The University of Texas at San Antonio (UTSA)

Klesse College of Engineering and Integrated Design

Department of Mechanical, Aerospace, and Industrial Engineering
ME 3293.002 Thermodynamics I

Final Exam, Spring 2025

Tu. 05/13/2025
Start: 03:00 PM, End: 04:50 PM

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[30 points total]
1) The pressure, temperature, and volume at the beginning of compression in an air-
standard Diesel cycle represented by the 𝑝 − 𝑣 diagram shown below are 86.1
kPa, 300 K, and 0.2 m3, respectively. At the end of the heat addition, the pressure
is 6,026 kPa and the temperature is 2,150 K. The pressure at the beginning of the
isometric heat transfer removal is 114.8 kPa. Determine:

a) The mass of air, in kg. [2.5 points]

Using the ideal gas equation 𝑝𝑉 = 𝑚𝑅𝑇 for state 1 (beginning of compression):

𝑝1 𝑉1 (86.1 kN/m2 )(0.2 m3 )

𝒎= = = 𝟎. 𝟐 𝐤𝐠
𝑅𝑇1 8.314 kJ
( ) (300 K)
28.97 kg ∙ K

b) The heat removal in the cycle, in kJ. [7.5 points]

The heat removal refers to process 4-1, which is an isometric process (i.e., a constant volume process). Applying the 1st
law of thermodynamics to this process:

∆𝐸41 = ∆𝐾𝐸41 + ∆𝑃𝐸41 + ∆𝑈41 = 𝑄41 − 𝑊41 → 𝑄41 = 𝑚(𝑢1 − 𝑢4 )

The temperature at state 4 is obtained with the ideal gas equation and considering that 𝑉4 = 𝑉1 = 0.2 m3 .

𝑝4 𝑉4 (114.8 kN/m2 )(0.2 m3 )

𝑇4 = = = 400 K
𝑚𝑅 8.314 kJ
(0.2 kg) ( )
28.97 kg ∙ K

The specific internal energies are obtained from Table A-22:

𝑢1 (𝑇1 = 300 K) = 214.07 kJ/kg 𝑢4 (𝑇4 = 400 K) = 286.16 kJ/kg

The heat removal is calculated as:

𝑸𝟒𝟏 = 𝑄𝑜𝑢𝑡 = 𝑚(𝑢1 − 𝑢4 ) = (0.2 kg)(214.07 kJ/kg − 286.16 kJ/kg) = −𝟏𝟒. 𝟒𝟏𝟖 𝐤𝐉

c) The heat addition in the cycle, in kJ. [10 points]

The heat addition refers to the isobaric process 2-3, where 𝑝2 = 𝑝3 . Applying the 1st law of thermodynamics:

∆𝐸23 = ∆𝐾𝐸23 + ∆𝑃𝐸23 + ∆𝑈23 = 𝑄23 − 𝑊23

𝑄23 = 𝑚(𝑢3 − 𝑢2 ) + 𝑊23

3
𝑄23 = 𝑚(𝑢3 − 𝑢2 ) + ∫ 𝑝𝑑𝑉
2

𝑄23 = 𝑚(𝑢3 − 𝑢2 ) + 𝑚(𝑝3 𝑣3 − 𝑝2 𝑣2 ) = 𝑚(ℎ3 − ℎ2 )

Process 1-2 is an isentropic process (∆𝑠12 = 0) for which it is possible to obtain the following using Table A-22 and
considering that 𝑇1 = 300 K (𝑝𝑟1 = 1.3860 from Table A-22):

𝑝2 𝑝𝑟2 𝑝 6,026 kPa

= → 𝑝𝑟2 = 𝑝𝑟1 ( 2 ) = 1.3860 ( ) = 97.00
𝑝1 𝑝𝑟1 𝑝1 86.1 kPa
The specific enthalpies are obtained from Table A-22 (no need for interpolation):

ℎ2 (𝑝𝑟2 = 97.00) = 1,000.55 kJ/kg ℎ3 (𝑇3 = 2,150 K) = 2,440.3 kJ/kg

The heat addition becomes:

𝑸𝟐𝟑 = 𝑄𝑖𝑛 = 𝑚(ℎ3 − ℎ2 ) = (0.2 kg)(2,440.3 kJ/kg − 1,000.55 kJ/kg) = 𝟐𝟖𝟕. 𝟗𝟓𝟎 𝐤𝐉

d) The net work done by the cycle, in kJ. [2.5 points]

The net work done by the cycle is given as:

𝑊𝑐𝑦𝑐𝑙𝑒 = 𝑄𝑐𝑦𝑐𝑙𝑒 = 𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 = |𝑄23 | − |𝑄41 |

𝑾𝒄𝒚𝒄𝒍𝒆 = |287.950 kJ| − |−14.418 kJ| = 𝟐𝟕𝟑. 𝟓𝟑𝟐 𝐤𝐉

e) The thermal efficiency of the cycle. [2.5 points]

By definition:

𝑊𝑐𝑦𝑐𝑙𝑒 273.532 kJ
𝜼= = = 𝟎. 𝟗𝟒𝟗𝟗 = 𝟗𝟒. 𝟗𝟗%
𝑄𝑖𝑛 287.950 kJ

f) The mean effective pressure, in kPa. [5 points]

By definition:

𝑊𝑐𝑦𝑐𝑙𝑒
𝑚𝑒𝑝 =
𝑉1 − 𝑉2

The temperature at state 2 is obtained from Table A-22 considering that 𝑝𝑟2 = 97.00 (calculated in part c):

𝑇2 (𝑝𝑟2 = 97.00) = 960 K

The volume in state 2 is obtained with the ideal gas equation:

8.314 kJ
𝑚𝑅𝑇2 (0.2 kg) (28.97 kg ∙ K) (960 K)
𝑉2 = = = 0.0091439 m3
𝑝2 (6,026 kPa)

Thus, the mean effective pressure becomes:

𝑊𝑐𝑦𝑐𝑙𝑒 273.532 kNm

𝒎𝒆𝒑 = = = 𝟏, 𝟒𝟑𝟑. 𝟏𝟖𝟒𝟓 𝐤𝐏𝐚
𝑉1 − 𝑉2 0.2 m3 − 0.0091439 m3
[25 points total]
2) An air compressor and a regenerative heat exchanger
operating at steady state are shown below. Air as an ideal gas
flows from the compressor to the regenerator (states 1, 2, and
3), and a separate stream of air as an ideal gas passes though
the regenerator in counterflow (states 4 and 5). Operating
data are provided on the figure. Stray heat transfer to the
surroundings and kinetic and potential energy effects can be
neglected. The compressor power input is 6750 kW, and the
mass flow rate at the inlet of the compressor (𝑚̇1 ) is the same
as the mass flow rate at the exit of the regenerator (𝑚̇5 ).

a) Determine the mass flow rate of air entering the

compressor, in kg/s. [5 points]

The mass flow rate of air flowing through the compressor, in kg/s, is obtained by applying the law of continuity and the
first law of thermodynamics in steady state to the compressor (one inlet and one outlet, neglecting changes in kinetic
and potential energy, as well as heat transfer to the surroundings):

𝑑𝐸𝑐𝑣 𝑉𝑖2 𝑉𝑒2

̇ ̇
= 𝑄𝑐𝑣 − 𝑊𝑐𝑣 + ∑ 𝑚̇𝑖 (ℎ𝑖 + + 𝑔𝑧𝑖 ) − ∑ 𝑚̇𝑒 (ℎ𝑒 + + 𝑔𝑧𝑒 )
𝑑𝑡 2 2
𝑖 𝑒

After simplifications:
𝑊̇𝑐 = 𝑚̇𝑎𝑖𝑟 (ℎ1 − ℎ2 )

Solving for the mass flow rate, and obtaining the specific enthalpies from Table A-22:

𝑊̇𝑐 −6,750 kJ/s 𝐤𝐠

𝒎̇𝒂𝒊𝒓 = = = 𝟐𝟎. 𝟓𝟖𝟔𝟖
ℎ1 − ℎ2 300.19 kJ − 628.07 kJ 𝐬
kg kg

b) Determine the temperature of the air exiting the regenerator at state 5, in K. [5 points]

The specific enthalpy at the exit of the regenerator is obtained by applying the law of continuity and the first law of
thermodynamics in steady state to the entire regenerator (one inlet and one outlet, neglecting changes in kinetic and
potential energy, as well as heat transfer to the surroundings):

𝑑𝐸𝑐𝑣 𝑉𝑖2 𝑉𝑒2

̇ ̇
= 𝑄𝑐𝑣 − 𝑊𝑐𝑣 + ∑ 𝑚̇𝑖 (ℎ𝑖 + + 𝑔𝑧𝑖 ) − ∑ 𝑚̇𝑒 (ℎ𝑒 + + 𝑔𝑧𝑒 )
𝑑𝑡 2 2
𝑖 𝑒

After simplifications:
0 = 𝑚̇𝑎𝑖𝑟 ((ℎ2 − ℎ3 ) + (ℎ4 − ℎ5 )) → ℎ5 = ℎ2 − ℎ3 + ℎ4

The specific enthalpies are obtained from Table A-22 for the given temperatures:

kJ kJ kJ 𝐤𝐉
𝒉𝟓 = ℎ2 − ℎ3 + ℎ4 = 628.07 − 778.18 + 800.03 = 𝟔𝟒𝟗. 𝟗𝟐
kg kg kg 𝐤𝐠

The temperature of air at the exit of the regenerator is obtained from Table A-22 using ℎ5 as a reference in interpolation:

kJ
𝑻𝟓 (ℎ5 = 649.92 ) ≈ 𝟔𝟒𝟎 𝐊
kg
c) Determine the rate of entropy production in the compressor, in kW/K. [7.5 points]

The entropy rate balance equation for a control volume that encloses the compressor, which operates adiabatically and
in steady state conditions is:

𝑑𝑆𝑐𝑣 𝑄̇𝑗
= ∑ + ∑ 𝑚̇𝑖 𝑠𝑖 − ∑ 𝑚̇𝑒 𝑠𝑒 + 𝜎̇𝑐𝑣
𝑑𝑡 𝑇𝑗
𝑗 𝑖 𝑒

Solving for the rate of entropy production:

𝜎̇𝑐𝑣 = 𝑚𝑎𝑖𝑟 (𝑠2 − 𝑠1 )

The change in specific entropy of an ideal gas is given as:

𝑝2
𝑠2 − 𝑠1 = 𝑠 0 (𝑇2 ) − 𝑠 0 (𝑇1 ) − 𝑅 ln ( )
𝑝1

Collecting data from Table A-1 and A-22 and substituting:

kg kJ kJ 8.314 kJ 12 bar 𝐤𝐖
𝝈̇ 𝒄𝒗 = (20.5868 ) (2.44356 − 1.70203 −( ) ln ( )) = 𝟎. 𝟓𝟖𝟒𝟓𝟕
s kg ∙ K kg ∙ K 28.97 kg ∙ K 1 bar 𝐊

d) Determine the rate of entropy production in the regenerator, in kW/K. [7.5 points]

The entropy rate balance equation for a control volume that encloses the compressor, which operates adiabatically and
in steady state conditions is:

𝑑𝑆𝑐𝑣 𝑄̇𝑗
= ∑ + ∑ 𝑚̇𝑖 𝑠𝑖 − ∑ 𝑚̇𝑒 𝑠𝑒 + 𝜎̇𝑐𝑣
𝑑𝑡 𝑇𝑗
𝑗 𝑖 𝑒

Solving for the rate of entropy production:

𝜎̇𝑐𝑣 = 𝑚𝑎𝑖𝑟 (𝑠3 − 𝑠2 ) + 𝑚𝑎𝑖𝑟 (𝑠5 − 𝑠4 )

The change in specific entropy of an ideal gas is given as:

𝑝 𝑝
𝑠3 − 𝑠2 = 𝑠 0 (𝑇3 ) − 𝑠 0 (𝑇2 ) − 𝑅 ln ( 3 ) 𝑠5 − 𝑠4 = 𝑠 0 (𝑇5 ) − 𝑠 0 (𝑇4 ) − 𝑅 ln ( 5 )
𝑝2 𝑝4

Given that 𝑝2 = 𝑝3 and 𝑝4 = 𝑝5 , the terms involving natural logarithms result in zero, as ln(1) = 0. Collecting data
from A-22 and substituting:

𝜎̇𝑐𝑣 = 𝑚𝑎𝑖𝑟 (𝑠 0 (𝑇3 ) − 𝑠 0 (𝑇2 ) + 𝑠 0 (𝑇5 ) − 𝑠 0 (𝑇4 ))

kg kJ kJ kJ kJ 𝐤𝐖
𝝈̇ 𝒄𝒗 = (20.5868 ) (2.66176 − 2.44356 + 2.47716 − 2.69013 ) = 𝟎. 𝟏𝟎𝟕𝟔𝟕
s kg ∙ K kg ∙ K kg ∙ K kg ∙ K 𝐊
[35 points total]
3) Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 10 MPa and 480 °C. The
condenser pressure is 6 kPa, and saturated liquid enters the pump. The turbine and pump have isentropic efficiencies of
80% and 70%, respectively. Ignore changes in kinetic and potential energy and consider steady state conditions.

a) Determine the specific enthalpy at the turbine exit (ℎ2 ), in kJ/kg. [10 points]

The properties of superheated vapor at state #1 (from table A-4):

𝑝1 = 10 MPa and 𝑇1 = 480 °C: ℎ1 = 3,321.4 kJ/kg, and 𝑠1 = 6.5282 kJ/kg · K.

Properties of the two-phase mixture of liquid and vapor at state #2s (considering isentropic expansion at the turbine
with 𝑠2𝑠 = 𝑠1 , from table A-3):

𝑝2 = 𝑝2𝑠 = 6 kPa 𝑠2𝑠 = 𝑠𝑓2𝑠 + 𝑥2𝑠 (𝑠𝑔2𝑠 − 𝑠𝑓2𝑠 )

𝑠2𝑠 = 𝑠1 = 6.5282 kJ/kg · K
𝑠2𝑠 − 𝑠𝑓2𝑠 6.5282 − 0.5210
𝑠𝑓2𝑠 (@ 6 kPa) = 0.5210 kJ/kg · K 𝒙𝟐𝒔 = = ≈ 𝟎. 𝟕𝟔𝟗𝟐𝟐𝟕
𝑠𝑔2𝑠 (@ 6 kPa) = 8.3304 kJ/kg · K 𝑠𝑔2𝑠 − 𝑠𝑓2𝑠 8.3304 − 0.5210
ℎ𝑓2𝑠 (@ 6 kPa) = 151.53 kJ/kg
ℎ𝑔2𝑠 (@ 6 kPa) = 2,567.4 kJ/kg ℎ2𝑠 = ℎ𝑓2𝑠 + 𝑥2𝑠 (ℎ𝑔2𝑠 − ℎ𝑓2𝑠 )
kJ kJ kJ 𝐤𝐉
𝒉𝟐𝒔 = 151.53 + (0.7692) (2,567.4 − 151.53 ) = 𝟐, 𝟎𝟎𝟗. 𝟖𝟖𝟐
kg kg kg 𝐤𝐠

The isentropic efficiency of the turbine is:

ℎ1 −ℎ2
𝜂𝑡 = → ℎ2 = ℎ1 − 𝜂𝑡 (ℎ1 − ℎ2𝑠 )
ℎ1 −ℎ2𝑠

kJ kJ kJ
ℎ2 = 3,321.4 − (0.80) (3,321.4 − 2,009.882 )
kg kg kg

𝐤𝐉
𝒉𝟐 = 𝟐, 𝟐𝟕𝟐. 𝟏𝟖𝟓𝟗𝟒𝟔
𝐤𝐠
b) Determine the specific enthalpy at the pump exit (ℎ4 ), in kJ/kg. [10 points]

The properties of saturated liquid water at state #3 (𝑝3 = 𝑝2 , 𝑥3 = 0, from table A-3):
𝑣3 = 𝑣𝑓3 + 𝑥3 (𝑣𝑔3 − 𝑣𝑓3 ) = 𝑣𝑓3 ≈ 0.0010064 m3 /kg
ℎ3 = ℎ𝑓3 + 𝑥3 (ℎ𝑔3 − ℎ𝑓3 ) = ℎ𝑓3 ≈ 151.53 kJ/kg
𝑠3 = 𝑠𝑓3 + 𝑥3 (𝑠𝑔3 − 𝑠𝑓3 ) = 𝑠𝑓3 ≈ 0.5210 kJ/kg · K

Properties of liquid water at state #4s (considering isentropic compression with 𝑠4𝑠 = 𝑠3 , and isobaric heat addition
with 𝑝4 = 𝑝1 , from table A-3): 𝑝4 = 10 MPa, 𝑠4𝑠 = 𝑠3 = 0.5210 kJ/kg · K. The isentropic efficiency of the pump is
given as:
(𝑊̇p ⁄𝑚̇) ℎ4𝑠 −ℎ3 𝑣3 (𝑝4 −𝑝3 ) 𝑣3 (𝑝4 −𝑝3 )
𝑠
𝜂𝑝 = = = → ℎ4 = ℎ3 +
(𝑊̇p ⁄𝑚̇) ℎ4 −ℎ3 ℎ4 −ℎ3 𝜂𝑝

m3 N
kJ (0.0010064
kg
)(10−0.006)MPa 106 2 1 kJ 𝐤𝐉
m
𝒉𝟒 = 151.53 + | || | = 𝟏𝟔𝟓. 𝟖𝟗𝟖𝟓𝟏𝟕
kg 0.7 1 MPa 103 Nm 𝐤𝐠

c) Determine the heat transfer to the water passing through the boiler, in kJ per kg. [5 points]

The heat transfer to the water passing through the boiler is calculated with the 1st law of thermodynamics:

𝑑𝐸𝑐𝑣 𝑉𝑖2 𝑉𝑒2

= 𝑄̇41 − 𝑊̇41 + 𝑚̇𝑖 (ℎ𝑖 + + 𝑔𝑧𝑖 ) − 𝑚̇𝑒 (ℎ𝑒 + + 𝑔𝑧𝑒 )
𝑑𝑡 2 2

0 = 𝑄̇41 − 0 + 𝑚̇(ℎ4 + 0 + 0) − 𝑚̇(ℎ1 + 0 + 0)

𝑸̇𝒊𝒏 𝑸̇𝟒𝟏 kJ kJ 𝐤𝐉
= = ℎ1 − ℎ4 = 3,321.4 − 165.898517 = 𝟑, 𝟏𝟓𝟓. 𝟓𝟎𝟏𝟒𝟖𝟑
𝒎̇ 𝒎̇ kg kg 𝐤𝐠

d) Determine the thermal efficiency. [5 points]

The thermal efficiency of the Rankine cycle is:

𝑊̇𝑐𝑦𝑐𝑙𝑒 𝑊̇𝑡 ⁄𝑚̇ + 𝑊̇𝑝 ⁄𝑚̇ (ℎ1 − ℎ2 ) + (ℎ3 − ℎ4 )

𝜂= = =
𝑄̇𝑖𝑛 𝑄̇𝑖𝑛 ⁄𝑚̇ ℎ1 − ℎ4

kJ kJ kJ kJ
(3,321.4 − 2,272.185946 ) + (151.53 − 165.898517 )
kg kg kg kg
𝜼= = 𝟎. 𝟑𝟐𝟕𝟗𝟓 (𝟑𝟐. 𝟕𝟗𝟓%)
kJ kJ
3,321.4 − 165.898517
kg kg

e) Determine the back work ratio. [5 points]

The back-work ratio is given as:

kJ kJ
|𝑊̇𝑝 ⁄𝑚̇| |ℎ3 − ℎ4 | |151.53 − 165.898517 |
kg kg
𝒃𝒘𝒓 = = = = 𝟎. 𝟎𝟏𝟑𝟔𝟗𝟓 (𝟏. 𝟑𝟔𝟗𝟓%)
|𝑊̇𝑡 ⁄𝑚̇| |ℎ1 − ℎ2 | |3,321.4 kJ − 2,272.185946 kJ |
kg kg
[10 points total]
4) Steam enters an adiabatic turbine steadily at 3 MPa and 400 °C, and leaves at
100 kPa and 100°C. The power output of the turbine is 3 MW. Ignoring kinetic
and potential energy effects, determine the isentropic efficiency of the turbine.

At both the inlet and the exit we have superheated water vapor. From Table A-4:

kJ kJ
ℎ1 = 3,230.9 ℎ2 = 2,676.2
kg kg
kJ kJ
𝑠1 = 6.9212 𝑠2 = 7.3614
kg∙K kg∙K

If isentropic operation is considered, then the entropy between states 1 and 2

remains the same 𝑠1 = 𝑠2 = 6.9212 kJ/kg ∙ K. It is necessary to find the
enthalpy at state 2s but we note that, for a pressure of 100 kPa, we don’t have an
entropy of 6.9212 kJ/kg∙K listed in the table for superheated water vapor. This means that we no longer have superheated
water vapor, but instead a two-phase mixture of liquid and vapor. Thus, we must first identify the quality of this mixture
and then the enthalpy at state 2s by examining Table A-3:
kJ kJ
ℎ𝑔2𝑠 = 2,675.5 𝑠𝑔2𝑠 = 7.3594
kg kg∙K
kJ kJ
ℎ𝑓2𝑠 = 417.46 𝑠𝑓2𝑠 = 1.3026
kg kg∙K

The quality is given as:

𝑠2 = 𝑠𝑓2𝑠 + 𝑥2𝑠 (𝑠𝑔2𝑠 − 𝑠𝑓2𝑠 )

𝑠2 − 𝑠𝑓2𝑠 6.9212 − 1.3026

𝑥2𝑠 = = = 0.92765
𝑠𝑔2𝑠 − 𝑠𝑓2𝑠 7.3594 − 1.3026

The enthalpy at state 2 is then given as:

ℎ2𝑠 = ℎ𝑓2𝑠 + 𝑥2𝑠 (ℎ𝑔2𝑠 − ℎ𝑓2𝑠 )

kJ kJ kJ kJ
ℎ2𝑠 = 417.46 + 0.92765 (2,675.5 − 417.46 ) = 2,512.13434
kg kg kg kg

The isentropic efficiency of the turbine is calculated as:

𝑊̇𝑡 /𝑚̇ ℎ1 − ℎ2 3,230.9 − 2,676.2

𝜼𝒕 = = = = 𝟎. 𝟕𝟕𝟏𝟕𝟒𝟎 (𝟕𝟕. 𝟏𝟕𝟒𝟎%)
(𝑊̇𝑡 /𝑚̇)
𝑠
ℎ1 − ℎ2𝑠 3,230.9 − 2,512.13434