Question

A chemical process facility has a pipe system with the inner surface coated with polypropylene lining. The ASME Code for Process Piping limits the maximum use temperature for polypropylene lining to \(107^{\circ} \mathrm{C}\) (ASME B31.3-2014, Table A323.4.3). A double-pipe counterflow heat exchanger is located upstream to reduce the hot fluid temperature before it flows into the pipe system coated with polypropylene lining. The inner tube of the heat exchanger has a negligible wall thickness, and its length and diameter are \(5 \mathrm{~m}\) and \(25 \mathrm{~mm}\), respectively. The convection heat transfer coefficients inside and outside of the heat exchanger inner tube are $1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and \)1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The hot fluid enters the heat exchanger at a heat capacity rate of \(1 \mathrm{~kW} / \mathrm{K}\) and \(120^{\circ} \mathrm{C}\). In the cold fluid stream, water enters the heat exchanger at $10^{\circ} \mathrm{C}\( and exits at \)30^{\circ} \mathrm{C}$. Determine the highest level of fouling factor that the heat exchanger can tolerate before it becomes unable to cool the hot fluid to \(107^{\circ} \mathrm{C}\) at the outlet.

Short answer

Based on the given information and calculations, the highest fouling factor the double-pipe counterflow heat exchanger can tolerate is 0.0013 m²·K/W.

Step-by-step solution

1. Overall Heat Transfer Coefficient (U)

To find the overall heat transfer coefficient, we need to use the heat transfer coefficients of the inner and outer surfaces, as well as the fouling factor. We'll use the formula: \(U=\left( \frac{1}{h_{i}}+\frac{1}{h_{o}}+R_{f} \right)^{-1}\) Since we know the inner and outer convection heat transfer coefficients (\(h_{i}\) and \(h_{o}\)) are \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, we can rewrite the above equation as: \(U=\left( \frac{1}{1000}+\frac{1}{1500}+R_{f} \right)^{-1}\)

2. Calculate Heat Transfer Rates

To determine the required heat transfer rates, we first need to calculate the heat capacity rate of the water (\(C_{c}\)) and the temperature difference (\(\Delta T_{m}\)). The heat capacity rate of the water can be calculated as: \(C_{c}= C_{h}\frac{T_{h,i} - T_{h,o}}{T_{c,o} - T_{c,i}}\) Where: - \(C_{h}\) is the heat capacity rate of the hot fluid, given as \(1 \mathrm{~kW} / \mathrm{K}\) - \(T_{h,i}\) is the inlet temperature of the hot fluid, given as \(120^{\circ} \mathrm{C}\) - \(T_{h,o}\) is the outlet temperature of the hot fluid, desired as \(107^{\circ} \mathrm{C}\) - \(T_{c,i}\) is the inlet temperature of the cold fluid, given as \(10^{\circ} \mathrm{C}\) - \(T_{c,o}\) is the outlet temperature of the cold fluid, given as \(30^{\circ} \mathrm{C}\) Plugging in the values, we get: \(C_{c} = 1\frac{120-107}{30-10} = 0.65 \mathrm{~kW} / \mathrm{K}\) Next, we'll calculate the required temperature difference. For a counterflow heat exchanger, the logarithmic mean temperature difference (\(\Delta T_{m}\)) is calculated as: \(\Delta T_{m} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln{(\frac{\Delta T_{1}}{\Delta T_{2}})}}\) Where: - \(\Delta T_{1} = T_{h,i} - T_{c,o} = 120 - 30 = 90 \mathrm{~K}\) - \(\Delta T_{2} = T_{h,o} - T_{c,i} = 107 - 10 = 97 \mathrm{~K}\) Plugging in the values, we get: \(\Delta T_{m} = \frac{90 - 97}{\ln{(\frac{90}{97})}} = -93.63 \mathrm{~K}\)

3. Determine Heat Transfer Area

To determine the heat transfer area (A), we will use the given length (L) and diameter (D) of the inner tube of the heat exchanger: \(A = \pi D L\) Where: - \(D = 25 \mathrm{~mm} = 0.025 \mathrm{~m}\) - \(L = 5 \mathrm{~m}\) Plugging in the values, we get: \(A = \pi\cdot0.025\cdot5 = 0.3927 \mathrm{~m}^{2}\)

4. Calculate the Fouling Factor (Rf)

Now that we have all the necessary values, we can use the formula for the overall heat transfer coefficient (U) and rearrange it to find the fouling factor (Rf): \(R_{f} = \frac{1}{U} - \frac{1}{h_{i}} - \frac{1}{h_{o}}\) Before we can use this, we need to first find the value of U. Since we know that the heat transfer rate (\(Q\)) is equal to the product of U, A, and \(\Delta T_{m}\), we can write: \(U = \frac{Q}{A\Delta T_{m}}\) Where \(Q = C_{c}(T_{c,o} - T_{c,i}) = 0.65(30 - 10) = 13 \mathrm{~kW}\) So, we can now calculate the value of U: \(U = \frac{13}{0.3927\cdot(-93.63)} = 0.3508 \mathrm{W} / \mathrm{m}^{2}\cdot\mathrm{K}\) Finally, we can now use this value to find the fouling factor (Rf): \(R_{f} = \frac{1}{0.3508} - \frac{1}{1000} - \frac{1}{1500} = 0.0013~\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\) The highest fouling factor that the heat exchanger can tolerate is \(0.0013~\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\).