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PROBLEM 6.42
PROBLEM 6.49
PROBLEM 6.53
A system consisting of 10 lb of air contained within a closed, rigid tank is initially at 1 atm and
600oR. Energy is transferred to the system by heat transfer from a reservoir at 900oR until the
temperature of the air is 800oR. During the process, the temperature of the system boundary
where the heat transfer occurs is 900oR. Using the ideal gas model, determine the amount of
energy transfer by heat, in Btu, and the amount of entropy produced, in Btu/oR.
KNOWN: A system consisting of a fixed volume of air is heated from a specified initial state to
a specified final temperature. Heat transfer occurs from a reservoir at a given temperature and
the boundary temperature where the heat transfer occurs is known.
FIND: Determine the amounts of energy transfer by heat and entropy produced.
SCHEMATIC AND GIVEN DATA:
Reservoir
ENGINEERING MODEL: (1) The air
900oR
is a closed system. (2) The air is modeled
as an ideal gas. (3) The volume is constant
Q
and W = 0. (4) The boundary temperature
o
where heat transfer occurs is Tb = 900 R.
(5) Kinetic and potential energy effects are
negligible.
Air
T1 = 600oR
p1 = 1 atm
10 lb
T2 = 800oR
Tb = 900 oR
ANALYSIS: The energy balance reduces to give: ∆U + ΔKE + ΔPE = Q – W . With data from
Table A-22
Q = m (u2 – u1) = (10 lb)(136.97 – 102.34)Btu/lb = 346.3 Btu (in)
With modeling assumption 4, the entropy balance gives
ΔS = Q/Tb + σ
→
σ = m (s2 – s1) – Q/Tb
For the air as an ideal gas; (s2 – s1) = so(T2) – so(T1) – R ln (p2/p1). Since the volume is constant,
p2/p1 = T2/T1. Thus
σ = m [so(T2) – so(T1) – R ln (T2/T1)] – Q/Tb
= (10 lb)[(0.69558 – 0.62607 – (1.986/28.97) ln(800/600)]Btu/lb∙oR - (346.3 Btu/900oR)
= 0.1131 Btu/ oR
PROBLEM 6.63
A rigid, well-insulated tank contains air. A partition in the tank separates 12 ft3 of air at 14.7
lbf/in.2, 40oF from 10 ft3 of air at 50 lbf/in.2, 200oF, as illustrated in fig. P6.63. The partition is
removed and air from the two sides mix until a final equilibrium state is attained. The air can be
modeled as an ideal gas, and kinetic and potential energy effects can be neglected. Determine
the final temperature, in oF and pressure, in lbf/in.2 Calculate the amount of entropy produced, in
Btu/oR.
KNOWN: Air is contained in a rigid, well-insulated tank on two sides of a partition. The initial
states of the air on each side are specified. The partition is removed and equilibrium is attained.
FIND: Determine the final temperature and pressure and the amount of entropy produced.
SCHEMATIC AND GIVEN DATA:
air
air
p2, T2
air
VL = 12 ft3
pL = 14.7 lbf/in.2
TL = 40oF = 500oR
VR = 10 ft3
PR = 50 lbf/in.2
TR = 200oF = 660oR
ENGINEERING MODEL: (1) The air on both sides of the partition are a closed system. (2)
The air is modeled as an ideal gas. (3) Kinetic and potential energy effects can be neglected. (4)
There is no heat transfer from the contents of the tank to the surroundings and W = 0.
ANALYSIS: First, we use the ideal gas equation of state to calculate the masses on the left and
right sides of the partition, respectively.
mL =
=
= 0.9526 lb
and
mtot = 2.9981 lb
mR =
=
= 2.0455 lb
Now, using the energy balance: ΔKE + ΔPE + ΔU = Q – W
Thus
U1 = mLuL + mRuR and U2 = mtotu2
So
u2 = (mLuL + mRuR)/mtot
→
ΔU = 0
PROBLEM 6.61 (CONTINUED)
With data from Table A-22: uL = 85.20 Btu/lb and uR = 112.67 Btu/lb
u2 = [(0.9526 lb)(85.20 Btu/lb) + (2.0455)(112.67)]/(2.9981 lb) = 103.94 Btu/lb
Interpolating in Table A-22: T2 ≈ 609.3oR = 149.3oF
The final pressure is
p2 =
= 30.75 lbf/in.2
=
Now, the entropy balance reduces to: ΔS =
+σ
→
σ = ΔS = mL(s2 – sL) + mR(s2 – sR)
σ = ΔS = mL[so(T2) – so(TL) – R ln (p2/pL)] + mR[so(T2) – so(TR) – R ln (p2/pR)]
From Table A-22: so(TL) = 0.59233 Btu/lb∙oR, so(TR) = 0.64902 Btu/lb∙oR, and so(T2) ≈ 0.62973 Btu/lb∙oR
Thus
σ = (0.9526 lb)[(0.62973 – 0.59233) Btu/lb∙oR – (1545/28.97) Btu/lb∙oR ln(30.75/14.7)]
(2.0455)[(0.62973 – 0.64902) – (1545/28.97) ln(30.75/50)]
= 15.533 Btu/oR
The entropy production indicates that the mixing process is irreversible.
PROBLEM 6.86
PROBLEM 6.87
PROBLEM 6.112
PROBLEM 6.112 (CONTINUED)
PROBLEM 6.138