HEAT EXCHANGERS 6.10) A condenser condenses steam (saturated vapour) into conden

Question

HEAT EXCHANGERS 6.10) A condenser condenses steam (saturated vapour) into condensate (saturated liquid) by passing it over a tube bank through which cold liquid water flows. The cooling water (specific heat capac CP-4200J/kgK) enters the condenser at 6°C and exits at 24°C (after being heated by the condensing steam). The steam and condensate are at saturation conditions of 0.2bar - What mass flow rate me of cooling water is required to condense m, -0.13kg's o - What is the rate of heat transfer from the steam to the cooling water (neglecting heat le the condenser to ambient)? ity age from (Ans. 4.06 kg/s; 307kW) Page 1 of 2

Explanation / Answer

The question talks about a condenser that condenses steam into condensate. Hence,it becomes evident to us that the only effect that the condensor does is change the state of steam, and not the temperature of the steam. So the heat transferred from the condensor to the cooling water is only responsible for the change of state of steam (Latent heat) and not change in temperature (sensible heat).

Total heat lost by the condensor will amount to the latent heat of vaporization of the steam at that pressure.

ie, Total heat lost (rate) = Qlost = Mass (flow rate) of hot substance ( in this case steam) * Latent heat of vaporization (of steam at 0.2bar)

From the steam tables we get the latent heat of vaporization of steam at 0.2bar will be 2358.4 kJ/kg.

So, heat lost (rate) Qlost = 0.13 * 2358.4 = 306.592 kJ/s or kW

Also, for heat exchangers without heat leakages or losses, heat supplied by the condenser (hot liquid) will be equal to the heat gained by the cool liquid.

Qlost = Qgained

So, Qgained = 306.592kW

Also, we know that the heat gained by the cooling water is responsible for the change in temperature of the cooling water (sensible heating).

So, Total heat gained (rate) = Qgained = Mass (flowrate) of cold substance * specific heat capacity * temperature difference. Substituting the values of Qgained, specific heat capacity and temperature difference,

306 .592 = mc * 4.2 * (24-6) [Cp = 4200J/kgK or 4.2kJ/kgK]

Solving for mc, we get mc= 4.055 kg/s.

So, mass flow rate of cooling water is 4.06kg/s and rate of heat transfer from steam to the cooling water = Qlost = 307kW.

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