Air is expanded reversibly and adiabatically in a nozzle from 13 bar and 150°C to a pressure of 6 bar. The inlet velocity of the nozzle is very small and the process occurs under steady state flow conditions. Calculate the exit velocity of the nozzle. (U.P.S.C., 1992)

Question

Air is expanded reversibly and adiabatically in a nozzle from 13 bar and 150°C to a pressure of 6 bar. The inlet velocity of the nozzle is very small and the process occurs under steady state flow conditions. Calculate the exit velocity of the nozzle. (U.P.S.C., 1992)

Accepted Answer

Given : [latex] p_{1} [/latex] = 13 bar ; [latex] T_{1} [/latex] = 150 + 273 = 423 K ; [latex] p_{2} [/latex] = 6 bar ; [latex] C_{1}[/latex] = 0

Exit velocity, [latex] C_{2} [/latex] :
Applying the energy equation at ‘1’ and ‘2’, we get

[latex] m [h_{1} + \frac{ C_{1} ²}{2} + Z_{1} g] + Q = [h_{2} + \frac{ C_{2} ²}{2} + Z_{2} g ] + W[/latex] ...(i)

Since the air is expanded reversibly and adiabatically in a nozzle from condition ‘1’ to ‘2’, therefore,

Q = 0 ; Also W = 0 and [latex] Z_{1} = Z_{2} [/latex]
∴ Eqn. (i) reduces to :

[latex] h_{1} + \frac{ C_{1} ²}{2} = h_{2} + \frac{ C_{2} ²}{2} [/latex]

But [latex] C_{1} [/latex] = 0

∴ [latex]C_{2} ² = 2 (h_{1} - h_{2})[/latex]

or [latex]C_{2} = \sqrt{ 2 (h_{1} - h_{2})} = \sqrt{ 2 × c_{p} (T_{1} - T_{2})}[/latex] ...(ii)

Now, [latex] \frac{ T_{2}}{T_{1}} = (\frac{ p_{2}}{p_{1}})^{ \frac{ γ - 1}{γ} } = (\frac{ 6}{13})^{ \frac{ 1.4 - 1}{1.4} }[/latex] = 0.8018

∴ [latex]T_{2}[/latex] = 423 × 0.8018 = 339.16 K
Substituting the values in eqn. (ii), we get

[latex]C_{2} = \sqrt{ 2 × 1.005 (423 - 339.16) } [/latex] = 12.98 m/s.

Question 5.23: Air is expanded reversibly and adiabatically in a nozzle fro...

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