/*! 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

Hot water at \(60^{\circ} \mathrm{C}\) is cooled to \(36^{\circ} \mathrm{C}\) through the tube side of a 1-shell pass and 2-tube passes heat exchanger. The coolant is also a water stream, for which the inlet and outlet temperatures are \(7^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The overall heat transfer coefficient and the heat transfer area are \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(15 \mathrm{~m}^{2}\), respectively. Calculate the mass flow rates of hot and cold water streams in steady operation.

In a 1-shell and 2-tube heat exchanger, cold water with inlet temperature of \(20^{\circ} \mathrm{C}\) is heated by hot water supplied at the inlet at \(80^{\circ} \mathrm{C}\). The cold and hot water flow rates are \(5000 \mathrm{~kg} / \mathrm{h}\) and \(10,000 \mathrm{~kg} / \mathrm{h}\), respectively. If the shelland-tube heat exchanger has a \(U A_{s}\) value of \(11,600 \mathrm{~W} / \mathrm{K}\), determine the cold water and hot water outlet temperatures. Assume \(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.

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.

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.
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