Answer to Question #234537 in Molecular Physics | Thermodynamics for Unknown346307

Question

The power developed by a turbine in a certain steam plant is 1200 kW. The

heat supplied to the steam in the boiler is 3360 kJ/kg, the heat rejected by the system to cooling

water in the condenser is 2520 kJ/kg and the feed pump work required to pump the condensate

back into the boiler is 6 kW.

Calculate the steam flow round the cycle in kg/s

Answer 1

The power developed by the turbine = 1200 kW

The heat supplied to the steam in the boiler = 3360 kJ/kg

The heat rejected by the system to cooling water = 2520 kJ/kg

Feed pump work = 6 kW

Figure shows the cycle. A boundary is shown which encompasses the entire plant. Strictly,

this boundary should be thought of as encompassing the working fluid only.

$\oint dQ = 3360- 2520 = 840 \;kJ/kg$

Let the system flow be in kg/s.

$\oint dQ = 840 \;m \;kJ/s \\ \oint dW = 1200-6 = 1194 \;kJ/s \\ \oint dQ = \oint dW \\ 840 \;m=1194 \\ m=\frac{1194}{840}=1.421 \;kg/s$

Steam flow round the cycle = 1.421 kg/s.