copper pipe of diameter 3.5 cm and wall thickness 4 mm is used in an antibiotic factory toconvey hot water a distance of 60 m. The inlet temperature of the water is 90°C. Due to heatlosses to the atmosphere, the outlet water temperature is 84°C. The ambient temperature is25°C. The inside of the pipe is covered with a layer of fouling with heat transfer coefficient7500 W m-2 K-1 . The rate of heat loss is estimated as 503 kW.(a) What proportion of the total resistance to heat transfer is provided by the fouling layer?

To solve this problem, we first need to understand the concept of thermal resistance. Thermal resistance is a measure of the resistance to heat flow, and in this case, it is made up of the resistance of the fouling layer and the resistance of the copper pipe.

The thermal resistance (R) can be calculated using the formula:

R = 1/(h*A)

where h is the heat transfer coefficient and A is the surface area.

1. Calculate the surface area (A) of the pipe:

The surface area of a cylinder can be calculated using the formula:

A = 2πrL

where r is the radius and L is the length. The radius of the pipe is half of the diameter, so r = 3.5 cm / 2 = 1.75 cm = 0.0175 m. The length of the pipe is 60 m.

So, A = 2π * 0.0175 m * 60 m = 6.63 m^2.

2. Calculate the thermal resistance of the fouling layer (Rf):

The heat transfer coefficient of the fouling layer is given as 7500 W/m^2K.

So, Rf = 1/(7500 W/m^2K * 6.63 m^2) = 0.0000201 K/W.

3. Calculate the total thermal resistance (Rt):

The rate of heat loss is given as 503 kW, and the temperature difference between the inlet and outlet is 90°C - 84°C = 6°C = 6 K.

So, Rt = ΔT/Q = 6 K / 503 kW = 0.0000119 K/W.

4. Calculate the proportion of the total resistance provided by the fouling layer:

The proportion is given by Rf / Rt.

So, the proportion = 0.0000201 K/W / 0.0000119 K/W = 1.69.

This means that the fouling layer provides approximately 169% of the total resistance to heat transfer. This value is greater than 100% because the fouling layer is actually increasing the total resistance to heat transfer, making it more difficult for heat to flow through the pipe.