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Mechanical Engineering

MECHENG 235

Donald Siegel

Winter

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ME235 EXAM II, NOVEMBER 30, 2010
C. BORGNAKKE and D. SIEGEL
1. [30%]
In a remote location you run a heat engine to provide the power to run a
refrigerator. The input to the heat engine is at 800 K and the low T is 400 K, it has an
actual efficiency equal to ½ of the corresponding Carnot unit. The refrigerator hasLa T =
-10C and T H T ambient 10, with a COP that is 1/3 of the corresponding carnot unit.
Assume the ambient is at 25C and that a cooling capcity of 2 kW is needed.
a. Find the rate of heat input to the heat engine
b. Find the entropy generation rate in the heat engine
c. Find the entropy generation rate in the refrigerator
2. [30%]
Air at 100 kPa, 20C is compressed in a piston cylinder to a volume 9 times
smaller in a polytropic process where n = 1.25. If heat transfer goes out it goes into the
ambient at 20C and if the heat transfer comes in it comes from a 250C energy reservoir.
a) Find the final T and P of the air
b) Find the specific work and specific heat transfer
c) Find the entropy generation per unit mass of air
3. [40%]
When large concrete pipes for storm drains are made they are placed in a sealed
perfectly insulated constant P flexible bag. Assume the bag contains 1000 kg concrete at
30C which releases 1.2 W/kg due to chemical reaction. This chemical heat release is like
a work term in and not heat transfer. The bag also contains 25 kg of saturated water
vapor at 100 kPa. The pipe cures for 6 hr and not all the steam condenses.
a) Find the final T (assumed uniform) and quality x of the water.
b) Find the total entropy generation. EXAM II, SOLUTION NOVEMBER 30, 2010
C. BORGNAKKE and D. SIEGEL
1. [30 points]
T = 800 K T = -10 C
H
The coolin.
Q H1 QL2 capacity isL2 .
W
HE REF.
The only term
Q Q that does the
L1 H2 cooling.
400 K 35 C
Definition of the heat engine efficiency
TL1 400
HE = 0.5 CARNOT = 0.5 (1 T ) = 0.5 ( 1 800) = 0.25
H
Definition of the refrigerator COP
1 1 T L 1 263.15
= = = = 1.949
3 CARNOT 3 T H T L 3 35 - (-10)
From the required cooling load
. .
W = Q / = 2 kW = 1.026 kW
L2 1.949
Heat engine output determined so find the input
. . 1.026
Q = W / = kW = 4.104 kW
H1 HE 0.25
Entropy equation for the two control volumes
. .
. QL1 Q H 4.104 - 1.0264.104
Sgen totT T = 400 800 = 0.002565 kW/K
L1 H
. .
. Q H2 Q L2 2 + 1.026 2
Sgen refT T = 308.15 263.15 = 0.00222 kW/K
H2 L2
Notice there is entropy generated outside of the heat engine (from the 400 K to the 25C
ambient redistribution of heat) and outside the refrigerator (from the 35C to the 25C
redistribution of heat). 2. [30 points]
C.V. Air of constant mass m 2 m = 1.
Energy Eq.5.11: m(u2 u 1 = 1 2 W 1 2
Entropy Eq.8.3 (37): m(s2 s1) = dQ/T + 1 2 gen
1.25
Process: Pv = constant, v2/v1= 1/9
State 1: P 1 100 kPa, T =1293.15 K
State 2:

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