Liquid water(H2O) at 200 KPa and 20oC is heated in a chmaber by mixing it with a

Question

Liquid water(H2O) at 200 KPa and 20oC is heated in a chmaber by mixing it with a steam (H20) at 200 KPa and 150oC. The liquid water enters the mixing chamber at a mass flow rate of 2.5 kg/s and the steam enters the mixing chamber at a mass of 0.2 kg/s. The mixture leaves the chmaber at 200 KPa and 60oC. The mixing chamber operates steadily and there is no work interaction involved in the mixing process. Assume that the change in kinetic energy is negligible. Determine the magnitude and direction of heat transfer rate during the process.

Explanation / Answer

assume T is final temp

heat gain by liquid water =Q=mcpdt=2.5 Cp(T-20) =2.5*4.184*10^3*(T-20)

heat loss by steam =Q=mcpdt=0.2 Cp(150-T)=0.2* 1.996*10^3*(150-T)

heat gain by liquid water= heat loss by steam

2.5*4.184*10^3*(T-20)= 0.2*1.996*10^3*(150-T)

10.46(T-20) = 0.3992(150-T)

10.46T - 209.2 = 59.88 - 0.3992 T

T = 24.78 C

Q= 2.5*4.184*10^3*(24.78-20)

Q = 49.998 KJ

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