Question

Two long copper rods of diameter D 10 mm are soldered together end to end, with solder having a melting point of 650 C. The rods are in air at 25 C with a convection coefficient of 10 W/m2 K. What is the minimum power input needed to effect the soldering?

Answer 1

Answer:

Qf 194.2 W

Explanation:

Given data:

Diameter of copper rod D is 0.01 ,

Temperature at junction is Tb = 650 + 273 = 923 K

Temperature of air is 15+273 = 288 K

$Tmean = \frac{923 + 288}{2} = 605.5 K$

For temperature 605.5 degree celcius thermal conductivity is K = 379 W/m K

Heat transfer is calculated as

$qf = \sqrt{hPkAc(Tb -T\infty)}$

$qf = \sqrt{h\pi D k\frac{\pi}{4} D^2 (Tb -T\infty)}$

$qf = \sqrt{25 \pi \times 0.01\times 379 \times \frac{\pi}{4} \times 0.01^2(920 -288)}$

qf = 97.1 W

Hence, the rate of heat is

$Qf = 2qf = 2\times 97.1 = 194.2 W$

Answer 2

Answer:

125pi  W/m

Explanation:

This is the formula for heat transfer by convection

Q=hA(T2-T1)

Q=Power

h=convection coefficient

A=area

T2= temperature of melting point

T1=temperature of air

Area of tow rods

A=2*pi*D=2*pi*0.01=pi/50

Q=10*(´pi/50)(650-25)

Q=125pi  W/m