/*! 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 26 A double-pipe heat exchanger is ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 26

A double-pipe heat exchanger is constructed of a copper \((k=380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) inner tube of internal diameter \(D_{i}=\) \(1.2 \mathrm{~cm}\) and external diameter \(D_{o}=1.6 \mathrm{~cm}\) and an outer tube of diameter \(3.0 \mathrm{~cm}\). The convection heat transfer coefficient is reported to be \(h_{i}=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the inner surface of the tube and \(h_{o}=1400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on its outer surface. For a fouling factor \(R_{f, i}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) on the tube side and \(R_{f, o}=\) \(0.0002 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) on the shell side, determine \((a)\) the thermal resistance of the heat exchanger per unit length and \((b)\) the overall heat transfer coefficients \(U_{i}\) and \(U_{o}\) based on the inner and outer surface areas of the tube, respectively.

Short Answer

Expert verified
Question: Calculate the thermal resistance of the heat exchanger per unit length and the overall heat transfer coefficients based on the inner and outer surface areas, given the convection coefficients, fouling factors, and tube dimensions. Answer: To calculate the thermal resistance of the heat exchanger per unit length (R_tot) and the overall heat transfer coefficients (U_i and U_o), follow these steps: 1. Calculate the tube wall's thermal resistance per unit length (R_t) using the formula: \[ R_{t} = \frac{\ln\frac{D_{o}}{D_{i}}}{2 \pi k} \] 2. Calculate inner and outer convective resistance, considering fouling factors (R_i and R_o) using the formulas: \[ R_{i} = \frac{1}{h_{i} \pi D_{i}} + R_{f, i} \] and \[ R_{o} = \frac{1}{h_{o} \pi D_{o}} + R_{f, o} \] 3. Determine the total thermal resistance per unit length (R_tot) by adding up all the resistances: \[ R_{tot} = R_{i} + R_{t} + R_{o} \] 4. Calculate the overall heat transfer coefficients based on the inner and outer surface areas (U_i and U_o) using the formulas: \[ U_{i} = \frac{1}{R_{tot}} \pi D_{i} \] and \[ U_{o} = \frac{1}{R_{tot}} \pi D_{o} \ ] Plug in the given values and perform the calculations to obtain the thermal resistance of the heat exchanger per unit length (R_tot), and the overall heat transfer coefficients (U_i and U_o).

Step by step solution

01

Calculate the tube wall's thermal resistance per unit length

Finding the thermal resistance in the tube wall between the inner and outer surfaces: $$R_{t} = \frac{\ln\frac{D_{o}}{D_{i}}}{2 \pi k}$$ Where \(R_t\) is the tube wall's thermal resistance, \(k\) is the copper thermal conductivity, \(D_i\) is the inner diameter, and \(D_o\) is the outer diameter.
02

Calculate inner and outer convective resistance, considering fouling factors

We are given the convection coefficients (\(h_{i}\) and \(h_{o}\)) and fouling factors (\(R_{f, i}\) and \(R_{f, o}\)) for the heat exchanger. To account for fouling, we need to calculate the thermal resistance for both the inner and outer surfaces. For the inner surface: $$R_{i} = \frac{1}{h_{i} \pi D_{i}} + R_{f, i}$$ For the outer surface: $$R_{o} = \frac{1}{h_{o} \pi D_{o}} + R_{f, o}$$
03

Determine the total thermal resistance per unit length

We will add up all the resistances: $$R_{tot} = R_{i} + R_{t} + R_{o}$$
04

Calculate the overall heat transfer coefficients based on inner and outer surface areas

We can find the overall heat transfer coefficients, \(U_i\) and \(U_o\), based on the inner and outer surface areas: For the inner surface: $$U_{i} = \frac{1}{R_{tot}} \pi D_{i}$$ For the outer surface: $$U_{o} = \frac{1}{R_{tot}} \pi D_{o}$$ Now we can plug in the given values and perform the calculations to obtain the thermal resistance of the heat exchanger per unit length \((R_{tot})\), and the overall heat transfer coefficients \((U_i\) and \(U_o)\).

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Key Concepts

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

Thermal Resistance
In a heat exchanger, thermal resistance is quite similar to electrical resistance, but in the context of heat flow instead of electricity. It represents how difficult it is for heat to flow through a material or interface. This concept is vital when designing and analyzing heat exchangers since it affects how efficiently heat can be transferred from one medium to another.
For our copper heat exchanger, thermal resistance refers to the resistance to heat flow through the tube wall. It depends on several factors including material properties and the geometry of the tube. In this case, we use the formula \( R_t = \frac{\ln\frac{D_{o}}{D_{i}}}{2 \pi k} \). Here, \( D_{i} \) and \( D_{o} \) are the inner and outer diameters of the tube, and \( k \) is the thermal conductivity of copper.
Calculating thermal resistance is crucial for determining how well a heat exchanger functions. Lower thermal resistance indicates more efficient heat transfer, allowing heat energy to pass through quickly and easily.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient is a measure of a heat exchanger's performance. It consolidates all the resistances to heat transfer鈥攃onductive through the wall, convective from fluid, and fouling鈥攊nto a single composite value.
When designing or analyzing a heat exchanger, determining the values of \( U_i \) and \( U_o \) helps in understanding the rate of heat transfer on both the inner and outer surfaces of the tube.
  • Inner surface coefficient, \( U_i \), is calculated as \( U_{i} = \frac{1}{R_{tot}} \pi D_{i} \)
  • Outer surface coefficient, \( U_o \), is calculated as \( U_{o} = \frac{1}{R_{tot}} \pi D_{o} \)
Both coefficients essentially tell us how efficiently the heat exchanger is operating by relating the total thermal resistance to the geometry of the heat exchanging surfaces.
Fouling Factor
The fouling factor is the extra resistance to heat transfer caused by deposits or "fouling" on the heat exchanger surfaces, either from the fluids being heated/cooled or other substances. Over time, these deposits accumulate, reducing heat exchanger efficiency by adding resistance.
The fouling factors \( R_{f, i} \) and \( R_{f, o} \) represent this additional thermal resistance on the inner and outer surfaces respectively. In our calculation:
  • Inner fouling resistance is added as \( R_{f, i} \) in \( R_{i} = \frac{1}{h_{i} \pi D_{i}} + R_{f, i} \)
  • Outer fouling resistance is added as \( R_{f, o} \) in \( R_{o} = \frac{1}{h_{o} \pi D_{o}} + R_{f, o} \)
Managing fouling is key for maintaining heat exchanger performance and efficiency over its operational life. Regular cleaning, choosing appropriate materials, and operating conditions help in mitigating fouling impact.

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Most popular questions from this chapter

Consider a recuperative cross flow heat exchanger (both fluids unmixed) used in a gas turbine system that carries the exhaust gases at a flow rate of \(7.5 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(500^{\circ} \mathrm{C}\). The air initially at \(30^{\circ} \mathrm{C}\) and flowing at a rate of \(15 \mathrm{~kg} / \mathrm{s}\) is to be heated in the recuperator. The convective heat transfer coefficients on the exhaust gas and air side are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Due to long term use of the gas turbine the recuperative heat exchanger is subject to fouling on both gas and air side that offers a resistance of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) each. Take the properties of exhaust gas to be the same as that of air \(\left(c_{p}=1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). If the exit temperature of the exhaust gas is \(320^{\circ} \mathrm{C}\) determine \((a)\) if the air could be heated to a temperature of \(150^{\circ} \mathrm{C}(b)\) the area of heat exchanger \((c)\) if the answer to part (a) is no, then determine what should be the air mass flow rate in order to attain the desired exit temperature of \(150^{\circ} \mathrm{C}\) and \((d)\) plot variation of the exit air temperature over a temperature range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\) with air mass flow rate assuming all the other conditions remain the same.

Under what conditions can a counter-flow heat exchanger have an effectiveness of one? What would your answer be for a parallel-flow heat exchanger?

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with ethylene glycol \(\left(c_{p}=2680 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Water enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(70^{\circ} \mathrm{C}\). Ethylene \(\mathrm{glycol}\) enters the shell at \(110^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the tube side is \(280 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area on the tube side.

Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hotand cold-water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same.

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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