Question

Consider a double-pipe heat exchanger with a tube diameter of 10 cm and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be 0.025 K/W when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors 0.00045 (K*m2)/W and 0.00015 (K*m2)/W respectively. The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is:

A. 2.3%
B. 6.8%
C. 7.1%
D. 7.6%
E. 8.5%

Answer 1

The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is approximately 2.4%.

None of the given options is correct

Step-by-step explanation:

To calculate the percentage decrease in the rate of heat transfer due to fouling in the heat exchanger, we need to find the new thermal resistance after fouling. The total thermal resistance of the clean heat exchanger is 0.025 K/W.

The fouling factors for the inner and outer surfaces are 0.00045 (K*m²)/W and 0.00015 (K*m²)/W respectively. We can calculate the new thermal resistance by adding the fouling factors:

New thermal resistance = total thermal resistance + fouling factor of inner surface + fouling factor of outer surface

• New thermal resistance = 0.025 K/W + 0.00045 (K*m²)/W + 0.00015 (K*m²)/W
• New thermal resistance = 0.025 K/W + 0.00045 (K*m²)/W + 0.00015 (K*m²)/W

Now we can calculate the percentage decrease in the rate of heat transfer using the formula:

• Percentage decrease = (New thermal resistance - Total thermal resistance) / Total thermal resistance * 100
• Percentage decrease = (0.025 K/W + 0.00045 (K*m²)/W + 0.00015 (K*m²)/W - 0.025 K/W) / 0.025 K/W * 100
• Percentage decrease = (0.0006 (K*m²)/W) / 0.025 K/W * 100
• Percentage decrease ≈ 2.4%

Therefore, the percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is approximately 2.4%.

None of the given options is correct