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m
1
and
m
2
can be found by using the formula:
v
V
m
State 1 is saturated vapor at 100kPa; the relevant properties can be found in table A-5.
v
1
= v
g at 100 kPa
= 1.6941 m
3
/kg
State 2 is at 300kPa and 200
°
C and is therefore a superheated vapor; the relevant
properties can be found in table A-6.
v
2
= 0.71643 m
3
/kg
The container is a rigid tank and the volume V stays constant at 2 m
3
during this
process.
kg
kg
m
m
v
V
m
791619558
.
2
/
0.71643
2
3
3
2
2
kg
kg
kg
m
m
m
i
8
1.61105155
180568
.
1
8
2.79161955
1
2
(b) By using the first law relation for a transient process, the enthalpy of the supplied
steam
h
sup
can be found.
𝑄
𝑐𝑣
− 𝑊
𝑐𝑣
= ∑ 𝑚
𝑒
ℎ
𝑒
𝑒𝑥𝑖??
− ∑ 𝑚
𝑖
ℎ
𝑖
𝑖𝑛𝑙𝑒??
+ (𝑚
2
𝑢
2
− 𝑚
1
𝑢
1
)
𝑐𝑣
(1)
Energy leaves the system during cooling, therefore,
Q
cv
=
–
700 kJ.
There is no work done on/by the system,
W
cv
= 0.
Steam only enters the tank, so
m
e
= 0 and
h
i
= h
sup
.
kg
kg
m
m
v
V
m
180568
.
1
/
6941
.
1
2
3
3
1
1

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