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Consider a heat exchanger that has an NTU of \(0.1\). Someone proposes to triple the size of the heat exchanger and thus triple the NTU to \(0.3\) in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?

Short Answer

Expert verified
Answer: Yes, tripling the NTU in a counter-flow heat exchanger leads to a higher effectiveness, indicating that more heat is transferred between the fluids. However, other factors such as initial investment, maintenance costs, and space constraints should also be considered before making a final decision.

Step by step solution

01

Understand the relationship between effectiveness and NTU

Effectiveness (\(\varepsilon\)) of a heat exchanger is defined as the ratio of the actual amount of heat transferred to the maximum possible heat transfer. The relationship between effectiveness and NTU is different for different types of heat exchangers (parallel-flow, counter-flow, etc.). To simplify the analysis, we will assume a counter-flow heat exchanger, where the relationship between effectiveness and NTU is given by the formula: \(\varepsilon = \frac{1 - e^{-(NTU)^{1-{C_r}}}}{1 - C_r e^{-(NTU)^{1-{C_r}}}}\) where \(C_r\) is the heat capacity rate ratio, which is the ratio of the minimum heat capacity rate to the maximum heat capacity rate.
02

Calculate the effectiveness for the original NTU

Since we are not given the value of \(C_r\), we will only have a general understanding of how increasing NTU affects effectiveness. Using the given NTU as \(0.1\), and assuming a counter-flow heat exchanger, let's calculate the effectiveness: \(\varepsilon_{0.1} = \frac{1 - e^{-(0.1)^{1-{C_r}}}}{1 - C_r e^{-(0.1)^{1-{C_r}}}}\)
03

Calculate the effectiveness for the proposed NTU

Let's also calculate the effectiveness when the NTU is tripled to \(0.3\): \(\varepsilon_{0.3} = \frac{1 - e^{-(0.3)^{1-{C_r}}}}{1 - C_r e^{-(0.3)^{1-{C_r}}}}\)
04

Compare the two effectiveness values

Now, we want to compare the two effectiveness values to determine if the proposal would indeed lead to increased effectiveness. As can be seen from the formulas derived in step 2 and step 3, it is evident that \(\varepsilon_{0.3} > \varepsilon_{0.1}\). This means that increasing the NTU will result in a higher effectiveness for the heat exchanger.
05

Conclusions and recommendations

Tripling the size of the heat exchanger and thus tripling the NTU to \(0.3\) in order to increase the effectiveness of the heat exchanger will indeed result in higher effectiveness, according to the counter-flow heat exchanger formula. However, keep in mind that increasing the size of the heat exchanger can also lead to increased initial investment and maintenance costs. Additionally, in real-world applications, there can be other factors limiting the size of heat exchangers, such as space constraints or weight limitations. Therefore, as a teacher, more information on these aspects needs to be considered before recommending or not supporting this proposal.

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Most popular questions from this chapter

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.

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