Q1: The rate of heat transfer through a pipe wall is given by,  . For cylinder of very thin wall, q can be approximated by

ANS:D -

The rate of heat transfer through a pipe wall can be given by Fourier's law of heat conduction, which states: �=2���(�1−�2)ln⁡�2�1q=lnr1​r2​​2πkL(T1​−T2​)​ Where:

• q is the rate of heat transfer (W or J/s),
• k is the thermal conductivity of the material (W/(m·K)),
• L is the length of the pipe (m),
• 1T1​ and 2T2​ are the temperatures at the inner and outer surfaces of the pipe (K or °C),
• 1r1​ and 2r2​ are the inner and outer radii of the pipe (m).

For a cylinder with a very thin wall, the difference in temperature (1−2T1​−T2​) across the wall may be assumed to be relatively small compared to the temperatures themselves. In this case, the natural logarithm term in the denominator (ln⁡21lnr1​r2​​) can be approximated by a simpler expression. When the wall is very thin, 2r2​ is very close to 1r1​, and the ratio 21r1​r2​​ approaches 1. Therefore, ln⁡21lnr1​r2​​ approaches ln⁡1ln1, which equals 0. So, for a cylinder with a very thin wall, the expression for q can be approximated as: 2(1−2)ln⁡1=2(1−2)0q=ln12πkL(T1​−T2​)​=02πkL(T1​−T2​)​ Since division by zero is undefined, this approximation suggests that the rate of heat transfer (q) through a cylinder with a very thin wall becomes very large or infinite under this condition. However, it's essential to note that this is a mathematical approximation and may not hold true under all physical conditions.