Determine the throat area, exit area and exit velocity for a steam nozzle to pass a mass flow of 0.2 kg/s when inlet conditions are 10 bar and 250°C and the final pressure is 2 bar. Assume expansion is isentropic and that the inlet velocity is negligible. Use pv^1.3 constant. Do not calculate from

Question

Determine the throat area, exit area and exit velocity for a steam nozzle to pass a mass flow of 0.2 kg/s when inlet conditions are 10 bar and 250°C and the final pressure is 2 bar. Assume expansion is isentropic and that the inlet velocity is negligible. Use [latex] pv^{1.3} [/latex] constant. Do not calculate from h-s chart. (N.U.)

Accepted Answer

Mass of steam flowing through the nozzle, [latex] \dot{m} _{s}[/latex] = 0.2 kg/s
Inlet pressure, [latex] p_{1}[/latex] = 10 bar
Inlet temperature, [latex] T_{1}[/latex] = 250 + 273 = 523 K
Specific volume at 10 bar, 250°C = 0.233 m³/kg (From steam tables)
Final pressure, [latex] p_{3}[/latex] = 2 bar

Throat area ([latex] A_{2}[/latex] ), exit area ([latex] A_{3}[/latex] ) and exit velocity ([latex] C_{3}[/latex] ) :

[latex] \frac{p_{2} }{ p_{1}} = (\frac{2}{n + 1}) ^{\frac{n}{n - 1}} = (\frac{2}{1.3 + 1}) ^{\frac{1.3}{1.3 - 1}} = 0.5457 [/latex]

∴ [latex] p_{2}[/latex] = 0.5457 × [latex] p_{1}[/latex] = 0.5457 × 10 = 5.457 bar

At throat :

Velocity [latex] C_{2} = \sqrt {\frac{2 n}{n - 1} p_{1} v_{1} [1 - ( \frac{p_{2} }{ p_{1}}) ^{\frac{n - 1}{n}} ]} [/latex]

= [latex] \sqrt {\frac{2 × 1.3}{1.3 - 1} × (10 × 10^{5}) × 0.233 [1 - ( \frac{5.457}{ 10}) ^{\frac{1.3 - 1}{1.3}} ]} [/latex]

= [latex] \sqrt{2019333.3 (1 - 0.8696)} [/latex] = 513.15 m/s

Also [latex] v_{2} = v_{1} ( \frac{p_{1} }{ p_{2}}) ^{\frac{ 1}{n}} = 0.233 ( \frac{10 }{ 5.457}) ^{\frac{ 1}{1.3}} [/latex] = 0.3712 m³ / kg

Mass flow rate, [latex] \dot{m}_{s} = \frac{A_{2} C_{2}}{v_{2}} or 0.2 = \frac{A_{2} × 513.15}{0.3712} [/latex]

or Throat area, [latex] A_{2} = \frac{ 0.2 × 0.3712 }{ 513.15}[/latex] = 1.446 × [latex]10^{–4}[/latex] m².

At exit :

Velocity, [latex] C_{3} = \sqrt {\frac{2 n}{n - 1} p_{1}v_{1} \left\{ 1 - (\frac{p_{3} }{ p_{1}}) ^{\frac{n - 1}{n}} \right\} } [/latex]

= [latex] \sqrt {\frac{2 × 1.3}{1.3 - 1} × (10 × 10^{5}) × 0.233 \left\{ 1 - ( \frac{2}{ 10}) ^{\frac{1.3 - 1}{1.3}} \right\} } [/latex]

= [latex] \sqrt{2019333.33 × 0.3102} [/latex] = 791.45 m /s

[latex] v_{3} = v_{1} ( \frac{p_{1} }{ p_{3}}) ^{1 / n} = 0.233 ( \frac{10 }{ 2}) ^{\frac{ 1}{1.3}} [/latex] = 0.8035 m³/kg

Mass flow rate, [latex] \dot{m} _{s} = \frac{A_{3} C_{3}}{v_{3}} or 0.2 = \frac{A_{3} × 791.45}{0.8035} [/latex]

or Exit area, [latex] A_{3} = \frac{ 0.2 × 0.8035 }{791.45}[/latex] = 2.03 × [latex]10^{–4}[/latex] m².

Question 5.19: Determine the throat area, exit area and exit velocity for a...

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