Question: Derive the expression for maximum discharge through steam nozzle.
0

Mumbai University > Mechanical Engineering > SEM 6 > Thermal and Fluid Power Engineering

Marks: 6M

ADD COMMENTlink
modified 19 months ago  • written 19 months ago by gravatar for Juilee Juilee2.3k
0

Let

$p_1$ and $p_2$= Pressure at inlet and exit respectively;

$v_1$ and $v_2$= Velocity at inlet and exit respectively;

$A_1$ and $A_2$= Area at inlet and exit respectively;

n = adiabatic index;

ṁ = mass flow rate or discharge

$\frac{ṁ}{A_2} = \sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\bigg\{\Big(\frac{p_2}{p_1}\Big)^{\frac{2}{n}}- \Big(\frac{p_2}{p_1}\Big)^{\frac{n+1}{2}}\bigg\}\Bigg]}$

On substituting the condition of maximum discharge,

$\Big(\frac{p_2}{p_1}\Big) = \Big(\frac{2}{n+1}\Big)^{\frac{n}{n-1}}$

$m_{max} = A_2\sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\bigg\{\Big(\frac{2}{n+1}\Big)^{\frac{n}{n-1}\times\frac{2}{n}}- \Big(\frac{2}{n+1}\Big)^{\frac{n}{n-1}\times\frac{n+1}{2}}\bigg\}\Bigg]}$

$m_{max} = A_2\sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\bigg\{\Big(\frac{2}{n+1}\Big)^{\frac{2}{n-1}}- \Big(\frac{2}{n+1}\Big)^{\frac{n+1}{n-1}}\bigg\}\Bigg]}$

$m_{max} = A_2\sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\Big(\frac{2}{n+1}\Big)^{\frac{n+1}{n-1}}\bigg\{ \Big(\frac{2}{n+1}\Big)^{\frac{2}{n-1}\frac{n+1}{n-1}}-1\bigg\}\Bigg]}$

$m_{max} = A_2\sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\Big(\frac{2}{n+1}\Big)^{\frac{n+1}{n-1}}\bigg\{ \Big(\frac{2}{n+1}\Big)^{-1}-1\bigg\}\Bigg]}$

$m_{max} = A_2\sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\Big(\frac{2}{n+1}\Big)^{\frac{n+1}{n-1}}\bigg\{ \Big(\frac{n+1}{2}\Big)^{1}-1\bigg\}\Bigg]}$

$m_{max} = A_2\sqrt{\Bigg[2\Big(\frac{n}{n-1}\Big)\frac{p_1}{v_1}\Big(\frac{2}{n+1}\Big)^{\frac{n+1}{n-1}}\bigg\{ \Big(\frac{n-1}{2}\Big)^{1}\bigg\}\Bigg]}$

$m_{max} = A_2\sqrt{\Bigg[n.\frac{p_1}{v_1}\Big(\frac{2}{n+1}\Big)^{\frac{n+1}{n-1}}}\Bigg]$

ADD COMMENTlink
written 19 months ago by gravatar for Juilee Juilee2.3k
Please log in to add an answer.