Answer to Question #244126 in Molecular Physics | Thermodynamics for kimmmmm

Question

A turbine operates under steady flow conditions receiving steam at the following state: pressure 1.2 MPa, temperature 188 C, enthalpy 2785 kJ/kg, velocity 33.3 m/s and elevation 3m. The steam leaves the turbine at the following state: pressure 20 kPa, enthalpy 2512 KJ/kg, velocity 100 m/s and elevation 0 m. . Heat loss to the surroundings is at a rate of 0.29 kJ/s. If the rate of steam flow through the turbine is 0.42 kg/s, what is the power output of the turbine in kW?

Answer 1

Step 1

given data

At the inlet of the turbine

P _i = 1200 kPa

T _i = 188 ^o C

h _i = 2785 kJ/kg

v _i = 33.3 m/s

z _i = 3 m

Q _r = 0.29 kW

m ˙ = 0.42 kg/s

And at the outlet of the turbine

P _o = 20 kPa

h _o = 2512 kJ/kg

v _o = 100 m/s

z _o = 0 m

To determine

Power output of the turbine in kW

Step 2

From the steady-state equation for turbine

$m[h_i + \frac{v_1^2}{2000} + \frac{gz_i}{1000} ] + Q_r = m[h_2 + \frac{v^2_0}{2000} + gz_0] + W \\ 0.42[2785 + \frac{33.3^2}{2000} + \frac{3 \times 9.81}{1000}] = 0.42[2512 + \frac{100^2}{2000} + \frac{9.81 \times 0}{1000}] + W \\ 169.655 =1057.14 +W \\ W = 112.5 \; kW$

Thus, output power of turbine is 112.5 kW