Example $3\mathrm{.}5.$ Mass flow in a converging nozzle

Air is allowed to flow from a large reservoir through a converging nozzle with an exit area of $50c{m}^2.$ The reservoir is large enough so that negligible changes in reservoir pressure and temperature occur as fluid is exhausted through the nozzle. Assume isentropic$,$ steady flow in the nozzle, with $p_r=500$kPa and $T_r=400K.$ Assume also that air behaves as a perfect gas with constant specific heats; $=1\mathrm{.}4.$ Determine the mass flow through the nozzle for back pressures of $0,125,250,$ and $375$ kPa $.$

Convergent Nozzle...

Solution

For $=1\mathrm{.}4,$ the critical pressure ratio is $0\mathrm{.}5283$ ; therefore, for all back pressures below $(500)(0\mathrm{.}5283)=264\mathrm{.}15$kPa, the nozzle is choked. Under these conditions, the Mach number at the exit plane is unity, the pressure at the exit plane $(p_e)$ is $264\mathrm{.}15$ kPa $($does not equal the back pressure if $p_b<mn>2</mn><mn>6</mn><mn>4</mn><mn>.</mn><mn>1</mn><mn>5</mn>$kPa $),$ and the temperature at the exit plane $(T_e)$ is $[(\frac{T}{T_o}{)}_e]T_r=(0\mathrm{.}8333)(400)=333\mathrm{.}3K,$ where $(\frac{T}{T_o}{)}_e$ is obtained from Eq$.(3\mathrm{.}12)$ at $M_e=1$ for $=1\mathrm{.}4($or from the tables in Appendix B$).$ The mass$-$flow rate for all back pressures below the critical pressure is

$m^{}={}_e{A}_e{V}_e$

$=(\frac{p_e}{R{T}_e})A_e(M_e\sqrt[\phantom{2}]{R{T}_e})$

$=[\frac{(264\mathrm{.}15k\frac{N}{m^2})}{(0\mathrm{.}287kN*\frac{m}{k}g)(333\mathrm{.}3(K))}](50{10}^{-4}{m}^2)$

$[(1\mathrm{.}0)\sqrt[\phantom{2}]{(1\mathrm{.}4)(287N*\frac{m}{k}g*K)(333\mathrm{.}3K)}]$

$=5\mathrm{.}053k\frac{g}{s}$

Convergent Nozzle...

This is the flow rate at back pressures of $0,125,$ and $250$ kPa $.($See Figure $3\mathrm{.}13.)$

For a back pressure of $375$ kPa $,$ which is above the critical pressure, the nozzle is not choked, and the exit$-$plane pressure is equal to the back pressure. For $\frac{p_d}{p_r}=\frac{375}{500}=0\mathrm{.}75=(\frac{p}{p_o}{)}_e,$ we find that $M_e=0\mathrm{.}654[$from Eq$.(3\mathrm{.}16)$ or from the tables in Appendix $B$ and use it to determine that $\frac{T_e}{T_r}=0\mathrm{.}921.$ So$,$

$m^{}=[\frac{375}{(0\mathrm{.}287)(0\mathrm{.}921)(400)}](50{10}^{-4})[(0\mathrm{.}654)\sqrt[\phantom{2}]{(1\mathrm{.}4)(287)(0\mathrm{.}921)(400)}]=4\mathrm{.}462k\frac{g}{s}$

$38$