Name: Discussion Section:

Size: px

Start display at page:

Download "Name: Discussion Section:"

Madeline Johns
5 years ago
Views:

1 CBE 141: Chemical Engineering Thermodynamics, Spring 2018, UC Berkeley Midterm 1 February 13, 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 Problem 7 8 Problem 8 20 Problem 9 21 Problem Problem Problem Extra Credit Q 9 5 Extra Credit Q Total (points) 155 Total (percentage) 100 1

2 Problem 1 (6 points) The plot below represents a gas going from state 1 to state 2 through three different processes. a) Circle the letter corresponding to the process in which the gas does the least amount of work: A B C D A B C D b) Circle the letter corresponding to the process in which the internal energy of the gas changes the most: Problem 2 (5 points) An ideal gas is at 1 atm in an insulated piston-cylinder (see figure). The piston is weightless and frictionless, and in a vacuum. When the mass is removed from the piston, the gas temperature A. Increases B. Decreases C. Does not change D. Need to know initial entropy E. Need to know initial and final entropy du = Q (=0 due to insulation) + W (zero due to vacuum) = 0. Therefore dt = 0 Vacuum 1 kg Gas Stopper Block Piston Problem 3 (5 points) For a Carnot Cycle, the following are true (you may circle more than one answer): A) The Carnot efficiency is highest when the Carnot cycle generates maximum entropy 2

B) The Carnot cycle is reversible C) The work output of a Carnot engine is largest when T hot and T cold are nearly the same D) The Carnot cycle extracts entropy from the hot reservoir, and deposits

3 B) The Carnot cycle is reversible C) The work output of a Carnot engine is largest when T hot and T cold are nearly the same D) The Carnot cycle extracts entropy from the hot reservoir, and deposits entropy in the cold reservoir E) The Carnot cycle is most efficient when the working fluid has a high heat capacity Problem 4: (5 points) Adiabatic expansions and compressions are shown for an ideal gas (see Figure). One curve is reversible and one is irreversible in each figure. Which ones are the irreversible curves? A. 1 & 3 B. 1 & 4 C. 2 & 3 D. 2 & 4 E. Need more information Problem 5 (5 points) Two containers are at the same temperature and pressure and have the same number of moles of the same non-ideal gas: Container A: sealed container, fixed volume Container B: constant pressure piston and cylinder. If 100 kj of heat are added to each container, container A is container B. a. Hotter b. Cooler 3

4 c. Same temperature d. Not enough information. Reason: Cv < Cp for the same fluid, and dt is larger for smaller specific heats. Problem 6 (5 points) A gas undergoes reversible expansion from 40 bar and 500 K to 5 bar and 500 K by two pathways. Which process creates more work? A. A constant pressure process to the final volume and then constant volume to 5 bar. B. A constant volume process to 5 bar and then constant pressure to the final volume. C. Same work for both Problem 7 (8 points) By volume, dry air contains 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.04% carbon dioxide. Can air be considered an ideal gas at 100 ºC? (Tc nitrogen = -147 C). Answer in 3 sentences or less. A: Air is composed primarily of nitrogen. The critical temperature of nitrogen is -147 ºC, well below the specified 100 ºC temperature. Therefore, air does behave as an ideal gas at 100 ºC Problem 8 (20 points) Considered a closed system filled with a gas. The path followed by the gas during a mechanically reversible process is described by the following hypothetical equation of state (that we have not covered in class): P + av = c where a and c are constants. In the initial state, P1= 30 bar and V1 = m 3. In the final state P2= 65 bar and V2= m 3. During the process, heat in the amount of 6000 J is transferred to the gas. Hint: solve for constants a and c, and then substitute a and c into the equation for W. 1 bar = 105 N/m2 and 1 J = 1 N m. a) Determine W and DU for the process (15 points) b) Suppose the gas followed a different path connecting the same initial and final states. Which of the quantities (Q, W, DU) must be unchanged? Why? (5 points) 4

5 Since the process is reversible, the work done is given by dw = - PdV. In order to integrate to find the total work W, we must find an expression for P in terms of V. This is given by the equation of state: P = c av Then 2 æ T 2 T T T av ö W =-ò( c- av ) dv =- ç cv - V 2 è ø To find the change in internal energy, we use the first law of thermodynamics: U = Q+W, where, Q is given and W is calculated from the above equation. To get a numerical answer, we have to find a and c from the data given for pressure and volume: P1 = 30 bar, V1 T = m 3, P2 = 65 bar, and V2 T = m 3. Then P1 + a V1 = c P2 + a V2 = c First, we solve for a by subtracting the two equations to eliminate c: (P1 - P2) + a(v1 - V2) = 0 P2 - P1 a = T T V -V 3 a = bar/m Then we calculate c by eliminating a: multiply the first equation by V2, multiply the second equation by V1, and subtract: V2 T * (P1 - a V1 = c) V1 T * (P2 - a V2 = c) (P2 V1 P1 V2 = c(v1 V2) PV - PV c = T T V1 -V2 c = 100 bar Why didn t we find c just by substituting values of a, P, and V into the equation of state? The reason is that if you somehow made an error in calculating a, your value of c would also be wrong. On principle, it is more reliable to determine all parameters independent of each other to minimize errors. Now we substitute a and c into the equation for W, and find W = bar m 3 = 14,250 J Where, we have used 1 bar = 10 5 N/m 2 and 1 J = 1 N m. Finally, U t = Q + W = ,250 U t = J V T T V V 2 1 5

6 b) In this problem, only the change in internal energy DU is a state function (path independent). Heat and work are path dependent. Problem 9 (21 points) The diagram below represents an ideal gas in a closed system. a) For each process below (1, 2, 4), what is the value of that process work, heat, and change in internal energy in going from initial to final state? You can write your answers in terms of molar volume. Fill in the blanks below: (18 points, 2 points/answer) Process 1 work = -P(Vf-Vi) = -R(Tf-Ti) Process 1 heat = Cp(Tf-Ti) Process 1 internal energy change = ΔU = Cv(Tf-Ti) Process 2 work = -RT ln(vf/vi) Process 2 heat = Q = -W = RT ln(vf/vi) Process 2 internal energy change = 0 6

7 Process 4 work = 0 Process 4 heat = Cv(Tf-Ti) Process 4 internal energy change = Cv(Tf-Ti) True b) True or false: These isotherms are all above the critical temperature of this ideal gas. c) EXTRA CREDIT Q9: For process #3, how are the system pressure and volume related? Hint: start with the ideal gas law, and show that the system pressure and volume can be related by a constant that incorporates Cp and Cv, but not temperature. (5 points final answer must be correct, no partial credit!): pv = nrt first law for an adiabatic process is: dq = CvdT + pdv = 0 we can write the ideal gas law as: PdV + Vdp = nrdt solving for dt: dt = [(PdV)/R] + [(VdP)/R] substituting this equation into the ideal gas law equation: dq = Cp PdV + Cv VdP since dq = 0: (students get full credit if they make it this far) (final answer ok if students state it this way also) 7

8 Problem 10 (30 points) The diagram here shows a reversible, three-reservoir cyclic heat engine. For a cycle the engine adsorbs 1200kJ of heat from the reservoir at 700K and performs 200 kj of mechanical work. a. Find Q2 and Q3, the amounts of heat exchanged with other reservoirs, and state whether each reservoir takes in or gives out heat. (20pts) b. Calculate the entropy change of each reservoir and of the engine. (10pts) 8

9 9

10 Problem 11 (30 points) Carbon dioxide (assumed to be an ideal gas) is fed into an adiabatic turbine. The carbon dioxide, initially at 10 bar and 400 o C is expanded to 1 bar. Because of irreversibilities and poor design of the turbine, the work obtained is found to be 25 % less than that of a well-designed (reversible and adiabatic) turbine. a. Compute the outlet temperature of the carbon dioxide and the work obtained to the turbine for both the reversible and irreversible cases. (26 points) b. Calculate the entropy change of the gas, the surroundings, and the universe. (4 points) Data: for CO 2: C " = 22.2 J/(mol K) 10

11 Problem 12 (15 points) The mixing tank shown below initially contains 50 kg of water at 25 C. Suddenly the two inlet valves and the single outlet valve are opened. Both inlet streams have a flow rate of 5 kg/min and the exit stream has a flow rate of 10 kg/min. The tank is insulated and well mixed so that the outlet temperature is always the same as the water temperature in the tank. Assume: C V = C P = C = constant for water Calculate the steady state water temperature obtained in the tank. P = constant T 1 = 80 o C T 2 = 50 o C T 3 (t)=? 11

12 EXTRA CREDIT Q 12 (10 points final answer must be correct, no partial credit!): Develop an expression for water temperature in the tank at any time (i.e. before steady state is reached). 12

13 13

14 14

15 Additional space provided for work. 15

16 Additional space provided for work. 16

17 Additional space provided for work. 17

Name: Discussion Section:

Name: Discussion Section: 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