Download - HW 03 Solutions
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PROBLEM 3.9
COMMENT: As the pressure increases, the difference in specific volume between saturated
vapor and saturated liquid decreases. At the critical pressure, the two states
coincide and the difference is zero.
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PROBLEM 3.37
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PROBLEM 3.55
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PROBLEM 3.63
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PROBLEM 3.85
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PROBLEM 3.98
Five kg of butane (C4H10) in a piston-cylinder assembly undergo a process from p1 = 5 MPa, T1 =
500 K to p2 = 3 MPa during which the relationship between pressure and specific volume is pv =
constant. Determine the work, in kJ.
KNOWN: Five kg of C4H10 undergo a process for which pv = constant in a piston-cylinder
assembly between two states.
FIND: Determine the work.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: 1. The Butane is a closed system. 2. The process is polytropic with
pv = constant. 3. Volume change is the only work mode.
ANALYSIS: The work is given by W = m
. With pv = constant = p1v1
W = m
= (p1v1) m ln
(*)
To evaluate W requires v1 and v2. The compressibility chart can be used to obtain v1: From
Table A-1; pc = 38 bar, Tc = 425 K, M = 58.12 kg/kmol.
pR1 = p1/pc = (50)/(38) = 1.32
Fig. A-2: Z1 ≈ 0.67
TR1 = T/Tc = (500)/(425) = 1.18
Accordingly, with v1 = Z1 (RT1/p1), we get
v1 = (0.67)
= 0.0096 m
3/kg
Now, with pv = constant,
= [(5 MPa)(0.0096 m3)]/(3 MPa) = 0.016 m
3
Now, inserting values into (*), we get
Butane
(C4H10)
m = 5 kg
p1 = 5 MPa = 50 bar
T1 = 500 K p1 = 3 MPa = 30 bar →
pv = constant
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PROBLEM 3.98 (CONTINUED) – PAGE 2
W = (5 MPa)(0.0096 m3/kg)(5 kg) ln [(0.016)/(0.0096)]
= 122.6 kJ (out)
Alternative Evaluation of
pR1 = p1/pc = (50)/(38) = 1.32
Fig. A-2: ≈ 0.6
TR1 = T/Tc = (500)/(425) = 1.18
v1 =
= (0.6)
= 0.0096 m
3/g
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PROBLEM 3.105
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PROBLEM 1.333
Two kg of nitrogen (N2) gas are contained in a closed, rigid tank surrounded by a 10-kg water
bath, as shown in Fig. P1.133. Data for the initial states of the nitrogen and water are shown on
the figure. The entire unit is well-insulated, and the nitrogen and water interact until thermal
equilibrium is achieved. The measured final temperature is 34.1oC. The water can be modeled
as an incompressible substance, with c = 4.179 kJ/kg∙K, and the nitrogen is an ideal gas with
constant cv. From the measured data, determine the average value of the specific heat cv, in
kJ/kg∙K.
KNOWN: A tank of nitrogen gas is surrounded by a well-insulated water bath. The gas and
water are initially at different temperatures and they interact until equilibrium is achieved.
FIND: Determine the average values of the specific heat cv for the nitrogen.
SCHEMATIC AND GIVEN DATA:
ANALYSIS: Reducing the energy balance
ΔKE + ΔPE + ΔUw + Δ = Q – W → ΔUw + Δ
= 0
The final equilibrium temperature is T2. With ΔUw = mwcw (T2 – T1,w) and Δ =
cv(T2 - )
mw(T2 – T1,w) + cv(T2 -
) = 0
Solving for cv
cv =
=
= 0.741 kJ/kg∙K
Nitrogen
10 kg
water bath
2 kg
Nitrogen
= 50
oC
Water
T1, w = 20oC
ENGINEERING MODEL: (1) As
shown on the accompanying sketch, the
closed system consists of the nitrogen
tank and the water bath. (2) W = 0 and
Q = 0. (3) Kinetic and potential energy
effects can be neglected. (4) The water
is an incompressible substance with cw =
4.179 kJ/kg∙K. (5) The nitrogen is
modeled as an ideal gas with constant cv.
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PROBLEM 3.135
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PROBLEM 3.142
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PROBLEM 3.142 (CONTINUED)