Air from a reservoir at 10 atm and 80^{\circ} C is discharged through a convergent–divergent nozzle fitted to the reservoir. The thrust exerted by the jet issuing from the nozzle is 11.12 kN. If the backpressure is 1 atm, calculate the nozzle throat and exit areas and Mach number of the jet issuing from the nozzle. Assume the flow through the nozzle to be isentropic and correctly expanded.
Question 2.3: Air from a reservoir at 10 atm and 80°C is discharged throug...
Given, p_{0}=10 atm , T_{0}=80+273.15=353.15 K , \text { and } p_{b}=1 atm . The flow is correctly expanded; therefore, p_{b}=p_{e}.
The pressure ratio across the nozzle is
\frac{p_{e}}{p_{0}}=\frac{1}{10}=0.1
For p_{e} / p_{0}=0.1, from the isentropic table we have the Mach number of the jet issuing from the nozzle as
M_{e}=2.16
Also,
\frac{T_{e}}{T_{0}}=0.5173\frac{A_{e}}{A_{\text {th }}}=1.935
Therefore,
T_{e}=0.5173 \times 353.15=182.68 Ka_{e}=\sqrt{1.4 \times 287 \times 182.68}=270.93 m s ^{-1}
The velocity at the nozzle exit is
V_{e}=M_{e} a_{e}=2.16 \times 270.93=585.2 m s ^{-1}
The thrust generated by the jet issuing from the nozzle is
F=11.12 \times 10^{3}=\dot{m} V_{e}
Therefore,
\dot{m}=\frac{11120}{585.2}=19 kg s ^{-1}
The mass flow rate \dot{m} is also given by
\dot{m}=\frac{0.6847}{\sqrt{R T_{0}}} p_{0} A_{\text {th }}
Therefore, the throat area becomes
A_{ th }=\frac{19 \sqrt{287 \times 353.15}}{0.6847 \times 10 \times 101325} =0.0087188 m ^{2}=87.19 cm ^{2}
The nozzle exit area is
A_{e}=1.935 \times 87.19=168.71 cm ^{2}
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