ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 13 June 2007 Midterm Examination R. Culham This is a 2 hour, open-book examination. You are permitted to use: course text book calculator There are 4 questions to be answered. Read the questions very carefully. Clearly state all assumptions. Do not assume properties are constant with respect to temperature unless clearly stated in the problem. It is your responsibility to write clearly and legibly. Question 1 (12 marks) An insulated rigid tank is initially divided into two equal volumes by a piston held in place by a pin, as shown in the figure below. Side A contains 1 kg of water (liquid and/or vapor) at 260 C and 2 MPa. Side B contains 5 kg of water (liquid and/or vapor) at 260 C and 4.688 MPa. The pin is then pulled and the piston moves until the pressures on each side of the piston equalizes at a value of 4.688 MPa. Heat transfer across the piston keeps the temperature at 260 C on both sides of the piston. a) Fill in the following table and show all calculations necessary to determine these unknown values. b) Show the complete process on a T v diagram. Property State 1 State 2 Side A Side B Side A Side B Liquid Volume [m 3 ] Vapor Volume [m 3 ] Pressure [MPa] 2 4.688 4.688 4.688
Assumptions 1. the process is quassi-equilibrium 2. isothermal 3. outer shell of the tank is adiabatic Part a) From Table A-5, the saturation temperature of side A at the initial conditions is therefore the water is superheated and T sat (2 MPa) 212.42 C<260 C v A1 0.11425 m 3 /kg V A1 m A v A1 (1kg)(0.11425 m 3 /kg) 0.11425 m 3 v B1 V B1 0.11425 m3 0.02285 m 3 /kg m B 5 kg At the initial conditions, Side B is under the vapor dome. When the pin is removed and the pressure equalizes, we know that volume A increases and volume B decreases such that the pressure and temperature are constant on both sides of the piston. This is achieves as follows: i) volume A2 decreases and state A2 becomes a saturated vapor. In this case, B2 will remain in the saturated region (under the dome) at 260 C and P 2 4.688 MPa V A2 m A v A2 (1kg)(0.04221 m 3 /kg) 0.04221 m 3 and V B2 V V A2 2 (0.11425 m 3 ) (0.04221 m 3 )0.18629 m 3 We note that v B2 region is correct. v B2 V B2 m B 0.18629 m3 5 kg 0.03726 m 3 /kg <v g (@ 260 C) therefore our assumption of state 2 being in the saturated State B1 x B1 v B1 v f v g v f 0.02285 0.001276 0.04221 0.001276 0.52704 V gb1 m gb1 v g x B1 m B v g (0.52704)(5 kg)(0.04221 m 3 /kg) 0.11123 m 3 V fb1 m fb1 v f (1 x B1 )m B v f (1 0.52704)(5 kg)(0.001276 m 3 /kg) 0.00302 m 3
State B2 x B2 v B2 v f v g v f 0.03726 0.001276 0.04221 0.001276 0.87907 V gb2 m gb2 v g x B2 m B v g (0.87907)(5 kg)(0.04221 m 3 /kg) 0.18553 m 3 V fb1 m fb1 v f (1 x B1 )m B v f (1 0.87907)(5 kg)(0.001276 m 3 /kg) 0.00077 m 3 Property State 1 State 2 Side A Side B Side A Side B Liquid Volume [m 3 ] 0 0.00302 0 0.00077 Vapor Volume [m 3 ] 0.11425 0.11123 0.04221 0.18553 Pressure [MPa] 2.0 4.688 4.688 4.688 Part b)
Question 2 (12 marks) A vortex tube is a steady-state device that splits a high pressure gas stream into two streams, one warm and one cool. During a controlled test, air entered the vortex tube at 19.3 C and 0.52 MPa. The warm air left the tube at 26.5 C and the cool air left at 21.8 C. Both exit streams were at 101.35 kp a. The ratio of the mass flow rate of the warm air to that of the cool air was 5.39. There was no work transfer. The temperature of the surroundings was 20 C. For a compressed air flow rate of 1.5 kg/s, determine: a) the rate of heat loss from the vortex tube to the surroundings, [kw ] b) the rate of entropy generation in the vortex tube, [kw/k] compressed air 1 o T 0 20 C cool air 3 warm air 2 Assumptions 1. steady state, steady flow 2. properties are calculated at T avg Part a) From conservation of mass we know ṁ 1 ṁ 2 + ṁ 3 From the given conditions we know ṁ 1 1.5 kg/s ṁ 2 5.39 ṁ 3 Therefore ṁ 1 6.39 ṁ 3 ṁ 3 0.2347 kg/s and
From conservation of energy we know ṁ 2 5.39 0.2347 1.2653 kg/s Q out ṁ 1 h 1 ṁ 2 h 2 ṁ 3 h 3 (ṁ 2 + ṁ 3 ) h 1 ṁ 2 h 2 ṁ 3 h 3 ṁ 2 (h 1 h 2 )+ṁ 3 (h 1 h 3 ) ṁ 2 c p (T 1 T 2 )+ṁ 3 c p (T 1 T 3 ) (1.2653 kg/s) (1.0048 /) (292.45 299.65) C +(0.2347 kg/s) (1.0039 /) (292.45 251.35) C 0.5299 kw part a) where specific heat was calculated from Table A-2b as follows at at T 1 + T 2 2 T 1 + T 3 2 296.05 K c p 1.0048 / 271.90 K c p 1.0039 / Part b) A second law balance gives or ṁ 1 s 1 + Ṡ gen ṁ 2 s 2 + ṁ 3 s 3 + Q out T 0 Ṡ gen ṁ 2 s 2 + ṁ 3 s 3 ṁ 1 s 1 + Q out T 0 ṁ 2 s 2 + ṁ 3 s 3 (ṁ 2 + ṁ 3 )s 1 + Q out T 0 where ṁ 2 (s 2 s 1 )+ṁ 3 (s 3 s 1 )+ Q out T 0 s 2 s 1 c p ln(t 2 /T 1 ) R ln(p 2 /P 1 ) 1.0048 ln(299.65/292.45) 0.286 ln(0.10135/0.52)
0.4921 s 3 s 1 c p ln(t 3 /T 1 ) R ln(p 3 /P 1 ) 1.0039 0.3156 ln(251.35/292.45) 0.286 ln(0.10135/0.52) Therefore the entropy production can be written as ( ) ( Ṡ gen (1.2653 kg/s) 0.4921 +(0.2347 kg/s) 0.3156 ) 0.5299 kw + (20 + 273.15) K 0.6985 kw/k part b)
Question 3 (12 marks) A well insulated piston-cylinder arrangement initially contains water vapor at 500 kp a and 300 C. The initial volume of water vapor is 1500 L asshowninthefigure. The insulation is suddenly removed and the steam is allowed to condense by heat rejection from the cylinder to the surrounding air. Two stops are installed inside the cylinder in order to prevent the downward motion of the piston. After sometime, the piston touches the stops and the volume inside the cylinderatthattimeis500 L. More heat is released to the surrounding air so that the final pressure inside the cylinder becomes 75 kp a. (Assume frictionless piston movement and neglect heat transfer through the piston). Find: a) the mass [kg] of the steam inside the cylinder, its initial specific volume, [m 3 /kg] and the specific internal energy [/kg] b) the specific volume [m 3 /kg], the quality of the steam (if it is a mixture) when the piston just touches the stops and the specific internal energy [/kg] c) the volume of liquid water [m 3 ], the volume of water vapor [m 3 ], and the specific internal energy [/kg] atthefinal state d) the total work transfer [/kg] for the entire process e) the amount of heat rejection [/kg] to the surrounding for the entire process. f) show the entire process on a T v diagram. Assumptions steady state quassi equilibrium frictionless piston no heat transfer through the piston
Part a) @ 500 kp a T sat 151.86 C<T 1 300 C Therefore the steam is superheated at state 1. From Table A-6 u 1 2802.9 /kg v 1 0.5226 m 3 /kg m (1500 L 10 3 m 3 /L) 0.5226 m 3 /kg 2.87 kg Part b) As the volume decreases and the piston moves but before it touches the stops, the process can be treated as a constant pressure process. The specific volume just as the piston touches the stops is v 2 (500 L 10 3 m 3 /L) 2.87 kg 0.1742 m 3 /kg From Table A-5, at a pressure, P 2 500 kp a, v f 0.001093 m 3 /kg v g 0.3749 m 3 /kg Therefore state point 2 is a saturated mixture (under the dome) and The specific internal energy is x v 2 v f v g v f 0.1742 0.001093 0.3749 0.001093 0.4631 u 2 u f + xu fg 639.68 + 0.4631 1921.6 1529.57 /kg Part c) Once the piston touches the stops, At 75 kp a, from Table A-5 we see v 3 v 2 0.1741 m 3 /kg v f 0.001037 m 3 /kg v g 2.217 m 3 /kg Therefore state point 3 is a saturated mixture (under the dome) and x v 2 v f v g v f 0.1742 0.001037 2.217 0.001037 0.0781
The volume of the liquid water is V liquid v f (1 x)m (0.001037 m 3 /kg)(1 0.0781)(2.87 kg) 0.002743 m 3 The volume of the water vapor is V vapor v g (x)m (2.217 m 3 /kg)(0.0781)(2.87 kg) 0.4969 m 3 The specific internal energy at state 3 is u 3 u f + xu fg 384.31 + 0.0781 2112.4 549.29 /kg Part d) The total work done is given as 2 W 13 W 12 + W 23 PdV +0 P 1 (V 2 V 1 ) 1 (500 kp a)(0.5 m 3 1.5 m 3 ) 500 Note: this will be 500 using Cengel s sign convention of work done by the system is positive. Part e) To find the heat rejection, apply the 1st law W in Q out + mu 1 mu 3 Q out (2.87 kg)(2802.9 549.29) /kg + 500 6967.9 Part f)
Question 4 (10 marks) A solar driven power plant uses water as the working fluid. During daytime operation, the turbine inlet state is 100 C at 50 kp a. The air cooled condenser operates at a pressure of 5 kp a. The turbine and the pump are assumed to be isentropic. 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 b) determine the mass flow rate of the working fluid, [kg/s] c) determine the required collector surface area, [m 2 ] State T [ C] P [kp a] v [m 3 /kg] x h[/kg] s [/] 1 5 0.001005 137.82 2 50 137.865 3 100 50 2682.5 7.6947 4 5 2348.2 7.6947 State Point 3 @50kP a T sat 81.33 C<T 3 100 C Therefore state point 3 is a superheated vapor and h 3 2682.5 /kg and s 3 7.6947 / State Point 4 Check to see is state point 4 is inside the dome. We know that s 4 s 3 7.6947 /
and at a pressure of 5 kp a the entropy of the saturated vapour is s g 8.3951 /. 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 7.6947 0.4764(1 x) +8.3951x Therefore x 4 0.912. The enthalpy at 4 is given as h 4 137.82(1 0.912) + 2561.5 (0.912) 2348.2 /kg Part a) Performing an energy balance over the turbine we get w t h 3 h 4 2682.5 2348.2 334.3 /kg We have a saturated liquid at the entry to the pump at P 1 5kP a, therefore v 1 0.001005 m 3 /kg and h 1 137.82 /kg The work input of the pump can be determined as w p v 1 (P 1 P 4 ) (0.001005 m 3 /kg)(50 5) kp a 1 /m3 1 kp a and 0.045 /kg h 2 h 1 + w p 137.82 /kg +0.045 /kg 137.865 /kg The heat input at the collector is given as q in h 3 h 2 2682.5 137.865 2544.635 /kg The first law efficiency is given as η 1 w t w p q in 334.3 0.045 2544.635 0.131 Part b) The mass flow rate of the working fluid is given as ṁ Ẇt w t 75 /s 334.3 /kg 0.224 kg/s Part c) The required collector area is A sc ṁ q in Q sc 0.224 kg/s 2544.635 /kg 0.65 kw/m 2 876.92 m 2