Question

Fluid enters a device at 4 m/s and leaves it at 2 m/s. If there is no change in the PE of tihe flow, and there is no heat and (non-flow) work across boundaries of the device, what is the increase in specific enthalpyof the fluid (hg-hi) in kJ/kg? Assume steady state operation of the device.

Answer 1

`h_2-h_1=6* 10^(-3)(KJ)/(Kg)`

Step-by-step explanation:

Now from first law for open system

`h_1+(V_1^2)/(2)+Q=h_2+(V_2^2)/(2)+w`

Here given Q=0 ,w=0

So
`h_1+(V_1^2)/(2)=h_2+(V_2^2)/(2)`

`V_1=4 m/s,V_2=2 m/s`

`h_1+(4^2)/(2000)=h_2+(2^2)/(2000)`

`h_2-h_1=6* 10^(-3)`

So increase in specific enthalpy

`h_2-h_1=6* 10^(-3)(KJ)/(Kg)`