# ME235_Exam2_F05.pdf

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University of Michigan - Ann Arbor

Mechanical Engineering

MECHENG 235

Donald Siegel

Winter

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ME 235 EXAM 2, NOVEMBER 22, 2005
C. BORGNAKKE, ME
Problem 1 (50 Points)
Air at 300 K, 100 kPa in a rigid box is heated to 1000 K by a 1500 K source.
a) Find the final pressure.
b) Find the specific heat transfer.
c) Find the total entropy generation per kg air.
Air at 300 K, 100 kPa is compressed in a reversible adiabatic process by a piston cylinder
so T = 1000 K.
2
d) Find the final pressure
e) Find the specific work.
Problem 2 (20 Points)
.
A large heat pump upgrades 5 MW of heat (Q ) at 85 C to be delivered at 185 C.o
L
Suppose the actual heat pump .as a COP of β HP = 2.5.
a) How much power W is required to do this?
b) What is the rate of entropy generation inside the heat pump?
Problem 3 (30 Points)
A 0.005 m 3tank is empty and filled with steam from a supply line at 500 kPa, 200 C.
The operator filled it slowly so the final state was measured to be 200 C, 500 kPa.
a) Find the final mass in the tank.
b) How much heat transfer took place?
c) If it was filled quickly so no heat transfer, how hot would it be? ME 235 EXAM 2, SOLUTION NOVEMBER 22, 2005
C. BORGNAKKE, ME
1) [30 Points] Take CV as the air out to the source surface.
The first process is a constant volume process 1o 2 = 0
Energy Eq: m(u –2u ) 1 Q 1 2 = 1 2 1 2 => q = u2- u1
Entropy Eq.: m(s – s ) = Q /T + S
2 1 1 2 s 1 2 gen
Process & ideal gas: P 1 1 mRT , 1 P2 2= mRT 2
P = P T /T = 100 × 1000 / 300 = 333.3 kPa
2 1 2 1
From energy eq.: 1 2= u 2 u 1 C v o(T2– T 1
= 0.717 (1000 – 300) = 501.9 kJ/kg
1000
From Eq.8.??: s2 – 1 = Cv o ln(T2/ T1) = 0.717 ln300 = 0.8632 kJ/kgK
From entropy eq.: 1 2 gen= s2– s1– q1 2 s
501.9
= 0.8632 – 1500 = 0.5286 kJ/kgK
[20 Points]
The second process is isentropic, q = 0, so with constant heat capacity we get:
k 1.4
k-1 1000 0.4
P 2= P1(T 2T )1 = 100 ( 300 )

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