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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.

Short Answer

Expert verified
Answer: The heat transfer rating of the heat exchanger is 1760 kW, and a suitable type of heat exchanger for cooling oil at the given conditions is the shell and tube heat exchanger.

Step by step solution

01

Identify the known parameters and formula for heat transfer

We are given the following information: - The specific heat capacity of oil: \(c_{p} = 2200 \mathrm{~J} / \mathrm{kg \cdot K}\) - The mass flow rate of oil: \(\dot{m} = 10 \mathrm{~kg} / \mathrm{s}\) - The initial temperature of oil: \(T_{1} = 120^{\circ} \mathrm{C}\) - The final temperature of oil: \(T_{2} = 40^{\circ} \mathrm{C}\) We can find the heat transfer rating using the formula: \(q = \dot{m} \cdot c_{p} \cdot \Delta T\) where \(q\) is the heat transfer rating \(\dot{m}\) is the mass flow rate \(c_{p}\) is the specific heat capacity \(\Delta T\) is the temperature difference
02

Calculate the temperature difference and heat transfer rating

First, determine the temperature difference: \(\Delta T = T_{1} - T_{2} = 120^{\circ} \mathrm{C} - 40^{\circ} \mathrm{C} = 80 \mathrm{K}\) Now, we can calculate the heat transfer rating: \(q = \dot{m} \cdot c_{p} \cdot \Delta T = 10 \mathrm{~kg} / \mathrm{s} \cdot 2200 \mathrm{~J} / \mathrm{kg \cdot K} \cdot 80 \mathrm{K}\) \(q = 1760000 \mathrm{~W}\) or \(1760 \mathrm{kW}\)
03

Propose a suitable type of heat exchanger

There are different types of heat exchangers, such as shell and tube, plate, and finned tube. For cooling oil, a plate or shell and tube heat exchanger is typically used due to their efficiency and ability to handle a large temperature difference. In this case, since the temperature difference is 80 K and we need high efficiency for cooling, a shell and tube heat exchanger could be a suitable choice. To summarize: Heat transfer rating of the heat exchanger: \(1760 \mathrm{kW}\) Proposed type of heat exchanger: Shell and tube heat exchanger

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Most popular questions from this chapter

Consider a water-to-water counterflow heat exchanger with these specifications. Hot water enters at \(90^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of the hot water is \(15^{\circ} \mathrm{C}\) greater than that of the cold water, and the mass flow rate of the hot water is 50 percent greater than that of the cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(2200 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be $c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, determine \)(a)\( the outlet temperature of the cold water, \)(b)$ the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

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 heat exchanger that has an NTU of 4 . Someone proposes to double the size of the heat exchanger and thus double the NTU to 8 in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?

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?

In a textile manufacturing plant, the waste dyeing water $\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)80^{\circ} \mathrm{C}$ is to be used to preheat fresh water $\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)10^{\circ} \mathrm{C}$ at the same flow rate in a double-pipe counterflow heat exchanger. The heat transfer surface area of the heat exchanger is \(1.65 \mathrm{~m}^{2}\), and the overall heat transfer coefficient is $625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the rate of heat transfer in the heat exchanger is \)35 \mathrm{~kW}$, determine the outlet temperature and the mass flow rate of each fluid stream.

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