Question

Steam at 4 MPa and 400C enters a nozzle steadily with a velocity of 60 m/s, and it leaves at 2 MPa and 300C. The inlet area of the nozzle is 50 cm2, and heatis being lost at a rate of 75 kJ/s. Determine(a)the mass flow rate of the steam,(b)the exit velocity of the steam, and(c)the exit area of the nozzle.

Answer 1

Answer:

(A) 4.09 kg/s

(B) 589.9 m/s

(C)   0.0008707 m^{3} =  8.71 cm^{2}

Explanation:

inlet pressure of steam (P1) = 4 MPa

inlet temperature of steam (T1) = 400 degree celcius

inlet velocity (V1) = 60 m/s

outlet pressure (P2) = 2 MPa

outlet temperature (T2) = 300 degree celcius

inlet area (A1) = 50 cm^{2} = 0.005 m^{2}

rate of heat loss (Q) = 75 kJ/s

(A) mass flow rate (m) = \frac{A1 x V1}{α1}

where the initial specific volume (α1) for the given temperature and pressure is gotten from tables A-6 = 0.07343 m^3/kg

m = \frac{0.005 x 60}{0.07343}

m = 4.09 kg/s

(B) we can get the outlet velocity using the energy balance equation

  E in = E out

   m(h1 + \frac{(V1)^{2}}{2}) =  m(h2 + \frac{(V2)^{2}}{2})

V2 = $\sqrt{2(h1 - h2) +(V1)^{2} - 2\frac{Q}{m}$

where h1 and h2 are the enthalpies and are gotten from table A-6

V2 = $\sqrt{2 x 1000 x(3214.5 - 3024.2) +(60)^{2} - 2\frac{75 x 1000}{4.09}$

V2 = 589.9 m/s  

(C) the outlet area is gotten from mass flow rate (m) = \frac{A2 x V2}{α}

  A2 = (α2 x m) / V2

where the initial specific volume (α2) for the given temperature and pressure is gotten from tables A-6 = 0.12552 m^3/kg

A2 = (0.12552 x 4.09) / 589.5 = 0.0008707 m^{3} =  8.71 cm^{2}