Question #123789

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


1
Expert's answer
2020-06-26T14:37:41-0400

Solution.

h1=3052kJ/kg=3.052106J/kg;h_1=3052kJ/kg=3.052\sdot10^6J/kg;

C1=60m/s;C_1=60m/s;

h2=2790kJ/kg=2.79106J/kg;h_2=2790kJ/kg=2.79\sdot10^6J/kg;

A1=0.1m2;A_1=0.1m^2;

v1=0.19m3/kg;v_1=0.19m^3/kg;

v2=0.5m3/kg;v_2=0.5m^3/kg;

h1+C122+Z1g+Q=h2+C222+Z2g+W;h_1+\dfrac{C_1^2}{2}+Z_1g+Q=h_2+\dfrac{C_2^2}{2}+Z_2g+W;

Q=0;W=0;Z1=Z2;Q=0; W=0; Z_1=Z_2;

h1+C122=h2+C222;h_1+\dfrac{C_1^2}{2}=h_2+\dfrac{C_2^2}{2};

C222=h1h2+C122;\dfrac{C_2^2}{2}=h_1-h_2+\dfrac{C_1^2}{2};

C2=2h12h2+C12;C_2=\sqrt{2h_1-2h_2+C_1^2};

C2=23.05210622.79106+602=726.36m/s;C_2=\sqrt{2\sdot3.052\sdot10^6-2\sdot2.79\sdot10^6+60^2}=726.36m/s;

m=A1C1v1;m=\dfrac{A_1C_1}{v_1};

m=0.1600.19=31.58kg;m=\dfrac{0.1\sdot60}{0.19}=31.58kg;

m=A2C2v2;m=\dfrac{A_2C_2}{v_2};

A2=mv2C2;A_2=\dfrac{mv_2}{C_2};

A2=31.580.5726.36=0.0217m2;A_2=\dfrac{31.58\sdot0.5}{726.36}=0.0217m^2;


Answer: C2=726.36m/s;C_2=726.36m/s;

m=31.58kg;m=31.58kg;

A2=0.0217m2.A_2=0.0217m^2.







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