A convergent–divergent nozzle of throat area 10 cm^2 and exit area 24 cm^2 is run from an air storage tank at 300 kPa and 300K. Calculate the range of backpressure for which (a) the entire divergent portion will be supersonic and (b) the exit Mach number is less than 1. (c) Are the mass flow and

Question

A convergent–divergent nozzle of throat area [latex]10 cm^{2}[/latex] and exit area [latex]24 cm^{2}[/latex] is run from an air storage tank at 300 kPa and 300K. Calculate the range of backpressure for which (a) the entire divergent portion will be supersonic and (b) the exit Mach number is less than 1. (c) Are the mass flow and exit pressure independent of the backpressure?

Accepted Answer

(a) After choking, if the pressure [latex]p^{\ast }[/latex] at the throat is larger than the backpressure [latex]p_{b},[/latex] the flow will expand further downstream as a supersonic stream in the divergent portion of the nozzle. This will continue as long as the backpressure [latex]p_{b},[/latex] is less than the pressure that will compel the flow to establish a normal shock at the nozzle exit. Therefore, this problem essentially becomes a determination of the backpressure required for the formation of a normal shock at the nozzle exit.

Given:

[latex]\frac{A_{e}}{A_{th}}=\frac{24}{10}=2.4[/latex]

For this area ratio, from the isentropic table, we have

[latex]M_{e}=2.4,[/latex] [latex]\frac{p_{e}}{p_{0}}=0.0684[/latex]

This becomes the upstream Mach number [latex]M_{1}[/latex] for the normal shock at the nozzle exit. Therefore, from the normal shock table (Table A.2 in the Appendix), for [latex]M_{1}=2.4[/latex],

[latex]\frac{p_{2}}{p_{1}}=6.5533[/latex]

Thus,

[latex]p_{2}=6.5533 p_{1}=6.5533 p_{e}, ext { since } p_{e}=p_{1}[/latex]

[latex]=6.5533 imes 0.0684 imes p_{0}[/latex]

[latex]=6.5533 imes 0.0684 imes 300[/latex]

[latex]=134.47 kPa[/latex]

Hence, the flow in the complete divergent portion of the nozzle will be supersonic in the range of backpressures

[latex]0 \leq p_{b} \leq 134.47 kPa[/latex]

(b) The nozzle exit Mach number will be subsonic when the flow in the entire divergent portion of the nozzle is subsonic or if there is a normal shock in the divergent portion. For this to happen, the given area ratio will have to give a subsonic Mach number at the exit. Thus, for [latex]A_{e} / A_{ th }=2.4[/latex] from the isentropic table, we get

[latex]M_{e}=0.25, \quad \frac{p_{e}}{p_{0}}=0.9575[/latex]

[latex]p_{e}=0.9575 imes 300=287.25 kPa[/latex]

When [latex]p_{b} \geq 287.25[/latex] the entire divergent portion of the nozzle is subsonic flow. When [latex]p_{b}=134.47[/latex] there will be a normal shock at the nozzle exit. Therefore, for the subsonic flow in the divergent portion the backpressure has to be in the range

[latex]134.47

(c) The throat will be choked for [latex]\frac{p_{ ext {th }}}{p_{0}} \leq 0.528,[/latex] thus

[latex]p_{ ext {th }} \leq 0.528 imes 300[/latex]

[latex]\leq 158.40 kPa[/latex]

The choked flow will make the flow in the convergent portion frozen for any change in the downstream condition. Therefore, [latex]\dot{m}[/latex] is independent of [latex]p_{b}[/latex] for

[latex]0 \leq p_{b} \leq 158.40 kPa[/latex]

The exit pressure is independent of [latex]p_{b}[/latex] for

[latex]0 \leq p_{b} \leq 134.97 kPa[/latex]

This is because when the entire divergent duct flow is supersonic, for any fixed area ratio [latex]A_{e} / A_{ th }, M_{e}[/latex] is constant and, therefore, for a fixed [latex]p_{0}[/latex] and [latex]A_{e} / A_{ th },p_{e}[/latex] becomes independent of [latex]p_{b}[/latex] as long as the above conditions are maintained.

Question 2.9: A convergent–divergent nozzle of throat area 10 cm^2 and exi...

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