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In the heat transfer relation \(\dot{Q}=U A_{s} \Delta T_{\mathrm{lm}}\) for a heat exchanger, what is \(\Delta T_{\mathrm{lm}}\) called? How is it calculated for a parallel-flow and a counterflow heat exchanger?

Short Answer

Expert verified
Answer: The term \(\Delta T_{\mathrm{lm}}\) is called the Logarithmic Mean Temperature Difference (LMTD) in the heat transfer equation for a heat exchanger. For both parallel-flow and counterflow heat exchangers, LMTD is calculated using the formula: \(\Delta T_{\mathrm{lm}} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})}\). In parallel-flow heat exchangers, \(\Delta T_1 = T_{h1} - T_{c1}\) and \(\Delta T_2 = T_{h2} - T_{c2}\), while in counterflow heat exchangers, \(\Delta T_1 = T_{h1} - T_{c2}\) and \(\Delta T_2 = T_{h2} - T_{c1}\).

Step by step solution

01

Definition of \(\Delta T_{\mathrm{lm}}\)

The \(\Delta T_{\mathrm{lm}}\) in the heat transfer equation of a heat exchanger is called the Logarithmic Mean Temperature Difference (LMTD).
02

Calculating LMTD for Parallel-flow Heat Exchanger

In a parallel-flow heat exchanger, the hot and cold fluids flow in the same direction. Let's denote the temperatures for the hot fluid as \(T_{h1}\) and \(T_{h2}\) for the inlet and outlet, respectively. For the cold fluid, the temperatures are denoted as \(T_{c1}\) and \(T_{c2}\) for the inlet and outlet. The LMTD for a parallel-flow heat exchanger is calculated using the following formula: $$ \Delta T_{\mathrm{lm}} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})} $$ where \(\Delta T_1 = T_{h1} - T_{c1}\) and \(\Delta T_2 = T_{h2} - T_{c2}\).
03

Calculating LMTD for Counterflow Heat Exchanger

In a counterflow heat exchanger, the hot and cold fluids flow in opposite directions. We use the same temperature notations as in the parallel-flow case. The LMTD for a counterflow heat exchanger is calculated using the same formula: $$ \Delta T_{\mathrm{lm}} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})} $$ However, in the counterflow case, the temperature differences are \(\Delta T_1 = T_{h1} - T_{c2}\) and \(\Delta T_2 = T_{h2} - T_{c1}\). In summary, the \(\Delta T_{\mathrm{lm}}\) is called the Logarithmic Mean Temperature Difference (LMTD), and it can be calculated for parallel-flow and counterflow heat exchangers using the same formula but with different temperature differences.

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Most popular questions from this chapter

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?

A heat exchanger is to cool oil $\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at a rate of \)10 \mathrm{~kg} / \mathrm{s}$ from \(120^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) by air. Determine the heat transfer rating of the heat exchanger and propose a suitable type.

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Water is boiled at \(150^{\circ} \mathrm{C}\) in a boiler by hot exhaust gases $\left(c_{p}=1.05 \mathrm{~kJ} / \mathrm{kg}^{\circ}{ }^{\circ} \mathrm{C}\right)\( that enter the boiler at \)540^{\circ} \mathrm{C}$ at a rate of \(0.4 \mathrm{~kg} / \mathrm{s}\) and leave at \(200^{\circ} \mathrm{C}\). The surface area of the heat exchanger is \(0.64 \mathrm{~m}^{2}\). The overall heat transfer coefficient of this heat exchanger is\(\mathrm{kg} / \mathrm{s}\) with cold water $\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( that enters the heat exchanger at \)20^{\circ} \mathrm{C}$ at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the heat transfer area of the heat exchanger is (a) \(0.745 \mathrm{~m}^{2}\) (b) \(0.760 \mathrm{~m}^{2}\) (c) \(0.775 \mathrm{~m}^{2}\) (d) \(0.790 \mathrm{~m}^{2}\) (e) \(0.805 \mathrm{~m}^{2}\)

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water $\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)$ enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools $5000 \mathrm{~kg} / \mathrm{h}\( of air \)\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( which enters the air handler at \)25^{\circ} \mathrm{C}$. If these streams are in counterflow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

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