# Chapter 1 - UniMasr.com • Home

```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.
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.
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?
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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
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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.
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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.
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.
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?
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.
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
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
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
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.
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
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.
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.
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.
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.
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.
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.
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
∴WT ,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.
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.
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.
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.
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.
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
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
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
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
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)
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
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
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?
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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
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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.
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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.
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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.
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