Air enters a frictionless adiabatic converging nozzle at 10 bar 500 K with negligible velocity. The nozzle discharges to a region at 2 bar. If the exit area of the nozzle is 2.5 cm², find the flow rate of air through the nozzle. Assume for air cp = 1005 J/kg K and cv = 718 J/kg K. (GATE)

Question

Air enters a frictionless adiabatic converging nozzle at 10 bar 500 K with negligible velocity. The nozzle discharges to a region at 2 bar. If the exit area of the nozzle is 2.5 cm², find the flow rate of air through the nozzle. Assume for air [latex] c_{p} [/latex] = 1005 J/kg K and [latex] c_{v} [/latex]= 718 J/kg K. (GATE)

Accepted Answer

Refer Fig. 26.
Given : [latex] p_{1}[/latex] = 10 bar ; [latex] T_{1}[/latex] = 500 K ; [latex] C_{1}[/latex] = 0 ; [latex] p_{2} [/latex] = 2 bar ; [latex] A_{2}[/latex] = 2.5 cm² ; [latex] c_{p} [/latex] = 1005 J/kg K ; [latex] c_{v} [/latex] = 718 J/kg K.

Flow rate of air, Q :

γ = [latex] \frac{c_{p}}{c_{v}} = \frac{1005}{718} [/latex] = 1.4

For the isentropic process 1-2 in the nozzle,

[latex] \frac{T_{1} }{T_{2} } = ( \frac{p_{1}}{p_{2}})^{\frac{γ - 1}{γ}} [/latex]

or [latex] \frac{500 }{T_{2} } = ( \frac{10}{2})^{\frac{1.4 - 1}{1.4}} [/latex] = 1.584

∴ [latex] T_{2} = \frac{500 }{1.584 } [/latex] = 315.6 K

Now, [latex] \frac{C_{2} ² - C_{1}² }{2} = h_{1} - h_{2}[/latex]

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

or [latex] C_{2} = \sqrt{2 (h_{1} - h_{2})} [/latex] (∵ [latex] C_{1} [/latex] = 0)

= [latex] \sqrt{2 c_{p} (T_{1} - T_{2})} [/latex]

= [latex] \sqrt{2 × 1005 (500 - 315.6)} [/latex] = 608.8 m/s

∴ Flow rate of air, [latex]Q = A_{2}C_{2} [/latex]

[latex]= 2.5 × 10^{–4} × 608.8[/latex]

= 0.1522 m³/s.

Question 5.24: Air enters a frictionless adiabatic converging nozzle at 10 ...

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