Question

Two long copper rods of diameter D = 10 mm are soldered together end to end, with solder having a melting point of ${650}^{\circ }C$. The rods are in air at ${25}^{\circ }C$ with a convection coefficient of $10W/m^{2}K$. Heat loss through the rods is (thermal conductivity of cooper is 379 W/mK)

• A. 116.05 W
• B. 11.68 W
• C. 60.45 W
• D. 120.9 W

Answer 1

The correct option is D 120.9 W



K = 379 W/m-K

$h=10W/m^2-K$

Heat transfer rate from the core side of the fin,

$Q=\sqrt{phk{A}_c}\times (T_o-T_{\infty })$

$Q=\sqrt{\pi d\times h\times k\times \frac{\pi }{4}{d}^2}\times (650-25)$

$Q=\sqrt{\pi \times 0.01\times 10\times 379\times \frac{\pi }{4}\times {0.01}^2}\times 625$

Q = 60.437 W

Minimum power required = Total heat transfer rates

$=2\times 60.437=120.87W$