CBE 141: Chemical Engineering Thermodynamics, Spring 2018, UC Berkeley Midterm 2 March 22, 2018 Time: 80 minutes, closed-book and closed-notes, one-sided 8 ½ x 11 equation sheet allowed Please show all work and clearly mark your answers. Write your name on any additional pages of scratch work. You must turn in your equation sheet with your exam. Point Totals: Problem 1-6 32 Problem 7 20 Problem 8 30 Problem 9 20 Problem 10 15 Extra credit 10b 5 Problem 11 20 Extra credit 12 10 Total (points) 137 Total (percentage) 100 This exam has 14 pages and 12 questions 1
Problem 1 (6 points) A turbine cannot be run with 100% efficiency due to a variety of losses, one of which includes friction at the upstream nozzle which reduces the available fluid enthalpy coming into the turbine. Which of the following is a consequence of this non-ideality? You may circle more than one, or none at all. A. The friction of the fluid in the nozzle increases the enthalpy available to the turbine B. The friction of the fluid in the nozzle reduces the enthalpy available to the turbine C. The velocity leaving the nozzle is less than the velocity calculated with isentropic expansion. D. The velocity leaving the nozzle is more than the velocity calculated with isentropic expansion. E. The flow coming into the turbine can be considered isothermal F. The flow coming into the turbine can be considered isentropic Rubric: 1 point for each a-f (i.e. did they circle or not-circle correctly) Problem 2 (5 points) When comparing the Otto and Diesel cycles, which of the following are true? You may circle more than one, or none at all. A. Otto cycle has higher efficiency for the same maximum (i.e. compression) pressure B. Otto cycle has higher efficiency for the same compression ratio C. The area enclosed by the Diesel cycle on a PV diagram (i.e. Work) is larger than the area enclosed by a real compression engine. D. The area enclosed by the Otto cycle on a PV diagram (i.e. Work) is larger than the area enclosed by a real spark-ignition engine. E. The largest contributor to non-ideality (i.e. inefficiency) of running an Otto or Diesel cycle is ignoring the intake and exhaust strokes. Rubric: 1 point for each a-e (i.e. did they circle or not-circle correctly) Problem 3 (5 points) Consider the following exact differential: dddd = TTTTTT + VVVVVV, which of the following is INCORRECT: a. PP = TT d. VV = PP b. = VV e. = SS SS PP c. = 2 HH SS Rubric: 1 point for each a-e (i.e. did they circle or not-circle correctly) 2
Problem 4: (5 points) Which will have the highest magnitude for the residual enthalpy (H R ) at 100 o C? The vapor pressure of water at 100 o C is 1 bar. A. Water at 2.0 bar B. Water at 0.5 bar C. Argon at 2.0 bar D. Argon at 0.5 bar E. Not enough information H R = H Hig Water at 2 bar is liquid Water at 0.5 bar is vapor Therefore, water at 2 bar has higher H R Water vs. Argon: Mol interaction for water >> Argon (argon vapor is close to ideal gas) Therefore answer A Problem 5 (5 points) Which of the following statements are true regarding an ideal gas mixing process. You may circle one, more than one, or none at all: A. Mixing is always a spontaneous process B. Enthalpy of mixing must be greater than zero C. Entropy of mixing must be greater than zero D. The change in entropy during a mixing is not temperature dependent E. A 50/50 binary mixture will achieve the highest change in entropy when mixed 3
Problem 6 (6 points) Label the on the given Txy graph of benzene and toluene at 1atm and an initial composition of z = 0.40 (3 points): A. Superheated vapor region B. Dew point C. Bubble point and calculate the vapor/liquid compositions and relative compositions of the vapor and liquid phases if a mixture with an initial composition of z = 0.40 at 75 C is heated to 100 C (3 points). (+1 pt for each correct label) 4
(3 pts for these calculations) xx 1 = 0.26 aaaaaa yy 1 = 0.45 (1pt) LL = yy 1 zz 1 = 0.45 0.40 = 0.26 (1pt) yy 1 xx 1 0.45 0.26 VV = zz 1 xx 1 = 1 LL = 0.74 (1pt) yy 1 xx 1 5
Problem 7 (20 points) Consider the following ideal Vapor-Compression cycle: Starting with the first law energy balance and second law entropy balance, prove that the maximum value of the coefficient of performance for a refrigeration cycle is equal to the following expression: 1 st law: For a cycle U = Q + W = 0 (returns to the initial state) (2 points) Q E + Q C + W C = 0 2 nd law: (3 points) Solve 1 st law for Q C and substitute into 2 nd law with Q E /W C =COP (5 points) on the left hand side: (5 points) Because the denominator term is always positive, and entropy generation decreases the COP, therefore the maximum COP occurs when Sgen = 0 (2 points) and therefore: (3 points) For grading, OK if students first set Sgen to zero and omit it from the energy balance altogether: 6
Problem 8 (30 points) A diesel cycle has a compression ratio of 18 and a cutoff ratio of 2.5. Initially (at state 1), the state is fixed by a pressure of 0.9 bar, and a temperature of 300 K. a) What are the temperatures of the states, with the initial state as state 1, following an isentropic compression to T 2, an isobaric heat addition to T 3, and an isentropic expansion to T 4. (30 points) b) Extra Credit What is the thermal efficiency of the cycle? You may assume Cp = 1.005 kj/kg K and Cp/Cv = 1.4 for the working fluid. (5 points no partial credit) 7
ηη = WW QQ iiii = QQ iiii QQ oooooo QQ iiii = 0.61 oooo 61% 8
Problem 9 (20 points) Using the ideal gas equation of state (PV=RT), calculate the three partial derivatives that appear in the triple product rule between P, T, and V. Hint: Show that their product equals -1. 9
Problem 10 (15 points) Pure ethanol boils at a temperature of 63.5 ºC at a pressure of 400 torr. It also boils at 78.4 ºC and 760 torr. a) What is the enthalpy of vaporization of pure ethanol? (15 points) b) Extra credit: Estimate the saturation pressure for ethanol at 100 ºC. (5 points no partial credit) a) Using the Clausius-Clapeyron equation: ddpp ssssss dddd = HH vvvvvv RRTT 2 / PP ssssss For a liquid to gas transition using the simplified form of the Clausius-Clapeyron equation: ln PP 2 ssssss HHvvvvvv ssssss = 1 1 PP 1 RR TT 2 TT 1 Plug in values to find H vap Remember to use temperature in Kelvin HH vvvvvv = RR ln PP 2 ssssss ssssss PP 760 tttttttt 1 JJ ln 1 1 = 8.314 KK mmmmmm 400 tttttttt 1 TT 2 TT 1 (78.4 + 273.15 KK) 1 (63.5 + 273.15 KK) HH vvvvvv = 42386 JJ = 42.386 kkkk/mmmmmm mmmmmm b) Now plug in our temperature of 100 ºC into the equation and solve for corresponding saturation pressure. ssssss PP TT=100 CC = 760 tttttttt exp HHvvvvvv 1 RR (100 + 273.15 KK) 1 78.4 + 273.15 KK) PP ssssss = 1759.47 tttttttt Or similarly ssssss PP TT=100 CC = 400 tttttttt exp HHvvvvvv 1 RR (100 + 273.15 KK) 1 63.5 + 273.15 KK) PP ssssss = 1759.47 tttttttt 10
Problem 11 (20 points) Consider a mixture of chemical species 1 and 2. Below is a plot of partial molar volumes in cm 3 /mol for both species (VV 1 and VV ) 2 with respect to x 1. For a mixture of 2 mol of species 1 and 3 mole of species 2, determine the following (approximate values based on interpretations of the graph above are okay): a) Total molar volume, V b) Pure component molar volumes, V 1 and V 1 c) Molar volume of an ideal mixture of 2 mol species 1 and 3 mol species 2 a) VV = xx 1 VV 1 + xx 2 VV 2 = (0.4) 48 ccmm3 ccmm3 ccmm3 + (0.6) 69 = 60.6 mmmmmm b) Look up VV 1 at x 1 = 1 and VV 2 at x 1 = 0 V 1 = 50 cm 3 /mol, V 2 = 70 cm 3 /mol c) V = x 1 V 1 + x 2 V 2 = 62 cm 3 /mol mmmmmm mmmmmm 11
EXTRA CREDIT Q 12 (10 points final answer must be correct, no partial credit!): What is the residual volume in the ideal gas state? Hint: Show that simultaneously applying both ideal gas limits of P 0 and Z 1 yields an indeterminate (i.e. 0 0). Hint: Calculate the ratio of V R /V as a function of Z. V R = V (RT/P) V R = RT/P (Z-1) V R /V = [RT(Z-1)]/PV => lim P->0 and Z->1 yields => [RT(1-1)]/0V = 0/0 The ratio of V R /V does go to zero as the ideal gas limit approaches. V R represents an infinitesimal fraction of the system volume, and is thus negligible. 12
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