A hot-water stream at 70°C enters an adiabatic mixing chamber with a mass flow rate of 3.6 kg/s, where it is mixed with a stream of cold water at 20°C. If the mixture leaves the chamber at 42°C, determine
(a) the mass flow rate of the cold water and
(b) the rate of entropy generation during this adiabatic mixing process. Assume all the streams are at a pressure of 200 kPa

Where min is the mass flow that is entering the system, cin is the specific heat of the substance that is entering, Tin is the temperature of the substance that is entering, and the out is from the substances that are leaving the system. Because there is only water in the system cin = cout. The energy balance will be then:

mh*Th + mc*Tc = (mh + mc)*Tmix

Where h is from the hot water, c is from the cold water, and mix for the mixture that leaves the chamber. Th = 70ºC + 273 = 343 K, Tc = 20ºC + 273 = 293 K, Tmix = 42ºC + 273 = 315 K

3.6*343 + mc*293 = (3.6 + mc)*315

293mc - 315mc = 3.6*315 - 3.6*343

- mc*(315 - 293) = -3.6*(343 - 315)

mc = 3.6*(343 - 315)/(315 - 293)

mc = 3.6*28/22

mc = 4.58 kg/s

b) The entropy balance is:

Sin + Sout - Sgen = ΔSsystem

Where Sin is the entropy of the entering substances, Sout the entropy that the leaving substances, and Sgen the entropy that is generated in the process.

The entropy variantion for an adiabatic process is 0, so ΔSsystem = 0

Sgen = Sin + Sout

The sum of entropy is:

m*cp*ln(Tfinal/Tinitial), where cp is the specific heat (4.184 kJ/kg.K)

Sgen = mc*cp*ln(Tmix/Tc) + mh*cp*ln(Tmix/th)

Sgen = 4.58*4.184*ln(315/293) + 3.6*4.184*ln(315/343)

Sgen = 1.3874 + (-1.2827)

Sgen = 0.105 kW/K