Question

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 300°C and 200 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm2, determine the velocity and the volume flow rate of the steam at the nozzle exit.

Answer 1

Answer:606 m/s

Explanation:

Given

Steam Inlet temperature $400^{\circ} C$ and pressure 800 KPa

$h_1=3267.7 kJ/kg$

initial Velocity 10 m/s

Steam Outlet Velocity is $300^{\circ} C$ and pressure is 200 KPa

$h_2=3072.1 kJ/kg$

$\nu _2=1.31632$

From steam table

Heat loss 25  KW

inlet area 800 cm^2

applying Steady Flow Energy Equation

$h_1+\frac{1}{2}v_1^2++Q=h_2+\frac{1}{2} v_2^2+W$

$3267.7+\frac{1}{2000}10^2-25=3072.1+\frac{1}{2000}v_2^2$

$3267.7-3072.1+0.05-25=\frac{1}{2}v_2^2$

$v_2=\sqrt{195.65\times 2}=606 m/s$

and volume flow rate is $m=\dot{m}\mu _2=2.082\times 1.31623=2.74 m^3/s$

Answer 2

Answer:

YOUR ANSWER IS 606 m/s

Explanation: