Question #123977
A nozzle is a device for increasing the velocity of a steadily flowing stream of fluid. At the inlet to a certain nozzle the enthalpy of the fluid is 3025kj/kg and the velocity is 60m/s. At the exit from the nozzles the enthalpy is 2790kj/kg.The nozzle is horizontal and there is negligible heat loss from it.
(i) Find the velocity at the nozzle exit.
(ii) If the inlet area is 0.1m^2 and the specific volume at the inlet is 0.19m^3/kg. Find the rate of flow of fluid.
(ii) If the specific volume at the nozzle exit is 0.5m^3/kg. Find the exit area of the nozzle
1
Expert's answer
2020-06-30T18:25:23-0400

i) According to Conservation Energy:



h1+0.5v12=h2+0.5v22h_1+0.5v^2_1=h_2+0.5v^2_2



3000.103+0.5(60)2=2762.103+0.5v223000 .10^3+0.5(60)^2=2762.10^3+0.5v^2_2

v2=692.5msv_2=692.5\frac{m}{s}


ii)Rate of flow of fluid:



dmdt=dvdtdvdm\frac{dm}{dt}=\frac{dv}{dt}\frac{dv}{dm}

dvdt=Av1\frac{dv}{dt}=Av_1

dmdt=60.0.10.19=31.6kgs\frac{dm}{dt}=\frac{60.0.1}{0.19}=31.6\frac{kg}{s}


ii) Exit area of nozzle;


dmdt=v2Advdm\frac{dm}{dt}={v_2A'}\frac{dv}{dm}

31.6=692.5A0.531.6=\frac{692.5A'}{0.5}

A=0.0228m2A'=0.0228m^2


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