/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 155 A single-pass cross-flow heat ex... [FREE SOLUTION] | 91影视

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A single-pass cross-flow heat exchanger uses hot air (mixed) to heat water (unmixed), flowing with a mass flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), from \(30^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). The hot air enters and exits the heat exchanger at \(220^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the required surface area.

Short Answer

Expert verified
Answer: The required surface area of the single-pass cross-flow heat exchanger is approximately 30.86 m虏.

Step by step solution

01

Calculate the heat transfer rate (Q)

To find the heat transfer rate (Q), we can use the specific heat capacity of water (Cpw = 4.18 kJ/kg.K) and the mass flow rate (m) of water, as well as the temperature difference in water from the inlet to the outlet, as follows: Q = m * Cpw * (T_outlet_water - T_inlet_water) where m = 3 kg/s (mass flow rate of water) Cpw = 4.18 kJ/kg.K (specific heat capacity of water) T_outlet_water = 80掳C T_inlet_water = 30掳C Q = 3 kg/s * 4.18 kJ/kg.K * (80掳C - 30掳C) Converting kJ to J: Q = 3 kg/s * 4180 J/kg.K * (50 K) Q = 627000 W The heat transfer rate at which water is being heated is 627000 W.
02

Calculate the log-mean temperature difference (螖Tm)

Next, we need to find the log-mean temperature difference (螖Tm) which can be used to determine the surface area. The log-mean temperature difference can be determined by using the inlet and outlet temperatures of both the hot air and the water: 螖T鈧 = T_inlet_hotair - T_outlet_water 螖T鈧 = T_outlet_hotair - T_inlet_water To find 螖Tm, use the formula: 螖Tm = (螖T鈧 - 螖T鈧) / ln(螖T鈧 / 螖T鈧) where T_inlet_hotair = 220掳C T_outlet_hotair = 100掳C T_inlet_water = 30掳C T_outlet_water = 80掳C 螖T鈧 = 220掳C - 80掳C = 140 K 螖T鈧 = 100掳C - 30掳C = 70 K 螖Tm = (140 K - 70 K) / ln(140 K / 70 K) 螖Tm 鈮 101.46 K The log-mean temperature difference is approximately 101.46 K.
03

Calculate the required surface area (A)

Now that we have found the heat transfer rate (Q) and the log-mean temperature difference (螖Tm), we can determine the required surface area (A) of the heat exchanger using the formula: Q = U * A * 螖Tm where U = 200 W/m虏.K (overall heat transfer coefficient) Rearranging the formula for A and substituting known values: A = Q / (U * 螖Tm) A = 627000 W / (200 W/m虏.K * 101.46 K) A 鈮 30.86 m虏 The required surface area of the single-pass cross-flow heat exchanger is approximately 30.86 m虏.

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Key Concepts

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

Log-Mean Temperature Difference
The Log-Mean Temperature Difference (LMTD) is an essential concept when analyzing heat exchangers. It serves as the driving force for heat transfer in these systems.
The LMTD accounts for the temperature difference between two fluids at different points within the heat exchanger. This method provides a more accurate measure than simply using the average temperature difference. Given its accuracy, it is widely preferred in heat exchanger calculations.
To calculate the LMTD, you need the temperature differences at the entry and exit points of the fluids. Let's break this down:
  • Calculate the temperature difference at the inlet and outlet of the hot and cold fluids.
  • Use these differences to calculate the LMTD with the formula: \[ \Delta T_{m} = \frac{(\Delta T_{1} - \Delta T_{2})}{\ln(\Delta T_{1}/\Delta T_{2})} \] where \( \Delta T_{1} \) and \( \Delta T_{2} \) are the temperature differences at the start and end of the heat exchange process.
Understanding LMTD simplifies complex temperature variations and helps predict the efficiency of the heat exchanger effectively.
Heat Transfer Rate Calculation
The Heat Transfer Rate is pivotal in determining how efficient a heat exchanger is. This rate helps determine how much energy is being transferred between fluids.
It is quantified by the formula:
  • First, identify the mass flow rate ( \( m \) ) and specific heat capacity (\( C_{p} \)) of the fluid being heated or cooled.
  • Calculate the temperature difference of the fluid across the heat exchanger.
  • The heat transfer rate ( \( Q \) ) is given by: \[ Q = m \cdot C_{p} \cdot \Delta T \] where \( \Delta T \) is the temperature change from inlet to outlet.
For example, in the given exercise, this translates to using the mass flow rate and specific heat capacity of water to calculate the required energy transfer. This rate tells us how much energy is being transferred at any given moment and can provide insights into whether the system meets the desired heat transfer performance.
Overall Heat Transfer Coefficient
The Overall Heat Transfer Coefficient (\( U \)) represents the heat transfer capability of the entire heat exchanger surface. It's a collective measure of how well the heat exchanger materials transfer heat between fluids.
Several factors can affect this coefficient:
  • The type and thickness of the heat exchanger wall materials.
  • The cleanliness or fouling of the heat exchange surfaces.
  • The convection heat transfer coefficients on both sides of the exchanger.
Combine these to find the overall efficiency of heat transfer within the system using this coefficient.
The coefficient is key in the formula:
\[ Q = U \cdot A \cdot \Delta T_{m}\] Where:
  • \( Q \) is the heat transfer rate, and
  • \( A \) is the heat exchange surface area.
Efficient heat exchangers aim to maximize \( U \), as this leads to better heat transfer with less surface area, ultimately reducing costs and energy consumption.

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Most popular questions from this chapter

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the \(\varepsilon-\mathrm{NTU}\) method.

The condenser of a room air conditioner is designed to reject heat at a rate of \(15,000 \mathrm{~kJ} / \mathrm{h}\) from refrigerant-134a as the refrigerant is condensed at a temperature of \(40^{\circ} \mathrm{C}\). Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows across the finned condenser coils, entering at \(25^{\circ} \mathrm{C}\) and leaving at \(35^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the refrigerant side is \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer area on the refrigerant side.

There are two heat exchangers that can meet the heat transfer requirements of a facility. One is smaller and cheaper but requires a larger pump, while the other is larger and more expensive but has a smaller pressure drop and thus requires a smaller pump. Both heat exchangers have the same life expectancy and meet all other requirements. Explain which heat exchanger you would choose and under what conditions.

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 \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2} .\)

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

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