Answer:
The rate of heat transfer; Q = 160512[W] ; the area = 14.7 [m^2]
Explanation:
The first law of thermodynamics requires that the rate of heat transfer from the hot fluid is equal to the heat transfer to the cold; that is to say:
[tex]Q=m_{cold}*Cp_{cold}*(Tc_{out} - Tc_{int} ) \\ \\and\\Q=m_{hot}*Cp_{hot}*(Th_{int} - Th_{out} ) \\\\where:\\\\m_{cold}, m_{hot} = mass flow\\Cp_{cold}, Cp_{hot} = specific heat\\\\Tc_{out}, Th_{out} = outlet temperatures\\Tc_{int}, Th_{int} = inlet temperatures[/tex]
Now taking the water as the cold fluid we can find the rate of heat transfer:
[tex]Q=0.8*4180*(70-22)\\Q=160512[watt][/tex]
By means of the logarithmic mean temperature difference we can find the heat transfer area using the following equation:
[tex]Q=U*A_{s}*DT_{ml} \\\\DT_{ml} =\frac{(DT_{1}-DT_{2} )}{ln\frac{DT_{1} }{DT_{2} } } \\[/tex]
[tex]DT1=Th_{inlet} - Tc_{out} \\DT2=Th_{out} - Tc_{inlet} \\\\DT1= 110-70 = 40 [C]\\DT2=60-22 =38[C][/tex]
Now replacing:
[tex]DT_{ml} = \frac{(40-38)}{ln\frac{40}{38} } \\DT_{ml} = 38.9[C][/tex]
[tex]A_{s}=\frac{Q}{U*DT_{lm} } \\A_{s}=\frac{160512}{280*38 } \\\\A_{s}=14,7[m^{2}][/tex]
