The condenser of a power plant has a two-phase liquid plus vapor mixture of wate

Question

The condenser of a power plant has a two-phase liquid plus vapor mixture of water enter at 10 kPa and a quality of 95% (state 1) and exit at 45 degree C (state 2). Heat is transferred from the high temperature stream to a cooler stream of water. The two streams do not mix. The cooling water stream enters at 20 degree C at 110 kPa (state 3) and exists at 35 degree C (state 4). Determine the ratio of the mass flow rate of cooling water stream over the mass flow rate of the high temperature stream (also known as the condensing stream). Determine the rate of heat leaving the condensing stream per mass flow rate of the condensing stream.

Explanation / Answer

a) From steam table for saturated liquid, enthalpy (hfg) = 2392.8 KJ/kg corresponding to pressure 10 kPa, specific heat of pressurized cold water stream is 4.221 KJ/kgK.

For condenser, heat balance equation is

mh*(hfg) = mc*Cc*(Tco-Tci)

Where, mh is mass flow rate of hot water stream, mc is mass flow rate of cooling water stream, Tco is tempearture of state 4, Tci is temperature of state 3.

so the ratio of mass flow rate of cooling water stream to that of hot water stream can be found out by rearranging above equation and its value comes to be 37.791.

b) Enthalpy at state 2 = hf=h2= 191.81 KJ/kg, from steam table , enthalpy at state 3 = h3.

h3 = hf + x*(hfg) = 2464.97 KJ/kg

where x is quality of hot stream.

So rate of heat leaving the condensing stream per mass flow rate of condensing stream = h3-h2 = 2273.16 KJ/kg.

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