PROBLEM 6.3 - Nick G. Glumac

PROBLEM 6.3
PROBLEM 6.4
Using the appropriate tables, determine the change in specific entropy between the specified
states, in Btu/lb∙oR. Show the states on a sketch of the T-s diagram.
(a) water, p1 = 10 lbf/in.2, saturated vapor; p2 = 500 lbf/in.2, T2 = 700oF.
(b) ammonia, p1 = 140 lbf/in.2, T1 = 160oF; T2 = - 10oF, h2 = 590 Btu/lb.
(c) air as an ideal gas, T1 = 80oF, p1 = 1 atm; T2 = 340oF, p = 5 atm.
(d) oxygen as an ideal gas, T1 = T2 = 520oR, p1 = 10 atm, p2 = 5 atm.
(a) water, p1 = 10 lbf/in.2, saturated vapor; p2 = 500 lbf/in.2, T2 = 700oF.
State 1: Table A-3E: s1 = 1.7877 Btu/ lb∙oR
State 2: Table A-4E: s2 = 1.6112 Btu/ lb∙oR
T
→
Δs = - 0.1765 Btu/ lb∙oR
500 lbf/in.2
(2)
.
700oF
10 lbf/in.2
.
(1)
s
2
o
(b) ammonia, p1 = 140 lbf/in. , T1 = 160 F; T2 = - 10oF, h2 = 590 Btu/lb.
State 1: Table A-15E: s1 = 1.3025 Btu/ lb∙oR
State 2: Table A-113E: h2 < [email protected] -10oF. The quality is x2 =
0.969
s2 = sf2 + x2(sg2 – sf2) = 0.0729 + (0.969)(1.3544 – 0.0729) = 1.3147 Btu/ lb∙oR
Δs = 0.0122 Btu/ lb∙oR
T
140 lbf/in.2
(1)
160oF
.
23.738 lbf/in.2
.
-10oF
(2)
s
PROBLEM 6.4 (CONTINUED)
(c) air as an ideal gas, T1 = 80oF = 540 oR, p1 = 1 atm; T2 = 340oF = 800 oR, p = 5 atm.
For air as an ideal gas; Δs = so(T2) – so(T1) – R ln (p2/p1). With data from Table A-22E
Δs = (0.69558) – (0.60078) – (1.986/28.97) ln(5/1) = - 0.01553 Btu/lb∙oR
T
5 atm
.
(2)
800oR
.(1)
1 atm
540oR
s
(d) oxygen as an ideal gas, T1 = T2 = 520oR, p1 = 10 atm, p2 = 5 atm.
For oxygen as an ideal gas; Δs =
– R ln (p2/p1). Since T1 = T2
Δs = - (1.986/32.00) ln(5/10) = 0.04302 Btu/lb∙oR
T
10 atm
5 atm
(1)
.
(2)
.
T1 = T2 = 520oR
s
PROBLEM 6.10
Propane undergoes a process from state 1, where p1 = 1.4 MPa, T1 = 60oC, to state 2,
where p2 = 1.0 MPa, during which the change in specific entropy is s2 – s1 = - 0.035
kJ/kg∙K. At state 2, determine the temperature, in oC, and the specific enthalpy, in kJ/kg.
= 1.81 kJ/kg∙K
PROBLEM 6.15
PROBLEM 6.22
A system consisting of 2 kg of water initially at 160oC, 10 bar, undergoes an internally
reversible, isothermal expansion during which there is energy transfer by heat into the system of
2700 kJ. Determine the final pressure, in bar, and the work, in kJ.
KNOWN: Two kg of water undergoes an isothermal process from a specified initial state. The
energy transfer by heat is given.
ENGINEERING MODEL: (1) The water is the
closed system. (2) The expansion takes place
FIND: Determine the final pressure and the work.
isothermally. (3) The process is internally
reversible. (4) Kinetic and potential energy effects
SCHEMATIC AND GIVEN DATA:
can be neglected.
T
10 bar
o
T = 160 C
W
Water
6.178 bar
p1 = 10 bar
m = 2 kg
(1)
.
.
(2)
160oC
Q = 2700 kJ
ANALYSIS: Using assumptions (2) and (3), and the definition of entropy change: dS=
= T(S2 – S1) = mT(s2 – s1)
or, on rearrangement and with data from Table A-2: s1 ≈ [email protected] C = 1.9427 kJ/kg∙K
/mT = 1.9427 kJ/kg∙K + (2700 kJ)/[(2 kg)(433 K)] = 5.0605 kJ/kg∙K
s2 = s1 +
Since sf < s2 < sg at 160oC, state 2 is in the two-phase liquid-vapor region, and the quality is
x2 =
=
= 0.649
and the pressure is the saturation pressure at 160oC: p2 = 6.178 bar
With modeling assumption (4), the energy balance reduces to give
W = Q + m(u1 – u2)
With data from Table A-2; u1 = uf(160oC) = 674.86 kJ/kg and
u2 = uf2 + x2(ug2- uf2) = 674.86 + (0.649)(2568.4 – 674.86) = 1903.77 kJ/kg
Finally
W = (2700 kJ) + (2 kg)(674.86 – 1903.77)kJ/kg = 242.2 kJ (out)
s
PROBLEM 6.26
A gas initially at 2.8 bar and 60oC is compressed to a final pressure of 14 bar in an isothermal
internally reversible process. Determine the work and heat transfer, each in kJ per kg of gas, if
the gas is (a) Refrigerant 134a, (b) air as an ideal gas. Sketch the process on p-v and T-s
coordinates.
KNOWN: A gas is compressed isothermally with no internal irreversibilities from a specified
initial state to a specified final pressure.
FIND: Determine the work and the heat transfer, per unit mass of gas, for (a) R-134a), (b) air as
an ideal gas. Sketch the process on p-v and T-s coordinates.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The gas is a
closed system. (2) The compression takes place
isothermally and with no internal
irreversibilities. (3) Kinetic and potential
energy effects are negligible. (4) For part (b),
the air behaves as an ideal gas.
T = 60oC
p1 = 2.8 bar
Gas
p2 = 14 bar
ANALYSIS: Using modeling assumption (2), the definition of entropy change: dS =
can be used
to obtain
= mT( s2 – s1)
Q=
→
Q/m = T( s2 – s1)
The energy balance reduces to give: ΔU = Q – W
→
(*)
W/m = Q/m + (u1 – u2)
(**)
(a) R-134a From Table A-12: u1 = 277.23 kJ/kg, s1 = 1.1079 kJ/kg∙K; u2 = 262.17 kJ/kg, s1 =0.9297kJ/kg∙K
Thus
Q/m = (60 + 273)K(0.9297 – 1.1079) kJ/kg∙K = -59.34 kJ/kg (out)
W/m = (-59.34 kJ/kg) + (277.23 – 262.17)kJ/kg = - 44.28 kJ/kg (in)
p
16.813bar
T
.
2
16.813bar
14 bar
2.8 bar
.
2
14 bar
.1
.1
2.8 bar
v
s
PROBLEM 6.26 (CONTINUED)
(b) Air Since T1 = T2, the specific entropy change of the air reduces to
s2 – s1 = - R ln (p2/p1) =
= -0.46139 kJ/kg∙K
ln
and
Q/m = T( s2 – s1) = (333 K)(- 0.46139 kJ/kg∙K) = -153.81 kJ/kg (out)
Since T1 = T2, the change in specific internal energy is zero, and
W/m = Q/m + (u1 – u2) = -153.81 kJ/kg (in)
p
T
.
2
14 bar
.1
2.8 bar
v
14 bar
.2
2.8 bar
.1
s
PROBLEM 6.28
PROBLEM 6.35