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Chapter 11: Problem 110

Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), where it is heated by a hot water stream \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(70^{\circ} \mathrm{C}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\). Determine the maximum heat transfer rate and the outlet temperatures of both fluids for that case.

Short Answer

Expert verified
Answer: The maximum heat transfer rate is 209500 J/s, with the outlet temperature of the hot water stream being 20°C and the outlet temperature of the air being 43.28°C.

Step by step solution

01

Use the conservation of energy principle

Since energy is conserved, for the maximum heat transfer rate, all the energy lost by the hot water stream must be equal to the energy gained by the air stream. Let \(Q\) be the heat transfer rate, \(m\) be the mass flow rate, and \(c_p\) be the specific heat capacity. The energy change for the hot water stream can be represented as follows: \((m_{hot} \cdot c_{p_{hot}}) (T_{in_{hot}} - T_{out_{hot}}) = Q\) Simultaneously, the energy change for the air can be represented as: \(Q = (m_{air} \cdot c_{p_{air}}) (T_{out_{air}} - T_{in_{air}})\)
02

Find the maximum heat transfer rate, Q

Since we want the maximum heat transfer rate, the outlet temperature of the hot water stream will be equal to the inlet temperature of the air. Therefore, \(T_{out_{hot}} = T_{in_{air}}\). Plugging in the provided values and solving for \(Q\), we get \(Q = (m_{hot} \cdot c_{p_{hot}}) (T_{in_{hot}} - T_{in_{air}})\) \(Q = (1\,\mathrm{kg/s}) \cdot (4190\,\mathrm{J/kg\cdot K}) \cdot (70\,^{\circ}\mathrm{C} - 20\,^{\circ}\mathrm{C})\) \(Q = 209500\,\mathrm{J/s}\)
03

Calculate the outlet temperature of the air

Now, we will use the heat transfer rate and the air stream's energy change equation to find the outlet temperature of the air: \(Q = (m_{air} \cdot c_{p_{air}}) (T_{out_{air}} - T_{in_{air}})\) Rearranging the equation, we get: \(T_{out_{air}} = T_{in_{air}} + \frac{Q}{(m_{air} \cdot c_{p_{air}})}\) Plugging in the provided values, we get: \(T_{out_{air}} = 20\,^{\circ}\mathrm{C} + \frac{209500\,\mathrm{J/s}}{(3\,\mathrm{kg/s}) \cdot (1005\,\mathrm{J/kg\cdot K})}\) \(T_{out_{air}} = 20\,^{\circ}\mathrm{C} + 23.28\,^{\circ}\mathrm{C}\) \(T_{out_{air}} = 43.28\,^{\circ}\mathrm{C}\)
04

Present the final results

We calculated the maximum heat transfer rate, Q, to be \(209500\,\mathrm{J/s}\) and the outlet temperatures for both fluids: - Outlet temperature of the hot water stream: \(T_{out_{hot}} = 20\,^{\circ}\mathrm{C}\) - Outlet temperature of the air: \(T_{out_{air}} = 43.28\,^{\circ}\mathrm{C}\)

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Most popular questions from this chapter

In a parallel-flow, water-to-water heat exchanger, the hot water enters at \(75^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) and cold water enters at \(20^{\circ} \mathrm{C}\) at a rate of \(09 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6.4 \mathrm{~m}^{2}\), respectively. The specific heat for both the hot and cold fluid may be taken to be \(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). For the same overall heat transfer coefficient and the surface area, the increase in the effectiveness of this heat exchanger if counter-flow arrangement is used is (a) \(0.09\) (b) \(0.11\) (c) \(0.14\) (d) \(0.17\) (e) \(0.19\)

Hot water \(\left(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with mass flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) at \(100^{\circ} \mathrm{C}\) enters a thin-walled concentric tube counterflow heat exchanger with a surface area of \(23 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with mass flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(20^{\circ} \mathrm{C}\), determine \((a)\) the heat transfer rate for the heat exchanger and \((b)\) the outlet temperatures of the cold and hot fluids. After a period of operation, the overall heat transfer coefficient is reduced to \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine (c) the fouling factor that caused the reduction in the overall heat transfer coefficient.

There are two heat exchangers that can meet the heat transfer requirements of a facility. Both have the same pumping power requirements, the same useful life, and the same price tag. But one is heavier and larger in size. Under what conditions would you choose the smaller one?

Can the temperature of the cold fluid rise above the inlet temperature of the hot fluid at any location in a heat exchanger? Explain.

By taking the limit as \(\Delta T_{2} \rightarrow \Delta T_{1}\), show that when \(\Delta T_{1}=\Delta T_{2}\) for a heat exchanger, the \(\Delta T_{\mathrm{lm}}\) relation reduces to \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2} .\)
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