Answer:
[tex]T_{out}=27.0000077[/tex] ºC
Explanation:
First, let's write the energy balance over the duct:
[tex]H_{out}=H_{in}+Q[/tex]
It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:
[tex]m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q[/tex]
The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.
[tex]m*Cp(T_{out}-T_{in})=Q[/tex]
So, let's isolate [tex]T_{out}[/tex]:
[tex]T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}[/tex]
The Cp of the air at 27ºC is 1007[tex]\frac{J}{kgK}[/tex] (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are [tex]T_{out}[/tex] and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:
[tex]P=2*H+2*A=2*0.004m+2*0.016m=0.04m[/tex]
Surface area:
[tex]A=P*L=0.04m*1m=0.04m^2[/tex]
Then, the heat Q is:
[tex]600\frac{W}{m^2} *0.04m^2=24W[/tex]
Finally, find the exit temperature:
[tex]T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077[/tex]
[tex]T_{out}[/tex]=27.0000077 ºC
The temperature change so little because:
