Question

Air enters a turbine operating at steady state at 440 K, 20 bar, with a mass flow rate of 6 kg/s, and exits at 290 K, 5 bar. The velocities at the inlet and exit are 18 m/s and 30 m/s, respectively. The air is modeled as an ideal gas, and potential energy effects can be neglected. If the power developed is 815 kW, determine the rate of heat transfer, in kW, for a control volume enclosing the turbine.

Answer 1

Answer:

The rate of heat transfer is - 935.392 kW

Explanation:

Given;

inlet pressure, P₁ = 20 bar

inlet temperature, T₁ = 440 k

outlet pressure, P₂ = 5 bar

outlet temperature, T₂ = 290 k

inlet velocity, v₁ = 18 m/s

outlet velocity, v₂ = 30 m/s

mass flow rate, Q = 6kg/s

developed power, W = 815 kW

From steam table;

at P₁ and T₁, interpolating between T = 400 and T = 450, h₁ = 3335.52 kJ/kg

at P₂ and T₂, interpolating between T= 250 and T= 300, h₂ = 3043.5 kJ/kg

Applying steady state energy balance equation;

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

substitute values above into this equation and evaluate the rate of heat transfer

$\frac{Q}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2}) +\frac{W}{m}\\\\\frac{Q}{m} = 3043.5 - 3335.52 +\frac{1}{2000} (30{^2} -18{^2}) + \frac{815}{m} \\\\\frac{Q}{m} = -291.732 + \frac{815}{m}\\\\\frac{Q}{m} = \frac{815}{m} -291.732\\\\Q = 815 - m(291.732)\\Q = 815 - 6*291.732\\\\Q = 815-1750.392 = -935.392 \ kW$

Therefore, the rate of heat transfer, in kW, for a control volume enclosing the turbine is - 935.392 kW