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Consider two double-pipe counterflow heat exchangers that are identical except that one is twice as long as the other one. Which heat exchanger is more likely to have a higher effectiveness?

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Answer: The double-pipe counterflow heat exchanger that is twice as long as the other one is more likely to have a higher effectiveness.

Step by step solution

01

Factors affecting heat exchanger effectiveness

In a double-pipe counterflow heat exchanger, the effectiveness (ε) depends on several factors like the heat capacity rates of the fluids involved, the heat transfer coefficient, and the surface area for heat transfer. The formula for the effectiveness of a counterflow heat exchanger is given by: ε = \frac{1 - e^{-NTU(1-C_r)}}{1 - C_r e^{-NTU(1-C_r)}} Here, NTU (Number of Transfer Units) measures the size of the heat exchanger relative to the capacity rates of the fluids and is defined as: NTU = \frac{UA}{C_{min}} where U is the overall heat transfer coefficient, A is the heat transfer surface area, and C_{min} is the minimum heat capacity rate of the fluids (C_{min} = min(C_h, C_c), where C_h and C_c are the hot fluid and cold fluid heat capacity rates, respectively). C_r is the capacity rate ratio, defined as C_r = \frac{C_{min}}{C_{max}}. From the given information, the heat exchangers are identical except for their length. We can assume that their heat capacity rates, heat transfer coefficients, and fluid properties are the same for both heat exchangers. Now, let's consider the impact of length on the surface area and effectiveness.
02

Impact of length on the surface area and NTU

The surface area (A) for heat transfer in a double-pipe heat exchanger is directly proportional to the length (L) of the exchanger: A = k*L where k is a constant factor that depends on the geometry and dimensions of the heat exchanger. The NTU of the longer heat exchanger (L2) will be twice that of the shorter heat exchanger (L1) as L2 = 2*L1: NTU_2 = \frac{UA_2}{C_{min}} = \frac{2kL_1U}{C_{min}} = 2 * NTU_1
03

Impact of increased NTU on the effectiveness

Since the heat capacity rates (C_h and C_c) and their ratio (C_r) are the same for both heat exchangers, the only difference in their effectiveness will be due to the difference in their NTUs, as shown in the formula: ε = \frac{1 - e^{-NTU(1-C_r)}}{1 - C_r e^{-NTU(1-C_r)}} As NTU_2 = 2 * NTU_1, the effectiveness of the longer heat exchanger will be higher than that of the shorter heat exchanger as the term e^{-NTU(1-C_r)} will decrease with increasing NTU, resulting in a higher numerator and lower denominator in the effectiveness formula.
04

Conclusion

Based on the analysis, we can conclude that the double-pipe counterflow heat exchanger that is twice as long as the other one is more likely to have a higher effectiveness. This is because the longer heat exchanger has a larger surface area for heat transfer, resulting in a higher NTU and, consequently, a higher effectiveness.

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Most popular questions from this chapter

A performance test is being conducted on a doublepipe counterflow heat exchanger that carries engine oil and water at a flow rate of $2.5 \mathrm{~kg} / \mathrm{s}\( and \)1.75 \mathrm{~kg} / \mathrm{s}$, respectively. Since the heat exchanger has been in service for a long time, it is suspected that fouling might have developed inside the heat exchanger that could affect the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of $450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and a surface area of \)7.5 \mathrm{~m}^{2}\(, the oil must be heated from \)25^{\circ} \mathrm{C}$ to \(55^{\circ} \mathrm{C}\) by passing hot water at $100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

Consider a condenser unit (shell and tube heat exchanger) of an HVAC facility where saturated refrigerant \(\mathrm{R}-134 \mathrm{a}\) at a saturation pressure of \(1318.6 \mathrm{kPa}\) and at a rate of $2.5 \mathrm{~kg} / \mathrm{s}$ flows through thin-walled copper tubes. The refrigerant enters the condenser as saturated vapor and we wish to have a saturated liquid refrigerant at the exit. The cooling of refrigerant is carried out by cold water that enters the heat exchanger at \(10^{\circ} \mathrm{C}\) and exits at \(40^{\circ} \mathrm{C}\). Assuming the initial overall heat transfer coefficient of the heat exchanger to be $3500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the surface area of the heat exchanger and the mass flow rate of cooling water for complete condensation of the refrigerant. In practice, over a long period of time, fouling occurs inside the heat exchanger that reduces its overall heat transfer coefficient and causes the mass flow rate of cooling water to increase. Increase in the mass flow rate of cooling water will require additional pumping power, making the heat exchange process uneconomical. To prevent the condenser unit from underperforming, assume that fouling has occurred inside the heat exchanger and has reduced its overall heat transfer coefficient by 20 percent. For the same inlet temperature and flow rate of refrigerant, determine the new flow rate of cooling water to ensure complete condensation of the refrigerant at the heat exchanger exit.

A shell-and-tube heat exchanger is used for heating $14 \mathrm{~kg} / \mathrm{s}\( of oil \)\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( from \)20^{\circ} \mathrm{C}\( to \)46^{\circ} \mathrm{C}$. The heat exchanger has one shell pass and six tube passes. Water enters the shell side at \(80^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). The overall heat transfer coefficient is estimated to be $1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Calculate the rate of heat transfer and the heat transfer area.

A two-shell-pass and four-tube-pass 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}\(. A water stream enters the shell side at \)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 the fouling factor.

Explain how the maximum possible heat transfer rate \(\dot{Q}_{\max }\) in a heat exchanger can be determined when the mass flow rates, specific heats, and inlet temperatures of the two fluids are specified. Does the value of \(\dot{Q}_{\max }\) depend on the type of the heat exchanger?

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