Question 5.19: Determine the throat area, exit area and exit velocity for a...
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 h-s chart. (N.U.)
Mass of steam flowing through the nozzle, \dot{m} _{s} = 0.2 kg/s
Inlet pressure, p_{1} = 10 bar
Inlet temperature, T_{1} = 250 + 273 = 523 K
Specific volume at 10 bar, 250°C = 0.233 m³/kg (From steam tables)
Final pressure, p_{3} = 2 bar
Throat area ( A_{2} ), exit area ( A_{3} ) and exit velocity ( C_{3} ) :
\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∴ p_{2} = 0.5457 × p_{1} = 0.5457 × 10 = 5.457 bar
At throat :
Velocity C_{2} = \sqrt {\frac{2 n}{n – 1} p_{1} v_{1} [1 – ( \frac{p_{2} }{ p_{1}}) ^{\frac{n – 1}{n}} ]}
= \sqrt {\frac{2 × 1.3}{1.3 – 1} × (10 × 10^{5}) × 0.233 [1 – ( \frac{5.457}{ 10}) ^{\frac{1.3 – 1}{1.3}} ]}
= \sqrt{2019333.3 (1 – 0.8696)} = 513.15 m/s
Also v_{2} = v_{1} ( \frac{p_{1} }{ p_{2}}) ^{\frac{ 1}{n}} = 0.233 ( \frac{10 }{ 5.457}) ^{\frac{ 1}{1.3}} = 0.3712 m³ / kg
Mass flow rate, \dot{m}_{s} = \frac{A_{2} C_{2}}{v_{2}} or 0.2 = \frac{A_{2} × 513.15}{0.3712}
or Throat area, A_{2} = \frac{ 0.2 × 0.3712 }{ 513.15} = 1.446 × 10^{–4} m².
At exit :
Velocity, C_{3} = \sqrt {\frac{2 n}{n – 1} p_{1}v_{1} \left\{ 1 – (\frac{p_{3} }{ p_{1}}) ^{\frac{n – 1}{n}} \right\} }
= \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\} }
= \sqrt{2019333.33 × 0.3102} = 791.45 m /s
v_{3} = v_{1} ( \frac{p_{1} }{ p_{3}}) ^{1 / n} = 0.233 ( \frac{10 }{ 2}) ^{\frac{ 1}{1.3}} = 0.8035 m³/kg
Mass flow rate, \dot{m} _{s} = \frac{A_{3} C_{3}}{v_{3}} or 0.2 = \frac{A_{3} × 791.45}{0.8035}
or Exit area, A_{3} = \frac{ 0.2 × 0.8035 }{791.45} = 2.03 × 10^{–4} m².