ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 12 June 2006

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1 ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 1 June 006 Midterm Examination R. Culham This is a hour, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side only, and the Property Tables booklet. There are 3 questions to be answered. Read the questions very carefully. Clearly state all assumptions. It is your responsibility to write clearly and legibly. Question 1 (1 marks Air enters an insulated diffuser at 100 kp a, 3 C and 60 m/s and exits at 130 m/s. You can assume that the device operates at steady state conditions and that the properties are constant based on the inlet conditions. Employing an ideal gas model and ignoring potential energy, determine: a the temperature of the air at the exit, [K]. b the maximum attainable air pressure [kp a]. c explain how you would solve for exit temperature if properties were not assumed constant Assume 1. diffuser operates at steady state, steady flow conditions. no heat transfer to the surroundings 3. PE =0 4. air is an ideal gas

2 Part a ṁ 1 = ṁ = ṁ mass balance and an energy balance over the diffuser gives or Q 0 +Ẇ 0 +ṁ(h 1 h + h = h 1 + ( V 1 V + g(z 1 z 0 =0 ( V 1 V Foranidealgash = C p T where from Table Ab, C p (70 K = / K h 1 = C p T 1 = K 70 K = h = C p T = K T T = = h 1 + ( V 1 V C p ( ( m /s ( 1 / / K 1000 m /s = K Part b The maximum air pressure at the exit is achieved when the process is isentropic, i.e. s =0.For an ideal gas and an isentropic process, we can write s s 1 = C p ln ( T T 1 R ln ( P P 1 =0 The gas constant R is independent of temperature and therefore at 70 K R = 0.87 / K C p = / K Note: if you do not believe that R is independent of temperature, calculate C p (70 K and C v (70 K and R = C p (70 K C v (70 K = = 0.87 / K

3 Solve for P and K ln ( K K P = kp a ( K ln P 100 kp a Part c If properties are not constant with respect to temperature we would have to iterate. 1. solve for T by assuming and initial guess for the exit temperature, perhaps T = T 1 as we didinparta.. once we calculate a value of T using the initial guess, go back and calculate a new value of C p with this new estimate of T 3. repeat this procedure until the updated value of T does not change significantly from the starting value of T 4. since R is independent of temperature, we would not have to update this value with a change in exit temperature =0

4 Question (16 marks Consider the piston/cylinder arrangement containing water, as shown in the figure below. The frictionless piston is free to move between the stops and because of its weight requires 175 kp a to move. When the piston rests on the lower stops, the enclosed volume is 0.4 m 3. When the piston reaches the upper stops, the volume is 0.6 m 3. The cylinder initially contains.5 of water at 175 kp a with a quality of 0%. It is heated until the water exists as saturated vapor. a For the initial state, determine the volume of the liquid water [m 3 ], the volume of the water vapor [m 3 ] and the internal energy [/]. b As the piston moves from the initial state to the final state, how would you expect the pressure to change before the piston contacts the upper stops, and after the piston contacts the upper stops. c Sketch the P v diagram for the actual process, from initial to final states. d For the final state, determine the pressure [kp a], specific volume [m 3 /] and the internal energy [/] e Find the work transfer for the entire process, []. f Find the heat transfer for the entire process, []. Assume 1. the piston is frictionless. KE = PE =0 Part a 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 = 175 kp a. v f1 v g1 = m 3 / = m 3 /

5 The mass of the liquid and gas can be determined based on the quality such that m f1 = 0..5 =0.45 m g1 = =1.8 Now the volume can be determined as V f1 = (1.8 ( m 3 / = m 3 V g1 = (0.45 ( m 3 / =0.45 m 3 The total volume is V 1 = V f1 + V g1 = m m 3 = m 3 Since the starting volume is greater than 0.4 m 3 the piston starts off the lower stops. The internal energy is u 1 = (1 xu f + xu g = ( = / Part b The pressure will be constant from the initial state until the piston reaches the upper stops. At this point the pressure will begin to rise (constant volume process until the final state of a saturated vapor is reached. Part c

6 Part d At the final state, x =1.0and v = V m = 0.6 m3.5 =0.667 m3 / From Table A-4 we see that the specific volume, v is between C, therefore we interpolate to find T = C P = kp a u = / Part e The only boundary work done is between state and 3. W 3 = PdV = 150 kp a ( m 3 ( kp a m 3 = (where the negative indicates work out of the system W out = Part f The heat transfer can be obtained by performing an energy balance. Q in + W in 0 + U + KE 0 + PE 0 = Q out 0 +W out Q in = (U 3 U 1 +W 3 =.5 ( / =

7 Question 3 (17 marks A steam power plant is proposed as shown in the sketch below. The boiler has a volume of 1 m 3 and initially contains 80% liquid and 0% vapor by volume at 100 kp a. The burner is turned on. When the pressure reaches 1 MPa, the pressure-regulating valve holds the pressure in the boiler constant at 1 MPa, and dry, saturated vapor at this pressure leaves the boiler. Dry, saturated vapor at a pressure of 100 kp a leaves the turbine and is discharged to the atmosphere. When the liquid in the boiler is depleted, the burner shuts off automatically. Assume that the boiler, the turbine and all connecting hardware are fully insulated. Hint: the mass through the turbine is determined by finding the difference between the mass initially in the boiler and the mass remaining when the boiler shuts off. a determine the work output for one charge of the boiler, [MJ]. b determine the energy input for one charge of the boiler, [MJ]. c determine the entropy generated across the turbine for one charge of the boiler, [MJ/K]. Assume 1. the entire system is adiabatic with respect to the surroundings. the process is complete when x finish =1 Part a The mass flow through the turbine can be determined by performing a mass balance over the boiler. m turbine =(m start m finish boiler To find the mass in the boiler before heat is applied m g = V g v g = 1 m m 3 / =0.1181

8 m f = V f v f = 1 m m 3 / = m start = m g + m f = = x start = m g = m f + m g = The process is complete when the liquid in the boiler is depleted, i.e. P quality is 1. = 1 MPa and the x finish = 1 m finish = V v g = 1 m m 3 / = The work out of the turbine is W turbine = m turbine (h h 3 = ( ( / = =78.MJ where h (@1 MP a, sat.vap. = / (Table A-5 h 3 (@100 kp a, sat.vap. = / (Table A-5 Part b The energy input to the boiler can be determined by performing and energy balance over the boiler. U start + Q in = U finish + m turbine h Q in = (m finish u finish (m start u start +m turbine h where u start = (1 x 1 u f1 + x 1 u g1 = ( ( = K u finish = (sat. 1 MPa

9 The heat input is then Q = ( = MJ ( ( Part c Since the turbine is assumed to be adiabatic, there is no heat transfer in or out of the system boundary. The flow is assumed steady over the duration of the charge. Therefore S gen = m turbine (s 3 s = ( ( / K where = K =0.589 MJ K s (@1 MP a, sat.vap. = / K (Table A-5 s 3 (@100 kp a, sat.vap. = / K (Table A-5

ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:

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