AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class For multiple choice, circle THE BEST answer (only one answer per question)

1. In a cycle consisting of a sequence of four processes, over the complete cycle, the system MUST

a. increase in b. decrease in energy

c. have no change in energy d. do no

e. increase in

2. At the critical point for any pure substance,

a. liquid-solid-vapor phases are in equilibrium b. Pv = RT

c. equals 22.064 MPa d. hf = hg

e. 0 < x < 1

3. In an open system,

a. mass crosses the system boundary b. h > u for the flowing mass

c. can cross the system boundary d. work can cross the system boundary

e. all of the above

4. Mercury is used in barometers more often than other liquids because

a. it is shiny b. it is inexpensive

c. it is liquid at room d. it doesn’t react with glass

e. it has very high density

5. transferred during an , Strans = Q/T where Q is the heat transferred between the system and surroundings and T is the temperature of the surroundings. The units of Strans are

a. kJ/K b. kw/K

c. kJ/oC d. kJ/kg

e. kN/oC

6. The specific properties of a pure compressed liquid

a. are not affected by pressure b. are not affected by temperature

c. are properties of the liquid at STP d. cannot be measured or calculated

e. depend on the mass of liquid

1 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class

o o 7. Steam at 50 kPa has Tsat = 81.3 C. After T increases to 100 C in an . the steam will be:

a. compressed liquid b. superheated vapor

c. saturated vapor d. saturated mixture

e. at the critical point

o 8. Steam at 100 C has Psat = 101.4 kPa. After P decreases in an to 50 kPa, the steam will be

a. compressed liquid b. superheated vapor

c. saturated vapor d. saturated mixture

e. at the critical point

9. All real gases deviate somewhat from behavior: PV = mRT. For which of the following conditions is the deviation the smallest?

a. Low temperature and high pressure b. High temperature and low specific volume

c. Low temperature and low pressure d. High temperature and low pressure

e. At the critical point

10. What is the change in of air (an ideal gas) cooled from 550 oC to 100 oC? given that R = 0.287 kJ/kg/K, Cp = 1.013 kJ/kg/K and Cv = 0.726 kJ/kg/K

a. - 470 kJ/kg b. - 400 kJ/kg u = Cv(T2-T1)

c. - 130 kJ/kg d. - 460 kJ/kg u = 0.726(100-550) = -327 kJ/kg

e. - 330 kJ/kg

11. The internal energy of refrigerant R-134a is reduced from 300 kJ/kg at 0.8 MPa to 150 kJ/kg in a condenser at constant pressure. What is the final quality of the refrigerant, given the following: o P (kPa) T ( C) uf (kJ/kg) ufg (kJ/kg) ug (kJ/kg) usuperheated (kJ/kg) 800 31.31 95.47 171.82 267.29 800 90 300

o a. 0.0 b. 0.50 x=(u2-uf)/ufg @ T = 150 C

c. 0.32 d. 0.36 x=(150-95.47)/171.82 = 0.32

e. not applicable, refrigerant is superheated at final state

2 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 12. Draw the processes or locate the points on the T-v graph for refrigerant, R-134a, below. For each part, label the line or point on the graph with the corresponding letter. No calculations should be necessary, although you may have to use Tables A-11, A-12, and A-13.

A. Refrigerant at T = 374.2 K and P = 4.38 MPa, the critical point for R-134a.

Critical state marked by star on graph

B. Isochoric heating of saturated vapor with specific volume = 0.035969 m3/kg to a final temperature of 80 oC. o v1 = vg @ 20 C = Tsat (A-11) isochoric process, v1 = v2

C. Isobaric compression of superheated vapor at 1 MPa and 60 oC to saturated liquid.

3 o v1 = 0.0231 m /kg, Tsat = 39.37 C @ 1 MPa (A-13, A-12)

D. Isothermal expansion a saturated mixture of R-134a with v = 0.035 m3/kg at 293.01 kPa, to a final pressure of 100 kPa. o T = Tsat @ 293.01 kPa = 0 C 3 v2 = 0.2163 m /kg (A-13)

E. R-134a when equilibrium pressure = 400 kPa and temperature = 253 K = -20 oC. o o 3 @ 400 kPa, Tsat = 8.9 C. For T < Tsat, compressed liquid, v ~ vf @ -20 C = 0.00074 m /kg State marked by triangle on diagram

R-134a Saturated Liquid-Vapor

120 A 100

80

60 B C 40 C) ( T 20 D 0 E

-20 -40 0.0001 0.0010 0.0100 0.1000 1.0000

3 v (m /kg)

3 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 13. A piston-cylinder contains 50 kg water of which 12.5 kg is saturated liquid. The equilibrium pressure of the system is 100 kPa at the initial state.

3 a) Find the system initial volume, V1, in m

3 x = 0.75; v1 = 0.75(1.6941-0.001043) + 0.001043 = 1.271 m /kg

3 3 V1 = mv1 = 5 kg (1.271 m /kg) = 63.5 m

b) The water undergoes an isothermal expansion until the pressure reaches 50 kPa. Find the change in the system volume, V, in m3, assuming that the process temperature is 100 oC.

o P < Psat (101.4 kPa) for 100 C, so water is superheated vapor.

3 v2 = 3.4187 m /kg (A-6) 3 3 V2 = 50 kg(3.4187 m /kg) = 170.9 m V = 170.9 – 63.5 = 107.4 m3

c) Calculate how much energy must be added to the system ( U in kJ) for the expansion.

U = m(u2 – u1) u1 = x(ufg) + uf = 0.75(2088.2) + 417.4 kJ/kg = 1,983.6 kJ/kg (A-5) u2 = 2511.5 kJ/kg (A-6)

U = 50 kg(2511.5 – 1983.6) kJ/kg = 26,395 kJ

H = m(h2 – h1) h1 = 0.75(2,257.5) + 417.51 kJ/kg = 2,110.6 kJ/kg (A-5) h2 = 2,682.4 kJ/kg (A-6) H = m(h2 – h1) = 50 kg (2,682.4-2,110.6) = 28,590 kJ

4 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class

d) After the expansion, the water undergoes an isobaric compression until V3 = V1. Find the quality at the final equilibrium state.

3 P3 = 50 kPa. v3 = 1.271 m /kg x3 = (1.271-0.00103)/(3.2403-0.00103) = 0.392

e) Draw the process on the P-v diagram for water below and show temperature and specific volume values at the equilibrium states in the table next to the graph. (5 points)

(oC)

T :__100______0.3 1

T :__100______2 0.25

T :__81.32___ 3

0.2

(m3/kg) 0.15

P (MPa) P v :_1.271__ 1 0.1

v :_3.419____ 2

0.05 v :_1.271_____ 3 0

0.0001 0.001 0.01 0.1 1 10 v (m3/kg)

5 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 14. An ideal gas and 365 ml of liquid water are contained in a 500-ml rigid vessel; the vessel itself weighs 20 grams. The water and gas are at 25 oC. The vessel and its contents weigh a total of 385.218 grams at equilibrium. A pressure gage reads Pg = 100 kPa. Assuming Patm = 100 kPa:

a. Calculate the molecular weight of the gas above the water.

PV = mRT, R = (PV/mT) -3 3 -3 (g) R = 200 kPa(0.135x10 m )/(0.218 x10 kg)(298K) 500 mL R = 0.4156 kJ/kg-K container 365 mL (l) Since m = 385.218 – 20 – 365 g = 0.218 g OR 0.218 x10-3 kg

385.218 gr scale MW = /R = 8.314 kJ/kmol-K/0.4156 kJ/kg-K = 20 kg/kmol

Gas is neon (FYI)

b. If the vessel is heated so that the temperature of the system reaches 99.5 oC, what is the pressure of the system?

Assume Vgas = constant since water does not boil

P1/T1 = P2/T2 and P2 = P1T2/T1 = 200 kPa (372.5/298) = 250 kPa

o Check assumption, T2 < Tsat for water @250 kPa (127.4 C), so okay c. Are the contents of the vessel a pure substance? Explain.

No, because system does not have homogeneous molecular composition (Neon and water). (Answer: because system is two is NOT CORRECT.) d. You neglected the vapor pressure of the water to do this problem. Explain why is that justified in both parts a and b.

Can neglect vapor pressure of water because water is compressed liquid at both states 1 and 2 and vapor pressure would be negligible.

o o State 1: T = 25 C << Tsat @ 200 kPa (120.2 C) o o State 2: T = 99.5 C < Tsat @ 250 kPa (127.4 C)

6 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 15. A rigid container initially holds water at 1.2 MPa and 250 oC. The container and its contents are allowed to cool to 120 oC.

a) the final pressure 3 o v1 = 0.19241 m /kg (Table A-6) = v2 . vf < v2 < vg @ 120 C so water is saturated mixture o P2 = Psat @ 120 C = 198.7 kPa

b) the system quality @ 120 oC x2 = (v2-vf)/vfg = (0.19241-0.00106)/(0.89133-0.00106) = 0.215

c) the final specific

o h2 = x(hfg) +hf @ 120 C = 0.215(2202.1) + 503.81 kJ/kg = 977.3 kJ/kg

d) Show this process on the T-v diagram below

Water Liquid-Vapor PhaseT-v Diagram

400 350 300 250 200 150

100

C) o

50 T ( T 0 0.0001 0.001 0.01 0.1 1 10 100 1000

v (m3/kg)

7 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 3 16. Goodyear designs steel belted tires for trucks, which initially contain 0.5 m of compressed air at 200 kPa. When the trucks are loaded, the tires compress and their effective volumes drop to 0.3 m3.

Answer the following questions, and draw the corresponding points and paths on the P-v diagram. Show your work, calculations and assumptions on a separate page. Label all the paths where WORK is done on or by the system with a capital W

PATH 1  2 Assuming the temperature inside the tire is the same as outside air (30 C), estimate the pressure increase inside the tire under load

2  3 Assume the loaded truck moves on hot pavement and the temperature increases from 30C to 46 C; estimate the pressure increase assuming the effective volume of 0.3 m3 is completely constrained by the steel belts

3  4 Assume the hot tires are then unloaded at 46 C, and the effective volume rebounds to the original 0.5 m3 1 2: P1V1 = P2V2 P2 = 200 kPa(0.5/0.3) = 333.3 kPa 2  3: P2/T2 = P3/T3 P3 = 333.3 kPa(319/303) = 350 kPa P 3 4: P4V1 = P3V2 P4 = 350 kPa(0.3/0.5) = 210 kPa

kPa 350 ---

333 ---

210 --- 200 ---

V

0.1 0.2 0.3 0.4 0.5 0.6

8 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 17. A 0.2 m3 container has 5 kg R-134a at 0 oC (273 K) in a closed system at state 1.

a) What is the pressure of the R-134a at state 1? 3 3 o v1 = 0.2 m /5 kg = 0.04 m /kg. vf < v1 < vg @ 0 C  system is saturated mixture o P = Psat @ 0 C = 293 kPa (A-11)

b) What is the mass of liquid refrigerant in the container at state 1? mf/mT = 1 - x x = (0.04-0.0007723)/(0.069255-0.0007723) = 0.573 mf = 5 kg (1-0.573) = 2.14 kg

Heat is added and the R-134a expands in an isothermal process until P = 100 kPa at state 2.

c) Calculate the change in enthalpy ( H) in kJ for process 1  2.

H = m(h2 – h1) o h1 = x(hfg) + hf @ 0 C = 0.573(198.6) + 51.86 = 165.5 kJ/kg @ P = 100 kPa, T = 0 oC, R-134a is superheated vapor h2 = 255.58 kJ/kg (A-13)

H = 5 kg (255.58 – 165.5) kJ/kg = 450 kJ

Then heat is removed from the R-134a in an isobaric process until T = - 32 oC (241 K) at state 3.

d) What phase is the refrigerant at state 3? Justify your answer. (5) o @ P = 100 kPa, Tsat = -26.37 C. T3 < Tsat, so R-134a is compressed liquid.

e) Calculate the change in enthalpy ( H) in kJ for process 2  3. (5)

o h1 ≈ hf @ -32 C = 10.1 kJ/kg (A-11)

H =5 kg (10.1 – 255.58) = - 1,227.4 kJ

f) Draw the processes on the P-v and on the T-v diagrams below

9 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class

Saturated Liquid-Vapor Phase Diagram R-134a

800 700 600 500

P (kPa) 400 300 200 100 0 0.0001 0.0010 0.0100 0.1000 1.0000

v (m3/kg)

R-134a Saturated Liquid-Vapor

120

100 80

60

40

C) ( T 20

0

-20 -40 0.0001 0.0010 0.0100 0.1000 1.0000 v (m3/kg)

10 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 18. 7 kg of air, T = 27 oC and P = 100 kPa is compressed in an isothermal process (1  2) such that:

V1 V 2 3 a) Calculate V2 and P2 at state 2

P1V1 = P2V2 P2 = P1V1/V2 = 100 kPa (3) = 300 kPa

b) What is the change in enthalpy ( H) for 1  2?

H = mcP (T2 – T1) = 0 kJ

c) What is the boundary work of the system for 1  2?

Wb = mRT(ln(V2/V1)) = 7 kg(0.287 kJ/kg-K)300K(ln(1/3)) = -662 kJ

Heat is then added to the air, which expands in an isobaric process (2  3) until V3 = V1.

d) What is the change in enthalpy ( H) for the process 2  3?

H = mcP (T3 – T2)

T3 = (V3/V2)T2 (K) = (V1/V2)T2 (K) = 3(300 K) = 900 K cP @ (300+900)/2 = 1.051 kJ/kg-K H = 7 kg(1.051 kJ/kg-K)(900-300)K = 4,414 kJ

e) What is the boundary work of the system for 2  3?

Wb = P2(V3 – V2) = P2(V1 – V2) = P2(3V2 – V2) = 2P2V2

3 V2 = 7 kg(0.287 kJ/kg-K)300K/300 kPa = 2 m

3 Wb = 2*300 kPa(2 m ) = 1,200 kJ

11 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class

Finally, heat is removed from the air in a constant volume (isochoric) process until P = P1.

f) Calculate the net boundary work of the cycle.

Wb31 = 0 (isochoric)

Wnet = 1,200 – 662 kJ = 538 kJ

g) Draw the cycle on the P-V diagram below. Use arrows to indicate direction of process.

Cycle P-V Diagram

350 300 P (kPa) 250 200 150 100 50 0 0 1 2 3 4 5 6 7 8

V (m3)

12 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 19. An autoclave is a rigid-tank device where wet steam (x = 0.7) is used to sterilize equipment or other material. The working temperature of an autoclave with a volume of 0.1 m3 (exclusive of material being sterilized) is 140 oC. a) What is the pressure in the autoclave vessel during operation?

o x = 0.7, P = Psat @ 140 C = 361.53 kPa (A-4) b) What is the mass of liquid water that must be added to the autoclave at the start of the sterilization to maintain the working conditions above? m = V/v

o 3 v = 0.7(vfg) + vf @ 140 C = 0.7(0.5085-0.00108)+0.00108 = 0.3563 m /kg (A-4) m = 0.1 m3/0.03563 m3/kg = 0.281 kg

c) What is the volume of the added water at 100 kPa and 25 oC? (5)

V = m/ where at atmospheric P and room temp = 1 kg/L

V = 0.281 kg/1 kg/L = 0.281 L

13 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class Autoclaves fail disastrously if the heater fails to shut off and the pressure rises beyond the capacity of the tank vessel. To prevent an explosion, a pressure relief valve is set to release steam when the pressure reaches 500 kPa. d) Will there still be liquid water in the autoclave when the steam pressure reaches 500 kPa? Justify your answer.

Rigid tank, v = 0.3563 m3/kg @ 500 kPa. vf < v < vg @ 500 kPa  system is saturated mixture (A-5) therefore there will be liquid water

In fact x = (0.3563-0.001093)/(0.37483-0.001093) = 0.95 so 5% of saturated mixture will be liquid.

e) What would the temperature of the steam in the autoclave be just before the valve releases at 500 kPa?

o T = Tsat @ 500 kPa = 151.83 C (A-5) f) Draw the process going from normal operating conditions for the autoclave to maximum pressure on the P-v diagram for water below.

0.6

0.5

0.4

0.3 P (MPa) P 0.2

0.1

0 0.0001 0.001 0.01 0.1 1 10 100 v (m3/kg)

14 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class

20. Complete the following table

substance P (kPa) T (oC) v (m3/kg) x* phase H2O 200 120.21 0.35495 0.4 Saturated mixture H2O 200 85 0.001032 na Compressed liquid air 200 354 0.9 na Ideal gas

R-134a 200 -10.09 0.0500 0.497 Saturated Mixture R-134a 200 0 0.10481 na Superheated vapor * use "na" for "not applicable" where quality does not apply

1. v = 0.4(0.88578-0.001061) + 0.001061 = 0.35495 o 3 2. 2. T < Tsat @ 200 kPa, so water is compressed liquid, v ≈ vf @ 85 C = 0.001032 m /kg 3. T = Pv/R = 200 kPa(0.9 m3/kg)/0.287 kJ/kg-K = 627.2K – 273 = 354 oC o 4. vf < v < vg @ 200 kPa, so R-134a is saturated mixture, T = Tsat @ 200 kPa = - 10.09 C x = (0.05-0.0007533)/(0.099867-0.0007533) = 0.497 o 5. T > Tsat @ 200 kPa (- 10.09 C) so R-134a is superheated vapor, A-13 for v

15 AREN 2110, Spring 2010 NAME______HOMEWORK 5: Practice Problems for Midterm 1 DUE Friday, 2/19 in class 21. Use the P-V diagram below to answer the following questions

P

4 3

1 2

Circle the correct answer V

a. The net work for the cycle is:

1. Zero

2. Positive

3. Negative

4. Cannot tell from the diagram

b. The processes from states 1  2 and 3  4 are:

1. Isothermal

2. Isobaric

3. Isochoric

4. Isenthalpic

c. The net enthalpy change for the cycle is

1. Zero

2. Positive

3. Negative

4. Cannot tell from the diagram

16