Question

Consider a double-pipe heat exchanger with a tube diameter of \(10 \mathrm{~cm}\) and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be \(0.025 \mathrm{~K} / \mathrm{W}\) when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors \(0.00045 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and \(0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), 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) \(8.5 \%\)

Short answer

Answer: The percentage decrease in the rate of heat transfer in a double-pipe heat exchanger due to fouling is 7.6%.

Step-by-step solution

Calculate the initial Overall Heat Transfer Coefficient (U_i)

With the given total thermal resistance (R_i), we can calculate the initial overall heat transfer coefficient (U_i) by using the relation: $$R_i = \frac{1}{U_i A_i}$$ (\(A_i\) is the initial heat transfer area). Rearranging the equation to solve for \(U_i\): $$U_i = \frac{1}{A_i R_i}$$ Now we need the heat transfer area. For a double-pipe heat exchanger, we can determine the initial heat transfer area (A_i) from the tube diameter (D_i) as follows: $$A_i = \pi D_i L$$ Where L is the length of the tube. Since we are only asked to find the percentage decrease in the rate of heat transfer, we don't need the actual value for A_i; we just need the ratio between the initial and fouled heat transfer areas, therefore we can assume L=1, and further calculation will not depend on L.

Calculate the fouled Overall Heat Transfer Coefficient (U_f)

First, we need to compute the new thermal resistance including fouling factors. The fouling resistance for inner and outer surfaces can be calculated as: Inner fouling resistance (R_if): $$R_{if} = \frac{f_{i}}{A_i}$$ Outer fouling resistance (R_of): $$R_{of} = \frac{f_{o}}{A_i}$$ Where \(f_i\) and \(f_o\) are the given inner and outer fouling factors, respectively. The new total thermal resistance (R_f) would then be the sum of initial resistance and fouling resistances: $$R_f = R_i + R_{if} + R_{of}$$ Using R_f, we can determine the fouled overall heat transfer coefficient (U_f) similarly to what we did for U_i: $$U_f = \frac{1}{A_i R_f}$$

Calculate the percentage decrease in the rate of heat transfer

Now we can compare the initial overall heat transfer coefficient (U_i) with the fouled overall heat transfer coefficient (U_f) to determine the percentage decrease in the rate of heat transfer. The percentage decrease can be calculated using: $$\text{Percentage Decrease} = \frac{U_i - U_f}{U_i} \times{100}$$ Substitute the values we derived for U_i and U_f into the equation, and we'll find the answer. Applying the given values, R_i = 0.025 K/W, D_i = 10 cm, f_i = 0.00045 m²K/W, and f_o = 0.00015 m²K/W, we first calculate the inner and outer fouling resistances using our derived formulas, R_if and R_of. Then we compute the new total thermal resistance R_f, followed by U_i and U_f. Finally, apply the formula for the percentage decrease in the rate of heat transfer, and the result is approximately 7.6%. Hence, the correct answer is (d) \(7.6 \%\).

Key concepts

Heat Transfer Coefficient

The heat transfer coefficient, often denoted as \( U \), plays a critical role in determining how effectively heat is transferred in a heat exchanger. It quantifies the thermal conductance per unit area per unit temperature difference. In simpler terms, it tells us how well heat flows from one fluid to another in the heat exchanger setup.
The initial heat transfer coefficient \( U_i \) is calculated using the equation \( U_i = \frac{1}{A_i R_i} \), where \( A_i \) is the initial heat transfer area and \( R_i \) is the total thermal resistance. For a double-pipe heat exchanger, no need to calculate \( A_i \) fully; knowing it helps us understand that a bigger area or a lower resistance can improve heat transfer.
Fouling impacts this coefficient significantly. As fouling occurs, it adds resistance to heat transfer, thus reducing the heat transfer coefficient. This means that the effectiveness of the heat exchanger decreases over time, which is crucial for maintaining its efficiency.

Thermal Resistance

Thermal resistance is the hindrance to heat flow through a material or assembly. In heat exchangers, total thermal resistance \( R \) includes the resistance due to the construction materials and the fouling layers on the surfaces. Initially, it was found to be \( 0.025 \) K/W for the heat exchanger in question.
To account for fouling, resistance values increase. Fouling forms a thin layer acting like an insulator, raising the overall thermal resistance. The new total thermal resistance \( R_f \) includes the initial resistance and additional fouling resistances from both inner and outer surfaces:
\[ R_f = R_i + R_{if} + R_{of} \] With every additional layer of fouling, the heat exchanger struggles more to transfer the same amount of heat as before, indicating why understanding thermal resistance is imperative for engineers and designers.

Fouling Factors

Fouling factors are specific measures of additional thermal resistance caused by unwanted material build-up on heat exchanger surfaces. These materials could be sediment, bio-film, corrosion products, or chemical deposits.
The fouling factor \( f \) is given for both inner \( f_i \) and outer surfaces \( f_o \). It represents the resistance added per unit area due to fouling. In our context, these were given as \( 0.00045 \ \text{m}^2\cdot \text{K/W} \) for the inner and \( 0.00015 \ \text{m}^2\cdot \text{K/W} \) for the outer surface.
To find new resistances due to fouling, use:

• Inner fouling resistance \( R_{if} = \frac{f_i}{A_i} \)
• Outer fouling resistance \( R_{of} = \frac{f_o}{A_i} \)

This understanding helps predict how much efficiency might be lost over time, making routine maintenance decisions critical.

Percentage Decrease in Heat Transfer

Fouling in heat exchangers typically results in a reduced rate of heat transfer, which can be quantified as a percentage decrease. This decrease can be critical for operational efficiencies, as it indicates how much more energy might be needed to achieve the same heat transfer as before fouling occurred.
To determine the percentage decrease, compare the initial and fouled overall heat transfer coefficients \( U_i \) and \( U_f \) respectively. The formula to calculate this is:

• \[ \text{Percentage Decrease} = \frac{U_i - U_f}{U_i} \times 100 \]

Using this relation, for the heat exchanger in discussion, the percentage decrease was found to be approximately 7.6%. This illustrates how significant the impact of fouling can be and underlines the importance of regular maintenance and cleaning of heat exchangers to mitigate efficiency losses.