Question

Steam is accelerated by a nozzle to a velocity of 200 m/s at a rate of 4 kg/s. If the steam at the nozzle exit is 300 °C and 2 MPa, then what is the cross-sectional area in cm2 of the exit opening of the nozzle? 2am icNiel Crews, 2013 cm2

Answer 1

In this problem, first of all Calculate the volume flow rate of steam at exit of nozzle with the help of pressure and temperature after that Calculate the exit area of nozzle with the help of mass flow rate formula as below:-

Given data :- At exit, in velocity, V = 200 mis ( ) may flow rate, in=4 legis (") Pressure P2 = 2 MPa = 2000 kPa. (iv) Temp., To - 300*C - 573 K for steam, R= 0.4615 kJ/kg-k belation : - We know that, S P V2 = RT2 .. volume flow rate (m?lng), V = ♡ = 0.1322 miling. f in=f. Au or in= Az Vz V2 RT2 0.4615x 523 2000 = 0.1322 m .: Exit cross section, Az = An 4x 0.1322 m2 800 A2 = 2.644+ x 103 m2 OR. I A2 = 26.444 cm2