Question

Assume that you need to transfer 4628320 J of energy in ten minutes between the two reservoirs. To enhance the rate of energy transfer, a steel rod (k = 43 W/m-K) of the same length is added between the two reservoirs. What should the cross-sectional area of the steel rod be in order to achieve the proper rate of energy transfer?

Answer 1

Answer:

$A = \frac{179.4}{\frac{dT}{dx}}$

Explanation:

As we know that thermal conduction of heat is given by the formula

$\frac{dQ}{dt} = \frac{kA}{L}(T_2 - T_1)$

now we have

$\frac{dQ}{dt} = \frac{4628320}{10(60)}$

$\frac{dQ}{dt} = 7713.87 Watt$

now we also know that

k = 43 W/m-k

now we have

$7713.87 = \frac{(43)A}{dT}{dx}$

so we have

$A = \frac{179.4}{\frac{dT}{dx}}$

here we know that dT/dx is temperature gradient of the rod which is ratio of temperature difference of two ends of the rod and length of the rod.