Answer:
[tex]Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J[/tex]
Explanation:
To solve this problem we use the expression for the temperature film
[tex]T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5[/tex]
Then, we have to compute the Reynolds number
[tex]Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}[/tex]
Re<5*10^{5}, hence, this case if about a laminar flow.
Then, we compute the Nusselt number
[tex]Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77[/tex]
but we also now that
[tex]Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\[/tex]
but the average heat transfer coefficient is h=2hx
h=2(8.48)=16.97W/m^{2}K
Finally we have that the heat transfer is
[tex]Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J[/tex]
In this solution we took values for water properties of
v=16.96*10^{-6}m^{2}s
Pr=0.699
k=26.56*10^{-3}W/mK
A=1*0.5m^{2}
I hope this is useful for you
regards
