/*! 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 53 A counter-flow heat exchanger is... [FREE SOLUTION] | 91Ó°ÊÓ

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A counter-flow heat exchanger is stated to have an overall heat transfer coefficient of \(284 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) when operating at design and clean conditions. Hot fluid enters the tube side at \(93^{\circ} \mathrm{C}\) and exits at \(71^{\circ} \mathrm{C}\), while cold fluid enters the shell side at \(27^{\circ} \mathrm{C}\) and exits at \(38^{\circ} \mathrm{C}\). After a period of use, built-up scale in the heat exchanger gives a fouling factor of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If the surface area is \(93 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer in the heat exchanger and \((b)\) the mass flow rates of both hot and cold fluids. Assume both hot and cold fluids have a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
Question: Determine the rate of heat transfer (Q) in clean and fouled conditions, and calculate the mass flow rates of the hot fluid (m_h) and cold fluid (m_c) using the given information. Given information: - Overall heat transfer coefficient (clean) = 284 W/m²K - Fouling factor = 0.0004 m²K/W - Input/output temperatures of hot fluid (Th1, Th2) = 100°C, 70°C - Input/output temperatures of cold fluid (Tc1, Tc2) = 20°C, 40°C - Specific heat capacity for both fluids = 4200 J/kgK

Step by step solution

01

1. Calculate initial overall heat transfer coefficient (U_fouled)

To find the overall heat transfer coefficient when the heat exchanger has developed some fouling, we can use the following equation: $$U_fouled = \frac{1}{(1/U_{clean}) + R_{fouling}}$$ where \(U_{fouled}\) is the overall heat transfer coefficient in fouled conditions, \(U_{clean}\) is the overall heat transfer coefficient in clean conditions (given as 284 W/m²K), and \(R_{fouling}\) is the fouling factor (given as 0.0004 m²K/W). Plugging in given values, we can compute the \(U_{fouled}\).
02

2. Calculate the LMTD (Log Mean Temperature Difference)

To determine the LMTD of the temperature profile in the heat exchanger, we'll use the expression: $$LMTD = \frac{(T_{h1} -T_{c2}) - (T_{h2} - T_{c1})}{\ln[(T_{h1} - T_{c2}) / (T_{h2} - T_{c1})]}$$ where \(T_{h1}\) is the temperature of the hot fluid entering, \(T_{h2}\) is the temperature of the hot fluid leaving, \(T_{c1}\) is the cold fluid entering, and \(T_{c2}\) is the cold fluid leaving. Plugging in the given values, we can compute the LMTD.
03

3. Calculate the rate of heat transfer (Q)

Now that we have the LMTD and the overall heat transfer coefficients (both clean and fouled), we can calculate the rate of heat transfer using the equation: $$Q = U \cdot A \cdot LMTD$$ where \(Q\) represents the rate of heat transfer, \(U\) is the overall heat transfer coefficient (both clean and fouled), \(A\) is the surface area of the heat exchanger, and \(LMTD\) is the log mean temperature difference. We can compute the rate of heat transfer under both clean and fouled conditions.
04

4. Calculate the mass flow rate of hot and cold fluids (m_h, m_c)

To find the mass flow rates of the hot and cold fluids, we can use the equation \(\Delta T = (Q / mC_p)\) for each fluid, and rearrange the equation as: \(m = Q / (C_p\Delta T)\). The specific heat capacity (\(C_p\)) is given as 4200 J/kgK. For the hot fluid: $$m_h = \frac{Q}{C_H (T_{h1} - T_{h2})}$$ and for the cold fluid: $$m_c = \frac{Q}{C_C (T_{c2} - T_{c1})}$$. Solving for both mass flow rates, we can find \(m_h\) and \(m_c\).

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Key Concepts

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

Overall Heat Transfer Coefficient
The overall heat transfer coefficient is a crucial factor in heat exchanger performance. It denotes the efficiency with which heat is transferred between fluids. It is symbolized by \(U\). This coefficient depends on factors such as the material's thermal conductivity, the fluid velocities, and the surface conditions of the heat exchanger.
During clean conditions, the coefficient has a higher value but as fouling develops, its value diminishes due to added thermal resistance. The formula to adjust for fouling is: \[U_{fouled} = \frac{1}{(1/U_{clean}) + R_{fouling}}\] Here, \(U_{clean}\) is the clean surface coefficient, and \(R_{fouling}\) is the fouling factor. Fouling mildly reduces \(U\), hence lowering effectiveness.
Fouling Factor
The fouling factor represents the resistance to heat transfer caused by deposits on heat exchanger surfaces. This occurs over time, especially in environments prone to scaling or muck formation. As fouling increases, the capacity of the heat exchanger to pass heat decreases, requiring more energy to accomplish the same level of heat transfer.
In calculations, it adds to the resistance in the heat transfer equation, thus lowering the effective heat transfer coefficient \(U\). It is a vital consideration in maintenance to ensure longevity and efficiency of heat exchangers.
Log Mean Temperature Difference
The Log Mean Temperature Difference (LMTD) is pivotal in quantifying the temperature driving force within a heat exchanger. Unlike simple averages, it considers the non-linear temperature gradient between the fluid streams. Calculating LMTD allows for more accurate determination of the heat exchanger’s performance: \[LMTD = \frac{(T_{h1} - T_{c2}) - (T_{h2} - T_{c1})}{\ln\left(\frac{T_{h1} - T_{c2}}{T_{h2} - T_{c1}}\right)}\] Where \(T_{h1}, T_{h2}, T_{c1},\) and \(T_{c2}\) denote inlet and outlet temperatures of the respective fluids. Accurate LMTD ensures precise energy transfer calculation.
Mass Flow Rate
The mass flow rate is an essential parameter indicating the amount of fluid mass moving through the heat exchanger per unit time. It affects the heat transfer rate significantly, given more mass can convey more heat energy. To determine the mass flow rates of hot \((m_h)\) and cold \((m_c)\) fluids, we'll use the energy balance: \[\Delta T = \frac{Q}{mC_p}\] Rearranging for \(m\) yields: \[m = \frac{Q}{C_p\Delta T}\] Where \(C_p\) is the specific heat. These calculations give insights into system dynamics and energy efficiency.
Specific Heat
Specific heat, denoted by \(C_p\), is an intrinsic material property that indicates how much energy a substance can store per unit mass and temperature change. It plays a vital role in heat exchanger calculations since it affects the heat transfer rate by influencing the \(\Delta T\) in the flow rate calculations: The specific heat value for both fluids in our exercise is given as \(4.2 \, \text{kJ/kg} \cdot \text{K}\). This knowledge helps engineers calculate energy needs and balance the exchanger's systems efficiently.

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Most popular questions from this chapter

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\). The rate of heat transfer between the two streams is (a) \(8.2 \mathrm{~kW}\) (b) \(23.7 \mathrm{~kW}\) (c) \(33.8 \mathrm{~kW}\) (d) \(44.8 \mathrm{~kW}\) (e) \(52.8 \mathrm{~kW}\)

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

Saturated liquid benzene flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(75^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) by using a source of cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(3.5 \mathrm{~kg} / \mathrm{s}\) and \(15^{\circ} \mathrm{C}\) through a \(20-\mathrm{mm}-\) diameter tube of negligible wall thickness. The overall heat transfer coefficient of the heat exchanger is estimated to be \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the specific heat of the liquid benzene is \(1839 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and assuming that the capacity ratio and effectiveness remain the same, determine the heat exchanger surface area for the following four heat exchangers: \((a)\) parallel flow, \((b)\) counter flow, \((c)\) shelland-tube heat exchanger with 2 -shell passes and 40-tube passes, and \((d)\) cross-flow heat exchanger with one fluid mixed (liquid benzene) and other fluid unmixed (water).

Hot oil \(\left(c_{p}=2.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(110^{\circ} \mathrm{C}\) and \(8 \mathrm{~kg} / \mathrm{s}\) is to be cooled in a heat exchanger by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(10^{\circ} \mathrm{C}\) and at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). The lowest temperature that oil can be cooled in this heat exchanger is (a) \(10.0^{\circ} \mathrm{C}\) (b) \(33.5^{\circ} \mathrm{C}\) (c) \(46.1^{\circ} \mathrm{C}\) (d) \(60.2^{\circ} \mathrm{C}\) (e) \(71.4^{\circ} \mathrm{C}\)

A performance test is being conducted on a double pipe counter flow 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 over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected 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?

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