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 closed system cycle consisting of a sequence of four processes, over the complete cycle, the system MUST a. increase in energy b. decrease in energy c. have no change in energy d. do no work e. increase in volume 2. At the critical point for any pure substance, a. liquid-solid-vapor phases are in equilibrium b. Pv = RT c. pressure 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. heat 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 temperature d. it doesn’t react with glass e. it has very high density 5. Entropy transferred during an irreversible process, 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 isobaric process. 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 isothermal process 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 ideal gas 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 internal energy 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 phase 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.