April Norton
Q1: (25%) Air at !! = #$# [℃] and )! = * [+,-] enters a steady flow-steady state energy (heat) exchanger
(system A) and leaves at !" = .# [℃] and )" = . [+,-]. Energy is transferred (q) to a heat engine (system
B) that interacts with environment at !" = .# [℃] and produces work. Assume air is ideal gas with variable
#$
specific heat with / = 0. $2#[#%&].
A
B
a) Determine amount of energy transferred per unit mass flow of state 1 (q) from system A in [34/36]
b) If all processes are reversible:
i.
Define your system as that energy exchanger (A) and the engine (B), write the entropy balance
for this combined system (A+B) and calculate q0.
ii.
How much work is done by the engine per kg of air?
iii.
Calculate thermal efficiency of the engine.
c) What is flow exergy of air at state 1?
d) Compare flow exergy of air at state 1 with the work of reversible engine of part b and comment.
e) In an actual case having an irreversible engine the work output is 40% of the reversible engine. For this
case, calculate:
i.
amount of energy transferred from the engine to environment,
ii.
entropy production
iii.
exergy destruction per unit mass of air.
HINT: For part ( e ), use the engine itself as your system.
1
Q
O
0
q
h
n1
Solution
frac
9
6
0
4
T
2
h
as1
right
cdot
frac
left
Q
C
9
0
K
3
T
2
7
ns1
w
0
q
4
8
9
6
0
5
q
g
2
7
k
J1
right
left
T
0
q
s1
right
frac
left
l9
6
0
4
8
R
S
3
5
T
P
2
q
g
7
o
Jk1n
w
4
8
9
3
5
g
2
7
k
J1
frac
eta
w
6
9
0
8
3
5
2
q
71
6
9
7
right
left
6
9
0
4
8
3
5
T
2
g
7
h
e
Jks1j
w
e
1f no
wactual w
4
0
rev
q actual
0
qactual w
sigma
frac
aa1ctual
0
0
5
2
g
7
J
k1
0
5
g
7
k
J
right
frac
left
9
6
8
0
3
5
2
7
s1
sigma
9
6
0
T
3
5
2
g
E
7
J
k1
2
Q2: (25%) Outside fresh moist air enters a mixing section of an air handling unit at 4oC and relative
humidity of 20% in fall season. The fresh moist air then mixes with 1kg/s return flow of moist air from
the building to make 22oC temperature and 80% relative humidity. The return moist air enters the mixing
section at 35oC and 50% humidity. Total pressure is kept at 1[bar].
a) Find the mass flow rate of required fresh air enters at inlet 1.
b) Is this process adiabatic or heat energy as heat should be exchanged? Assume the dry air is perfect gas
#$
#$
with 7' = . [#% &] and / = 0. $2# [#% &].
3
Solution
phi
0
2
1
0
5
2i
cdot phi
circ
C
4
3
5
T
2
c1 phi803i
n
1
frac
g
2
n
a
k
s1
6
P
a
1r
omega
frac
V
6
0
P
2
Psat right
left
8
T
0
3
1 bar
right
left
8
6
T
0
5
2 bar
Psat
right
left
4
6
T
0
3
5
2 bar
Psat
phi
V
P
1 Psat
4
8
P
0
2
v
1
V
6
P
0
3
2
1
6
0
2
7
1
omega
4
0
1
omega
8
0
2
1
omega
4
0
3
5
1
3
2
n
a
1
Dryair
watervap
w
3
2
n
a
1
2
n
a
1
3
n
a
1
right
left
m
w
6
0
4
8
3
5
2
g
7
n
aks1
Q
0
3
2
h
n
a
1
4
3
g
2
n
a
k
s1
w
T
p
g
h
c
6
0
4
8
3
5
2
g
7
h1
right
left
6
0
4
8
3
5
2
g
7
h
Jk1
right
left
W
Q
6
4
8
0
3
5
2
7
k1
Q
0
Notadiabatic
4
Q3: (25%) 1[kg/sec] of fresh moist air enters an air handling unit, as shown below. The unit has two stages of
cooling and heating inline. Fresh moist air at 35[oC] and 90% relative humidity enters the cooling stage where
its temperature is decreased to 15[oC]. Then the saturated cooled air (i.e. its relative humidity is 100%) is
heated up to 25[oC] for serving the purpose of human being comfort zone. Assume dry air is perfect gas with
#$
#$
7' = . [#% &] and / = 0. $2# [#% &] and the total pressure is kept at 1[bar].
a) Fill the empty cells in the table below.
b) Find the rate of water condensation (8̇) [kg/s]) in cooling section.
c) Find the cooling load required in tons of refrigeration.
d) Find the heating load required in [kW].
e) On the T-ʋ diagram show the states of water vapor and on psychrometric chart draw the process
schematically and carefully labeled each state on the charts.
State
1
2
3
Temperature
Relative Humidity
o
(%)
90
(kPa)
(kgv/kga)
6
0
5
0
3
2
0
1
0
5
7
1
8
0
1
8
3
5
0
5
7
1
8
0
1
( C)
35
15
25
Vapor Partial pressure Humidity ratio
6
a
1r
1
cdot
3
cdot
2
2
5
3
Solution
phi
9
0
1i
cdot
circ
C
3
5
T
2
1
phi
0
2
1
Psat
Psat
Psat
cdot
rho
phi
V
P
1 sat1
right
left
8
6
T
0
5
2
1
rho
V
6
0
3
5
P
2
7
v1
right
left
T
0
5
2
7
1
right
left
9
6
T
0
3
1
cdot
phi
9
6
0
3
5
7
1
w
T
p
g
h
c
1
omega
right
frac
left
w
6
9
0
8
3
2
g
7
n
aks1
6
9
8
R
3
5
2
g
h1f
omega
frac
6
0
3
5
2
1
omega
9
0
2
7
1
omega
3
2
phi
8
0
3
5
8
3
5
9
0
4
5
2
g
h
Jk1
right
left
W
Q
w
6
9
4
0
3
5
2
h
7
n
c
a
k1j
w
4
8
3
T
5
2
p
g
h
c
J
k
right
left
W
Q
H
6
9
4
8
0
3
5
2
h
n
7
a
k
6
8
6
3
5
2
1
tons
Q4: (25%) Figure shows a vapor compression refrigeration cycle with refrigerant R134a circulating through
the components. Compressor and expansion valve are well insulated, kinetic, and potential energy effects are
negligible for all components. Some of the properties are on the sketch, for others you can use the table at the
back. The steady‐ state refrigeration capacity of the cycle is 7 kW (i.e. Q̇ *+ ). Let the T, = 27[, C] and
p, = 1[bar].
Note: This problem requires reading properties from the table in the equation sheet
a) Sketch the processes on the T‐ s diagram indicating all states
and process direction as well as isobars passing through all
the states.
Determine:
b) Specific enthalpy of refrigerant at the exit of the expansion
valve, ℎ- , in kJ/kg,
c) mass flow rate of refrigerant, in kg/s,
d) Power consumed by the compressor, in kW,
e) Coefficient of performance for refrigeration,
f) Isentropic efficiency of the compressor.
g) Find the quality at state 4.
h) Exergy rate destroyed in Expansion valve.
T
abar2is31.6bar2nN9l4
s
7
Solution
4
8
9
3
g
h
7
k
J
Value is
enthropic
frac
lg
Q
9
4
0
8
3
2
7
h
n
s1
right
left
Q
4
h
n
1i
right
left
W
4
9
8
0
3
2
h
n
7
c
1l
beta
frac
W
Qi
n
c
frac
eta
2
h
cs1
frac
9
3
5
2
7
1
at1.6bar
S
2
1
9
S
0
5
g
2
1
at 4 bar S betweenTsat21.6and30C
2160.097
0.9295
0
3 o 8
9
3
matrix
begin
end
eta
9
6
0
8
3
5
2
7
c1
4
3
h
27.3
30267.89
frac
9
0
4
8
2
g
7
h
ux1f
4
0
8
3
2
x
right
left
4
0
3
T
E
n
sj
T
0
3
2
7
has
21.6 zsa.la
right
left
6
9
0
4
8
S
3
5
2
a1rf
right
left
9
4
0
3
2
E
7
1
partial
W
8
0
E
k
1
8
circ
C
T
3
2
7
a
6
5
2
h
s1
Extra
9
ME 2380 THERMODYNAMICS
Fall 2023
EQUATION SHEET
General Balance Equations:
!
# = % #̇# − % #̇&
!" !"
#$%&'(
&)#'(
!
1 *
1
̇
̇
(!" = )!" − *!" + % #̇# ,ℎ# + 0# + 12# 3 − % #̇& ,ℎ& + 0& * + 12& 3
!"
2
2
#$%&'(
-
&)#'(
)̇+
!
4!" = % 5 7 + % #̇# 8# − % #̇& 8& + 9̇!"
!"
6+ ,
+./
#$%&'(
&)#'(
!
60
!<!"
Ε!" = % 51 − 7 )̇+ − ,*̇!" − ;0
3 + % #̇# =1# − % #̇& =1& − Ε̇2 ((@=A1B CDEFG=)
!"
6+
!"
+
#$%&'(
&)#'(
Ε̇2 = 60 9̇!" (IE"= JK (@=A1B !=8"ALG"MJF)
Energy balance for closed system: ∆F = G − I, where I = I./012345 + I676894:837 + I;<3=9 + ⋯
I. Boundary Work: I = ∫ MNO
a) Constant volume process: I = 0
b) Constant pressure process: I = M(O> − O? )
c) Polytropic process where MO 1 = STUVWXUW: I =
(A3 B3 CA4 B4 )
?C1
B3
d) If U = 1, MO = STUVWXUW : I = M? O? YU
B4
II. Electrical Work: I = O (ZTYWX[\) ∙ ^(S_``\UW) ∙ ∆W
E
Thermal Efficiency of a power cycle: a = F567
85
F
Coefficient of performance of a refrigeration cycle: b = E85
F
9:7
Coefficient of performance of a heat(energy) pump cycle: c = E
̇
(Iv)
B
Mass & Volume Flow rates: 8̇ = dȮ = H = H
Some Property Relations: e =
HCH;
H< CH;
=
ℎ = _ + fZ g = F + MO Z̅ = Zi
0C0;
0< C0;
=
<C<;
<< C<;
=
;C;;
;< C;;
LF
_j = _i V̅ = Vi G :19. = ∫ k NV V> − V? = ∫ l M m :19.
46H.
46H.
Compressed Liquid Approximation: Z(k, f) ≈ Z= (k) _(k, f) ≈ _= (k) V(k, f) ≈ V= (k)
ℎ(k, f) ≈ _= (k) + fZ= (k) ≈ ℎ= (k) + (f − f;39 )Z= (k)
H3
Specific exergy: \ = (_ − _N ) + fN (Z − ZN ) − kN (V − VN ) + > + [p
H3
Specific flow exergy: \= = (ℎ − ℎN ) − kN (V − VN ) + > + [p
8
P
Ideal Gas Relations: fZ = qk SO = SH + q r = 8= SO = PC? q
>
M
?
SH = PC? q
_> − _? = ∫M 3 SH (k)Nk, when SH is constant, _> − _? = SH (k> − k? )
4
M
ℎ> − ℎ? = ∫M 3 SO (k)Nk, when SO is constant, ℎ> − ℎ? = SO (k> − k? )
4
M3 8= (M)
O
O
M
O
M
V> − V? = ∫M M Nk − q ln O3 = (V>/ − V?/ ) − q ln O3 , when SO is constant, V> − V? = SO ln M3 − q ln O3 = SH ln M?3 +
4
4
4
4
4
H3
q ln H
4
If SH and SO are not constant, please check the table.
O3
O? (M3 )
@
M @A4
For isentropic process, when SO is not constant, O = O (M ); when SO is constant, O = (H ) = lM3 m
4
? 4
4
3
4
EB
<8 C<6
E8
<6C C<8
Isentropic turbine efficiency: a9 = E = < C<
Isentropic compressor efficiency: a8 = E = < C<
8
8
6C
O3
H4 P
B
10
6
8
Isentropic pump efficiency: aO =
E8
EB
H (O6 CO8 )
= 8
EB
Work of reversible open system: I46H. = − ∫ Z Nf for reversible pump, I 46H. = −Z(f6 − f: )
O0QO
MD
Carnot power cycles efficiency: aQ3R = 1 − M
E
M
D
Carnot refrigeration cycles efficiency: bQ3R = M CM
E
ME
Carnot heat (energy) pump cycles efficiency: cQ3R = M CM
E
D
D
Moist Air properties:
1
1
uH = 1> u3 = 1B fS = f;39 (k) k2 = k;39 (fH ) 8 = 83 + 8H = (1 + v)83 ℎQ = ℎ3 + vℎH
5>
5B
O
= O>
B
O
>
w = O (M)
<
Q
O
B
B
O
OCO
>
v = Q > = 0.622 O> = 0.622 OCO
= 0.622 O B
>
B
O
fH = ?T(N.U>>/W)
Approximations for Water Vapor at Low pressures:
X M
ZH ≅ O>
>
PY
qH ≅ 0.461 PS⋅[
_H ≅ _S (k)
ℎH ≅ ℎS (k)
Unit Conversions:
1 N = 1 kg-m/s2 1 kPa = 1 kN/m2 1 J = 1 N-m
1 Ton of Refrigeration = 3.51685 kW
1 bar = 100 kPa
11
O
VH ≅ VS (k) − qH ln {O> | = VS (k) − qH ln w
<
12
13
14
15
16
17