Given,
The enthalpy of the fluid = 3000 KJ/kg
Velocity of the fluid =60 m/s
At the time of discharge, specific enthalpy =2762 KJ/ Kg
a) Applying Bernoulli's Energy Equation,
⇒L1+2v12+gz1+Q=L2+2v22+gz2+W
⇒v22=2(h1−h2+Q)+v12
⇒v22=2(3000−2762−0.1)103+602
⇒v2=632.38m/s
b) A1=0.1m2
v1=0.19
m=ρ2A1V1
⇒m=0.190.1×60=31.578kg/s
c) m=ρ2A2V2
⇒A2=692.3831.578×0.5
⇒A2=0.0228m2
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