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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.

Respuesta :

Answer:

[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]

Explanation:

Now from first law for open system

[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]

Here given  Q=0 ,w=0

So [tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]

[tex]V_1=4 m/s,V_2=2 m/s[/tex]

[tex]h_1+\dfrac{4^2}{2000}=h_2+\dfrac{2^2}{2000}[/tex]

[tex]h_2-h_1=6\times 10^{-3}[/tex]

So increase in specific enthalpy

[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]

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