Question

A turbine operating under steady-flow conditions receives steam at the following state: pressure, 13.8 bar; specific volume 0.143 m^3/kg specific internal energy 2590 kj/kg, velocity 30 m/s. The state of the steam leaving the turbine is as follows: pressure 0.35 bar, specific volume 4.37 m^3/kg, specific internal energy 2360 kj/kg, velocity 90 m/s. Heat is rejected to the surroundings at the rate of 0.25 kW and the rate of steam flow through the turbine is 0.38 kg/s. Calculate the power developed by the turbine.

Answer 1

$$m\left(u_1+p_1v_1+\frac{C_1^2}{2}\right)-Q=\left(u_2+p_2v_2+\frac{C_2^2}{2}\right)+W$$

$$W=m\left((u_1-u_2)+\left(p_1v_1+\frac{C_1^2}{2}-p_2v_2-\frac{C_2^2}{2}\right)\right)-Q$$

$$W=0.38(2590-2360)10^3-0.25\cdot 10^3+\\+0.38\left((13.8\cdot10^5)(0.143)+\frac{30^2}{2}-(0.35\cdot10^5)(4.37)-\frac{90^2}{2}\right)$$

$$W=102.65\ kJ$$