Question

Circular copper rods of diameter D= 1.75 mm and length L= 25 mm are used to enhance heat transfer from a surface that is maintained at Ts,1= 100°C. One end of the rod is attached to this surface (at x= 0), while the other end (x= 25 mm) is joined to a second surface, which is maintained at Ts,2= 0°C. Air flowing between the surfaces (and over the rods) is also at a temperature of [infinity] T[infinity]= 0°C, and a convection coefficient of h= 125 W/m2·K is maintained. (1) What is the rate of heat transfer by convection from a single copper rod to the air, in W? (2) What is the total rate of heat transfer from a 1 m × 1 m section of the surface at 100°C, if a bundle of the rods is installed on 4-mm centers, in W?

Answer 1

Answer:

a. Rate of heat transfer = 1.273 W

b. Total rate of heat transfer = 285,933.8 W

Explanation:

Given

$\\\\\\\\D = 1.75mm\\L=25mm\\T_{s,1}=100^0C\\T_{s,2}=0^0C\\h=125W/m^2KT_{\infty}=0^0C$

a. Applying the conservation of energy

$q_{conv}=q_{cond,i}-q_{cond, o}$

The general heat conduction equation for the temperature distribution is given as

$T_{(x)}-T_{\infty}= C_1e^{mx}+C_2e^{-mx}\\\theta_{(x)} = C_1e^{mx}+C_2e^{-mx}$

Using boundary conditions, we find the constant values

$x=0:T=T_{s,1}$

Substituting the first boundary condition in the temperature distribution, we get

$T_{s,1}-T_{\infty}= C_1e^{mx*0}+C_2e^{-mx*0}\\T_{s,1}-T_{\infty}= C_1+C_2$

$100-0=C_1+C_2$

Using the second boundary conditions

$x=0.025m:T_x=T_{s,2}$

Substituting the second boundary condition in the temperature distribution, we get

$T_{x}-T_{\infty}= C_1e^{mx}+C_2e^{-mx}\\T_{s,2}-T_{\infty}= C_1e^{mx*0.025}+C_2e^{-mx*0.025}\\0-0=C_1e^{mx*0.025}+C_2e^{-mx*0.025}\\C_1e^{mx*0.025}=-C_2e^{-mx*0.025}\\C_1e^{mx*0.025}=\frac{-C}{e^{mx*0.025}}\\C_2=-C_1e^{2*mx*0.025}\\C_2=-C_1e^{0.05m}$

Since

$100=C_1+C_2\\100=C_1+C_1e^{0.05m}\\C_1=\frac{100}{1-e^{0.05m}}$

We therefore substitute to find $C_2$

$C_2=C_1e^{0.05m}\\=-\frac{100}{1-e^{0.05m}} e^{0.05m}$

Substituting into the heat conduction equation, we obtain

$\theta_{(x)} = C_1e^{mx}+C_2e^{-mx}\\\theta_{(x)} = \frac{100}{(1-e^{0.05m})} e^{mx}-\frac{100e^{0.05m}}{(1-e^{0.05m})} e^{-mx}\\=\frac{100}{(1-e^{0.05m})} [e^{mx}-e^{0.05m-mx}]$

We proceed to differentiate the temperature distribution with respect to x

$\frac{d}{dx}(\theta_{(x)}) = \frac{d}{dx}(\frac{100}{(1-e^{0.05m})}[e^{mx}-e^{0.05m-mx}] )\\\frac{d\theta_{(x)}}{dx} = \frac{100}{(1-e^{0.05m})}(me^{mx}-(-m)e^{0.05m-mx})\\\frac{d\theta_{(x)}}{dx} = \frac{100}{(1-e^{0.05m})}m(e^{mx}+e^{0.05m-mx})$

The thermal conductivity of copper from the table A.1, "Thermophysical properties of selected metallic solids" at an average temperature of 323K is 400W/m.K

Solving for m

$m=\sqrt{\frac{hP}{kA_c} }=\sqrt{\frac{h(\pi D)}{k(\frac{\pi D^2}{4} )} } =\sqrt{\frac{4h}{kD} } \\=\sqrt{\frac{4*125}{400*0.00175} }=26.73m^{-1}$

Using Fourier's law, we solve for the heat transfer rate by conduction

$q_{cond}=-kA_c\frac{d \theta}{dx} =-k* \frac{\pi D^2}{4}*\frac{100}{(1-e^{0.05m})}m(e^{mx}+e^{0.05m-mx})\\=\frac{-100k(\frac{\pi D^2}{4} )m}{(1-e^{0.05m})}[e^{mx}+e^{0.05m-mx}]\\=\frac{-(25 \pi)kmD^2}{(1-e^{0.05m})}[e^{mx}+e^{0.05m-mx}]$

Calculating the conductive heat transfer rate at x = 0, we get

$q_{cond,i}=\frac{-(25 \pi)kmD^2}{(1-e^{0.05m})}[e^{mx}+e^{0.05m-mx}]\\=\frac{-(25 \pi)*(400)*(26.73)*(0.00175)^2}{(1-e^{0.05*26.73})}[e^{26.73*0}+e^{(0.05*26.73)-(26.73*0)}]\\=\frac{-2.572}{-2.806}[1+3.806}]=4.405W$

Calculating the conductive heat transfer rate at x = 0.025 m, we get

$q_{cond,o}=\frac{-(25 \pi)kmD^2}{(1-e^{0.05m})}[e^{mx}+e^{0.05m-mx}]\\=\frac{-(25 \pi)*(400)*(26.73)*(0.00175)^2}{(1-e^{0.05*26.73})}[e^{(26.73*0.025)}+e^{(0.05*26.73)-(26.73*0.025)}]\\=\frac{-2.252}{-2.806} (3.903)=3.13W$

$q_{conv}=q_{cond,i}-q_{cond,o}=4.405-3.132=1.273W$

b. The rods are placed 4 mm apart on a 1 m by 1 m section

To find the number of rods on each side with 1 m being equal to 1000 mm,

$n = \frac{1000}{4} = 250$

We therefore calculate the total number of rods, N in 1 m by 1 m section

$N = 250*250=62,500rods$

To calculate the bare surface area, A

$A=1m^2-NA_c=1m^2-N(\frac{\pi D^2}{4} )\\=1-(62500*\frac{\pi*0.00175^2}{4} )=0.8497m^2$

Total rate of heat transfer can be calculated thus

$q=N*q_{cond,i}+hA(T_{s,1}-T_{\infty})\\=(62500*4.405)+(125*0.8497*(100-0))\\=275,312.5+10,621.3=285,933.8W$