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 = [tex]\sqrt{2(h1 - h2) +(V1)^{2} - 2\frac{Q}{m}[/tex]
where h1 and h2 are the enthalpies and are gotten from table A-6
V2 = [tex]\sqrt{2 x 1000 x(3214.5 - 3024.2) +(60)^{2} - 2\frac{75 x 1000}{4.09}[/tex]
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}
