Question 5.24: Air enters a frictionless adiabatic converging nozzle at 10 ...
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 c_{p} = 1005 J/kg K and c_{v} = 718 J/kg K. (GATE)
Refer Fig. 26.
Given : p_{1} = 10 bar ; T_{1} = 500 K ; C_{1} = 0 ; p_{2} = 2 bar ; A_{2} = 2.5 cm² ; c_{p} = 1005 J/kg K ; c_{v} = 718 J/kg K.
Flow rate of air, Q :
γ = \frac{c_{p}}{c_{v}} = \frac{1005}{718} = 1.4
For the isentropic process 1-2 in the nozzle,
\frac{T_{1} }{T_{2} } = ( \frac{p_{1}}{p_{2}})^{\frac{γ – 1}{γ}}or \frac{500 }{T_{2} } = ( \frac{10}{2})^{\frac{1.4 – 1}{1.4}} = 1.584
∴ T_{2} = \frac{500 }{1.584 } = 315.6 K
Now, \frac{C_{2} ² – C_{1}² }{2} = h_{1} – h_{2}
or C_{2} ² – C_{1}² = 2 (h_{1} – h_{2})
or C_{2} = \sqrt{2 (h_{1} – h_{2})} (∵ C_{1} = 0)
= \sqrt{2 c_{p} (T_{1} – T_{2})}
= \sqrt{2 × 1005 (500 – 315.6)} = 608.8 m/s
∴ Flow rate of air, Q = A_{2}C_{2}
= 2.5 × 10^{–4} × 608.8= 0.1522 m³/s.
