Question 2.3: Air from a reservoir at 10 atm and 80°C is discharged throug...

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.

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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 K

 

a_{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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