Solution;
Given;
m˙=1kg/s
p1=200kPa
V=0.4m3/kg
v1=100m/s
p2=3p1=600kPa
V2=3V1=1.2m3/kg
v2=3v1=300m/s
(a)
p1V1n=p2V2n
600200=(0.41.2)n
31=3n
3−1=3n
Therefore;
n=−1
(b)
From;
PV=mRT
Mco2=44
Hence;
R=448.314=0.18895
Therefore;
T1=mRpV=1×0.18895200×0.4=423.38K
T2=mRp2V2=1×0.18895600×1.2=3810.53K
(c)
Δh=cpΔT
cp=0.756kJ/kgK
Δh=0.756×(3810.53−423.38)
Δh=2560.69kJ/kg
(d)
Power = Change in Kinetic Energy
P=21m(v22−v12)
P=21×1(3002−1002)
P=40000W
P=40kW
(e)
Heat
Assuming the device is horizontal;
z1=z2
From steady flow energy equation;
Q=Δh+m˙(2000v12−v22)+W
By direct substitution;
Q=2560+40+40=2640.69kW
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