Ar ingressa num compressor a uma velocidade de 0.01 m/s, pressão de 1 atm e temperatura de 25 °C e deixa-o a pressão de 3 atm e a uma temperatura de 50 °C, através de uma abertura de 30 cm2 . Calcular a taxa de transferência de calor se a vazão mássica é de 10 kg/s e a potência do compressor é de 600 HP.
vi
0.01 m/s
pi
1 atm
Ti
25 °C
pf
3 atm
Tf
50 °C
Af
30 cm2
Φm
10 kg/s
P
600 HP
dQ/dt
a calcular
cp = 1000 J.kg-1 .K-1
Re = 290 J.kg-1 .K-1
Como neste problema temos um eixo que realiza trabalho, deve-se usar a seguinte forma da equação da primeira lei da termodinâmica:
d
Q
d
t
+
d
W
Ω
d
t
+
d
W
τ
d
t
+
d
W
o
u
t
r
o
d
t
=
∂
∂
t
∫
C
ρ
(
u
+
v
2
2
)
d
V
+
∫
S
[
ρ
(
u
+
v
2
2
)
−
σ
]
(
v
→
⋅
d
S
→
)
{\displaystyle {\frac {dQ}{dt}}\;+\;{\frac {dW_{\Omega }}{dt}}\;+\;{\frac {dW_{\tau }}{dt}}\;+\;{\frac {dW_{outro}}{dt}}\;=\;{\frac {\partial }{\partial t}}\int _{C}\rho (u\;+\;{\frac {v^{2}}{2}})dV\;+\;\int _{S}[\rho (u\;+\;{\frac {v^{2}}{2}})\;-\;\sigma ]({\vec {v}}\cdot d{\vec {S}})}
d
Q
d
t
+
P
+
0
+
0
=
0
+
∫
S
[
ρ
(
u
+
v
2
2
)
+
p
]
(
v
→
⋅
d
S
→
)
{\displaystyle {\frac {dQ}{dt}}\;+\;P\;+\;0\;+\;0\;=\;0\;+\;\int _{S}[\rho (u\;+\;{\frac {v^{2}}{2}})\;+\;p]({\vec {v}}\cdot d{\vec {S}})}
d
Q
d
t
+
P
=
∫
S
(
u
+
p
ρ
+
v
2
2
)
ρ
(
v
→
⋅
d
S
→
)
{\displaystyle {\frac {dQ}{dt}}\;+\;P\;=\;\int _{S}(u\;+\;{\frac {p}{\rho }}\;+\;{\frac {v^{2}}{2}})\rho ({\vec {v}}\cdot d{\vec {S}})}
A energia interna u é em parte termodinâmica e em parte gravitacional (u = ut + gz). Introduzindo a grandeza entalpia específica
h
=
H
m
=
U
t
+
p
V
m
=
U
t
m
+
p
ρ
=
u
t
+
p
ρ
{\displaystyle h\;=\;{\frac {H}{m}}\;=\;{\frac {U_{t}\;+\;pV}{m}}\;=\;{\frac {U_{t}}{m}}\;+\;{\frac {p}{\rho }}\;=\;u_{t}\;+\;{\frac {p}{\rho }}}
d
Q
d
t
+
P
=
∫
S
(
h
+
v
2
2
+
g
z
)
ρ
(
v
→
⋅
d
S
→
)
{\displaystyle {\frac {dQ}{dt}}\;+\;P\;=\;\int _{S}(h\;+\;{\frac {v^{2}}{2}}\;+\;gz)\rho ({\vec {v}}\cdot d{\vec {S}})}
d
Q
d
t
=
−
P
+
(
h
+
v
2
2
+
g
z
)
(
ρ
v
A
)
|
i
f
{\displaystyle {\frac {dQ}{dt}}\;=\;-\;P\;+\;\left.(h\;+\;{\frac {v^{2}}{2}}\;+\;gz)(\rho vA)\right|_{i}^{f}}
d
Q
d
t
=
−
P
+
(
h
f
+
v
f
2
2
+
g
z
f
)
(
ρ
f
v
f
A
f
)
−
(
h
i
+
v
i
2
2
+
g
z
i
)
(
ρ
i
v
i
A
i
)
{\displaystyle {\frac {dQ}{dt}}\;=\;-\;P\;+\;(h_{f}\;+\;{\frac {v_{f}^{2}}{2}}\;+\;gz_{f})(\rho _{f}v_{f}A_{f})\;-\;(h_{i}\;+\;{\frac {v_{i}^{2}}{2}}\;+\;gz_{i})(\rho _{i}v_{i}A_{i})}
∂
∂
t
∫
C
ρ
d
V
+
∫
S
ρ
v
→
⋅
d
s
→
=
0
⇒
∂
m
∂
t
+
ρ
v
A
|
i
f
=
0
{\displaystyle {\frac {\partial }{\partial t}}\int _{C}\rho dV\;+\;\int _{S}\rho {\vec {v}}\cdot d{\vec {s}}\;=\;0\Rightarrow \;\;\;{\frac {\partial m}{\partial t}}\;+\;\left.{\frac {}{}}\rho vA\right|_{i}^{f}\;=\;0}
0
+
ρ
f
v
f
A
f
−
ρ
i
v
i
A
i
=
0
⇒
ρ
f
v
f
A
f
=
ρ
i
v
i
A
i
{\displaystyle 0\;+\;\rho _{f}v_{f}A_{f}-\rho _{i}v_{i}A_{i}\;=\;0\Rightarrow \;\;\;\rho _{f}v_{f}A_{f}\;=\;\rho _{i}v_{i}A_{i}}
d
Q
d
t
=
−
P
+
ρ
f
v
f
A
f
(
h
f
+
v
f
2
2
+
g
z
f
−
h
i
−
v
i
2
2
−
g
z
i
)
{\displaystyle {\frac {dQ}{dt}}\;=\;-\;P\;+\;\rho _{f}v_{f}A_{f}\left(h_{f}\;+\;{\frac {v_{f}^{2}}{2}}\;+\;gz_{f}\;-\;h_{i}\;-\;{\frac {v_{i}^{2}}{2}}\;-\;gz_{i}\right)}
d
Q
d
t
=
−
P
+
Φ
m
(
h
f
−
h
i
+
v
f
2
2
−
v
i
2
2
)
{\displaystyle {\frac {dQ}{dt}}\;=\;-\;P\;+\;\Phi _{m}\left(h_{f}\;-\;h_{i}\;+\;{\frac {v_{f}^{2}}{2}}\;-\;{\frac {v_{i}^{2}}{2}}\right)}
p ΔT, onde Cp é a capacidade calorífica a pressão constante, e Δh = cp ΔT, onde cp é a capacidade calorífica específica a pressão constante
d
Q
d
t
=
−
P
+
Φ
m
(
c
p
(
T
f
−
T
i
)
+
v
f
2
−
v
i
2
2
)
{\displaystyle {\frac {dQ}{dt}}\;=\;-\;P\;+\;\Phi _{m}\left(c_{p}(T_{f}\;-\;T_{i})\;+\;{\frac {v_{f}^{2}\;-\;v_{i}^{2}}{2}}\right)}
p
f
V
f
=
n
R
T
f
⇒
p
f
m
ρ
f
=
n
R
T
f
⇒
ρ
f
=
p
f
m
n
R
T
f
=
p
f
m
e
R
T
f
=
p
f
R
e
T
f
{\displaystyle p_{f}V_{f}\;=\;nRT_{f}\Rightarrow \;\;\;p_{f}{\frac {m}{\rho _{f}}}\;=\;nRT_{f}\Rightarrow \;\;\;\rho _{f}\;=\;{\frac {p_{f}m}{nRT_{f}}}\;=\;{\frac {p_{f}m_{e}}{RT_{f}}}\;=\;{\frac {p_{f}}{R_{e}T_{f}}}}
e é o peso molecular do gás e Re = R/me é a constante específica para o gás.
f :
Φ
m
=
v
f
ρ
f
A
f
⇒
v
f
=
Φ
m
ρ
f
A
f
=
Φ
m
R
e
T
f
p
f
A
f
{\displaystyle \Phi _{m}\;=\;v_{f}\rho _{f}A_{f}\Rightarrow \;\;\;v_{f}\;=\;{\frac {\Phi _{m}}{\rho _{f}A_{f}}}\;=\;{\frac {\Phi _{m}R_{e}T_{f}}{p_{f}A_{f}}}}
d
Q
d
t
=
−
P
+
Φ
m
(
c
p
(
T
f
−
T
i
)
+
(
Φ
m
R
e
T
f
p
f
A
f
)
2
−
v
i
2
2
)
{\displaystyle {\frac {dQ}{dt}}\;=\;-\;P\;+\;\Phi _{m}\left(c_{p}(T_{f}\;-\;T_{i})\;+\;{\frac {({\frac {\Phi _{m}R_{e}T_{f}}{p_{f}A_{f}}})^{2}\;-\;v_{i}^{2}}{2}}\right)}
=
−
600
H
P
+
10
k
g
/
s
⋅
(
1000
J
⋅
k
g
−
1
⋅
K
−
1
⋅
(
50
o
C
−
25
o
C
)
+
(
10
k
g
/
s
⋅
290
J
⋅
k
g
−
1
⋅
K
−
1
⋅
50
o
C
3
a
t
m
⋅
30
c
m
2
)
2
−
(
0.01
m
/
s
)
2
2
)
{\displaystyle \;=\;-\;600HP\;+\;10\;kg/s\cdot \left(1000\;J\cdot kg^{-1}\cdot K^{-1}\cdot (50^{o}C\;-\;25^{o}C)\;+\;{\frac {({\frac {10\;kg/s\cdot 290\;J\cdot kg^{-1}\cdot K^{-1}\cdot 50^{o}C}{3\;atm\cdot 30\;cm^{2}}})^{2}\;-\;(0.01\;m/s)^{2}}{2}}\right)}
=
−
600
⋅
750
W
+
10
k
g
/
s
⋅
(
1000
J
⋅
k
g
−
1
⋅
K
−
1
⋅
25
K
+
(
10
k
g
/
s
⋅
290
J
⋅
k
g
−
1
⋅
K
−
1
⋅
(
50
+
273
)
K
300000
P
a
⋅
0.0030
m
2
)
2
−
(
0.01
m
/
s
)
2
2
)
{\displaystyle \;=\;-\;600\cdot 750\;W\;+\;10\;kg/s\cdot \left(1000\;J\cdot kg^{-1}\cdot K^{-1}\cdot 25\;K\;+\;{\frac {({\frac {10\;kg/s\cdot 290\;J\cdot kg^{-1}\cdot K^{-1}\cdot (50\;+\;273)\;K}{300000\;Pa\cdot 0.0030\;m^{2}}})^{2}\;-\;(0.01\;m/s)^{2}}{2}}\right)}
=
−
450000
W
+
250000
W
+
150000
W
=
−
50000
W
=
−
67
H
P
{\displaystyle \;=\;-\;450000\;W\;+\;250000\;W\;+\;150000\;W\;=\;-\;50000\;W\;=\;-\;67\;HP}
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