Midterm review

Mid-Term Review
Topics
1) Energy Analysis of Closed System (Lecture 4, Chap. 4
of textbook).
2) Mass and Energy Analysis Of Control Volumes
(Lecture 5, Chap. 5 of textbook).
3) Second Law of Thermodynamics (Lecture 6, Chap. 6
of textbook).
1

2
1) Energy Analysis of Closed System
Know how to interpret condition at final or initial
state, given the nature of the process; e.g. constant
pressure process, isothermal process, process inside a
closed rigid tank, process inside a piston-cylinder device.
Know how to obtain boundary work wb for the constant
pressure process, the isothermal process and the
polytropic process.
Know and understand the various forms of the energy
balance equation for a closed system.

3
Examples (Energy Analysis of Closed System)
1. An insulated piston–cylinder device contains 5 L of saturated
liquid water at a constant pressure of 175 kPa. Water is stirred by
a paddle wheel while a current of 8 A flows for 45 min through a
resistor placed in the water.
If one-half of the liquid is evaporated during this constant
pressure process and the paddle-wheel work amounts to 400
kJ, determine the voltage of the source. Also, show the process on
a P-v diagram.

4
Solution
We take the contents of the cylinder as the system. This is a
closed system since no mass enters or leaves. The energy
balance for this stationary closed system can be expressed as
since ΔU + Wb = ΔH during a constant pressure quasi-
equilibrium process.

5
The properties of water are
Substituting,

6
2. A mass of 5 kg of saturated liquid–vapor mixture of water is
contained in a piston–cylinder device at 150 kPa. Initially, half of
the water is in the liquid phase and the rest is in the vapor phase.
Heat is now transferred to the water, and the piston, which is
resting on a set of stops, starts moving when the pressure inside
reaches 300 kPa. Heat transfer continues until the total volume
increases by 20 percent. Determine (a) the initial and final
temperatures, (b) the mass of liquid water when the piston first
starts moving, and (c) the work done during this process.
Also, show the process on a P-v diagram.

7
Solution
Initially the system is a saturated mixture at 150 kPa pressure, and
thus the initial temperature is
Ti = Tsat@150 kPa = 111.35oC
The specific volumes at 150 kPa are vf = 0.001053 m3/kg and vg =
1.1594 m3/kg; thus total initial volume is
V1 = mf vf + mg vg
= (2.5 kg)( 0.001053 m3/kg) + (2.5 kg)(1.1594 m3/kg)
= 2.901 m3
Then the total and specific volumes at the final state are
V3 = 1.2 V1 = 1.2 2.901 = 3.481 m3
v3 = V3/m = 3.481 m3 / 5 kg = 0.69627 m3/k
Thus at the final state, P3 = 300 kPa, v3 = 0.69627 m3/kg.

8
At 300 kPa, vg = 0.60582 m3/kg. Hence since v3 > vg@300 KPa, state
3 is superheated vapour. Interpolating from Table A-6
(b) When the piston first starts moving, P2 = 300 kPa and V2 = V1
= 2.901 m3. The specific volume at this state is
V2 = V2/m = 2.901 m3 / 5 kg = 0.5802 m3/kg
which is less than vg = 0.60582 m3/kg at 300 kPa. Thus the quality
at state 2 is
Thus mass of the liquid at state 2 = (5 kg)(1 0.9576) = 0.212 kg.
3
3
150 0.69627 0.63402 0.69627 0.63402
150 200 150 187.8 C
200 150 0.71643 0.63402 0.71643 0.63402
T
T 
2 @300 kPa
2
@300 kPa
0.5802 0.001073
0.9576
0.60582 0.001073
f
fg
v v
x
v

9
(c) No work is done during process 1-2 since V1 = V2. The
pressure remains constant during process 2-3 and the work done
during this process is
3 3
2 3 2 300 kPa 3.481 2.901 m 206.4 kPa m 174 kJbW P V V

10
Example
Steam enters a turbine operating at steady state with a mass flow
rate of 4600 kg/h. The turbine develops a power output of 1000
kW. At the inlet, the pressure is 60 bar, the temperature is
400C, and the velocity is 10 m/s. At the exit, the pressure is 0.1
bar, the quality is 0.9 (90%), and the velocity is 50 m/s.
Calculate the rate of heat transfer between the turbine and
surroundings, in kW.

11
(assuming net heat transfer out of
the system and pe = 0.)

12
From the tables and using the usual procedures we obtain
h1 = 3178.3 kJ/kg and h2 = 2344.7 kJ/kg. Also
Substituting,
Thus heat is transferred out of the system at a rate of 63.6 kW.

13
Expansion or compression process where
pressure is not constant - Examples

14
FIGURE 4–10 Schematic and P-V diagram for Example 4–4.

22
A water
heater in
steady
operation.
Mass
balance
Energy
balance
2) Mass and Energy Analysis Of
Control Volumes

23
Turbines and
Compressors
Turbine drives the electric generator In steam,
gas, or hydroelectric power plants.
As the fluid passes through the turbine, work is
done against the blades, which are attached to
the shaft. As a result, the shaft rotates, and the
turbine produces work.
Compressors, as well as pumps and fans, are
devices used to increase the pressure of a
fluid. Work is supplied to these devices from
an external source through a rotating shaft.
A fan increases the pressure of a gas slightly
and is mainly used to mobilize a gas.
A compressor is capable of compressing the
gas to very high pressures.
Pumps work very much like compressors
except that they handle liquids instead of
gases.
Energy balance for the
compressor in this figure:

24
Example 1
The power output of an
adiabatic steam turbine is 5
MW, and the inlet and the
exit conditions of the steam
are as indicated in the
diagram.
(a) Compare the magnitudes
of h, ke, and pe.
(b) Determine the work done
per unit mass of the steam
flowing through the turbine.
(c) Calculate the mass flow
rate of the steam.

25
Solution
We take the turbine as the system. This is a control volume since
mass crosses the system boundary during the process. We
observe that there is only one inlet and one exit and thus
(a) At the inlet, steam is in a superheated vapor state, and its
enthalpy is
At the turbine exit, we have a saturated liquid–vapor mixture at
15 kPa. The enthalpy at this state is
.21 mm 

27
Single inlet and outlet, subs. 1 inlet, 2 outlet

29
A steam turbine has an inlet of 2 kg/s steam at 1000 kPa, 350 C
and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low
velocity. Determine the power output of the turbine in kW.
Solution
Properties of water
Example 2

30
The energy balance equation for the steady flow turbine is
Neglecting change in PE and given that V2 0, we then have
Or
Substituting values, we have

31
At 2 kg/s steam flow rate, the power output

32
A well-insulated turbine operating at steady conditions develops 23
MW of power for a steam flow rate of 40 kg/s. The steam enters at
360°C with a velocity of 35 m /s and exits as saturated vapor at 0.06
bar with a velocity of 120 m /s. Neglecting potential energy
effects, determine the inlet pressure, in bar.
Solution
Properties of water
Example 3

33
Neglecting change in PE and heat transfer (as turbine is well
insulated), the energy balance equation for the steady flow process
on a unit mass basis is,
Thus
Substituting values,
= 3147.6 kJ/kg

34
Hence
Checking Table A-4, we find hg@360 C = 2481.6 kJ/kg. Since h1 is
higher, we have superheated vapour state at 1.
(Note: Steam at turbine inlet is always superheated.)
Interpolating from Table A-6 (see below), we find P1 = 2.6 MPa or
26 bar.

35
Refrigerant-134a enters an adiabatic compressor as saturated vapor
at 24°C and leaves at 0.8 MPa and 60°C. The mass flow rate of
the refrigerant is 1.2 kg/s. Determine (a) the power input to the
compressor and (b) the volume flow rate of the refrigerant at the
compressor inlet.
Solution
From the refrigerant tables
Example 4

36
The energy balance for this steady-flow system can be expressed in
the rate form as

37
Air enters the compressor of a gas-turbine plant at ambient
conditions of 100 kPa and 25 C with a low velocity and exits at
1 MPa and 347 C with a velocity of 90 m/s. The compressor is
cooled at a rate of 1500 kJ/min, and the power input to the
compressor is 250 kW. Determine the mass flow rate of air
through the compressor.
Example 5

38
Solution
The energy balance for this steady-flow system can be expressed
in the rate form as
Substituting, the mass flow rate is determined to be

39
Throttling valves
Throttling valves are any kind of flow-restricting devices that
cause a significant pressure drop in the fluid.
What is the difference between a turbine and a throttling
valve?
The pressure drop in the fluid is often accompanied by a large
drop in temperature, and for that reason throttling devices are
commonly used in refrigeration and air-conditioning
applications.
The temperature of an ideal gas does not
change during a throttling (h =
constant) process since h = h(T).
During a throttling process, the enthalpy of a
fluid remains constant. But internal and flow
energies may be converted to each other.
Energy
balance

40
Example – Expansion of R-134a in a Refrigerator
Refrigerant-134a enters the capillary tube of a refrigerator as
saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12
MPa. Determine the quality of the refrigerant at the final state
and the temperature drop during this process.

41
Solution
A capillary tube is a simple flow-restricting device that is
commonly used in refrigeration applications to cause a large
pressure drop in the refrigerant. Flow through a capillary tube is a
throttling process; thus, the enthalpy of the refrigerant remains
constant.
Since hf < h2 < hg, the refrigerant exists as a saturated mixture at
the exit. The quality of this mixture is

42
Since the exit state is a saturated mixture at 0.12 MPa, the exit
temperature must be the saturation temperature at this
pressure, which is 22.32°C. Then the temperature change for this
process is

Part of the heat received by a heat
engine is converted to work, while the
rest is rejected to a sink.
The devices that convert heat to work.
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 in the form of a
rotating shaft.)
3. They reject the remaining waste
heat to a low-temperature sink (the
atmosphere, rivers, etc.).
4. They operate on a cycle.
Heat engines and other cyclic devices
usually involve a fluid to and from which
heat is transferred while undergoing a
cycle. This fluid is called the working
fluid.
Work can be converted to heat directly
and completely, but converting heat to
work requires the use of heat engines.
43
3) Second Law of Thermodynamics
Heat Engines

44
Thermal efficiency
Some heat engines perform better
than others (convert more of the
heat they receive to work).
Schematic of a
heat engine.
Even the most
efficient heat
engines reject
almost one-half
of the energy
they receive as
waste heat.

45
Example 1
Heat is transferred to a heat
engine from a furnace at a rate
of 80 MW. If the rate of waste
heat rejection to a nearby river
is 50 MW, determine the net
power output and the thermal
efficiency for this heat engine.

46
Solution
A schematic of the heat engine is shown in the previous slide.
The furnace serves as the high-temperature reservoir for this
heat engine and the river as the low-temperature reservoir. The
given quantities can be expressed as

47
Example 2
A car engine produces 136 hp on the
output shaft. The thermal efficiency of
the engine is 30%. Find the rate of heat
rejected to the ambient and the rate of
fuel consumption if the engine uses a
fuel with a heating value of 35,000
kJ/kg.
Note: 1 hp = 745.7 W = 0.7457 kW

48
Solution
s
kg
.
kJ/kg
kW.
HV
nconsumptiofuelofRate
kW.hp..
kWhp.
.
00970
35000
2338
723633171363453
23383453
30
136
H
L
t
HHt
Q
Q
.
W
QQW





49
An automobile engine consumes fuel at a rate of 28 L/h and delivers
60 kW of power to the wheels. If the fuel has a heating value of
44,000 kJ/kg and a density of 0.8 kg/L, determine the thermal
efficiency of this engine.
Example 3

50
Solution
The mass consumption rate of the fuel is
The rate of heat supply to the car is
The thermal efficiency of the car is then

51
REFRIGERATORS AND HEAT PUMPS
• The transfer of heat from a low-
temperature medium to a high-
temperature one requires special
devices called refrigerators.
• Refrigerators, like heat engines, are
cyclic devices.
• The working fluid used in the
refrigeration cycle is called a
refrigerant.
• The most frequently used
refrigeration cycle is the vapor-
compression refrigeration cycle.
Basic components of a
refrigeration system and
typical operating conditions.
In a household refrigerator, the freezer compartment
where heat is absorbed by the refrigerant serves as the
evaporator, and the coils usually behind the refrigerator
where heat is dissipated to the kitchen air serve as the
condenser.

52
Coefficient of Performance
The objective of a refrigerator is to
remove QL from the cooled space.
The efficiency of a refrigerator is expressed in
terms of the coefficient of performance (COP).
The objective of a refrigerator is to remove heat
(QL) from the refrigerated space.

53
Heat
Pumps
The objective of
a heat pump is
to
supply heat QH
into the
warmer space.
The work
supplied to a
heat pump is
used to extract
energy from the
cold outdoors
and carry it into
the warm
indoors.
for fixed values of QL and QH

54
The food compartment of a
refrigerator, shown at right, is
maintained at 4°C by removing
heat from it at a rate of 360
kJ/min. If the required power
input to the refrigerator is 2
kW, determine
(a) the COP of the refrigerator and
(b) the rate of heat rejection to
the room that houses the
refrigerator.
Example 1

55
Solution
(a) The coefficient of performance of the refrigerator is
That is, 3 kJ of heat is removed from the refrigerated space for each
kJ of work supplied.
(b) The rate at which heat is rejected to the room that houses the
refrigerator is determined from the conservation of energy relation:

56
A heat pump is used to meet the
heating requirements of a house and
maintain it at 20°C. On a day when the
outdoor air temperature drops to 2°C,
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 absorbed
from the cold outdoor air.
Example 2

57
(a) The power consumed by this heat pump is determined from the
definition of the coefficient of performance to be
(b) The house is losing heat at a rate of 80,000 kJ/h. If the house is
to be maintained at a constant temperature of 20°C, the heat pump
must deliver heat to the house at the same rate, that is, at a rate of
80,000 kJ/h. Then the rate of heat transfer from the outdoor
becomes

58
When a man returns to his well-sealed house, he finds that the
house is at 32°C. He turns on the air conditioner, which cools the
entire house to 20°C in 15 min. If the COP of the air-conditioning
system is 2.5, determine the power drawn by the air conditioner.
Assume the entire mass within the house is equivalent to 800 kg of
air for which cv = 0.72 kJ/kg · °C and cp = 1.0 kJ/kg · °C.
Solution
We first have to find the amount of heat that need to be removed.
Since the house is well-sealed, we can assume a constant volume
cooling process so that
This heat is removed in 15 minutes. Thus the average rate of heat
removal from the house is
Example 3

59
Using the definition of the coefficient of performance, the power
input to the air-conditioner is determined to be

60
Refrigerant-134a enters the condenser of a residential
heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and
leaves at 800 kPa as a saturated liquid. If the compressor consumes
1.2 kW of power, determine
(a) the COP of the heat pump and
(b) the rate of heat absorption from the outside air.
Example 4

61
Solution
(a) The enthalpies of R-134a at the condenser inlet and exit are
An energy balance on the condenser gives the heat rejected in
the condenser
The COP of the heat pump is

62
(b) The rate of heat absorbed from the outside air

63
THE CARNOT HEAT ENGINE
The Carnot
heat engine
is the most
efficient of
all heat
engines
operating
between the
same high-
and low-
temperature
reservoirs.
No heat engine can have a higher efficiency
than a reversible heat engine operating
between the same high- and low-
temperature reservoirs.
Any heat
engine
Carnot heat
engine

64
A Carnot heat engine receives 500 kJ
of heat per cycle
from a high-temperature source at
652°C and rejects heat to a low-
temperature sink at 30°C. Determine
(a) the thermal efficiency of this
Carnot engine and
(b) the amount of heat rejected to
the sink per cycle.
Example

65
(a) The Carnot heat engine is a reversible heat engine, and so its
efficiency is given by
(b) The amount of heat rejected QL by this reversible heat engine
is
Solution

66
THE CARNOT REFRIGERATOR
AND HEAT PUMP
No refrigerator can have a higher COP
than a reversible refrigerator operating
between the same temperature limits.
Any refrigerator or heat pump
Carnot refrigerator or heat pump

67
Example 1
An inventor claims to have developed a refrigerator that
maintains the refrigerated space at 2°C while operating in a room
where the temperature is 24°C and that has a COP of 13.5. Is this
claim reasonable?
Solution
The maximum COP is the COP of a reversible refrigerator
operating between these temperature limits.
Since the COPR,rev is lower than the claimed COP, the claim cannot
be justified.
512
1227324273
1
1
1
.
//
COPCOP max,Rrev,R
LH TT

68
Example 2
A heat pump is to be used to
heat a house during winter.
The house is to be
maintained at 21°C 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 5°C.
Determine the minimum
power required to drive this
heat pump.

69
Solution
The heat pump must supply heat to the house at a rate of QH =
135,000 kJ/h = 37.5 kW. The power requirements are minimum
when a reversible heat pump is used to do the job. The COP of a
reversible heat pump operating between the house and the
outside air is
The required power input to this reversible heat pump is then
311
2127352731
1
1
1
.
//
COP rev,HP
HL TT
kW.
.
kW.
COPHP
323
311
537H
,innet
Q
W

70
THE CARNOT
CYCLE
Reversible Isothermal Expansion (process 1-2, TH = constant)
Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)
Reversible Isothermal Compression (process 3-4, TL = constant)
Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)
Most efficient
cycles are
reversible
cycles. Most
familiar
reversible
cycle is the
Carnot cycle
Execution of the
Carnot cycle in a
closed system.

71
P-V diagram of the Carnot cycle. P-V diagram of the reversed
Carnot cycle.
The Reversed Carnot Cycle
The Carnot heat-engine cycle is a totally reversible cycle.
Therefore, all the processes that comprise it can be reversed, in which
case it becomes the Carnot refrigeration cycle.

72
THE 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.
2. The efficiencies of all
reversible heat engines
operating between the
same two reservoirs are
the same.

73
Proof of the First Carnot principle.
Let reversible HE run as refrigerator (dashed
line). If heat from refrigerator is supplied directly
to irrev. HE, can eliminate HT reservoir. But what
do we have then?
A HE operating from a single reservoir and
producing work amounting to Wirrev Wrev
If Wirrev > Wrev, net work is produced and this
violate the 2nd Law. Hence Wirrev < Wrev.

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