 89.1% 34 Problems on second law 1. State the expressions of Kelvin-Plank and Clausius for the second law and then prove that they are equivalent. 2. What are the two statements known as the Carnot principles? 3. Heat is transferred to a heat engine from a furnace at a rate of 800 MW. If the rate of waste heat rejection to nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine. (Answer: 30 MW, 0.375) 4. A heat pump is used to meet the heating requirements of a house and maintain it at 20oC. On a day when the outdoor air temperature drops to -2oC, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is extracted from the cold outdoor air. (Answer: 8.9 MW, 48000 kJ/h) 5. A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature source at 652oC and rejects heat to a low-temperature sink at 30oC. Determine (a) the thermal efficiency of this Carnot engine and (b) the amount of heat rejected to the sink per cycle. (Answer: 0.672, 163.8 kJ) 6. A heat pump is to be used to heat a house during the winter. The house is to be maintained at 21oC at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5oC. Determine the minimum power required to drive this heat pump unit. (Answer: 3.32 kW) 7. Consider two Carnot heat engines operating in series. The first engine receives heat from a reservoir at 1000 K and rejects the waste heat to another reservoir at temperature T. The second engine receives the energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at 300 K. If the thermal efficiencies of both engines are the same, determine the temperature T. (Answer: 547.7 K) 8. A Carnot heat engine receives heat from a reservoir at 900oC at a rate of 700 kJ/min and rejects the waste heat to the ambient air at 27oC. The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at -5oC and transfers it to the same ambient air at 27oC. Determine (a) the maximum rate of heat removal from the refrigerated space and (b) the total rate of heat rejection to the ambient air. (Answer: (a)4359 kJ/min) 9. A heat engine operates between two reservoirs at 727oC and 17oC. One-half of the work output of the heat engine is used to drive a Carnot heat pump which removes heat from the cold surroundings at 2oC and transfers it to a house maintained at 22oC. If the house is losing heat at a rate of 80,000 kJ/h, determine the minimum rate of heat supply to the heat engine required to keep the house at 22oC. 10. A Carnot cycle receives heat from a constant-temperature reservoir at 700 k and rejects heat to 5 kg of water which is initially saturated liquid at 100 kPa. As the engine operates, the water is heated at constant pressure until its temperature reaches 150oC. Calculate the work output of the Carnot engine. 35 Chapter 5: Entropy Clausius Inequality: This inequality is used to check that any cycle (engine or refrigerator) violates the second law of thermodynamic or not. • Consider a reversible power cycle (heat engines shown in Fig 5.1): and for • If the power cycle (heat engine) is irreversible and for the same QH: • Calusius Inequality: For all Cycles (Power/Ref/AC/H.P.) That is, the cyclic integral of ∆Q/T is always less than or equal to zero this inequality is valid for all cycles, reversible or irreversible. The symbol ∫ means that the integration is done over the entire cycle Example 5-1 A heat engine receives 600 KJ heat from a high-temperature source at 1000 K during a cycle. It converts 150 KJ of this heat to net work and rejects the remaining 450 KJ to a law of thermodynamics on the basis of (a) the Clausius inequality and (b) the Carnot principle. Solution (a) One way of the determining whether this cycle violates the second law is to check if it violates the Clausius inequality. A cycle that violates the Clausius inequality also violates the second law. By assuming that the temperature at location where heat is crossing the boundaries of the heat engine is equal to the temperature of the reservoirs, the cyclic integral of δ Q/T for the heat-engine cycle under consideration is determined to be 36 ∫ δQ T = QH QL 600 KJ 450 KJ − = − = −0.9 KJ K TH TL 1000 K 300 K Since the cyclic satisfied the Clausius inequality and thus the second law of thermodynamics. (b) Another way of determining whether this heat engine violates the second law is to check if the Carnot principle is satisfied. This is done by comparing the thermal efficiency of this heat engine to the thermal efficiency of a reversible heat engine, such as the Carnot engine, operating between the same temperature limits: η th = 1 − QL 450 KJ = 1− = 0.25 600 KJ QH η th.rev = 1 − TL 300 K = 1− = 0.70 1000 K TH or 25% or 70% Again, this heat engine is in complete compliance with the second law of thermodynamics since η th < η th.rev. Note that a cycle that violates the Clausius inequality will also violate the Carnot principle. Consider a closed system undergoing a reversible cyclic process (1-A-2-B-1) This means that the integration on the process or path A equals the integration on path B. A special case is the isothermal heat transfer process • • The values of the entropy (in kJ/kg.k) of pure substances such as steam, R-12……etc are obtained form tables. For ideal gases, Δs are obtained form some relation that are given in the next section. Form the above relations, we can write: or ΔQ = ∫Tds)rev δQ = T ds)rev. this means that the heat transfer during a process is represented by the area under the process curve on a T-S diagram 37 It has been shown that, for a reversible process For an irreversible process: The Entropy Increase Principle: The total energy change Δstotal or entropy generation Sgen and the entropy increase principle for any process is expressed as Sgen = Δstotal = Δssys + Δssurr ≥ 0 This principle is applicable to both closed (control mass) and open (control volume) system. It states that the total entropy change associated with a process must be positive or zero. The equality holds for reversible processes and the inequality for irreversible ones. Sine no actual process is truly reversible, we can conclude that the net entropy change for any process that takes place is positive, and therefore the entropy of the universe this means that the disorder or “mixed-upness” in the universe increases with the time. The entropy change of a system or its surroundings can be negative during a process, but their sum cannot. The entropy increase principle can be summarized as follows: The Entropy Increase Principle for a closed systems: Sgen = Δstotal = Δssys + Δssurr ≥ 0 Where Δssys = m (s2 –s1) = The difference between the final and initial entropies of the system. Δssurr = Qsurr / Tsurr = The entropy change of the surrounding caused by the heat transfer from the system. Qsurr = - Qsys For an adiabatic process Qsurr = 0 and Sgen = Δstotal = m (s2 – s1) ≥ 0 That is, the entropy of a closed system can never decrease during an adiabatic process. The entropy increase priniple for steady state Open system: At steady state conditions, the entropy change of a control volume is zero and Sgen =ΔSsurr = Se – Si + Qsurr ≥ο Tsurr Which can be expressed in the rate form as: Sgen = m(S e − S i ) + Or sgen = se – si + Qsurr ≥ 0 kW/K Tsurr q surr ≥ 0 KJ/kg.k Tsurr Thus the entropy of a fluid will increase as it flows through an adiabatic steady-flow device as a result of irreversibilities. 38 Causes of entropy change: The entropy of system changes because of - Heat transfer : if the heat is transferred from the system the entropy decreases and vise versa. - Irreversibility: • such as friction, fast expansion or compression. The entropy of a system can not decrease during an adiabatic process. If a process involves no heat transfer (adiabatic) and no irreversibility within the system (internally reversible), the entropy of a system must remain constant during that process. Such a process is called an internally reversible adiabatic or isentropic process. • The steady-flow devices deliver the highest work (such as turbines ) and consumes the minimum work (such as compressors and pumps ) when the process is reversible and adiabatic. Remarks an entropy: - Entropy can be viewed as a measure of molecular disorder, or molecular randomness. The concept of entropy as a measure of disorganized energy is useful in the calculation of the performance of engineering systems. The system that has high irreversibility’s gets greater entropy generation and becomes less efficient or becomes for away from the ideal system. - A process can occur in a certain direction only. This direction must give Δstotal ≥ 0. A process that violates this principle is impossible and cannot happen in nature. Entropy change of pure substances: For a pure substance such as steam & refrigerant – 12 the entropy values are obtained from tables and the entropy change during a process is simply the difference between the entropy values at the final and initial states: ΔS = m (S2 – S1) (kJ/k) (kJ/kg.k) or Δs = (s2 – s1) Isentropic process of pure substances: The entropy of a system will not change during an internally reversible, adiabatic process, which is called an isentropic process appears as a vertical line an a diagram. S2 = S1 (KJ/kg.k) Tds Equations 39 • Entropy change (ds) of solids and liquids Solids and liquids are incompressible since their volumes remain constant during a process. Thus dv = 0 for solids and liquids and hence. Where Carg is the average specific heat between T1 & T2. Entropy Change (ds) of an Ideal Gas For an ideal gas: • Ideal Gas Isentropic Process: ds = 0, i.e., s1 = s2 (Prove of Pvγ = c represent isentropic process). P T 0 = C p ln 2 − R ln 2 P1 T1 ∴ γ T R ln 2 γ −1 T1 T2 T1 P2 = P1 P = R ln 2 P1 γ γ −1 or P2 P1 v1 =40 v2 γ or P2 v 2γ = P1 v1γ = C Example 5.2: A steady-state compressor (Fig.5.2) changes air polytropically from 25 ◌C ْ and 1 bar to a final state of 4 bars. The content n = 1.3 for the process. Changes in kinetic and potential energies are negligible. If air is modeled as an ideal gas, determine (a) the exit temperature, in kelvins, and (b) the work and the heat transfer, both in kJ/kg, and (c) check the principle of entropy increase. Fig 5.2 Solution: (a) the final temperature is found from the polytropic-property relation (b) the work for a polytropic of air (R=287 J/kg.k) which is assumed as an ideal gas ( n −1) / n (1.3−1) / 1.3 P 4 T 2 = T 1 2 = 298 K = 410 K 1 P1 nR(T1 − T2 ) 1.3(287 )(410 − 298) kJ = = −139.3 kJ / kg wsf ,rev = (1.3 − 1) n −1 kg The heat transfer is found from the steady-state energy balance Δke = Δpe =0. 0 = q + w + h 1 – h2 Solving for q, the energy equation reduces to q =h2 – h1 –w = Cp (T2 - T1 ) = 1.005 ( 410 – 298) – 139.3 = -26.4 kJ/kg This means that heat is transferred from the compressor to the surroundings or q) surrounding = 26.9 KJ/kg (positive quantity). (c) Entropy = m se − m i si + σ = se − si + Q T surrounding q T surrounding T2 T1 P q − R ln 2 + P1 T surrounding 296 440 σ = 1.035 ln − 0.287 ln (4 ) + (273 + 25) 298 σ = 0.33 − 0.398 + 0.09 = 0.022 KJ kg. K σ = Cp ln As thus σ is positive which means that the entropy increase principle is satisfied. Example 5.3: One kilogram if liquid water is heated from 20 to 90 ºC, Calculate the entropy change Solution: Assume that water is an in compressible fluid T 363.2 ∆ = s2 − s1 = C ln 2 = 4.18 ln 293.2 T1 = 0.8949 KJ Kg . K = 0.8949 41 Problems on entropy A frictionless piston-cylinder device contains a saturated mixture of water at 100oC. During a constant-pressure process, 600 kJ of heat is transferred to the surrounding air which is at 25oC. As a result, part of the water vapor contained in the piston-cylinder device condenses. Determine (a) the entropy change of the water, (b) the entropy change of the surrounding air during the process, and (c) whether this process is reversible, irreversible, or impossible. 2. Steam at 7 MPa and 450oC is throttled in a valve to a pressure of 3 MPa during a steady-flow process. Determine the entropy generation for this process, and check if the increase-in-entropy principle is satisfied. 3. Water at 137.9 kPa and 10oC enters a mixing chamber at a rate of 2.268 kg/s, where it is mixed steadily with steam entering at 137.9 kPa and 115oC. The mixture leaves the chamber at 137.9 kPa and 130oC, and heat is lost to the surrounding air at 21oC at a rate of 3.16 kW. Neglecting the changes in kinetic and potential energies, determine the rate of entropy generation for this process. 4. In a system which undergoes an irreversible process, the work done is 5 kJ and the heat rejected is 7 kJ. The change in entropy is: (a) Positive (b) Negative (c) Zero (d) Can't tell 5. 2 kg of water at 65oC are mixed with 5 kg of water at 40oC in an adiabatic process. Calculate the change in entropy of the universe. 6. 10 kg of air at 350 kPa and 420 K are mixed with 5 kg of air at 200 kPa and 530 K. The mixing takes place in an adiabatic chamber. Calculate the change in entropy if the final pressure is 170 kPa. 7. A Carnot cycle operates between temperature limits of 900 and 300 K. The heat supplied at the high temperature is 4.0 MJ. Calculate the change in entropy during the heat-addition and rejection processes, the heat rejection and the work output. 8. A steady-flow compressor operates in an adiabatic manner to compress air from 1 atm, 20oC, to 5 atm, 210oC. Calculate: (a) The work required per kilogram of air. (b) The change in entropy for the process. 9. Steam expands in a turbine steadily at a rate of 25.000 kg/h, entering at 8 MPa and 450 ºC and leaving at 50 kPa as saturated vapor. If the power generation for this process. Assume the surrounding medium is at 25 ºC. 10. Air is campressed steadily by a 5 – kw compressor from 100 kPa and 17 ºC to 6000 kPa and 167 ºC at a rate of 16 kg/min. During this process. Some heat transfer takes place between the compressor and the surrounding medium at 17 ºC. Detrmine (a) the rate of entropy change of air, (b) the rate of entropy change of the surrounding medium (c) the rate of entropy generation during this process. 11. A mass- and atmosphere-loaded piston/cylinder contains 2 kg of water at 5 Mpa, 100 ºC. Heat is added from a reservoir at 700 ºC to the water until it reaches 700 ºC. Find the work, heat transfer, and total entropy production for the system and surroundings. 12. A rigid tank contains 2 kg of air at 200 kPa and ambient temperature, 20 ºC. An electric current now passes through a resistor inside the tank. After a total of 100 kJ of electrical work has crossed the boundary, the air temperature inside is 70 ºC. Is this possible? 13. An inventor proposes to build a non-flow device that will compress air isothermally at a temperature of 400 ºC from a pressure of 4.0 MPa to a final pressure of 15.0 MPa. If the device requires 150 kJ/kg of work input, is it reversible, irreversible, or impossible? 1. 42 Gas Power Cycles Our study of gas power cycles will involve the study of those heat engines in which the working fluid remains in the gaseous state throughout the cycle. We often study the ideal cycle in which internal irreversibilities and complexities (the actual intake of air and fuel, the actual combustion process, and the exhaust of products of combustion among others) are removed. We will be concerned with how the major parameters of the cycle affect the performance of heat engines. The performance is often measured in terms of the cycle efficiency. Air-Standard Assumptions In our study of gas power cycles, we assume that the working fluid is air, and the air undergoes a thermodynamic cycle even though the working fluid in the actual power system does not undergo a cycle. To simplify the analysis, we approximate the cycles with the following assumptions: • The air continuously circulates in a closed loop and always behaves as an ideal gas. • All the processes that make up the cycle are internally reversible. • The combustion process is replaced by a heat-addition process from an external source. • A heat rejection process that restores the working fluid to its initial state replaces the exhaust process. • The cold-air-standard assumptions apply when the working fluid is air and has constant specific heat evaluated at room temperature (25oC or 77oF). Terminology for Reciprocating Devices The following is some terminology we need to understand for reciprocating engines—typically pistoncylinder devices. Let’s look at the following figures for the definitions of top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure. The compression ratio r of an engine is the ratio of the maximum volume to the minimum volume formed in the cylinder. The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same mount of net work as that produced during the actual cycle. 43 Otto Cycle: The Ideal Cycle for SparkIgnition Engines Consider the automotive spark-ignition power cycle with the following processes: - Intake stroke - Compression stroke - Power (expansion) stroke - Exhaust stroke Often the ignition and combustion process begins before the completion of the compression stroke. The number of crank angle degrees before the piston reaches TDC on the number one piston at which the spark occurs is called the engine timing. What are the compression ratio and timing of your engine in your car, truck, or motorcycle? The air-standard Otto cycle is the ideal cycle that approximates the spark ignition combustion engine with the following processes: 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection Thermal Efficiency of the Otto cycle: Apply first law closed system to process 2-3, V = constant. 44 Thus, for constant specific heats, Apply first law closed system to process 4-1, V = constant. Thus, for constant specific heats, The thermal efficiency becomes Recall processes 1-2 and 3-4 are isentropic, so Since V3 = V2 and V4 = V1, we see that Is this the same as the Carnot cycle efficiency? Since process 1-2 is isentropic, where the compression ratio is r = V1/V2 and 45 We see that increasing the compression ratio increases the thermal efficiency. However, there is a limit on r depending upon the fuel. Fuels under high temperature resulting from high compression ratios will prematurely ignite, causing knock. Example 8-1 An Otto cycle having a compression ratio of 9:1 uses air as the working fluid. Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added. Determine all T's, P's, th, the back work ratio, and the mean effective pressure. Process Diagrams: Review the P-v and T-s diagrams given above for the Otto cycle. Assume constant specific heats with C v = 0.718 kJ/kg ⊕K, k = 1.4. (Use the 300 K data from Table A-2) Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9, The first law closed system for process 2-3 was shown to reduce to (your homework solutions must be complete; that is, develop your equations from the application of the first law for each process as we did in obtaining the Otto cycle efficiency equation) 46 Then, Using the combined gas law Process 3-4 is isentropic; therefore, Process 4-1 is constant volume. So the first law for the closed system gives, on a mass basis, For the cycle, u = 0, and the first law gives 47 The thermal efficiency is The mean effective pressure is The back work ratio is (can you show that this is true?) Air-Standard Diesel Cycle The air-standard Diesel cycle is the ideal cycle that approximates the Diesel combustion engine Process Description 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection The P-v and T-s diagrams are: 48 Thermal efficiency of the Diesel cycle Now to find Qin and Qout. Apply the first law closed system to process 2-3, P = constant. Thus, for constant specific heats Apply the first law closed system to process 4-1, V = constant (just like the Otto cycle) Thus, for constant specific heats The thermal efficiency becomes 49 where rc is called the cutoff ratio, defined as V3 /V2, and is a measure of the duration of the heat addition at constant pressure. Since the fuel is injected directly into the cylinder, the cutoff ratio can be related to the number of degrees that the crank rotated during the fuel injection into the cylinder. Recall processes 1-2 and 3-4 are isentropic, so Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain Therefore, What happens as rc goes to 1? Sketch the P-v diagram for the Diesel cycle and show rc approaching1 in the limit. 50 51 Brayton Cycle The Brayton cycle is the air-standard ideal cycle approximation for the gasturbine engine. This cycle differs from the Otto and Diesel cycles in that the processes making the cycle occur in open systems or control volumes. Therefore, an open system, steady-flow analysis is used to determine the heat transfer and work for the cycle. We assume the working fluid is air and the specific heats are constant and will consider both open and closed gas-turbine cycles. The closed cycle gas-turbine engine Process Description 1-2 Isentropic compression (in a compressor) 2-3 Constant pressure heat addition 3-4 Isentropic expansion (in a turbine) 4-1 Constant pressure heat rejection The T-s and P-v diagrams are Thermal efficiency of the Brayton cycle Now to find Qin and Qout. Apply the conservation of energy to process 2-3 for P = constant (no work), steady-flow, and neglect changes in kinetic and potential energies. For constant specific heats, the heat added per unit mass flow is 52 The conservation of energy for process 4-1 yields for constant specific heats (let’s take a minute for you to get the following result) The thermal efficiency becomes Recall processes 1-2 and 3-4 are isentropic, so Since P3 = P2 and P4 = P1, we see that The Brayton cycle efficiency becomes Is this the same as the Carnot cycle efficiency? Since process 1-2 is isentropic, 53 where the pressure ratio is rp = P2/P1 and Extra Assignment Evaluate the Brayton cycle efficiency by determining the net work directly from the turbine work and the compressor work. Compare your result with the above expression. Note that this approach does not require the closed cycle assumption. Example 8-2 The ideal air-standard Brayton cycle operates with air entering the compressor at 95 kPa, 22oC. The pressure ratio r p is 6:1 and the air leaves the heat addition process at 1100 K. Determine the compressor work and the turbine work per unit mass flow, the cycle efficiency, the back work ratio, and compare the compressor exit temperature to the turbine exit temperature. Assume constant properties. Apply the conservation of energy for steady-flow and neglect changes in kinetic and potential energies to process 1-2 for the compressor. Note that the compressor is isentropic. The conservation of mass gives For constant specific heats, the compressor work per unit mass flow is Since the compressor is isentropic 54 The conservation of energy for the turbine, process 3-4, yields for constant specific heats (let’s take a minute for you to get the following result) Since process 3-4 is isentropic 55 We have already shown the heat supplied to the cycle per unit mass flow in process 2-3 is The cycle efficiency becomes The back work ratio is defined as Note that T4 = 659.1 K > T2 = 492.5 K, or the turbine outlet temperature is greater than the compressor exit temperature. Can this result be used to improve the cycle efficiency? What happens to th, win /wout, and wnet as the pressure ratio r p is increased? Let's take a closer look at the effect of the pressure ratio on the net work done. 56 Note that the net work is zero when For fixed T3 and T1, the pressure ratio that makes the work a maximum is obtained from: This is easier to do if we let X = r p (k-1)/k Solving for X Then, the r p that makes the work a maximum for the constant property case and fixed T3 and T1 is For the ideal Brayton cycle, show that the following results are true. When rp = rp, max work, T4 = T2 When rp < rp, max work, T4 > T2 When rp > rp, max work, T4 < T2 Regenerative Brayton Cycle For the Brayton cycle, the turbine exhaust temperature is greater than the compressor exit temperature. Therefore, a heat exchanger can be placed between the hot gases leaving the turbine and the cooler gases leaving the compressor. This heat exchanger is called a regenerator or recuperator. The sketch of the regenerative Brayton cycle is shown below. 57 We define the regenerator effectiveness ∑regen as the ratio of the heat transferred to the compressor gases in the regenerator to the maximum possible heat transfer to the compressor gases. For ideal gases using the cold-air-standard assumption with constant specific heats, the regenerator effectiveness becomes Using the closed cycle analysis and treating the heat addition and heat rejection as steady-flow processes, the regenerative cycle thermal efficiency is Notice that the heat transfer occurring within the regenerator is not included in the efficiency calculation because this energy is not a heat transfer across the cycle boundary. Assuming an ideal regenerator ∑regen = 1 and constant specific heats, the thermal efficiency becomes (take the time to show this on your own) 58 When does the efficiency of the air-standard Brayton cycle equal the efficiency of the air-standard regenerative Brayton cycle? If we set th, Brayton = th, regen then Recall that this is the pressure ratio that maximizes the net work for the simple Brayton cycle and makes T4 = T2. What happens if the regenerative Brayton cycle operates at a pressure ratio larger than this value? For fixed T3 and T1, pressure ratios greater than this value cause T4 to be less than T2, and the regenerator is not effective. What happens to the net work when a regenerator is added? What happens to the heat supplied when a regenerator is added? The following shows a plot of the regenerative Brayton cycle efficiency as a function of the pressure ratio and minimum to maximum temperature ratio, T1/T3. Example 8-3: Regenerative Brayton Cycle Air enters the compressor of a regenerative gas-turbine engine at 100 kPa and 300 K and is compressed to 800 kPa. The regenerator has an effectiveness of 65 percent, and the air enters the turbine at 1200 K. For a compressor efficiency of 75 percent and a turbine efficiency of 86 percent, determine (a) The heat transfer in the regenerator. (b) The back work ratio. (c) The cycle thermal efficiency. Compare the results for the above cycle with the ones listed below that have the same common data as required. (a) The actual cycle with no regeneration, ∑ = 0. (b) The actual cycle with ideal regeneration, ∑ = 1.0. (b) The ideal cycle with regeneration, ∑ = 0.65. (d) The ideal cycle with no regeneration, ∑ = 0. (e) The ideal cycle with ideal regeneration, ∑ = 1.0. We assume air is an ideal gas with constant specific heats, that is, we use the cold-air-standard assumption. Summary of Results 59 Compressor analysis The isentropic temperature at compressor exit is To find the actual temperature at compressor exit, T2a, we apply the compressor efficiency Since the compressor is adiabatic and has steady-flow Turbine analysis The conservation of energy for the turbine, process 3-4, yields for constant specific heats (let’s take a minute for you to get the following result) Since P3 = P2 and P4 = P1, we can find the isentropic temperature at the turbine exit. 60 To find the actual temperature at turbine exit, T4a, we apply the turbine efficiency. The turbine work becomes The back work ratio is defined as Regenerator analysis To find T5, we apply the regenerator effectiveness. To find the heat transferred from the turbine exhaust gas to the compressor exit gas, apply the steady-flow conservation of energy to the compressor gas side of the regenerator. 61 Using qregen, we can determine the turbine exhaust gas temperature at the regenerator exit. Heat supplied to cycle Apply the steady-flow conservation of energy to the heat exchanger for process 5-3. We obtain a result similar to that for the simple Brayton cycle. Cycle thermal efficiency The net work done by the cycle is 62 The cycle efficiency becomes You are encouraged to complete the calculations for the other values found in the summary table. Other Ways to Improve Brayton Cycle Performance Intercooling and reheating are two important ways to improve the performance of the Brayton cycle with regeneration. Intercooling When using multistage compression, cooling the working fluid between the stages will reduce the amount of compressor work required. The compressor work is reduced because cooling the working fluid reduces the average specific volume of the fluid and thus reduces the amount of work on the fluid to achieve the given pressure rise. To determine the intermediate pressure at which intercooling should take place to minimize the compressor work, we follow the standard approach. For the adiabatic, steady-flow compression process, the work input to the compressor per unit mass is 63 Can you obtain this relation another way? Hint: apply the first law to processes 1-4. For two-stage compression, let’s assume that intercooling takes place at constant pressure and the gases can be cooled to the inlet temperature for the compressor, such that P3 = P2 and T3 = T1. The total work supplied to the compressor becomes ] To find the unknown pressure P2 that gives the minimum work input for fixed compressor inlet conditions T1, P1, and exit pressure P4, we get or, the pressure ratios across the two compressors are equal. Intercooling is almost always used with regeneration. During intercooling the compressor exit temperature is reduced; therefore, more heat must be supplied in the heat addition process. Regeneration can make up part of the required heat transfer. 64 To supply only compressed air, using intercooling requires less work input. The next time you go to a home supply store where air compressors are sold, check the larger air compressors to see if intercooling is used. For the larger air compressors, the compressors are made of two piston-cylinder chambers. The intercooling heat exchanger may be only a pipe with a few attached fins that connects the large piston-cylinder chamber with the smaller pistoncylinder chamber. Extra Assignment Obtain the expression for the compressor total work by applying conservation of energy directly to the lowand high-pressure compressors. Reheating When using multistage expansion through two or more turbines, reheating between stages will increase the net work done (it also increases the required heat input). The regenerative Brayton cycle with reheating is shown above. The optimum intermediate pressure for reheating is the one that maximizes the turbine work. Following the development given above for intercooling and assuming reheating to the high-pressure turbine inlet temperature in a constant pressure steady-flow process, we can show the optimum reheat pressure to be or the pressure ratios across the two turbines are equal. Vapor and Combined Power Cycles We consider power cycles where the working fluid undergoes a phase change. The best example of this cycle is the steam power cycle where water (steam) is the working fluid. The heat engine may be composed of the following components. Steam Power Cycle The working fluid, steam (water), undergoes a thermodynamic cycle from 1-2-3-4-1. The cycle is shown on the following T-s diagram. Carnot Vapor Cycle Using Steam The thermal efficiency of this cycle is given as 65 Note the effect of TH and TL on th, Carnot. : The larger the TH the larger the th, Carnot and the smaller the TL the larger the th, Carnot To increase the thermal efficiency in any power cycle, we try to increase the maximum temperature at which heat is added. Reasons why the Carnot cycle is not used: 1− Pumping process 1-2 requires the pumping of a mixture of saturated liquid and saturated vapor at state 1 and the delivery of a saturated liquid at state 2. 2− To superheat the steam to take advantage of a higher temperature, elaborate controls are required to keep TH constant while the steam expands and does work. To resolve the difficulties associated with the Carnot cycle, the Rankine cycle was devised. Rankine Cycle The simple Rankine cycle has the same component layout as the Carnot cycle shown above. The simple Rankine cycle continues the condensation process 4-1 until the saturated liquid line is reached. Ideal Rankine Cycle Processes Process Description 1-2 Isentropic compression in pump 2-3 Constant pressure heat addition in boiler 3-4 Isentropic expansion in turbine 4-1 Constant pressure heat rejection in condenser The T-s diagram for the Rankine cycle is given below. Locate the processes for heat transfer and work on the diagram. Example 9-1 Compute the thermal efficiency of an ideal Rankine cycle for which steam leaves the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa. We use the power system and T-s diagram shown above. P2 = P3 = 6 MPa = 6000 kPa T3 = 350oC P1 = P4 = 10 kPa Pump The pump work is obtained from the conservation of mass and energy for steady-flow but neglecting potential and kinetic energy changes and assuming the pump is adiabatic and reversible. Since the pumping process involves an incompressible liquid, state 2 is in the compressed liquid region, we use a second method to find the pump work or the h across the pump. Recall the property relation: Since the ideal pumping process 1-2 is isentropic, ds = 0. 66 The incompressible liquid assumption allows The pump work is calculated from Using the steam tables Now, h2 is found from Boiler To find the heat supplied in the boiler, we apply the steady-flow conservation of mass and energy to the boiler. If we neglect the potential and kinetic energies, and note that no work is done on the steam in the boiler, then We find the properties at state 3 from the superheated tables as 67 The heat transfer per unit mass is Turbine The turbine work is obtained from the application of the conservation of mass and energy for steady flow. We assume the process is adiabatic and reversible and neglect changes in kinetic and potential energies. We find the properties at state 4 from the steam tables by noting s4 = s3 and asking three questions. The turbine work per unit mass is 68 The net work done by the cycle is The thermal efficiency is Ways to improve the simple Rankine cycle efficiency: Superheat the vapor Average temperature is higher during heat addition. Moisture is reduced at turbine exit (we want x4 in the above example > 85 percent). Increase boiler pressure (for fixed maximum temperature) Availability of steam is higher at higher pressures. Moisture is increased at turbine exit. Lower condenser pressure Less energy is lost to surroundings. Moisture is increased at turbine exit. Extra Assignment For the above example, find the heat rejected by the cycle and evaluate the thermal efficiency from 69 Reheat Cycle As the boiler pressure is increased in the simple Rankine cycle, not only does the thermal efficiency increase, but also the turbine exit moisture increases. The reheat cycle allows the use of higher boiler pressures and provides a means to keep the turbine exit moisture (x > 0.85 to 0.90) at an acceptable level. The thermal efficiency is given by Example 9-2 Compare the thermal efficiency and turbine-exit quality at the condenser pressure for a simple Rankine cycle and the reheat cycle when the boiler pressure is 4 MPa, the boiler exit temperature is 400oC, and the condenser pressure is 10 kPa. The reheat takes place at 0.4 MPa and the steam leaves the reheater at 400oC. 70 Regenerative Cycle To improve the cycle thermal efficiency, the average temperature at which heat is added must be increased. One way to do this is to allow the steam leaving the boiler to expand the steam in the turbine to an intermediate pressure. A portion of the steam is extracted from the turbine and sent to a regenerative heater to preheat the condensate before entering the boiler. This approach increases the average temperature at which heat is added in the boiler. However, this reduces the mass of steam expanding in the lowerpressure stages of the turbine, and, thus, the total work done by the turbine. The work that is done is done more efficiently. The preheating of the condensate is done in a combination of open and closed heaters. In the open feedwater heater, the extracted steam and the condensate are physically mixed. In the closed feedwater heater, the extracted steam and the condensate are not mixed. Cycle with an open feedwater heater Rankine Steam Power Cycle with an Open Feedwater Heater Rankine Steam Power Cycle with an Open Feedwater Heater Cycle with a closed feedwater heater with steam trap to condenser Rankine Steam Power Cycle with a Closed Feedwater Heater Cycle with a closed feedwater heater with pump to boiler pressure 71 Consider the regenerative cycle with the open feedwater heater. To find the fraction of mass to be extracted from the turbine, apply the first law to the feedwater heater and assume, in the ideal case, that the water leaves the feedwater heater as a saturated liquid. (In the case of the closed feedwater heater, the feedwater leaves the heater at a temperature equal to the saturation temperature at the extraction pressure.) Conservation of mass for the open feedwater heater: Let y m m = ! / ! 6 5 be the fraction of mass extracted from the turbine for the feedwater heater. Conservation of energy for the open feedwater heater: Example 9-3 An ideal regenerative steam power cycle operates so that steam enters the turbine at 3 MPa, 500oC, and exhausts at 10 kPa. A single open feedwater heater is used and operates at 0.5 MPa. Compute the cycle thermal efficiency. Using the software package the following data are obtained. 72 The work for pump 1 is calculated from Now, h2 is found from The fraction of mass extracted from the turbine for the open feedwater heater is obtained from the energy balance on the open feedwater heater, as shown above. This means that for each kg of steam entering the turbine, 0.163 kg extracted for the feedwater heater. The work for pump 2 is calculated from Now, h4 is found from the energy balance for the pump. 73 Apply the steady-flow conservation of energy to the isentropic turbine. The net work done by the cycle is Apply the steady-flow conservation of mass and energy to the boiler. The heat transfer per unit mass entering the turbine at the high pressure, state 5, is 74 The thermal efficiency is If these data were used for a Rankine cycle with no regeneration, then th = 35.6 percent. Thus, the one open feedwater heater operating at 0.5 MPa increased the thermal efficiency by 5.3 percent. However, note that the mass flowing through the lower-pressure stages has been reduced by the amount extracted for the feedwater and the net work output for the regenerative cycle is about 10 percent lower than the standard Rankine cycle. Below is a plot of cycle thermal efficiency versus the open feedwater heater pressure. The feedwater heater pressure that makes the cycle thermal efficiency a maximum is about 400 kPa. Below is a plot of cycle net work per unit mass flow at state 5 and the fraction of mass y extracted for the feedwater heater versus the open feedwater heater pressure. Clearly the net cycle work decreases and the fraction of mass extracted increases with increasing extraction pressure. Why does the fraction of mass extracted increase with increasing extraction pressure? 75 Placement of Feedwater Heaters The extraction pressures for multiple feedwater heaters are chosen to maximize the cycle efficiency. As a rule of thumb, the extraction pressures for the feedwater heaters are chosen such that the saturation temperature difference between each component is about the same. Example 9-4 An ideal regenerative steam power cycle operates so that steam enters the turbine at 3 MPa, 500oC, and exhausts at 10 kPa. Two closed feedwater heaters are to be used. Select starting values for the feedwater heater extraction pressures. Deviation from Actual Cycles Piping losses--frictional effects reduce the available energy content of the steam. Turbine losses--turbine isentropic (or adiabatic) efficiency. The actual enthalpy at the turbine exit (needed for the energy analysis of the next component) is 76 Pump losses--pump isentropic (or adiabatic) efficiency. The actual enthalpy at the pump exit (needed for the energy analysis of the next component) is Condenser losses--relatively small losses that result from cooling the condensate below the saturation temperature in the condenser. 77 Problems on power cycles ----------------------------------------------------------------------------------------1. An ideal Otto cycle has a compression ratio of 8 . At the beginning of the compression process, air is at 95 kPa and 27 oC, and 750 kJ / kg of heat is transferred to air during the constant volume heat addition process. Determine (a) the pressure and temperature at the end of the heat addition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle. Answers: (a) 3898, 1 kPa, 1538 K, (b) 392 kJ/kg, (c) 52%, (d) 495 kPa 2. A four – cylinder spark – ignition engine has a compression ratio of 8, and each cylinder has a maximum volume of 0.6 L. At the beginning of the compression process, the air is at 98 kPa and 17 oC, and the maximum temperature in the cycle is 1800 K. Assuming the engine to operate on the ideal Otto cycle, determine (a) the amount of heat supplied per cylinder, (b) the thermal efficiency, and 9 (c) the number of revolutions per minute required for a net power output of 80 kW. 3. The compression ratio of an air – standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is at 100 kPa, 17 oC, and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K Determine (a) the highest temperature and pressure in the cycle, (b) the amount of heat added in kJ, (c) the thermal efficiency, and (d) the mean effective pressure. Answers: (a) 1987 K, 649 kPa, (b) 0.65 kJ/kg, (c) 59%, (d) 719 kPa 4. An air – standard Diesel cycle has a compression ratio of 18 .2 Air is at 27 oC and 0.1 MPa at the beginning of the compression process and at 2000 K at the end of the heat addition process. Accounting for the variation of specific heats with temperature, determine (a) the cutoff ratio (b)the heat rejection per unit mass, and (c) the thermal efficiency. 5. An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kPa and 20 oC. If the maximum temperature in the cycle is not to exceed 2200 K, determine (a) the thermal efficiency and (b) the mean effective pressure. Assume constant heats for air at room temperature. Answers: (a) 63.5 percent (b) 933 kPa. 6. An ideal dual cycle has a compression ratio of 12 and uses air as the working fluid. At the beginning of the compression process, air is 100 kPa and 30 oC and occupies a volume of 1.2 L. During the heat addition process, 0.3 kJ of heat is transferred to air at constant volume and 1.1 kJ at constant pressure. Determine the thermal efficiency of the cycle. 7. The compression ratio of an ideal dual cycle is 14. Air is at 100 kPa and 300 K at the beginning of the compression process and at 2200 K at the end of the heat addition process. Heat transfer to air takes place partly at constant volume and partly at constant pressure, and it amounts to 1520.4 kJ/kg. Determine (a) the fraction of heat transferred at constant volume and (b) the thermal efficiency of the cycle. 8. A simple Brayton cycle using air as the working fluid has a pressure ratio of 8. The minimum and maximum temperatures in the cycle are 310 and 1160 K. Assuming an adiabatic efficiency of 75 percent for the compressor and 82 percent for the turbine, determine (a) the air temperature at the turbine exit, (b) the net work output, and (c) the thermal efficiency. 9. Air enters the compressor of a gas – turbine engine at 300 K and 100 kPa, where it is compressed to 700 kPa and 580 K. Heat is transferred to air in the amount of 950 kJ/kg before it enters the turbine. For a turbine to efficiency of 86 percent, determine (a) the fraction of the turbine work output used to drive the compressor and (b) the thermal efficiency. Answers: (a) 64.7 percent, (b) 16.4 percent. 10. A gas turbine power plant operates on the simple Brayton cycle with air as the working fluid and delivers 15 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K and the pressure of air at the compressor exit is 8 times the value at the compressor inlet. Assuming an adiabatic efficiency of 80 percent for the compressor and 86 percent for the turbine, determine the mass flow rate of air through the cycle. 11. A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 8. The minimum and maximum temperatures in the cycle are 310 and 1150 K. Assuming an adiabatic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator; determine (a) the air temperature at the turbine exit, (b) the net work output, and (c) the thermal efficiency. Answers: (a) 763.07 kg/kg (b) 101.64 kg/kg, (c) 21.0 percent. 12. Air enters the compressor of a regenerative gas – turbine engine at 300 k and 100 kPa where it is compressed to 800 kPa and 580 K The regenerator has an effectiveness of 65 percent, and the air enters the turbine at 1200 K For a turbine efficiency of 86 percent, determine (a) the amount of heat transfer in the regenerator and (b) the thermal efficiency. Answers: (a) 137.7 kJ/kg (b) 35.0 percent. 13. Consider an ideal gas turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each stage of compressor and turbine is 3. The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K. Determine the back work ratio and the thermal efficiency of the cycle, assuming (a) no regenerator is used and (b) a regenerator with 75 percent effectiveness is used. 78 14. Consider a regenerative gas – turbine power plant with two stages of compression and two stages of expansion. The overall pressure ratio of the cycle is 9. The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K. Determine the minimum mass flow rate of air needed to develop a net power output of 50 MW. Answer: 113.4 kg/s. 15. A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3 MPa and 50 kPa. The temperature of the steam at the turbine inlet is 400 oC, and the mass flow rate of steam through the cycle is 40 kg/s. Show the cycle on a T– s diagram with respect to saturation lines, and determine (a) the thermal efficiency of the cycle and (b) the net power output of the power plant. 16. Steam enters the turbine of a steam power plant which operates on a simple ideal Rankine cycle at a pressure of 6 MPa, and it leaves as a saturated vapor at 7.5 kPa. Heat is transferred to the steam in the boiler at a rate of 105 kJ/s. Steam is cooled in the condenser by the cooling water from a nearby river which enters the condenser at 18 oC. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the turbine inlet temperature (b) the net power output and the thermal efficiency and (c) the minimum mass flow rate of the cooling water required. 17. A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 9 MPa and 10 kPa. The mass flow rate of steam through the cycle is 25 kg/s. The moisture content of the steam at the turbine exit is not to exceed 10 percent. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the minimum turbine inlet temperature, (b) the rate of heat input in the boiler (c) the thermal efficiency of the cycle. 18. A steam power plant operates on the ideal reheat Rankine cycle. Steam enters the high pressure turbine at 8 MPa and 500 oC and leaves at 3 MPa. Steam is then reheated at constant pressure to 500 oC before it expands to 20 kPa in the low pressure turbine. Determine the turbine work output, in kJ/kg, and the thermal efficiency of the cycle. Also show the cycle on a T-s diagram with respect to saturation lines. 19. Steam enters the high – pressure turbine of a steam power plant which operates on the ideal reheat Rankine cycle at 6 MPa and 450 oC and leaves as saturated vapor. Steam is then reheated to 400 oC before it expands to a pressure of 7.5 kPa. Heat is transferred to the steam in the boiler at a rate of 10 kJ/s. Steam is cooled in the condenser by the cooling water from a nearby river, which enters the condenser at 18 oC. Show the cycle on T-s diagram with respect to saturation lines and determine (a) the pressure at which reheating takes place (b) the net power output and thermal efficiency and (c) the minimum mass flow rate of the cooling water required. 20. A steam power plant operates on an ideal regenerative Rankine cycle. Steam enters the turbine at 6 MPa and 450 oC and is condensed in the condenser at 20 kPa. Steam is extracted from the turbine at 0.4 MPa to heat the feed water in an open feed water heater. Water leaves the feed water heater as a saturated liquid. Show the cycle on a T-s diagram and determine (a) the net work output per kilogram of steam flowing through the boiler and (b) the thermal efficiency of the cycle. Answers: (a) 1016 kJ/kg (b) 37.8 percent. 21. A steam power plant operates on an ideal reheat regenerative Rankine cycle and has a net power output of 80 MW. Steam enters the high pressure turbine at 10 MPa and 550 oC and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the feed water in an open feed water heater. The rest of the steam is reheated to 500 oC and is expanded in the low pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through the boiler and (b) the thermal efficiency of the cycle. Answers: (a) 54.56 kg / s (b) 44.4 percent. 79

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