ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION June 19, 2015 2:30 pm - 4:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closed-book examination. 2. Permitted aids include: one 8.5 in. 11 in. crib sheet (one side only), Property Tables Booklet for Introduction to Thermodynamics and Heat Transfer, Yunus Cengel, 2nd ed calculator 3. Answer all questions in the space provided. If additional space is required, use the back of the pages or the blank pages included. 4. It is your responsibility to write clearly and legibly. Clearly state all assumptions. Part marks will be given for part answers, provided that your methodology is clear. Question Marks Grade 1 19 2 15 3 17 TOTAL 51
ECE309 Midterm Exam S15 2 Question 1 (19 marks) Consider the piston/cylinder arrangement containing R-134a refrigerant, as shown in the figure below. The frictionless piston is free to move between the stops and because of its weight requires 380 kp a to move. When the piston rests on the lower stops, the enclosed volume is 20 L. When the piston reaches the upper stops, the volume is 60 L. The cylinder initially contains 2.5 kg of R134a at 380 kp a with a quality of 0.15. It is heated until the R134a exists as saturated vapor. a) for the initial conditions, determine if the piston is in contact with the lower stops or if it is positioned somewhere between the lower and upper stops b) for the initial state, determine the volume of the liquid R134a [m 3 ], the volume of the R134a vapor [m 3 ] and the internal energy [kj/kg] c) for the final state, determine the pressure [kp a], specific volume [m 3 /kg] and the internal energy [kj/kg] d) find the work transfer for the entire process, [kj] e) find the heat transfer for the entire process, [kj] f) sketch the P v diagram for the actual process, from initial to final states. Label the axes, state points and all relevant property values at the state points g) property tables for saturated R-134a (Table A-11) use 40 C as the reference value. Show how this data could be converted if 0 C was the new reference value. As an example, calculate the value of enthalpy for saturated liquid at 20 C (h f (@20 C) if the reference value is 0 C. R134a Q in Assumptions 1. the piston is frictionless 2. KE P E 0 3. steady state 4. quassi equilibrium
ECE309 Midterm Exam S15 3 Part a) and b) Since the initial pressure of the cylinder is equivalent to the pressure required to lift the piston, i.e. P 1 P cyl the piston is free to move off the bottom stops. To see if it is initially above the bottom stops, determine the volume of the initial state as follows: The specific volume of the liquid and the gas can be determined at the initial state of P 1 380 kp a. From Table A-12 v f1 v g1 0.000787 m 3 /kg 0.053970 m 3 /kg The mass of the liquid and gas can be determined based on the quality such that Now the volume can be determined as The total volume is m f1 0.85 2.5 kg 2.125 kg m g1 0.15 2.5 kg 0.375 kg V f1 (2.125 kg)(0.000787 m 3 /kg) 0.001672 m 3 Part b) V g1 (0.375 kg)(0.053970 m 3 /kg) 0.020239 m 3 Part b) V 1 V f1 + V g1 0.001672 m 3 + 0.020239 m 3 0.021911 m 3 Since the starting volume is greater than 0.02 m 3 the piston starts above the lower stops. Part a) The internal energy is u 1 (1 x)u f + xu g (1 0.15) 61.53 + 0.15 234.23 87.44 kj/kg Part b) Part c) At the final state, the R134a is a saturated vapour with x 3 1.0 and a volume of 0.06 m 3. v 3 v g3 V 3 m 0.06 m3 2.5 kg 0.024 m3 /kg
ECE309 Midterm Exam S15 4 From Table A-11 we see that the specific volume, v 3 is between 32 34 C, therefore we interpolate to find T 3 33.56 C P 3 852.81 kp a u 3 247.91 kj/kg Part d) The only boundary work done is between state 1 and 2 since the piston does not move when it hits the upper stops. W 1 2 P dv 380 kp a (0.06 0.021911) m 3 ( kj kp a m 3 14.47 kj (where the negative indicates work out of the system) W out 14.47 kj ) Part e) The heat transfer can be obtained by performing an energy balance. Q in + W in 0 + U + KE 0 + P E 0 Q out 0 +W out Q in (U 3 U 1 ) + W out 2.5 kg (247.91 87.44) kj/kg + 14.47 kj 415.65 kj
ECE309 Midterm Exam S15 5 Part f) P (kpa) P 852.81 3 3 P 380 1 1 2 v 0.024 2 piston stops 3 v (m /kg) Part g) The change in enthalpy between a reference state of 40 C and 0 C can be calculated as h C p (T T ref ) Alternate solution 1: From Table A-3 we can find the specific heat at 40 C and 0 C C p @ 40 C 1.24 kj/(kg K) C p @ 0 C 1.34 kj/(kg K) C p 1.29 kj/(kg K) Therefore h 40 C 0 C 1.29 kj/(kg K) (0 ( 40) 51.6 kj/kg With a reference value of 0 C h f (@20 C) (79.32 51.6) 27.72 kj/kg Alternate solution 2: The change in enthalpy from 40 C and 0 C in Table A-11 is h 40 C 0 C h f ((@0 C) h f ((@ 40 C) 51.86 0 51.86 kj/kg With a reference value of 0 C h f (@20 C) (79.32 51.86) 27.46 kj/kg
ECE309 Midterm Exam S15 6 Question 2 (15 marks) Liquid water at 90 C and 350 kp a steadily enters an engine at 5 kg/s. Superheated steam at 200 C and 350 kp a also flows steadily into the engine through another pipe at 10 kg/s. The two flows combine inside the engine and then leave the engine together at 350 kp a. The engine can be assumed to be adiabatic. Determine the maximum rate of work transfer [kw ] that can be produced by this engine. In general, what are the three different mechanisms that can cause the entropy in a control volume to change? Assumptions 1. steady state, steady flow 2. adiabatic conditions 3. KEPE0 Part a) For maximum work transfer, the process must be reversible. Conservation of Mass ṁ 3 ṁ 1 + ṁ 2 Entropy Balance Since the engine is steady state (ds/dt 0), adiabatic ( Q 0), and reversible (Ṡ gen 0) ṁ 1 s 1 + ṁ 2 s 2 ṁ 3 s 3 State 1: T sat (@350 C) 138.86 > T 1 90 C, therefore state 1 is sub cooled. Since the sub cooled tables are out of range we will use the approximation for s 1 s 1 s f@90 C 1.1929 kj/kg K Table A-4 s 2 7.24275 kj/kg K Table A-6
ECE309 Midterm Exam S15 7 s 3 ṁ1s 1 + ṁ 2 s 2 m 3 ṁ1s 1 + ṁ 2 s 2 ṁ 1 + ṁ 2 (5 kg/s)(1.1929 kj/kg K) + (10 kg/s)(7.24275 kj/kg K) (5 kg/s) + (10 kg/s) 5.2261 kj/kg K At state point 3 we can calculate the quality as x 3 s 3 s f s g s f Now we can calculate the enthalpy at state point 3 as h 3 h f + x 3 (h g h f ) 5.2261 1.7274 6.9402 1.7274 0.6712 584.26 + 0.6712 (2732.0 584.26) 2025.82 kj/kg Performing an energy balance over the engine and assuming the work produced is out of the engine ṁ 1 h 1 + ṁ 2 h 2 ṁ 3 h 3 + Ẇ The enthalpy at the various states can be determined as follows: h 1 h f@90 C + v f@90 C (P P sat@@90 C) 377.04 kj/kg + 0.001036 m 3 /kg (350 70.183)kP a 1 kj/m3 1 kp a 377.33 kj/kg use approximation since the sub cooled tables are out of range h 2 2863.4 kj/kg Table A-6 Ẇ ṁ 1 h 1 + ṁ 2 h 2 ṁ 3 h 3 (5 kg/s)(377.33 kj/kg) + (10 kg/s)(2863.4 kj/kg) (15 kg/s)(2025.82 kj/kg) 133.35 kw
ECE309 Midterm Exam S15 8 For a control volume system where there is mass crossing the boundary, the 3 mechanisms that can cause entropy to change are: 1. heat transfer ( Q/T T ER ) 2. irreversibilities ( Ṡ gen ) 3. entropy transport with mass ( ṁs)
ECE309 Midterm Exam S15 9 Question 3 (17 marks) A solar driven power plant uses water as the working fluid. During daytime operation, the inlet state of the boiler is 35 C at 50 kp a and the inlet state of the turbine is 100 C at 50 kp a. The turbine is assumed to be isentropic. The air cooled condenser operates at a pressure of 5 kp a. The work output is 75 kw when the solar collector input is 650 W/m 2. The water leaves the condenser as a saturated liquid. a) determine the first law efficiency of the system b) determine the mass flow rate of the working fluid, [kg/s] c) determine the required collector surface area, [m 2 ] d) are there any restrictions on this system based on the magnitude of the ambient temperature? If yes, what is the maximum and/or minimum ambient temperature where this system will function properly? 2 Q' sc650 W/m Turbine W 75 kw t Solar Collector/ Boiler W p Pump Q L Condenser (air cooled) Assumptions 1. steady state, steady flow 2. quassi equilibrium 3. turbine is isentropic 4. incident radiation on the solar cooler is uniformly distributed State T [ C] P [kp a] x h [kj/kg] s [kj/kg K] comments 1 5 137.75 sat. liq. 2 35 50 146.69 sub. cooled 3 100 50 2682.4 7.6953 super heated 4 5 0.9118 2347.0 7.6953 isentropic
ECE309 Midterm Exam S15 10 State Point 1 From Table A-5 @P 5 kp a h 1 h f 137.75 kj/kg State Point 2 The exit of the pump is in the sub cooled region since T sat (50 kp a) 81.32 C > T 2 50 C h 2 h f (T ) + v f (T )[P P sat (T )] 146.64 kj/kg + 0.001006 m 3 /kg[50 5.6291]kP a 146.69 kj/kg ( 1 kj 1 kp a m 3 ) State Point 3 @ 50 kp a T sat 81.33 C < T 3 100 C Therefore state point 3 is a superheated vapor and h 3 2682.4 kj/kg and s 3 7.6953 kj/kg K State Point 4 Check to see if state point 4 is inside the dome. We know that s 4 s 3 7.6953 kj/kg K From Table A-5 at a pressure of 5 kp a the entropy of the saturated vapour is s g 8.3938 kj/kg K. Therefore state point 4 is inside the dome and we can find the quality as follows: s 4 s f (1 x) + s g x 7.6953 0.4762(1 x) + 8.3938x Therefore x 4 0.9118. The enthalpy at 4 can be determined as h 4 137.75(1 0.9118) + 2560.7 (0.9118) 2347.00 kj/kg
ECE309 Midterm Exam S15 11 Part a) Performing an energy balance over the turbine we get w t h 3 h 4 2682.4 2347.00 335.4 kj/kg Performing an energy balance over the pump we get w p h 2 h 1 146.69 137.75 8.94 kj/kg The heat input at the collector is given as q in h 3 h 2 2682.4 146.69 2535.71 kj/kg The first law efficiency is given as η 1 w t w p q in 335.4 8.94 2535.71 0.129 Part b) The mass flow rate of the working fluid is given as ṁ Ẇt w t 75 kj/s 335.4 kj/kg 0.2236 kg/s Part c) The required collector area is A sc ṁ q in Q sc 0.2236 kg/s 2535.71 kj/kg 0.65 kw/m 2 872.28 m 2 Part d) Minimum Ambient Temperature If the piping and the solar collector are uninsulated then an ambient temperature below 0 C could potentially allow the working fluid of water to freeze. Maximum Ambient Temperature The condenser pressure is 5 kp a which leads to a saturation temperature of T sat (@ 5 KP a) 32.87 C (Table A-5). If the ambient temperature exceeds 32.87 C then the condenser will be unable to transfer heat to the ambient surroundings.