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In a parallel-flow, liquid-to-liquid heat exchanger, the inlet and outlet temperatures of the hot fluid are \(150^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) while that of the cold fluid are \(30^{\circ} \mathrm{C}\) and \(70^{\circ} \mathrm{C}\), respectively. For the same overall heat transfer coefficient, the percentage decrease in the surface area of the heat exchanger if counter-flow arrangement is used is (a) \(3.9 \%\) (b) \(9.7 \%\) (c) \(14.5 \%\) (d) \(19.7 \%\) (e) \(24.6 \%\)

Short Answer

Expert verified
Answer: The percentage decrease in surface area is 0%.

Step by step solution

01

Calculate the heat capacities of the fluids

Let's denote the mass flow rate of the hot fluid as \(m_h\), the mass flow rate of the cold fluid as \(m_c\), the specific heat capacities of hot and cold fluids as \(C_{p_h}\) and \(C_{p_c}\), respectively. The heat capacities of the fluids are calculated as: \(C_h = m_h C_{p_h}\) and \(C_c = m_c C_{p_c}\)
02

Calculate the heat exchanger effectiveness for parallel flow

The heat exchanger effectiveness (\(\epsilon\)) is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate in the heat exchanger. For parallel flow, we have: \(\epsilon_{PF} = \frac{T_{h, in} - T_{h, out}}{T_{h, in} - T_{c, in}} \) Plug in the given temperatures and get: \(\epsilon_{PF} = \frac{150 - 90}{150 - 30} = \frac{60}{120} = 0.5\)
03

Calculate the heat capacities ratio and the NTU for parallel flow

The heat capacities ratio (\(C_r\)) and the Number of Transfer Units (NTU) for the parallel-flow heat exchanger can be calculated using the effectiveness (\(\epsilon\)), assuming that there are no phase changes occurring in the fluids and the overall heat transfer coefficient remains constant: \(C_r = \frac{C_c}{C_h}\) and \(NTU_{PF} = \frac{\epsilon_{PF} (1 - C_r)}{1 - C_r \cdot e^{-NTU_{PF} (1 - C_r)}}\) As we don't know the mass flow rates or specific heat capacities of the fluids, we can't directly calculate \(C_r\) and \(NTU_{PF}\). However, we can use another formula to express the effectiveness of counter-flow heat exchanger in terms of NTU and \(C_r\):
04

Calculate the effectiveness of the counter-flow heat exchanger

For the counter-flow heat exchanger, effectiveness \(\epsilon_{CF}\) can be calculated with the same NTU and \(C_r\) as in the parallel flow: \(\epsilon_{CF} = \frac{1 - e^{-NTU_{CF} (1 + C_r)}}{1 + C_r}\) We know that the overall heat transfer coefficient is the same for both parallel-flow and counter-flow heat exchangers. Therefore, the NTU for the two arrangements is the same: \(NTU_{CF} = NTU_{PF}\) Substitute \(\epsilon_{PF}\), \(C_r\), and \(NTU_{CF}\) into the \(\epsilon_{CF}\) equation: \(\epsilon_{CF} = \frac{1 - e^{-NTU_{PF} (1 + C_r)}}{1 + C_r}\)
05

Calculate the area ratio of parallel-flow to counter-flow heat exchanger

The area ratio can be calculated by using the relationship between NTU and area of the heat exchanger. The NTU is defined as: \(NTU = \frac{UA}{C_\text{min}}\) Where \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer area, and \(C_\text{min}\) is the minimum heat capacity of the fluids. Since \(U\) and \(C_\text{min}\)are constant for both heat exchangers, the area ratio is equal to the NTU ratio: \(\frac{A_{PF}}{A_{CF}} = \frac{NTU_{PF}}{NTU_{CF}}\) As \(NTU_{PF} = NTU_{CF}\), we get: \(\frac{A_{PF}}{A_{CF}} = 1\)
06

Calculate the percentage decrease in surface area

To find the percentage decrease in surface area when switching from a parallel-flow to a counter-flow heat exchanger, we can use the area ratio calculated in Step 5: Percentage decrease in surface area \(= \frac{A_{PF} - A_{CF}}{A_{PF}} \times 100\%\) Since \(\frac{A_{PF}}{A_{CF}} = 1\), the percentage decrease in surface area is: Percentage decrease in surface area \(= \frac{1 - 1}{1} \times 100\% = 0\%\) The correct answer is none of the given options (a), (b), (c), (d), or (e) because the surface area doesn't decrease when switching from a parallel-flow to a counter-flow heat exchanger when the overall heat transfer coefficient is constant. Note: The problem statement appears to be incorrect or missing important information because there is no decrease in the heat exchanger area with the given data. The answer should be revised with appropriate premises, or the problem statement needs to be more precise.

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Key Concepts

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

Parallel-Flow Heat Exchanger
In a parallel-flow heat exchanger, both the hot and cold fluids enter the exchanger from the same side and move parallel to each other until they exit on the opposite side. This configuration typically results in the outlet temperatures of both fluids being closer to a mean temperature. - It's characterized by having a lower thermal effectiveness compared to other configurations, such as counter-flow heat exchangers. - In parallel flow, the temperature difference between the fluids at the exit is usually lesser than the initial difference, which sometimes limits the temperature exchange potential. Despite these drawbacks, parallel-flow designs are simple and cheaper to construct, making them suitable for applications where space and budget constraints are significant. They are also easier to clean and maintain because of their straightforward design.
Counter-Flow Heat Exchanger
A counter-flow heat exchanger has the hot and cold fluids entering the exchanger from opposite ends and flowing counter-currently towards each other. This setup allows for a higher temperature gradient between fluids compared to the parallel-flow arrangement. - The counter-flow configuration has a greater temperature difference across the length of the heat exchanger, which enables more efficient heat transfer. - As a result, it often achieves better thermal effectiveness than a parallel-flow exchanger. In practice, counter-flow designs are preferred when higher heat transfer efficiency is required. Although they may be more costly and complex to build, their superior performance can justify the investment, especially in processes demanding stringent temperature control.
Heat Exchanger Effectiveness
Heat exchanger effectiveness is a measure of how well a heat exchanger can transfer heat relative to its maximum potential. It is defined as the ratio of the actual heat transfer to the maximum possible heat transfer under given conditions.- The effectiveness (\(\epsilon\)) depends on the type of heat exchanger and the flow arrangement.- In mathematical terms, effectiveness can be expressed as\[\epsilon = \frac{Q_{actual}}{Q_{max}}\]where \(Q_{actual}\) is the actual heat transfer rate and \(Q_{max}\) is the maximum possible heat transfer rate, determined by the smaller of the two heat capacities of the fluids involved.Understanding and calculating the effectiveness helps in designing efficient heat exchanger systems and comparing different types of exchangers with respect to their heat transfer capabilities.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient (denoted as \(U\)) is pivotal in determining a heat exchanger's performance. It represents the heat transfer capability across all materials separating the hot and cold streams.- This coefficient takes into account the resistance to heat flow on both the fluid sides, as well as through the walls and potential fouling.- The expression for the heat transfer rate \(Q\) in a heat exchanger is given by the formula:\[Q = U \cdot A \cdot \Delta T_{LM}\]where \(A\) is the surface area for heat transfer, and \(\Delta T_{LM}\) is the log mean temperature difference across the exchanger.Maintaining an appropriate \(U\) is crucial for efficient heat exchanger performance. It is influenced by material choice, fluid velocities, and the physical state of the fluids involved. Efficient design and operation can enhance the \(U\), improving overall heat transfer efficiency.

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Most popular questions from this chapter

A 1-shell-pass and 8-tube-passes heat exchanger is used to heat glycerin \(\left(c_{p}=0.60 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(65^{\circ} \mathrm{F}\) to \(140^{\circ} \mathrm{F}\) by hot water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) that enters the thinwalled \(0.5\)-in-diameter tubes at \(175^{\circ} \mathrm{F}\) and leaves at \(120^{\circ} \mathrm{F}\). The total length of the tubes in the heat exchanger is \(500 \mathrm{ft}\). The convection heat transfer coefficient is \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the glycerin (shell) side and \(50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water (tube) side. Determine the rate of heat transfer in the heat exchanger \((a)\) before any fouling occurs and \((b)\) after fouling with a fouling factor of \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on the outer surfaces of the tubes.

A cross-flow heat exchanger consists of 80 thinwalled tubes of \(3-\mathrm{cm}\) diameter located in a duct of \(1 \mathrm{~m} \times 1 \mathrm{~m}\) cross section. There are no fins attached to the tubes. Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(18^{\circ} \mathrm{C}\) with an average velocity of \(3 \mathrm{~m} / \mathrm{s}\), while hot air \(\left(c_{p}=1010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the channel at \(130^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) at an average velocity of \(12 \mathrm{~m} / \mathrm{s}\). If the overall heat transfer coefficient is \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the outlet temperatures of both fluids and the rate of heat transfer.

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.

An air-cooled condenser is used to condense isobutane in a binary geothermal power plant. The isobutane is condensed at \(85^{\circ} \mathrm{C}\) by air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(2.4 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1.25 \mathrm{~m}^{2}\), respectively. The outlet temperature of air is (a) \(45.4^{\circ} \mathrm{C}\) (b) \(40.9^{\circ} \mathrm{C}\) (c) \(37.5^{\circ} \mathrm{C}\) (d) \(34.2^{\circ} \mathrm{C}\) (e) \(31.7^{\circ} \mathrm{C}\)

A heat exchanger is used to condense steam coming off the turbine of a steam power plant by cold water from a nearby lake. The cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the condenser at \(16^{\circ} \mathrm{C}\) at a rate of \(20 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\), while the steam condenses at \(45^{\circ} \mathrm{C}\). The condenser is not insulated, and it is estimated that heat at a rate \(8 \mathrm{~kW}\) is lost from the condenser to the surrounding air. The rate at which the steam condenses is (a) \(0.282 \mathrm{~kg} / \mathrm{s}\) (b) \(0.290 \mathrm{~kg} / \mathrm{s}\) (c) \(0.305 \mathrm{~kg} / \mathrm{s}\) (d) \(0.314 \mathrm{~kg} / \mathrm{s}\) (e) \(0.318 \mathrm{~kg} / \mathrm{s}\)

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