/*! 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 79 Consider a concentric tube heat ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a concentric tube heat exchanger characterized by a uniform overall heat transfer coefficient and operating under the following conditions: \begin{tabular}{lccrc} \hline & \(\dot{m}\) \((\mathbf{k g} / \mathbf{s})\) & \(c_{p}\) \((\mathbf{J} / \mathbf{k g} \cdot \mathbf{K})\) & \(T_{i}\) \((\boldsymbol{C})\) & \(T_{o}\) \((\mathbf{C})\) \\ \hline Cold fluid & \(0.125\) & 4200 & 40 & 95 \\ Hot fluid & \(0.125\) & 2100 & 210 & \(-\) \\ \hline \end{tabular} What is the maximum possible heat transfer rate? What is the heat exchanger effectiveness? Should the heat exchanger be operated in parallel flow or in counterflow? What is the ratio of the required areas for these two flow conditions?

Short Answer

Expert verified
The maximum possible heat transfer rate for the given concentric tube heat exchanger is \(Q_{max} = 44550 \ \, W\). The heat exchanger effectiveness is 0.648. Based on the given data, we cannot determine if the heat exchanger should be operated in parallel flow or counterflow. The ratio of required areas for parallel flow and counterflow arrangements is approximately 1.364.

Step by step solution

01

Find the Maximum Possible Heat Transfer Rate (Q_max)

We can determine the maximum heat transfer rate by using the equation: \(Q_{max} = C_{min}\)(\(T_{h,i} - T_{c,i}\), where \(C_{min}\) represents the minimum capacity rate of the two fluids, and \(T_{h,i}\) and \(T_{c,i}\) are the initial temperatures of the hot and cold fluids respectively. Calculate the capacity rates, \(C_h\) and \(C_c\), of the hot and cold fluids: \( C_h = m_h c_{p,h} \) \( C_c = m_c c_{p,c} \) Plug in the given values: \( C_h = (0.125 \ \, kg/s)(2100 \ \, J/kg \cdot K) = 262.5 \ \, W/K \) \( C_c = (0.125 \ \, kg/s)(4200 \ \, J/kg \cdot K) = 525 \ \, W/K \) Since \(C_h < C_c\), the minimum capacity rate \(C_{min} = C_h = 262.5 \ \, W/K\). Now, calculate the maximum heat transfer rate: \(Q_{max} = C_{min}(T_{h,i} - T_{c,i}) = 262.5 (210 - 40) = 44550 \ \, W\)
02

Determine the Heat Exchanger Effectiveness

Find the actual heat transfer rate, Q_actual, using the equation: \(Q_{actual} = C_c(T_{c,o} - T_{c,i})\) Plug in the given values for the cold fluid: \(Q_{actual} = 525(95 - 40) = 525 \cdot 55 = 28875 \ \, W\) With the actual heat transfer rate, we can now calculate the heat exchanger effectiveness as the ratio of the actual heat transfer rate to the maximum heat transfer rate: \(Effectiveness = \frac{Q_{actual}}{Q_{max}} = \frac{28875}{44550} = 0.648 \)
03

Determine the Optimal Flow Arrangement

Since the specific heat capacities and mass flow rates of both fluids are fixed, the heat exchanger effectiveness does not provide any information about the flow arrangement. Therefore, we cannot determine if the heat exchanger should be operated in parallel flow or counterflow based on the given data.
04

Calculate the Ratio of Required Areas

To find the ratio of the required areas for parallel flow and counterflow arrangements, we can use the following formula derived from the heat exchanger effectiveness: \( \frac{A_{\mathrm{parallel}}}{A_{\mathrm{counterflow}}} = \frac{\ln(\frac{1 - \mathrm{Effectiveness}}{1 - C_{\mathrm{ratio}} \cdot \mathrm{Effectiveness}})}{\ln(\frac{1 - \mathrm{Effectiveness} \cdot C_{\mathrm{ratio}}}{1 - C_{\mathrm{ratio}}})} \) Where \(C_{\mathrm{ratio}} = \frac{C_{\mathrm{min}}}{C_{\mathrm{max}}}\) Calculate \(C_{\mathrm{ratio}}\): \(C_{\mathrm{ratio}} = \frac{C_{\mathrm{min}}}{C_{\mathrm{max}}} = \frac{262.5}{525} = 0.5 \) Now, calculate the ratio of the required areas: \( \frac{A_{\mathrm{parallel}}}{A_{\mathrm{counterflow}}} = \frac{\ln(\frac{1-0.648}{1-0.5\cdot0.648})}{\ln(\frac{1-0.648\cdot0.5}{1-0.5})} = \frac{\ln\(0.352)}{\ln\(0.676)} = 1.364 \) Therefore, the ratio of required areas for parallel flow and counterflow is approximately 1.364.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Counterflow
In heat exchangers, **counterflow** is considered one of the most efficient flow arrangements. In a counterflow heat exchanger, the two fluids flow in opposite directions. One fluid enters the system on one side, while the second fluid enters from the opposite side. This setup allows the temperature gradient between the fluids to remain relatively high across the entire length of the exchanger.

The advantage of counterflow is that it maximizes the temperature difference between the fluids. As a result, heat transfer can occur more efficiently compared to other arrangements, like parallel flow. This means it can transfer more heat with a smaller surface area.

In practical terms, this means a counterflow heat exchanger can often accomplish the same degree of thermal exchange with less external resources. It's a preferred method in situations where space and cost constraints are significant considerations.
Heat Transfer Rate
The **heat transfer rate** is a measure of the energy transferred from one fluid to another, typically expressed in watts (W). It is an essential factor in determining the efficacy of a heat exchanger.

To compute this rate, you need to know several variables:
  • The mass flow rate of the fluid (\( \dot{m} \)).
  • The specific heat capacity (\( c_p \)) which describes how much heat the fluid can hold.
  • Temperature differences across the system.
For example, the rate can be calculated using the formula: \( Q = \dot{m} \times c_p \times \Delta T \), where \( \Delta T \)is the temperature change that occurs in the fluid as it passes through the exchanger.

Understanding the heat transfer rate is crucial for designing an efficient system. It helps engineers determine the amount of energy being transferred and optimize the system's performance by adjusting flow rates and temperatures.
Effectiveness
**Effectiveness** in the context of heat exchangers measures how well a heat exchanger performs compared to its maximum potential. It is expressed as a ratio or percentage, showing the fraction of maximum possible energy transfer achieved by the system.

The formula for effectiveness is:\( Effectiveness = \frac{Q_{actual}}{Q_{max}} \),where\( Q_{actual} \)is the actual heat transfer rate and\( Q_{max} \)is the maximum possible heat transfer rate based on the minimum capacity rate of the fluids involved.

Higher effectiveness indicates a more efficient heat exchanger. This measure is vital because it provides insight into how well the system is utilizing its potential capacity. However, keep in mind that even with a high effectiveness, operational and environmental factors can still affect overall system performance. Effectiveness does not provide a complete picture of heat exchanger efficiencies regarding flow arrangements like parallel or counterflow.
Capacity Rate
The **capacity rate** of a fluid in a heat exchanger is a crucial concept used to understand its thermal performance. It is the product of the fluid's mass flow rate and its specific heat capacity. Mathematically, it is represented as:
\( C = \dot{m} \times c_p \), where
  • \(\dot{m}\) is the mass flow rate.
  • \(c_p\)is the specific heat capacity.
This rate measures the heat-carrying capability of the fluid.

The smallest of the capacity rates between the two fluids in a heat exchanger becomes the limiting factor for heat exchange and determines the system's \(C_{min}\). This affects the maximum potential heat transfer. Understanding capacity rates helps design efficient heat exchange systems by ensuring the flow rates and specific heats are matched to maximize energy transfer.
Parallel Flow
**Parallel flow** is a configuration in a heat exchanger where both fluids enter the system from the same end and move parallel to each other. As both fluids travel in the same direction, the temperature gradient between them diminishes along the length of the exchanger.

Although easy to design and implement, parallel flow often results in a less effective heat transfer process compared to counterflow. This is because the temperature difference, which drives the heat transfer, decreases as the fluids progress through the exchanger.

A significant downside of the parallel flow arrangement is that it can never achieve complete thermal equilibrium between the fluids. This results in less overall heat being transferred and typically requires a larger area to accomplish the same amount of thermal exchange as a counterflow setup. In conclusion, while simpler, parallel flow is less favorable for applications requiring high thermal efficiency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Cooling of outdoor electronic equipment such as in telecommunications towers is difficult due to seasonal and diurnal variations of the air temperature, and potential fouling of heat exchange surfaces due to dust accumulation or insect nesting. A concept to provide a nearly constant sink temperature in a hermetically sealed environment is shown below. The cool surface is maintained at nearly constant groundwater temperature \(\left(T_{1}=5^{\circ} \mathrm{C}\right)\) while the hot surface is subjected to a constant heat load from the electronic equipment \(\left(q_{2}=50 \mathrm{~W}, T_{2}\right)\). Connecting the surfaces is a concentric tube of length \(L=10 \mathrm{~m}\) with \(D_{i}=100 \mathrm{~mm}\) and \(D_{o}=150 \mathrm{~mm}\). A fan moves air at a mass flow rate of \(m=0.0325 \mathrm{~kg} / \mathrm{s}\) and dissipates \(P=10 \mathrm{~W}\) of thermal energy. Heat transfer to the cool surface is described by \(q_{1}^{N}=\bar{h}_{1}\left(T_{h_{1} o}-T_{1}\right)\) while heat transfer from the hot surface is described by \(q_{2}^{\prime \prime}=\bar{h}_{2}\left(T_{2}-T_{f_{0}}\right)\) where \(T_{f_{0}}\) is the fan outlet temperature. The values of \(\bar{h}_{1}\) and \(h_{2}\) are 40 and \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. To isolate the electronics from ambient temperature variations, the entire device is insulated at its outer surfaces. The design engineer is concerned that conduction through the wall of the inner tube may adversely affect the device performance. Determine the value of \(T_{2}\) for the limiting cases of (i) no conduction resistance in the inner tube wall and (ii) infinite conduction resistance in the inner tube wall. Does the proposed device maintain maximum temperatures below \(80^{\circ} \mathrm{C}\) ?

Water at a rate of \(45,500 \mathrm{~kg} / \mathrm{h}\) is heated from 80 to \(150^{\circ} \mathrm{C}\) in a heat exchanger having two shell passes and eight tube passes with a total surface area of \(925 \mathrm{~m}^{2}\). Hot exhaust gases having approximately the same thermophysical properties as air enter at \(350^{\circ} \mathrm{C}\) and exit at \(175^{\circ} \mathrm{C}\). Determine the overall heat transfer coefficient.

Consider Problem 11.36. (a) For \(\dot{m}_{c \mathrm{CA}}=\dot{m}_{\mathrm{h}, \mathrm{B}}=10 \mathrm{~kg} / \mathrm{s}\), determine the outlet air and ammonia temperatures, as well as the heat transfer rate. (b) Plot the outlet air and outlet ammonia temperatures versus the water flow rate over the range \(5 \mathrm{~kg} / \mathrm{s} \leq \dot{m}_{c, \mathrm{~A}}=m_{h, \mathrm{~B}} \leq 50 \mathrm{~kg} / \mathrm{s}\).

11.3 A shell-and-tube heat exchanger is to heat an acidic liquid that flows in unfinned tubes of inside and outside diameters \(D_{i}=10 \mathrm{~mm}\) and \(D_{\mathrm{o}}=11 \mathrm{~mm}\), respectively. A hot gas flows on the shell side. To avoid corrosion of the tube material, the engineer may specify either a Ni-Cr-Mo corrosion-resistant metal alloy \(\left(\rho_{m}=8900 \mathrm{~kg} / \mathrm{m}^{3}, k_{\mathrm{w}}=8\right.\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) or a polyvinylidene fluoride (PVDF) plastic \(\left(\rho_{p}=1780 \mathrm{~kg} / \mathrm{m}^{3}, k_{p}=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The inner and outer heat transfer coefficients are \(h_{j}=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{v}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) Determine the ratio of plastic to metal tube surface areas needed to transfer the same amount of heat. (b) Determine the ratio of plastic to metal mass associated with the two heat exchanger designs. (c) The cost of the metal alloy per unit mass is three times that of the plastic. Determine which tube material should be specified on the basis of cost. 11.4 A steel tube \((k=50 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) of inner and outer diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{o}=26 \mathrm{~mm}\), respectively, is used to transfer heat from hot gases flowing over the tube \(\left(h_{\mathrm{h}}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) to cold water flowing through the tube \(\left(h_{c}=8000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\). What is the cold-side overall heat transfer coefficient \(U_{c}\) ? To enhance heat transfer, 16 straight fins of rectangular profile are installed longitudinally along the outer surface of the tube. The fins are equally spaced around the circumference of the tube, each having a thickness of \(2 \mathrm{~mm}\) and a length of \(15 \mathrm{~mm}\). What is the corresponding overall heat transfer coefficient \(U_{c}\) ?

The hot and cold inlet temperatures to a concentric tube heat exchanger are \(T_{h i}=200^{\circ} \mathrm{C}, T_{c, i}=100^{\circ} \mathrm{C}\), respectively. The outlet temperatures are \(T_{k, o}=110^{\circ} \mathrm{C}\) and \(T_{\omega_{0}}=125^{\circ} \mathrm{C}\). Is the heat exchanger operating in a parallel flow or in a counterflow configuration? What is the heat exchanger effectiveness? What is the NTU? Phase change does not occur in either fluid.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.