Water steam enters a conical nozzle at 2 MPa, 300 o C, with the inlet velocity b
ID: 2994917 • Letter: W
Question
Water steam enters a conical nozzle at 2 MPa, 300 oC, with the inlet velocity being 100 m/s. The nozzle operates at a steady state. The mass flow rate is 100 kg/s. The exit pressure and temperature are 0.6 MPa bar and 160 oC, respectively. The effect of the kinetic energy and heat transfer can be neglected. Please determine:- The inlet and outlet radii of the nozzle
- The exit flow velocity
- The inlet and outlet radii of the nozzle
- The exit flow velocity
- The inlet and outlet radii of the nozzle
- The exit flow velocity
- The inlet and outlet radii of the nozzle
- The exit flow velocity
Explanation / Answer
From water properties,
At P1 = 2 MPa and T1 = 300 deg C, we get h1 = 3020 kJ/kg, v1 = 0.126 m^3/kg
At P2 = 0.6 MPa and T2 = 160 deg C we get h2 = 2760 kJ/kg, v2 = 0.317 m^3/kg
Now m = A1*V1 / v1
100 = A1*100 / 0.126
A1 = 0.126 m^2
3.14 * r1 ^2 = 0.126
Inlet radius r1 = 0.2 m
Energy conservation:
h1 + V1 ^2 /2 = h2 + V2 ^2 /2
3020*10^3 + 100^2 /2 = 2760*10^3 + V2 ^2 /2
Solving, V2 = 728 m/s
m = A2*V2 / v2
100 = A2*728 / 0.317
A2 = 0.0435 m^2
3.14*r2 ^2 = 0.0435
r2 = 0.11776 m
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