Respuesta :
The local convection heat transfer coefficient at 1 m from the leading edge is [tex]0.44 \frac{W}{m^{2} \times K}[/tex] , the average convection heat transfer coefficient over the entire plate is [tex]0.293 \frac{W}{m^{2} \times K}[/tex]and the total heat flux transfer to the plate is [tex]61.6 KJ[/tex].
Explanation:
It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.
Hot carbon dioxide exhaust gas
physical properties
[tex]r= 1.05 \frac{kg}{m^{3}}[/tex]
[tex]c_p = 1.02 \frac{kJ}{Kg \times K}[/tex]
[tex]m= 231 \times 10^{7} \frac{N \times s }{m^2}[/tex]
υ = [tex]21.8 \times 10^{6} \frac{m^2}{s}[/tex]
[tex]k = 32.5 \times 10^{3} \frac{W}{m \times K}[/tex]
[tex]\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}[/tex]
[tex]Pr = 0.725[/tex]
Apart from these other data arr given below,
[tex]v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C[/tex]
To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,
[tex]Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })[/tex]
where h= Average heat transfer coefficient
L= Length of a plate
k= Thermal Conductivity of carbon dioxide
Re = Reynold's Number
Pr = Prandtle Number
(a) Convection heat transfer coefficient at 1 m from the leading edge
is referred as local convection heat transfer coefficient.
To find convection heat transfer coefficient at 1 m from leading edge,
[tex]Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })[/tex]
Here, first we have to find Re and Pr,
[tex]Re = \frac{r \times v \times L}{m}[/tex]
[tex]Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}[/tex]
[tex]Re = 20.63 \times 10^{-10}[/tex]
Pr number is take from physical property data and Pr is 0.725.
Putting value of Re and Pr in main equation,
we get
[tex]Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })[/tex]
[tex]h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })[/tex]
[tex]h_local = 0.44 \frac{W}{m^{2} \times K}[/tex]
(b) To find average convection heat transfer coefficient,
it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,
Therefore,
[tex]Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })[/tex]
Finally,
[tex]h = \frac{0.44}{1.5}[/tex]
[tex]h = 0.293 \frac{W}{m^{2} \times K}[/tex]
(C) Total heat flux transfer to the plate is found out by,
[tex]Q = h \times (T_g - T_s)[/tex]
[tex]Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ[/tex]
