Faculté de génieGénie mécanique

Faculty of EngineeringMechanical Engineering

MCG 2175 - THERMODYNAMICS II

Final Examination Time: 3 hours19 April 2007 Page 1 of 7Prof. W. Hallett

Closed book. Non-programmable calculators only allowed. Steam tables and a psychrometric chartare provided; other data are given with the questions.

1. (6 marks total) Descriptive questions - give brief answers in words. No calculations are required.You may write your answers in point form if you wish.(a) (2 marks) For an ideal Diesel cycle, sketch (i) a P-v diagram; (ii) a T-s diagram. Label the eventsin the cycle on each of these diagrams.(b) (3 marks) A student lives in a hot adiabatic apartment with no windows. He decides to put anelectrically-operated window-type air conditioner in the middle of the apartment to cool it. Whatwill happen to the temperature in the apartment and why? Will the student pass thermodynamics?(c) (1 mark) What does a regenerator do in a gas turbine cycle and how does it improve the cycleefficiency?

2. (11 marks total) Hydrocarbon fuels have the general chemical formula CnHm. To identify anunknown hydrocarbon gas, 0.001 kmol of this fuel are burned and the combustion products are foundto have a composition of 13.63% CO2, 5.58% H2O, 7.43% O2, and 73.36% N2 BY MASS (note: notthe usual volume percent). (a) (4 marks) If the total mass of the products produced from 0.001 kmol of fuel is 0.646 kg, identifythe hydrocarbon - i.e. solve for the unknowns n and m. (b) (7 marks) If this reaction is carried out in an adiabatic combustion chamber at constant pressure,the products reach a temperature around 1950 K. Using this information, find the Higher HeatingValue (HHV) for this fuel, in kJ/kg. The reactants (fuel and air) enter at 25°C.

Product Molar mass

(kg/kmol)(kJ/kmol) (kJ/kmol)

at 1900K

(kJ/kmol)

at 2000K

(kJ/kmol)

CO2 44 -393522 - - 85429 91450

H2O (g) 18 -241827 43 961 67613 72689

O2 32 0 - - 55434 59199

N2 28 0 - - 52551 56141

MCG 2175 - THERMODYNAMICS IIFinal Examination Time: 3 hours19 April 2007 Page 2 of 7

3. (13 marks total)

Some engineering students are taking a thermodynamics exam in a crowded auditorium. This largeclassroom receives conditioned air at state 3 (see diagram), and the breathing and sweating of the

occupants adds saturated vapour to the room at a mass flow rate = 0.027 kg/s and heat at the rate

of = 10 kW. The outside weather (stream 1 feeding the air conditioning unit) is a typical rainy

spring day with T1 = 15°C and N1 = 80%. The operation of the air conditioning unit involves

subtracting heat ( ) in the dehumidifying section and then reheating ( ) to deliver air at T3 =

20°C and N3 = 40% to the classroom. Throughout this problem, assume a mass flow rate of dry air

= 10 kg/s and a constant pressure of 100 kPa. Steam tables and a psychrometric chart are given

at the end of the paper, and the following relation might prove useful: .

(a) (2 marks) Find the specific humidity of states 1 and 3, T1 and T3, without using thepsychrometric chart.

For the following questions, the psychrometric chart must be used. However, you must showrelevant equations and specify how you got each reading on the chart (ex. “using TDRY and TWET,read T on the chart”)

(b) (4 marks) Describe state 4 (i.e. give T4, N4 and T4). Assume that the vapour added to the roomby the occupants is at 35°C.

(c) (4 marks) Describe state 2 (i.e. give T2, N2 and T2) and find the mass flow rate of condensate, ,

in kg/s.

(d) (1 marks) Calculate how much heat must be added in the reheating section, , in kW.

(e) (1 mark) Clearly draw these three processes (1-2, 2-3 and 3-4) on a sketch psychrometric chart.

(f) (1 mark) Write an equation to determine the heat removed from the dehumidifier. Explain

why it is acceptable to neglect the condensate enthalpy in this equation.

MCG 2175 - THERMODYNAMICS IIFinal Examination Time: 3 hours19 April 2007 Page 3 of 7

4. (15 marks total) The sketch shows thecycle of the new General ElectricLMS100 industrial gas turbine, which hasan intercooled compressor. Thecompressor turbine drives both the highpressure (HP) and low pressure (LP)compressors, while the power turbineproduces the output power. Bothcompressor stages have the same pressureratio rP = 6.5, and both have an isentropicefficiency of 0C = 0.92, while both of theturbine stages have an isentropicefficiency of 0T = 0.93. The working fluidis air, and the following cycle states aregiven:

1: T1 = 20°C, P1 = 100 kPa2: P2 / P1 = P7 / P6 = 6.53: T3 = 1380°C5: P5 = P1

6: T6 = T1

Properties for air: CP = 1.1 kJ/kg K, R = 0.287 kJ/kg K, k = 1.35.Isentropic relation for a perfect gas with constant specific heat:

(a) (3 marks) Sketch a T-s diagram of this cycle.(b) (3 marks) Determine the work required to drive the HP compressor in kJ/kg air and the exittemperature from it T7.(c) (2 marks) Knowing that the work of the compressor turbine is used entirely to drive the twocompressor stages, determine the temperature T4 at the compressor turbine outlet.(d) (4 marks) Calculate the pressure P4 at the compressor turbine outlet.(e) (3 marks) Calculate the efficiency of the cycle.

MCG 2175 - THERMODYNAMICS IIFinal Examination Time: 3 hours19 April 2007 Page 4 of 7

5. (15 marks total) The sketch shows arefrigeration cycle with a regenerator, verysimilar to the one in your lab. Theregenerator (heat exchanger) cools the liquidentering the valve using cold vapour leavingthe evaporator at 4. The refrigerant is R134a,and the following states are given:

1: saturated liquid at 30°C4: saturated vapour at -26.54°C

(a) (2 marks) Sketch a T-s diagram for thiscycle(b) (3 marks) If the effectiveness of the heatexchanger is g = 0.75, determine the temperature and the enthalpy of the liquid at point 2. Theeffectiveness is defined in exactly the same way as it is for a regenerative gas turbine cycle: it is theratio of the actual change in temperature of the stream going from 1 to 2 divided by the change intemperature that the stream would experience if heat transfer were ideal.(If you are unable to solvethis part of the question, assume T2 = T1 and T5 = T4 for the rest of the question.)

(c) (2 marks) Determine the heat transfer to the evaporator per kg of refrigerant flowing through

the cycle.(d) (3 marks) Calculate the enthalpy h5 and specific volume v5 at the exit from the heat exchanger(point 5).(e) (2 marks) The compression process may be described by a polytropic process Pvn = constant,

where n = 1.05. Determine the compressor power per kg of refrigerant and the coefficient of

performance of the cycle. The equation for the work done in a polytropic process is

(f) (3 marks) The compressor is cooled to reduce the work of compression. Assuming a polytropic

process, determine the enthalpy h6 at the compressor exit and the heat removed from the compressor

in kJ/kg. Do not make use of the perfect gas law in doing this.

Properties tables are appended to this paper.

Total marks for this paper: 60

MCG 2175 - THERMODYNAMICS IIFinal Examination Time: 3 hours19 April 2007 Page 5 of 7

Properties of saturated R134a

T (°C) P (kPa) vf (m3/kg) vg (m3/kg) hf (kJ/kg) hg (kJ/kg)

-26.54 100 .0 0.000728 0.19257 165.50 381.98

-20 133 .7 0.000738 0.14649 173.74 386.08

-10 201 .7 0.000755 0.09921 186.72 392.28

0 294 .0 0.000773 0.06919 200.00 398.36

10 415 .8 0.000794 0.04945 213.58 404.23

20 572 .8 0.000817 0.03606 227.49 409.84

30 771 .0 0.000843 0.02671 241.79 415.08

40 1017.0 0.000873 0.02002 256.54 419.82

50 1318.1 0.000908 0.01512 271.83 423.91

60 1681.8 0.000951 0.01146 287.79 427.13

Properties of superheated R134a

100 kPa 600 kPa 800kPa

T (°C) v (m3/kg) h (kJ/kg) v (m3/kg) h (kJ/kg) v (m3/kg) h (kJ/kg)

sat. 0.19257 381.98

- 20 0.19860 387.22

- 10 0.20765 395.27

0 0.21652 403.41

10 0.22527 411.67

20 0.23392 420.05

30 0.24250 428.56 0.03609 419.09

40 0.25101 437.22 0.03796 428.88 0.02711 424.86

50 0.25948 446.03 0.03974 438.59 0.02861 435.11

60 0.26791 454.99 0.04145 448.28 0.03002 445.22

70 0.27631 464.10 0.04311 457.99 0.03137 455.27

80 0.28468 473.36 0.04473 467.76 0.03268 465.31

90 0.04632 477.61 0.03394 475.38

100 0.04788 487.55 0.03518 485.50

MCG 2175 - THERMODYNAMICS IIFinal Examination Time: 3 hours19 April 2007 Page 6 of 7

Saturated Steam - Temperature Table

T (°C) P (kPa) hf (kJ/kg) hg (kJ/kg)

0.01 0.6113 0.01 2501.4

5 0.8721 20.98 2510.6

10 1.2276 42.01 2519.8

15 1.7051 62.99 2528.9

20 2.339 83.96 2538.1

25 3.169 104.89 2547.2

30 4.246 125.79 2556.3

35 5.628 146.68 2565.3

40 7.384 167.57 2574.3

MCG 2175 - THERMODYNAMICS IIFinal Examination Time: 3 hours19 April 2007 Page 7 of 7

Psychrometric Chart for a total pressure of 100 kPa