Unified Thermodynamics Quiz 1 November 22, 2006 Calculators allowed. No books or notes allowed. A list of equations is provided. Put your ID number on each page of the exam. Read all questions carefully. Do all work for each problem on the pages provided. Show intermediate results. Explain your work --- don t just write equations. Partial credit will be given (unless otherwise noted), but only when the intermediate results and explanations are clear. Please be neat. It will be easier to identify correct or partially correct responses when the response is neat. Show appropriate units with your final answers. Box your final answers. Exam Scoring #1 (5%) #2 (10%) #3 (25%) #4a (15%) #4b (15%) #5 (30%) Total 1
1) (5%, partial credit given, LO#2, LO #5) Is it possible to have an adiabatic process with a temperature change? Why or why not? Yes, it is possible. Consider the first law of thermodynamics for an ideal gas with constant specific heats, u = c v T = q w. If q=0, the temperature will still change if energy transfers are made to the system as work. 2) (10%, partial credit given, LO#1, LO#4, LO#6) An inventor claims that a new heat engine develops 2 kw of work when heat at the rate of 5 kj/s is transferred to the engine. The cycle operates between a maximum temperature of 1400 K and a minimum temperature of 430 K. Evaluate this claim. Is it possible? The most efficient engine operating between two different temperature reservoirs is a Carnot engine. For the Carnot engine Carnot = 1 T L /T H = 69%. The inventor claims an efficiency of W cycle /Q in = 2/5 = 40%. So yes, it is possible. 2
3) (25%, partial credit given, LO#1, LO#2, LO#4) The sketch below shows an ideal experiment done in a perfectly insulated, rigid container with compartments separated by a frictionless piston. The two compartments contain different amounts of the same gas (m A =1.2 m B ). The piston is nonadiabatic (heat can be transferred) and moves very slowly. Compartment A is initially at a higher temperature than compartment B (T A > T B ) but the pressure in the two compartments is the same. Assume an ideal gas with constant c p and c v. a) Which way, if at all, does the piston move? Indicate by an arrow on the drawing and give an explanation (6 pts). b) What are the final temperatures and volumes in compartments A and B? Express your answer in terms of given properties (17 pts). c) What is the net work done in the experiment (6 pts)? a) The piston moves to the left as shown on the diagram. Energy is transferred via heat from the high temperature side A to the low temperature side B. This causes the high temperature side A to contract and the low temperature side B to expand. b) Since the whole system is insulated and rigid, the total energy of the system (side A plus side B) is constant; there is no interaction with the surroundings. The work from side A = - work from side B, heat from side A = - heat from side B. In the final state, the temperature in both sides is the same, the pressure in both sides is the same, so the specific volume in both sides is the same. Since they differ in mass by a ratio of 1.2, the final volumes will differ by a ratio of 1.2 (more volume for side A since it has more mass). V Afinal = 1.2V Bfinal 3
-- BLANK PAGE FOR EXTRA WORK -- Therefore we can write: U system = Q W = 0 ( c v m A T final + c v m B T final )= ( c v m A T A + c v m B T B ) or T final = m A T A + m B T B m A + m B = 1.2T A + T B 2.2 c) The box is rigid and thermally-insulated. Therefore, there is no energy exchange between the system and the surroundings. The net work for the experiment is zero. 4
4a) (15%, partial credit given, LO#4, LO#6) Assume all processes are quasi-static. Neatly draw and label a thermodynamic cycle on a p-v diagram consisting of Leg 1-2: adiabatic compression Leg 2-3: isothermal expansion Leg 3-4: constant pressure compression Leg 4-1: constant volume heat extraction 300 250 2 Pressure (kpa) 200 150 100 4 3 50 1 0.5 0.75 1.0 1.25 1.5 Specific Volume (m 3 /kg) lines of constant temperature For 1-2, pv =constant adiabat is steeper than pv=constant (an isotherm) 4b) (15%, partial credit given, LO#4, LO#6) Determine the sign of the heat and work transfers and change in internal energy for each leg and the cycle as a whole. q (+, - or zero) w (+, - or zero) u (+, - or zero) Leg 1-2 0 - + Leg 2-3 + + 0 Leg 3-4 - - - Leg 4-1 - 0 - Sum for cycle + + 0 5
5) (30%, partial credit given, LO#1, LO#4, LO#6) A closed thermodynamic system employs the cycle shown in the figure below: a) If the system traverses the cycle in the counterclockwise sense, is the work transfer positive or negative? If the system traverses the cycle in the clockwise sense, is the work transfer positive or negative (5 pts)? b) What is the net quasi-static work transfer from this cycle (10 pts)? c) If the system performs as a heat engine and the heat transfer to the low temperature heat reservoir is 50 MJ, determine the thermal efficiency of the cycle (7 pts). d) If the system performs as a refrigerator and the heat transfer to the high temperature heat reservoir is 50 MJ, determine the coefficient of performance of the system, COP (8 pts). a) The work is negative for counter clockwise and positive for clockwise. b) Work for cycle is equal to the area enclosed by the ellipse = (0.68MPa)(2m 3 ) = 4.27MJ c) Thermal efficiency = W cycle /Q in from 1 st Law, W cycle = Q cycle = Q H +Q L therefore 4.27MJ = Q H -50MJ, so Q = 54.27MJ = W cycle /Q in = 4.27MJ/54.27MJ = 0.079 = 7.9% d) COP = Q L /(-W cycle ) from 1 st Law, W cycle = Q cycle = Q H +Q L therefore 4.27MJ = -50MJ + Q L so Q L =45.7MJ COP = 45.7MJ/4.27MJ = 10.7 6
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