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Consider an oil-to-oil double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold oil enters at \(20^{\circ} \mathrm{C}\) and leaves at \(55^{\circ} \mathrm{C}\), while the hot oil enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counter-flow heat exchanger? Why? Assuming the mass flow rates of both fluids to be identical, determine the effectiveness of this heat exchanger.

Short Answer

Expert verified
Answer: The flow arrangement of the given heat exchanger is counter-flow, and its effectiveness is approximately 0.583 or 58.3%.

Step by step solution

01

Identify the type of flow arrangement

In both parallel-flow and counter-flow heat exchangers, the temperature difference between the hot and cold fluids always allows heat transfer. For a parallel-flow heat exchanger, the hot and cold fluids enter at the same end and flow in the same direction, whereas, for a counter-flow heat exchanger, the hot and cold fluids enter at opposite ends and flow in opposite directions. Given temperature measurements: - Cold oil inlet temperature: \(T_{c, in} = 20^{\circ} \mathrm{C}\) - Cold oil outlet temperature: \(T_{c, out} = 55^{\circ} \mathrm{C}\) - Hot oil inlet temperature: \(T_{h, in} = 80^{\circ} \mathrm{C}\) - Hot oil outlet temperature: \(T_{h, out} = 45^{\circ} \mathrm{C}\) Comparing the temperatures, we can conclude the following: - \(T_{c, out} > T_{h, out}\), so the cold fluid outlet temperature is higher than the hot fluid outlet temperature. - \(T_{h, in} > T_{c, in}\), so the hot fluid inlet temperature is higher than the cold fluid inlet temperature. These temperature differences indicate that heat is being transferred from the hot fluid to the cold fluid, as expected in a heat exchanger. Since the hot fluid is cooling down and the cold fluid is heating up, it implies that the fluids are flowing in opposite directions, which is characteristic of a counter-flow heat exchanger.
02

Calculate the effectiveness of the heat exchanger

Assuming identical mass flow rates for both hot and cold fluids, we can calculate the effectiveness of the heat exchanger using the formula: Effectiveness \(= \dfrac{Q_{actual}}{Q_{max}}\) where \(Q_{actual}\) is the actual heat transfer rate and \(Q_{max}\) is the maximum possible heat transfer rate. To calculate \(Q_{actual}\), we can use the energy balance equation: \(Q_{actual} = m \times c_p \times (T_{h, in} - T_{h, out})\) where \(m\) is the mass flow rate, and \(c_p\) is the specific heat capacity of the fluid. Since the mass flow rates and specific heat capacities are identical for both fluids, this equation simplifies to: \(Q_{actual} = (T_{h, in} - T_{h, out})\) Next, we need to calculate the maximum possible heat transfer rate, \(Q_{max}\). The maximum possible heat transfer occurs when the cold fluid reaches the inlet temperature of the hot fluid. Then, \(Q_{max}\) can be calculated as: \(Q_{max} = (T_{h, in} - T_{c, in})\) Now, we can calculate the effectiveness of the heat exchanger: Effectiveness \(= \dfrac{Q_{actual}}{Q_{max}} = \dfrac{T_{h, in} - T_{h, out}}{T_{h, in} - T_{c, in}}\) Effectiveness \(= \dfrac{80 - 45}{80 - 20} = \dfrac{35}{60} \approx 0.583\) The effectiveness of this counter-flow heat exchanger is approximately 0.583, or 58.3%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Counter-flow Heat Exchanger
In a counter-flow heat exchanger, the hot and cold fluids flow in opposite directions. This setup allows for greater heat transfer efficiency compared to other types of heat exchangers. As the hot fluid runs from one end, the cold fluid moves from the opposite end, ensuring maximum temperature gradient between the fluids across the entire length of the heat exchanger. This large temperature gradient helps in achieving more effective heat transfer.

Due to the opposite flow directions, the cold fluid can leave the exchanger at a higher temperature than the outlet temperature of the hot fluid. In the given exercise, this specific configuration is indicated by the cold fluid exiting at a temperature of \(55^{\circ} \mathrm{C}\), which is higher than the hot fluid's exit temperature of \(45^{\circ} \mathrm{C}\). This confirms that the heat exchanger in question is of the counter-flow type.
Parallel-flow Heat Exchanger
A parallel-flow heat exchanger has both fluids entering from the same end and flowing in the same direction. This means that at the entry point, the temperature difference between the hot and cold fluids is at its maximum, but it reduces along the length of the exchanger.

Unlike the counter-flow setup, in a parallel-flow arrangement, the maximum temperature of the fluids at any point within the exchanger is limited. Typically, the cold fluid's exit temperature cannot exceed the hot fluid's outlet temperature. If this were a parallel-flow heat exchanger, the cold outlet would not be able to achieve a \(55^{\circ} \mathrm{C}\) temperature when the hot fluid exit is \(45^{\circ} \mathrm{C}\). Hence, a parallel-flow setup is less efficient in cases where high temperature change and effectiveness are desired.
Heat Transfer Rate
Determining the heat transfer rate in a heat exchanger involves calculating the energy transferred between fluids. The heat transfer rate, denoted as \(Q\), depends on the temperature difference between the inlet and outlet flows and the specific heat capacity of the fluids.

In the exercise, the heat transfer rate \(Q_{actual}\) for both the hot and cold oils can be expressed using the equation:
\(Q_{actual} = m \times c_p \times (T_{h, in} - T_{h, out})\)

This formula highlights how the heat transferred is directly related to the change in temperature of the hot fluid. It shows that the effectiveness of the heat exchanger impacts how much energy is actually being transferred from one fluid to the other during operation.
Energy Balance Equation
An essential principle in studying heat exchangers is the energy balance equation. This equation helps in understanding the heat exchange process by equating the energy lost by the hot fluid to the energy gained by the cold fluid.

The energy balance can be expressed as:
\(Q_{actual} = m \times c_p \times (T_{hot in} - T_{hot out}) = m \times c_p \times (T_{cold out} - T_{cold in})\)

This equation ensures that the total energy is conserved during the exchanging process. In the given exercise, this principle supports calculating the heat exchanger's effectiveness, thereby helping quantify how well the device performs in real-world scenarios. Understanding this concept is crucial for improving heat exchanger designs and increasing their efficiency.

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Most popular questions from this chapter

The mass flow rate, specific heat, and inlet temperature of the tube-side stream in a double-pipe, parallel-flow heat exchanger are \(2700 \mathrm{~kg} / \mathrm{h}, 2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), and \(120^{\circ} \mathrm{C}\), respectively. The mass flow rate, specific heat, and inlet temperature of the other stream are \(1800 \mathrm{~kg} / \mathrm{h}, 4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), and \(20^{\circ} \mathrm{C}\), respectively. The heat transfer area and overall heat transfer coefficient are \(0.50 \mathrm{~m}^{2}\) and \(2.0 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Find the outlet temperatures of both streams in steady operation using (a) the LMTD method and \((b)\) the effectiveness-NTU method.

Water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) is to be heated by solarheated hot air \(\left(c_{p}=0.24 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) in a double- pipe counterflow heat exchanger. Air enters the heat exchanger at \(190^{\circ} \mathrm{F}\) at a rate of \(0.7 \mathrm{lbm} / \mathrm{s}\) and leaves at \(135^{\circ} \mathrm{F}\). Water enters at \(70^{\circ} \mathrm{F}\) at a rate of \(0.35 \mathrm{lbm} / \mathrm{s}\). The overall heat transfer coefficient based on the inner side of the tube is given to be \(20 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine the length of the tube required for a tube internal diameter of \(0.5 \mathrm{in}\).

A counterflow double-pipe heat exchanger with \(A_{s}=\) \(9.0 \mathrm{~m}^{2}\) is used for cooling a liquid stream \(\left(c_{p}=3.15 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a rate of \(10.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(90^{\circ} \mathrm{C}\). The coolant \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the heat exchanger at a rate of \(8.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(10^{\circ} \mathrm{C}\). The plant data gave the following equation for the overall heat transfer coefficient in W/m \({ }^{2} \cdot \mathrm{K}: U=600 /\left(1 / \dot{m}_{c}^{0.8}+2 / \dot{m}_{h}^{0.8}\right)\), where \(\dot{m}_{c}\) and \(\dot{m}_{h}\) are the cold-and hot-stream flow rates in kg/s, respectively. (a) Calculate the rate of heat transfer and the outlet stream temperatures for this unit. (b) The existing unit is to be replaced. A vendor is offering a very attractive discount on two identical heat exchangers that are presently stocked in its warehouse, each with \(A_{s}=5 \mathrm{~m}^{2}\). Because the tube diameters in the existing and new units are the same, the above heat transfer coefficient equation is expected to be valid for the new units as well. The vendor is proposing that the two new units could be operated in parallel, such that each unit would process exactly one-half the flow rate of each of the hot and cold streams in a counterflow manner; hence, they together would meet (or exceed) the present plant heat duty. Give your recommendation, with supporting calculations, on this replacement proposal.

A company owns a refrigeration system whose refrigeration capacity is 200 tons ( 1 ton of refrigeration = \(211 \mathrm{~kJ} / \mathrm{min}\) ), and you are to design a forced-air cooling system for fruits whose diameters do not exceed \(7 \mathrm{~cm}\) under the following conditions: The fruits are to be cooled from \(28^{\circ} \mathrm{C}\) to an average temperature of \(8^{\circ} \mathrm{C}\). The air temperature is to remain above \(-2^{\circ} \mathrm{C}\) and below \(10^{\circ} \mathrm{C}\) at all times, and the velocity of air approaching the fruits must remain under \(2 \mathrm{~m} / \mathrm{s}\). The cooling section can be as wide as \(3.5 \mathrm{~m}\) and as high as \(2 \mathrm{~m}\). Assuming reasonable values for the average fruit density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for the quantities related to the thermal aspects of the forced-air cooling, including (a) how long the fruits need to remain in the cooling section, \((b)\) the length of the cooling section, \((c)\) the air velocity approaching the cooling section, \((d)\) the product cooling capacity of the system, in \(\mathrm{kg}\) fruit/h, \((e)\) the volume flow rate of air, and \((f)\) the type of heat exchanger for the evaporator and the surface area on the air side.

Oil in an engine is being cooled by air in a cross-flow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). Determine \((a)\) the heat transfer effectiveness and \((b)\) the outlet temperature of the oil.

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