Overall Heat Transfer Coefficient Calculation

To calculate the overall heat transfer coefficient based on the outer area, we need to consider the individual resistances to heat transfer in the heat exchanger.

1. Inner Convective Resistance
The inner convective resistance is caused by the fluid flowing inside the tube. The heat transfer coefficient based on the inner area can be calculated using the formula:

h_inner = (Q / A_inner) / ΔT

Where:
- h_inner is the heat transfer coefficient based on the inner area (W/m2.C)
- Q is the heat transfer rate (W)
- A_inner is the inner surface area of the tube (m2)
- ΔT is the temperature difference between the fluid and the tube wall (C)

2. Inner Conductive Resistance
The inner conductive resistance is caused by the conduction of heat through the tube wall. The thermal conductivity of the tube material can be used to calculate this resistance.

3. Outer Convective Resistance
The outer convective resistance is caused by the fluid flowing outside the tube. The heat transfer coefficient based on the outer area can be calculated using the formula:

h_outer = (Q / A_outer) / ΔT

Where:
- h_outer is the heat transfer coefficient based on the outer area (W/m2.C)
- A_outer is the outer surface area of the tube (m2)

4. Outer Conductive Resistance
The outer conductive resistance is caused by the conduction of heat through the tube wall.

Overall Heat Transfer Coefficient
The overall heat transfer coefficient can be calculated by considering both the convective and conductive resistances. Since the question provides the overall heat transfer coefficient based on the inner area, we can assume that the conductive resistances are negligible compared to the convective resistances.

Since the inner and outer surface areas of the tube are different, the heat transfer rates based on the inner and outer areas will also be different. However, the total heat transfer rate remains the same.

Q_inner = Q_outer

Using the equations for h_inner and h_outer, we can write:

(Q / A_inner) / ΔT = (Q / A_outer) / ΔT

Rearranging the equation, we get:

h_inner / A_inner = h_outer / A_outer

Since the total heat transfer rate is the same, the product of the heat transfer coefficient and the surface area is constant.

h_inner * A_inner = h_outer * A_outer

Since the tube is cylindrical, the inner and outer surface areas can be related as follows:

A_outer = A_inner + π * (D_outer - D_inner) * L

Where:
- D_outer and D_inner are the outer and inner diameters of the tube, respectively
- L is the length of the tube

Replacing the value of A_outer, we can solve for h_outer:

h_inner * A_inner = h_outer * (A_inner + π * (D_outer - D_inner) * L)

Simplifying the equation, we get:

h_outer = h_inner * (A_inner / (A_inner + π * (D_outer - D_inner) * L))

Substituting the given values, we have:

h_outer = 360 * (π * (0.025^2) / (π * (0.025^