Answer to Question #123789 in Molecular Physics | Thermodynamics for Clara ogu

Question

A nozzle is a device for increasing the velocity of a steady flowing steam of fluid.at the inlet to a certain nozzle the enthalpy of the fluid is 3052kj/kg and the velocity is 60m/s. At the exit from the nozzle the enthalpy is 2790kj/kg the nozzle is horizontal and there is negligible heat loss from it

Find the velocity at the nozzle exit

If the inlet area is 0.1m² and specific volume at inlet is 0.19m³/kg find the rate of flow of fluid

If the specific volume at the nozzle exit is 0.5m³/kg find the exit area of the nozzle

Answer 1

Solution.

$h_1=3052kJ/kg=3.052\sdot10^6J/kg;$

$C_1=60m/s;$

$h_2=2790kJ/kg=2.79\sdot10^6J/kg;$

$A_1=0.1m^2;$

$v_1=0.19m^3/kg;$

$v_2=0.5m^3/kg;$

$h_1+\dfrac{C_1^2}{2}+Z_1g+Q=h_2+\dfrac{C_2^2}{2}+Z_2g+W;$

$Q=0; W=0; Z_1=Z_2;$

$h_1+\dfrac{C_1^2}{2}=h_2+\dfrac{C_2^2}{2};$

$\dfrac{C_2^2}{2}=h_1-h_2+\dfrac{C_1^2}{2};$

$C_2=\sqrt{2h_1-2h_2+C_1^2};$

$C_2=\sqrt{2\sdot3.052\sdot10^6-2\sdot2.79\sdot10^6+60^2}=726.36m/s;$

$m=\dfrac{A_1C_1}{v_1};$

$m=\dfrac{0.1\sdot60}{0.19}=31.58kg;$

$m=\dfrac{A_2C_2}{v_2};$

$A_2=\dfrac{mv_2}{C_2};$

$A_2=\dfrac{31.58\sdot0.5}{726.36}=0.0217m^2;$

Answer: $C_2=726.36m/s;$

$m=31.58kg;$

$A_2=0.0217m^2.$