Question

Steam flows steadily into a turbine with a mass flow rate of 26 Kg/s and a negligible velocity at 6 MPa and 600°C. The steam leaves the turbine at 0.5 MPa and 200°C with a velocity of 180 m/s. The rate of work done by the steam in the turbine is measured to be 20 MW. If the elevation change between the turbine inlet and exit is negligible, Determine: (a) The rate of heat transfer associated with this process.

Answer 1

`Q_(out)=456.8kW`

Step-by-step explanation:

Hello,

To solve this problem, an energy balance is proposed as follows:

`mh_(in)=m[h_(out)+(V_(out)^2)/(2)]+W_(out)+Q_(out)`

The inlet conditions correspond to an oversaturated steam and the outlet conditions to an oversaturated steam as well, so the inlet and outlet enthalpies are extracted from the tables of oversaturated steam and have the following values:

`h_(in)=3658.8kJ/kg;h_(out)=2855.8kJ/kg`

Now, solving for the outgoing heat, we've got:

`Q_(out)=m[h_(in)-h_(out)-(V_(out)^2)/(2)]-W_(out) \\Q_(out)=26(kg)/(s)(3658.8(kJ)/(kg)-2855.8(kJ)/(kg)-((180(m)/(s))^2)/(2) ) -20MW*(1000kW)/(1MW) \\Q_(out)=26(kg)/(s)(803(kJ)/(kg)-16200(J)/(kg)*(1kJ)/(1000J) ) -20000kW\\Q_(out)=456.8kW`

This heat is positive due to the balance, but if we have used the classic notation it'll be negative indicating that is a lost heat.

Best regards.