Compressed air at 40°C and 145 kPa from a large tank is discharged to ambient atmosphere at 101 kPa, through a convergent nozzle of 10 cm^2 exit area. Calculate the mass flow rate through the nozzle when the ambient atmosphere is at (a) 101 kPa, (b) 50 kPa, and (c) 30 kPa.

Question

Compressed air at [latex]40^{\circ} C[/latex] and 145 kPa from a large tank is discharged to ambient atmosphere at 101 kPa, through a convergent nozzle of [latex]10 cm ^{2}[/latex] exit area. Calculate the mass flow rate through the nozzle when the ambient atmosphere is at (a) 101 kPa, (b) 50 kPa, and (c) 30 kPa.

Accepted Answer

Given, [latex]p_{0}=145 kPa ext { and } T_{0}=40^{\circ} C =40+273.15=313.15 K[/latex]

(a) When pa =101 kPa, the pressure ratio across the nozzle is

[latex]\frac{p_{0}}{p_{a}}=\frac{145}{101}=1.44[/latex]

The pressure ratio has to be above 1.89 for the flow to choke at the nozzle exit.Therefore, for pressure ratio 1.44 the flow at the nozzle exit is only subsonic. By isentropic relation,

[latex]\frac{p_{0}}{p_{a}}=\left(1+\frac{\gamma-1}{2} M^{2} ight)^{\frac{\gamma}{\gamma-1}}[/latex]

[latex]1.44=\left(1+0.2 M^{2} ight)^{3.5}[/latex]

This gives M=0.74. From the isentropic table (Table A.1 in the Appendix), for M=0.74, we get

[latex]\frac{ ho}{ ho_{0}}=0.7712, \frac{T}{T_{0}}=0.9013[/latex]

Therefore,

[latex] ho=0.7712 ho_{0}=0.7712 \frac{p_{0}}{R T_{0}}[/latex]

[latex]=\frac{0.7712 imes 145 imes 10^{3}}{287 imes 313.15}=1.244 kg m ^{-3}[/latex]

[latex]T=0.9013 T_{0}=0.9013 imes 313.15=282.24 K[/latex]

[latex]a=\sqrt{\gamma R T}[/latex]

[latex]=\sqrt{1.4 imes 287 imes 282.24}=336.75 ms ^{-1}[/latex]

[latex]V=M a=0.74 imes 336.75=249.2 ms ^{-1}[/latex]

The mass flow rate is

[latex]\dot{m}= ho A V[/latex]

[latex]=1.244 imes 10 imes 10^{-4} imes 249.2[/latex]

[latex]=0.31 kgs ^{-1}[/latex]

(b) Given: [latex]p_{a}=50 kPa[/latex] Therefore, [latex]\frac{p_{0}}{p_{a}}=\frac{145}{50}=2.9[/latex] and hence the flow is choked and the nozzle exit Mach number is 1.

Fromthe isentropic table, for M=1, we get

[latex]\frac{ ho}{ ho_{0}}=0.6339, \quad \frac{T}{T_{0}}=0.8333[/latex]

[latex] ho=0.6339 ho_{0}=0.6339 \frac{P_{0}}{R T_{0}}[/latex]

[latex]=\frac{0.6339 imes 145 imes 10^{3}}{287 imes 313.15}=1.023 kg m ^{-3}[/latex]

[latex]T=0.8333 T_{0}=0.8333 imes 313.15=260.95 K[/latex]

[latex]a=\sqrt{\gamma R T}[/latex]

[latex]=\sqrt{1.4 imes 287 imes 260.95}=323.8 m s ^{-1}[/latex]

[latex]V=M a=1.0 imes 323.8=323.8 m s ^{-1}[/latex]

[latex]\dot{m}= ho A V[/latex]

[latex]=1.023 imes 10 imes 10^{-4} imes 323.8[/latex]

= 0.331 kg s [latex]^{−1}[/latex]

(c) Given: [latex]p_{a} =30 kPa[/latex]. Here again M=1. After choking, any change in the conditions downstream of the nozzle cannot influence the flow in the nozzle. This is because, for a given stagnation condition, the nozzle flow is frozen once it is choked. Only changes in the stagnation condition can influence the nozzle flow and not the downstream condition of the nozzle.Therefore, for [latex]p_{a} =30 kPa[/latex] also,

[latex]\dot{m}= 0.331 kg s^{-1}[/latex]

Question 2.7: Compressed air at 40°C and 145 kPa from a large tank is disc...

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