ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 20 June 2005 Midterm Examination R. Culham & M. Bahrami This is a 90 minute, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side only), Conversion Factors (inside cover of text) and the Property Tables and Figures from your text book. There are 4 questions to be answered. Read the questions very carefully. Clearly state all assumptions. It is your responsibility to write clearly and legibly. Question 1 (12 marks) Air at 300 K, 0.14 MPa is contained in a piston-cylinder device as shown below. The initial volume is 0.3 m 3. The piston is then slowly pushed upward until the volume reaches 0.06 m 3. Heat transfer with the surroundings allows the temperature of the air inside the cylinder to remain at 300 K throughout the process. The pressure relief valve at the top of the cylinder opens when the pressure reaches 0.6 MPa to allow mass to escape and thus prevent the pressure from ever exceeding 0.6 MPa. a) Determine the specific volume [m 3 /kg] of the air in the cylinder at the final pressure and temperature. b) Determine the mass [kg] of the air that flows through the valve. c) If for some reason the valve failed to open, what would the final pressure [MPa] be? d) Determine the final temperature of the air, if the piston-cylinder is insulated (no heat transfer) and a final pressure of 0.6 MPa is achieved through a reversible compressive process while the valve remains closed.
Assume: air is an ideal gas the process is quassi-equilibrium Part a) We know the valve opens at 0.6 MPa and 300 K, therefore v 2 R T 2 P 2 (0.287 kj/kg K) (300 K) ( 10 3 kj/m 3 ) 0.1435 m 3 /kg (0.6 MPa) MPa Part b) From conservation of mass we know m m 1 m 2 m 1 P 1V 1 RT 1 m 2 P 2V 2 RT 2 ( 10 3 kj/m 3 ) (0.14 MPa) (0.3 m 3 ) MPa (0.287 kj/kg K) (300 K) ( 10 3 kj/m 3 ) (0.6 MPa) (0.06 m 3 ) MPa (0.287 kj/kg K) (300 K) 0.4878 kg 0.4181 kg The mass that escapes through the valve is then m (0.4878 kg) (0.4181 kg) 0.0697 kg Part c) Since air behaves as an ideal gas, we know mrt 1 P 1 V 1 mrt 2 P 2 V 2 Since the process is isothermal and a control mass, mrt 1 mrt 2 and we can write P 1 V 1 P 2 V 2 P 2 V 1 V 2 P 1 0.3 m3 0.14 MPa 0.70 MPa 0.06 m3
If the valve remains closed the pressure will rise to 0.7 MPa. Part d) Assume isentropic compression, i.e. process. adiabatic and reversible and no mass escapes during the T 2 T 1 ( ) (k 1)/k P2 P 1 where k 1.4 for air. T 2 (300 K) ( ) 0.6 MPa (1.4 1)/1.4 454.9 K 0.14 MPa
Question 2 (7 marks) An amusement park at the bottom of Niagara Falls wants to install a water turbine to produce 100 kw of power. Water (assumed to be incompressible) would enter the pipeline leading to the turbine at 12.5 C and 101.325 kp a at the top of the falls, 51 m above the turbine exit, with a velocity of 3 m/s. The water would leave the turbine at 12.5 C and 101.325 kp a. The pipeline and the turbine are both adiabatic. a) Determine the mass flow rate [kg/min] of the water. b) Determine the diameter [m] of the pipeline. Assume a circular cross section and uniform diameter throughout the system. Assumptions: steady state, steady flow Part a) Choose the control volume to include the inlet at the top of the falls and the outlet at the exit of the turbine. Performing an energy balance where: ṁe 1 ẇ + ṁe 2 The average temperature of the water is T avg (12 + 13)/2 12.5 C. The specific heat at this temperature is
C p (@12.5 C)4.205 kj/(kg K) ṁ ẇ e in e out ẇ (h 1 h 2 ) 0 +(pe 1 pe 2 )+(ke 1 ke 2 ) 0 ẇ g(z 1 z 2 ) ( 100 kw 1000 m 2 /s 2 ) (9.80665 m/s 2 ) (51 m) kj/kg 199.94 kg/s Part b) The mass flow rate can be written as ṁ ρva ρ (πd 2 /4) V At T avg 12.5 C the density of water is ρ 998.5 kg/m 3 D 4ṁ πρv 4 (199.94 kg/s) π (998.5 kg/m 3 ) (3 m/s) 0.2915 m
Question 3 (8 marks) An inventor has developed an engine that operates on the thermal gradients in the ocean. The surface waters at the proposed location are 29.5 C and those at a reasonable depth are at 10 C. The engine is claimed to produce 74.6 kw and rejects 300 kw of heat. a) What is the thermal efficiency of this engine? b) Clearly demonstrate if this process is or is not possible by comparing it with the Carnot cycle operating between T H and T L. c) Find the rate of entropy generation, Ṡ gen [kw/k], for the inventor s engine. Part a) From the first law for a heat engine, one can write: The thermal efficiency is: Q H Q L + Ẇ net η 1 Q L Q H 1 Q L Ẇ net + Q L 1 300 kw 74.6 kw + 300 kw 19.9% Part b) The maximum possible efficiency of a heat engine is the Carnot efficiency which works between the same thermal reservoirs. η Carnot 1 T L T H 1 283.15 K 302.65 K 6.4%
Since the heat engine s efficiency is greater than the Carnot cycle efficiency, the cycle is impossible. Part c) An entropy balance over the heat engine gives or Ṡ gen ( ) dscv dt }{{} 0 (steady) Q H T H + Q L T L +( S) MER 0 Ṡ gen Q L T L Q H T H 300 kw 283 K 374.6 kw 302.5 K 0.178 kw/k impossible since Ṡ gen < 0
Question 4 (8 marks) An adiabatic flash evaporator is used to make a small amount of clean water from dirty water. The dirty water enters as saturated liquid at 30 C andatarateof20 kg/min. Clean water leaves as a saturated vapour at 25 C. Dirty water leaves the evaporator as saturated liquid at 25 C. a) Determine the mass flow rate [kg/min] of the clean water vapour. b) Determine the rate of entropy generation [kw/k] within the evaporator. From Table A-4 we note that State T ( C) Phase h (kj/kg) s (kj/kg K) 1 30 sat. liquid 125.79 0.4369 2 25 sat. vapour 2547.2 8.558 3 25 sat. liquid 104.89 0.3674 Part a) Conservation of mass over the evaporator gives: ṁ 1 ṁ 2 + ṁ 3 Conservation of energy over the evaporator yields: Collecting common terms ṁ 1 h 1 ṁ 2 h 2 + ṁ 3 h 3 ṁ 2 h 2 +(ṁ 1 ṁ 2 )h 3
ṁ 2 ṁ1(h 1 h 3 ) (h 2 h 3 ) ( 20 kg )( ) 125.79 104.89 min 2547.2 104.89 0.17115 kg/min Part b) From the second law we know that Solving for the entropy production term 1 s 1 + Ṡ gen 2 s 2 + 3 s 3 Ṡ gen 2 s 2 + 3 s 3 2 s 2 +( 1 1 s 1 2 )s 3 1 s 1 ṁ 1 (s 3 s 1 )+ṁ 2 (s 2 s 3 ) ( 20 kg ) (0.3674 0.4369) min 0.01182 kj K min 1 min 60 s 0.000197 kw K ( kj kg K + 0.17115 kg ) (8.558 0.3674) min kj kg K