ME 235 EXAM I , Oct. 20, 2010
BORGNAKKE AND SIEGEL ME Dept.
1. (20 points) Determine the phase of the following substances and find the values of the
unknown quantities.
(a) Nitrogen, P = 1000 kPa, 120 K, v = ?, Z = ?
P = 1000 kPa, 300 K, v = ?, Z = ?
(b) Air, T = 500 K, v = 0.500 m /kg, P = ?
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(c) R-410a, T = 20°C, v = 0.01 m /kg, P = ?, h = ?
(d) Ammonia, T = 20°C, u = 1000 kJ/kg, P = ?, x = ?
2. (20 points) A rigid tank holds 0.75 kg ammonia at 800 kPa as saturated vapor. The tank is
now heated to 260°C by heat transfer.
(a) Which two properties determine the final state.
(b) Determine the amount of work and heat transfer during the process.
3. (30 points) Water in a tank A is at 250 kPa with a quality of 10% and mass 0.5 kg. It is
connected to a piston cylinder with a constant piston float pressure of 200 kPa initially with 0.5
kg water at 400°C. The valve is opened and enough heat transfer takes place to have a final
uniform temperature of 150°C.
(a) Find the initial volume of the tank
and piston cylinder. A B
(b) Find the final P and V
(c) Find the process work
(d) Find the process heat transfer
4. (30 points) A rigid tank contains 0.2 kg air at 300 K, 100 kPa and connected to a
piston/cylinder with 0.1 kg air at 2000 K, 200 kPa through a pipe with a valve. The piston keeps
constant pressure and now the valve is opened and all the air come to a single uniform state.
Assume we cool the system until the piston hits the bottom of the cylinder.
(a) Find the final pressure and final temperature of the air.
(b) Find the work
(c) Find the heat transfer
Do not use constant specific heats. ME 230 EXAM I SOLUTION, OCT 20, 2010
BORGNAKKE and SIEGEL ME Dept.
1. (20 points) Determine the phase of the following substances and find the values of the
unknown quantities.
(a) Nitrogen, P = 1000 kPa, 120 K, v = ?, Z = ?
P = 1000 kPa, 300 K, v = ?, Z = ?
(b) Air, T = 500 K, v = 0.500 m /kg, P = ?
(c) R-410a, T = 20°C, v = 0.01 m /kg, P = ?, h = ?
(d) Ammonia, T = 20°C, u = 1000 kJ/kg, P = ?, x = ?
Solution:
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(a) B.6.2 at 1000 kPa, 120 K: v = 0.03117 m /kg; A.5: R = 0.2968 kJ/kgK
Pv 1000 0.03117 kPa m /kg
Z = RT = 0.2968 120 (kJ/kgK) K = 0.875
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At 1000 kPa, 300 K: v = 0.08889 m /kg
Pv 1000 0.08889 kPa m /kg
Z = = = 0.998
RT 0.2968 300 (kJ/kgK) K
(b) Ideal gas, P = RT/v = 0.287 kJ/kgK 500 K/0.5 m /kg = 287 kPa
(c) B.4.1 at 20°C, v f 0.000923, v g 0.01758, v

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