1

Last Name: First Name: Purdue ID: Please write your name in BLOCK letters. Otherwise Gradescope may not recognize your name. CIRCLE YOUR LECTURE BELOW (Fail to circle is with 2-point penalty): MWF 10:30 am MWF 3:30 pm TR 8:30 am Prof. Chen Prof. Meyer Prof. Jain

FINAL EXAM ME 300 SP2018

INSTRUCTIONS 1. Please place all your electronics, including but not limited to cell phones, computers, watches,

tablets, etc., into your bag and store your bag under your seat. 2. The only calculator that is permitted to be used during this exam is the TI-30X IIS basic scientific

calculator. 3. This is a closed book and closed notes examination. You are provided with an equation sheet

and all the necessary property tables. 4. Do not hesitate to ask the instructor if you do not understand a problem statement. 5. Start each problem on the same page as the problem statement. Write on only the front side of

the page. Materials on the back side of the page will not be graded. 6. Put only one problem on a page. Another problem on the same page will not be graded. 7. Identify your system(s), list relevant assumptions, and start with basic equations for Problems 2,

3, 4, and 5. 8. If you give multiple solutions, you will receive only a partial credit although one of the solutions

might be correct. Cross out any solutions that you do not want to be graded. 9. For your own benefit, please write clearly and legibly. Maximum credit for each problem is

indicated below. 10. After you have completed the exam, put your papers in order. This may mean that you have to

remove the staple and then re-staple the exam. Do not turn in loose pages. 11. If you use additional blank pages, do not stable on your exam but use clip to hold it together. 12. Once time is called you will have three minutes to turn in your exam. Points will be subtracted

for exams turned in after these three minutes. 13. Keep your eyes on your own paper. If you are caught cheating you will get a zero for the exam

and your name will be turned over to the Dean of Students.

Problem Possible score1 45 2 35 3 25 4 40 5 55

Total 200

Solution Key

2

Last Name: First Name: Purdue ID: Problem 1 (45 points). Each part is 3 points. Please fill only ONE circle for your answer to each problem. (1) Which of the two heat transfer processes shown below has greater entropy generation? Process A Process B Same When σ = 0, the temperature difference between the source and sink must be indefinitely small. The larger is the temperature difference, the more entropy generation is. If you calculate σ σA = -Σ(Q/T) = - (1000/800 – 1000/300)=2.08, σB = -Σ(Q/T) = - (1000/400 – 1000/300)=0.83 (2) Is the exergy of a system different in an environment with a different temperature and pressure?

Yes No Insufficient information Yes, exergy is a function of T0 and P0. Please see the definition. (3) Engine A has a thermal efficiency of 38% and Engine B has a second law efficiency of 72%. For the same provided to each engine, which one produces more power? Engine A Engine B insufficient information The second law efficiency is calculated using a different definition than thermal efficiency. (4) In the figure below, the arrow represents a psychrometric process of _________. Heating and humidification Heating and dehumidification Cooling and humidification Increasing Tdb and humidity ratio. (5) To determine the properties of a superheated steam, which of the following is most accurate? Van der Waals equation Steam table Generalized compressibility chart Steam table does not use any assumptions/approximations.

Solution Key

CM CM

3

Last Name: First Name: Purdue ID: Problem 1 (Continued).

(6) In the enthalpy departure equation:* *

* *2 1 2 1 c

c c2 1

h h h hh h h h RT

RT RT

, the second

underlined term is a correction to the first underlined term, which is the enthalpy difference of a real gas at 298K a real gas at its critical point an ideal gas See page 694 of the textbook. (7) For chemical reaction equilibrium of 2 2 2CO H CO H O , log10K = -5.018 at 298 K and

log10K = -2.139 at 500 K. At which temperature does the equilibrium contain more H2? 298 K 500 K Insufficient information As K increases with temperature, more products (the chemical compounds on the right side of the formula) would be found at higher temperature. (8) A mixture of CO2 and N2O have the same mass fraction for the two gasses. The mole fraction of CO2 is that of N2O. much greater than much smaller than almost the same as Because the molar mass of the two gases is almost the same. (9) If moist air is heated to a very high temperature and the equilibrium composition consists of H2O, O2, N2, OH, H2, and NO, how many equilibrium constant relations are needed to determine the equilibrium composition of the mixture? 2 3 4 Since the mixture contains 6 compounds, it needs 6 equations. The 3 chemical elements have 3 atom balance equations so 3 equilibrium constant relations are needed. (10) Which cycle is the one depicted in the T-s diagram below? Regenerative vapor power (Rankine) cycle with an open feedwater heater Vapor power (Rankine) cycle with reheat Regenerative vapor refrigeration cycle with open feedwater heater Refer to the textbook on regenerative vapor power cycle with open feedwater heater.

T

Solution Key

4

Last Name: First Name: Purdue ID: Problem 1 (Continued). (11) To increase the efficiency of the Rankine cycle shown on the figure to the right, what should be done with the boiling and condensing temperatures? Increase boiling T, increase condensing T Decrease boiling T, increase condensing T Increase boiling T, decrease condensing T Increasing boiling temperature will increase efficiency but increasing condensing temperature will decrease efficiency. (12) Heat pumps are often difficult to use in very cold climates because the refrigerant temperature in the evaporator must be _____________ the ambient temperature. higher than lower than equal to The refrigerant temperature must be less than the ambient temperature to achieve the correct temperature gradient. (13) An ideal vapor compression refrigeration cycle is shown on a T-s diagram below. What does the area enclosed by the cycle represent? Net heat rejected per unit mass of flow Net work input per unit mass of flow No physical meaning Since 3-4 is throttling, the area below the line is not heat transfer. (14) Can regeneration be added to the Brayton cycle shown in the figure below? Yes No Insufficient information No, since T4 < T2 (15) In a gas turbine used for propulsion, the power generated by the turbine is idealy _____________ the power consumed by the compressor. much greater than much less than approximately equal to The turbine only needs to generate enough power to power the compressor; the goal of the turbo jet engine is to generate propulsive thrust, not shaft work.

T 2Tcond

Tevap

3

4 1

s

T

s

Solution Key

5

Last Name: First Name: Purdue ID: Problem 2 (35 points) Given: A compressed-air energy storage system fills an underground cavern with pressurized air.

The cavern has a fixed volume of 105 m3 and initially contains air at 285 K and 1 bar that is the same as the environment. After the filling process is complete, the cavern contains air at 800 K and 20 bar. The system boundary (control volume) includes the compressor, piping and cavern. Molar mass of air is 28.97 kg/kmol.

Find: (a) The initial and final mass of air in the cavern, in kg. (b) The total work done by the compressor, in kJ. (c) The total change in exergy contained within the cavern assuming that the filling process

is reversible, in kJ. Sketch of system: Assumptions:

Basic equations: Solution: (a)

Solution Key

Air behaves as an ideal gas Open system KE & PE negligible Adiabatic process

∑ ∑    

 

∑ ∑

1 ∑ , ∑ ,

CV

6

Last Name: First Name: Purdue ID: Problem 2 (Continued)

Solution Key

Using  , 

..   

..

 (b) Mass balance:

∑ ∑      

 Integrating,  ∆    Energy balance: 

   ∑ ∑  

∑  Integrating,  ∆ and ∆ ∆ ∆ ∆ Thus,

∆ = (m2 – m1)hi – (m2u2 – m1u1)

8.71 10 1.22 10 285.14

8.71 10 592.30 1.22 10 203.33

. (c) Exergy balance: 

1 ∑ , ∑ ,

, = 0 kJ/kg, because and ,

Integrating, ∆ . By stating that exergy destroyed for a reversible process is zero, thus exergy change is equal to the work done on the system is also acceptable.

Where   @285 @800 @285

7

Last Name: First Name: Purdue ID: Problem 2 (Continued)

Solution Key

8

Last Name: First Name: Purdue ID: Problem 3 (25 points) Given: Moist atmospheric air enters a cooling tower as shown below at 1 bar with a dry bulb

temperature of 30.0oC and 30% relative humidity and exits at a dry bulb temperature of 25.0oC and relative humidity of 90%. The mass flow rate of the dry air is 200 kg/s. Cooling water with a flow rate of 30 kg/s at 80oC enters the cooling tower, some evaporation occurs when it interacts with air. Assume constant specific heat for water to be c = 4.19 kJ/kg·K.

Find: (a) Show the air change process from state 3 to state 4 on the psychrometric chart on the

following page. (b) Calculate the amount of water evaporated to the air, in kg/s. (c) Determine the temperature of the water leaving the cooling tower at state 2, in oC. Sketch of system/process:

Assumptions: Basic equations:

1

2

4

2

1

1

30 /

80o

liquid H O

m kg s

T C

2

2

? o

liquid H O

T C

,3

,1

, 1

200 /

30

30%

dry air

oDB

moist air p bar

m kg s

T C

,2

4

, 1

25

90%

oDB

moist air p bar

T C

3

Solution Key

Moist air as an ideal gas KE & PE negligible Open system No work Steady state, steady flow

∑ ∑     

∑ ∑

CV

TDB,4 = 25oC

TDB,3 = 30oC

9

Last Name: First Name: Purdue ID: Problem 3 (Continued) Solution: (a)

Solution Key

10

Last Name: First Name: Purdue ID: Problem 3 (Continued)

Solution Key

(b)  

0.0183 0.008

200 . /

(c)

Energy balance: ∑ ∑

0

200 50.5

30 4.19 80 200

70.5

30 2.06 4.19 0

.

11

Q

Last Name: First Name: Purdue ID: Problem 4 (40 points) Given: One kmol of gaseous methane CH4 and 200% theoretical air at 298 K and 1.0 bar fills the

left hand side of a rigid vessel as shown below. This mixture is ignited, and complete combustion occurs with no dissociation of the nitrogen and oxygen. During the combustion, the brittle pins holding the massless piston in place fail, and this piston is driven to the right such that V2 = 2V1. Energy absorption by the pins, friction, and any boundary work are negligible. The final product temperature is 990 K. The ambient conditions are: To = 298 K and Po = 1 bar. Note that:

2

2

2

, 4

@990

@990

@990

74,850 /

216, 252 /

21,134 /

22,359 /

of CH

H O K

N K

O K

h kJ kmol

h kJ kmol

h kJ kmol

h kJ kmol

Find:

(a) The chemical reaction formula for the combustion process, including the relative number of moles of each reactant and product .

(b) The heat transfer during the entire combustion and expansion process, in kJ. (c) The entropy of the CO2, , in the product gases, in kJ/kmol·K.

Sketch of system/process: Assumptions: Basic equations:

Vacuum CH4 and 200% Theor. air

Solution Key

Ideal gas Complete combustion Closed system Quasi-equilibrium at the initial and final states No work KE & PE negligible ∆  

CM

12

Last Name: First Name: Purdue ID: Problem 4 (Continued) Solution:

Solution Key

(a) Stoichiometric: CH 2 O 3.76N ⟶ 2H O 1CO 2 3.76N Actual: CH 4 O 3.76N ⟶ 2H O 1CO 2O 4 3.76N (b) Energy balance: ∆ ∆ ∆ ∆ ∑ ∑

393,520 42,226 9,364 360,658 /

∑ 2 216,252 1 360,658

2 22,359 4 3.76 21,134

2 1 2 4 3.76 8.314 990

595,597.6 ∑ 1 74,850 4 0

4 3.76 0

1 4 4 3.76 8.314 298

124,500.5 Thus, , . , . , .

(c) ° ln

To determine ,

.

.

1.66 , .

0.05 ,

Thus,

° ln 268.670∙

8.314∙ln . .

.

.∙

13

Last Name: First Name: Purdue ID: Problem 4 (Continued)

Solution Key

14

Last Name: First Name: Purdue ID: Problem 4 (Continued)

Solution Key

15

Last Name: First Name: Purdue ID: Problem 5 (55 points) Given: An organic Rankine cycle follows the behavior of a regular Rankine cycle but uses

ammonia as the working fluid as opposed to water. Here we consider an organic Rankine cycle using ammonia as the working fluid that undergoes the cycle shown below with two turbine stages having efficiencies of T1 = T2 = 0.85. The pump efficiency is 100%. Assume that the specific volume, , is constant in the internally reversible pump. The environment temperature is 298 K.

Find:

(a) Fill in the table on the next page (Ignore the darkened cells. Do not interpolate property table values and instead use the closest values!)

(b) Draw the processes for the cycle on the T-s diagram on the next page. (c) Write an equation for the thermal efficiency of the cycle, th, in terms of enthalpy, h,

at the various states. (d) Compute the thermal efficiency of the cycle, th, in %. (e) Calculate the specific exergy destruction for the first-stage turbine, in kJ/kg.

Sketch of system/process:

Assumptions: Basic Equations:

Open system Steady state, steady flow KE & PE negligible Adiabatic turbines and pump The pump is internally reversible. No pressure loss in heat exchangers The specific volume is constant – - in the pump

0 ∑ ∑    

 

∑ ∑

∑ ∑ ∑

0   

16

Last Name: First Name: Purdue ID: Problem 5 (Continued) Solution: (a) Fill in table below. Please use the following abbreviations for “Phase”: SVLM – saturated vapor/liquid mixture, SHV – superheated vapor, CL – compressed liquid, SV – saturated vapor, and SL – saturated liquid. (Some values should be calculated. Please show your work to receive full credit.)

State T (C) P (bar) h (kJ/kg) s (kJ/kgK) v (m3/kg) Phase x

1 60 20 1509.5 4.8838 SHV

2s 12 1442.6 4.8838 SVLM 0.9790

2 12 1452.6 4.9168 SVLM 0.9878

3 60 12 1553.1 5.2347 SHV

4s 6 1453.16 5.2347 SHV

4 6 1468.15

5 6 223.32 0.0015982 SL

6 20 225.56 0.0015982 CL

(b) Draw the cycle processes on the following T-s diagram:

2s 4s

17

Last Name: First Name: Purdue ID: Problem 5 (Continued) Last Name: First Name: Purdue ID: Problem 5 (Continued)

State 2s

12 , 4.8838 SVLM (table A-14)

. .

. .0.979

0.979 1139.52 327.01 1442.6

State 2

0.85 .

. . 1452.6 SVLM (table A-14)

. .

.

.

. . . 4.9168

State 6 By assumption,

0.0015982 20 6 10 2.237

223.32 2.237 225.56

20 , 225.56 CL

(c)

⁄ ⁄ ⁄

⁄

(d)

1509.5 1452.6 1553.1 1468.15 2.2371509.5 225.56 1553.1 1452.6

0.101 10.1%

18

Last Name: First Name: Purdue ID: Problem 5 (Continued) (e)

Entropy balance: ∑ ∑ ∑

4.9168 4.8838 0.033

Exergy destruction:

. .