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

`A = (179.4)/((dT)/(dx))`

Step-by-step explanation:

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

`(dQ)/(dt) = (kA)/(L)(T_2 - T_1)`

now we have

`(dQ)/(dt) = (4628320)/(10(60))`

`(dQ)/(dt) = 7713.87 Watt`

now we also know that

k = 43 W/m-k

now we have

`7713.87 = ((43)A)/(dT){dx}`

so we have

`A = (179.4)/((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.