INTRODUCTION ECE309 TO THERMODYNAMICS & HEAT TRANSFER 16 June 2004 Midterm Examination R. Culham
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 (20 marks) Initially saturated water at 200 C is contained in a piston-cylinder device as shown below. The water is then heated isothermally until its volume is 100 times larger than its initial volume. a) Determine the increase in energy [kJ/kg] of the water b) Determine the work transfer [kJ/kg] and indicate whether it is into or out of the water. c) Determine the heat transfer [kJ/kg] and indicate whether it is into or out the water.
Part a) From Table A-4, the specic volume of saturated liquid water at 200 C is v1 = 0.001157 m3 /kg The volume at state 2 is 100 larger than state 1 v2 = 100 (0.001157 m3 /kg) = 0.1157 m3 /kg
The quality at state 2 is x2 = v2 vf vg vf = 0.1157 0.001157 0.12736 0.001157 = 0.9076
The internal energy at state 2 is u2 = uf + x2 (ug uf ) = 850.65 + 0.9076 (2595.3 850.65) = 2434.09 The internal energy at state 1 is u1 = uf = 850.65 kJ/kg The increase in energy of the water is u2 u1 = (2434.09 850.65) kJ/kg = 1583.44 kJ/kg Part b) The work transfer by way of expansion is given as W =
V2 V1
kJ kg
P dV = P (V1 V2 ) = P m (v1 v2 )
Noting that the saturation pressure at T = 200 C is P = 1.5538 M P a The work per unit mass is given as w= W m = P (v1 v2 ) = (1553.8 kP a) 1 kJ/m3 1 kP a (0.001157 0.1157) m3 /kg
= 177.98 kJ/kg where negative work implies the work is out of the water. Part c) From an energy balance on the water, we can write Q = W + U Q m = W m + (u2 u1 )
= (177.98 + 1583.44) kJ/kg = 1761.42 kJ/kg where positive heat transfer implies energy addition to the water.
Question 2 (20 marks) Air at 40 C and 0.6 M P a enters a 25 mm diameter pipeline at a mass ow rate of 0.01 kg/s. The air ows through several valves and leaves the pipeline at a pressure of 110 kP a. The pipeline and the valves can be assumed to be adiabatic. a) Determine the velocity of the air [m/s] entering the pipeline. b) Neglecting changes in kinetic energy, determine the exit state, i.e. T and v of air. c) Based on your answer to part b), determine the exit velocity [m/s] of the air. d) Determine the exit state, T and v of the air if the change in kinetic energy is accounted for by assuming the exit velocity is the value found in part c). How much difference is there between your new state and the state found in part b)? What conclusion do you reach regarding the importance of including kinetic energy terms in your analysis? Part a) Assume that the air behaves as an ideal gas and use the gas constant for air at a constant temperature of 300 K Rair = 0.287 kJ/(kg K) The specic volume of the air at the entrance can be determined as Rair T1 P1 0.287 = kJ (kg K) 600 kP a (40 + 273.15) K 1 kJ/m3 1 kP a = 0.1498 m3 /kg
v1 =
The velocity at the entrance is V1 = m v1 A = (0.01 kg/s) (0.1498 m3 /kg) = 3.05 m/s (0.025 m)2 4
Part b) If kinetic energy is considered negligible and the pipe and valves are considered adiabatic, the energy balance between the inlet and the outlet can be written as mh1 = mh2 h1 = h2
Since the air is considered an ideal gas we can write h1 = h2 T1 = T2 = 40 C
The specic volume at the exit can then be written as
v2 =
Rair T2 P2
0.287 =
kJ (kg K) 110 kP a
(40 + 273.15) 1 K kJ/m3 1 kP a = 0.8170 m3 /kg
Part c) The velocity at the exit is given as V2 = m v2 A = (0.01 kg/s) (0.8170 m3 /kg) = 16.64 m/s (0.025 m)2 4
Part d) If we now include the effects of kinetic energy, the energy balance equation is written as m h1 +
2 V1
2
= m h2 +
2 V2
2
Cp (T2 T1 ) = T2 T1 =
2 2 V1 V 2
2
2 2 V1 V 2
2Cp [(3.05)2 (16.64)2 ] m2 /s2 1 kJ/kg = 0.133 K 1000 m2 .s2 2 1.005 kJ/(kg K)
= Therefore
T2 = (40 + 273.15) K 0.133 K = 313.017 K and Rair T2 P2 0.287 = 313.017 K (kg K) = 0.81669 m3 /kg 3 1 kJ/m 110 kP a 1 kP a kJ
v2 =
Both the temperature and the specic volume are with 0.04% of the solution when kinetic energy was ignored. Therefore kinetic energy is insignicant in this calculation.
Question 3 (15 marks) Air is contained in a rigid, adiabatic 0.3 m3 container at 20 C and 101.325 kP a. A paddle wheel sticking into the container then does 50 kJ of work. Determine the entropy produced [kJ/K]. Since we are working with air we can assume ideal gas and therefore (0.3 m3 ) 1 kP a = 0.3613 kg = (0.287 kJ/(kg K))(20 + 273.15) K (101.325 kP a) 1 kJ/m3
m=
P1 V1 R T1
From a 1st law energy balance we know W = U = m Cv (T2 T1 ) Therefore we can solve for T2 as T2 = W m Cv + T1 = 50 kJ (0.3613 kg) (0.718 kJ/(kg K)) + 293.15 K
= 485.89 K = 212.74 C The change in entropy is given as s2 s1 = Cv ln T2 T1 + R ln v2 v1
Since the container is rigid, v1 = v2 and the second term in the above equation goes to zero. s2 s1 = Cv ln The entropy produced is S = m (s2 s1 ) = (0.3613 kg) 0.3628 kJ kg K = 0.131 kJ/K T2 T1 = 0.718 kJ/(kg K) ln 485.89 K 293.15 K = 0.3628 kJ kg K
Question 4 (20 marks) One kilogram of superheated water vapour (steam) at 700 C and 2.0 M P a is contained in an ejection tube behind a 50 kg projectile that is initially held in place by a pin. The pin is then removed and the vapour pushes the projectile forward into the reference atmosphere at Tatm = 20 C and Patm = 100 kP a. The process is adiabatic and occurs without friction. determine the expansion ratio V2 /V1 necessary to obtain the maximum possible projectile velocity. Hint: a process with no irreversibilities. b) determine the maximum possible proje...