Question

A 2-shell passes and 4-tube passes heat exchanger is used for heating a hydrocarbon stream \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) steadily from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). A water stream enters the shellside at \(80^{\circ} \mathrm{C}\) and leaves at \(40^{\circ} \mathrm{C}\). There are 160 thin-walled tubes, each with a diameter of \(2.0 \mathrm{~cm}\) and length of \(1.5 \mathrm{~m}\). The tube-side and shell-side heat transfer coefficients are \(1.6\) and \(2.5 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) Calculate the rate of heat transfer and the mass rates of water and hydrocarbon streams. (b) With usage, the outlet hydrocarbon-stream temperature was found to decrease by \(5^{\circ} \mathrm{C}\) due to the deposition of solids on the tube surface. Estimate the magnitude of fouling factor.

Short answer

Question: Calculate the fouling factor of the heat exchanger when the outlet temperature of the hydrocarbon stream decreases by 5ºC due to deposition of solids. Answer: The fouling factor of the heat exchanger is approximately 0.00022 m²·K/kW.

Step-by-step solution

Calculate the rate of heat transfer

To calculate the rate of heat transfer, we need to find the LMTD and multiply it with surface area (A) and overall heat-transfer coefficient (U): $$q=U \cdot A \cdot LMTD$$ To find the LMTD, it's necessary to determine the inlet and outlet temperature differences: \(\Delta T_{1}= T_{h1} - T_{c2} = 80^{\circ} C - 50^{\circ} C = 30^{\circ} C\) \(\Delta T_{2}= T_{h2} - T_{c1} = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C\) Now we can calculate LMTD: \(LMTD= \frac{\Delta T_{1} - \Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}}} = \frac{30-20}{\ln(\frac{30}{20})} \approx 24.36^{\circ}\mathrm{C}\) Next, we need to determine the surface area (A) and overall heat-transfer coefficient (U). First, let's calculate the tube-side surface area for one tube (A_t): \(A_{t}= \pi d L = \pi (0.02m)(1.5m) = 0.094 \mathrm{m}^2\) Since we have 160 tubes, the total surface area (A) is: \(A= 160 \cdot A_{t} = 160 \cdot 0.094 = 15.04 \mathrm{m}^2\) To calculate the overall heat-transfer coefficient (U), we can use the resistance equation: \(\frac{1}{U}= \frac{1}{h_{t}} + \frac{1}{h_{s}}\) \(U = \frac{1}{\frac{1}{1.6} + \frac{1}{2.5}} \approx 0.938 \mathrm{kW}/\mathrm{m^2}\cdot\mathrm{K}\) Finally, we can calculate the rate of heat transfer (q): \(q= U \cdot A \cdot LMTD = 0.938 \cdot 15.04 \cdot 24.36 = 343.79 \mathrm{kW}\)

Calculate the mass rate for both streams

To find the mass rate of the hydrocarbon and water streams, we can use the heat-transfer equation: $$q= m \cdot c_{p} \cdot \Delta T$$ For hydrocarbon stream: $$\dot{m}_{h} = \frac{q}{c_{p_h} \cdot (T_{c2} - T_{c1})} = \frac{343.79}{2.0 \cdot (50-20)} = 5.73 \mathrm{kg/s}$$ For water stream: $$\dot{m}_{w} = \frac{q}{c_{p_w} \cdot (T_{h1} - T_{h2})} = \frac{343.79}{4.18 \cdot (80-40)} = 2.05 \mathrm{kg/s}$$

Determine the overall heat transfer coefficient and surface area

We already calculated the overall heat-transfer coefficient (U) and the surface area (A) of the heat exchanger in step 1. So we can move forward to the next step.

Calculate the fouling factor

Since the outlet temperature of the hydrocarbon stream decreases by \(5^\circ\mathrm{C}\) due to the deposition of solids, the new rate of heat transfer must be recalculated. We can recalculate the LMTD, considering the new outlet temperature (\(45^\circ\mathrm{C}\)) for the hydrocarbon stream: \(\Delta T_1 = T_{h1} - T_{c2} = 80^{\circ} C - 45^{\circ} C = 35^{\circ} C\) (keeping the same, as this temperature doesn't change) \(\Delta T_2 = T_{h2} - T_{c1} = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C\) (keeping the same, as this temperature doesn't change) \(LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})} \approx 26.36^{\circ}\mathrm{C}\) Now we need to calculate the new rate of heat transfer (q'): $$q' = U' \cdot A \cdot LMTD = 343.79 \mathrm{kW}$$ $$U' = \frac{q'}{A \cdot LMTD} = \frac{343.79}{15.04 \cdot 26.36} \approx 0.89 \mathrm{kW}/\mathrm{m^2}\cdot\mathrm{K}$$ To calculate the fouling factor (R), we can subtract U' from the original overall heat transfer coefficient (U) and then divide it by U^3: $$ R = \frac{\frac{1}{U} - \frac{1}{U'}}{U^3} = \frac{\frac{1}{0.938} - \frac{1}{0.89}}{0.938^3} \approx 0.00022 \mathrm{m^2\cdot K/ kW}$$ This means that the fouling factor for this particular case is approximately equal to \(0.00022 \mathrm{m^2\cdot K/ kW}\).

Key concepts

Log Mean Temperature Difference (LMTD)

Understanding the Log Mean Temperature Difference (LMTD) is vital for anyone studying heat exchangers, as it represents the driving force behind heat transfer. LMTD calculates the average temperature difference between the hot and cold fluids across the exchanger. The formula for LMTD is given by
\[LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}\]
where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences at each end of the heat exchanger. The effectiveness of the heat exchanger largely depends on maximizing this temperature difference. However, it's important to note that LMTD is ideally used for heat exchangers with a constant flow rate and temperature difference; in practice, variations can occur which might require a more specific approach.

Overall Heat-Transfer Coefficient

The overall heat-transfer coefficient (U) measures how well a heat exchanger can transfer heat between the two fluids. It incorporates the conductive and convective heat transfer abilities of the materials as well as the geometrical design of the exchanger. To calculate U, we use the formula:
\[U = \frac{1}{\frac{1}{h_{t}} + \frac{1}{h_{s}}}\]
where \(h_{t}\) is the tube-side heat transfer coefficient and \(h_{s}\) is the shell-side heat transfer coefficient. The higher the U value, the more efficient the heat exchanger is at transferring heat. When the U value is known, combined with the heat exchanger surface area and the LMTD, we can estimate the rate of heat transfer (
\[q = U \cdot A \cdot LMTD\]
). It's crucial to keep in mind that the U value can decrease over time due to factors like fouling, which retards the heat transfer process, leading to inefficiency.

Fouling Factor

Fouling is a common problem in heat exchangers where deposits accumulate on the heat transfer surfaces, affecting the operation. The fouling factor (R) quantifies the thermal resistance introduced by this fouling. It's calculated by subtracting the new overall heat transfer coefficient (
\(U'\)
) from the original one (U) and then dividing by the cube of the original coefficient:
\[R = \frac{\frac{1}{U} - \frac{1}{U'}}{U^3}\]
This additional resistance caused by fouling reduces the overall heat transfer coefficient and thus the heat exchanger’s efficiency. Regular cleaning and maintenance are important to minimize the effects of fouling and maintain heat exchanger performance.

Hydrocarbon Stream Heating

Heating hydrocarbon streams is a common process in industrial applications such as refining, where heat exchangers play an integral role. Optimizing this process requires careful consideration of the hydrocarbon's specific heat capacity (
\(c_{p}\)
), the desired temperature increase, and the efficiency of the heat exchanger involved. For instance, the temperature increase from 20°C to 50°C in our exercise is achieved by balancing the rate of heat transfer with the heat capacity of the hydrocarbon stream. Efficient heating ensures that there is minimal energy waste and that the process remains cost-effective. Additionally, we have to be mindful of the impact of fouling on the efficiency of hydrocarbon stream heating, as it can lead to higher operational costs and reduced throughput.