/*! 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 65 Water is used for both fluids (u... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Water is used for both fluids (unmixed) flowing through a single-pass, cross- flow heat exchanger. The hot water enters at \(90^{\circ} \mathrm{C}\) and \(10,000 \mathrm{~kg} / \mathrm{h}\), while the cold water enters at \(10^{\circ} \mathrm{C}\) and \(20,000 \mathrm{~kg} / \mathrm{h}\). If the effectiveness of the exchanger is \(60 \%\), determine the cold water exit temperature.

Short Answer

Expert verified
To determine the cold water exit temperature in the single-pass, cross-flow heat exchanger with the given data, follow these steps: 1. Calculate heat capacity rates for hot (\(C_{h}\)) and cold (\(C_{c}\)) water using \(C = mc_p\). 2. Determine minimum (\(C_{min}\)) and maximum (\(C_{max}\)) heat capacity rates and calculate capacity ratio (\(\textit{CR}\)). 3. Calculate the dimensionless Effectiveness-NTU parameter using the given effectiveness and capacity ratio. 4. Determine the heat transfer (\(Q\)) on both fluid streams using effectiveness, NTU, and minimum heat capacity rate. 5. Calculate the cold water exit temperature (\(T_{c,o}\)) using the formula: \(T_{c,o} = T_{c,i} + \frac{Q}{C_{c}}\). By following these steps and plugging in the provided values, you will find the cold water exit temperature.

Step by step solution

01

Calculate Heat Capacity Rates

The heat capacity rate of a fluid can be calculated using the formula: \(C = mc_p\), where \(m\) is the mass flow rate of the fluid, and \(c_p\) is its specific heat capacity. Assuming constant specific heat capacities for both fluids, we will use the values of specific heat capacities for water at \(c_p=4.18\, kJ/(kg\cdot K)\). Hot water heat capacity rate: \(C_{h} = m_{h}c_{p}\) Cold water heat capacity rate: \(C_{c} = m_{c}c_{p}\)
02

Determine Minimum and Maximum Heat Capacity Rates and Calculate Capacity Ratio

The minimum and maximum heat capacity rates are \(C_{min}\) and \(C_{max}\), respectively. These values are found by comparing the hot and cold water heat capacity rates, and then we will calculate the capacity ratio with the formula: \(\textit{CR} = \frac{C_{min}}{C_{max}}\). Minimum heat capacity rate: \(C_{min} = min(C_{h}, C_{c})\) Maximum heat capacity rate: \(C_{max} = max(C_{h}, C_{c})\) Capacity ratio: \(\textit{CR} = \frac{C_{min}}{C_{max}}\)
03

Calculate Dimensionless Effectiveness-NTU parameter

Since the heat exchanger is a single-pass, cross-flow heat exchanger with unmixed fluids, the dimensionless parameter can be calculated using the given effectiveness and the capacity ratio: \(\textit{NTU} = -\ln \left( \frac{1 - \textit{CR}^{-1}(1 - \\ eff)}{\textit{CR}^{-1} - 1} \right)\)
04

Determine Heat Transfer on Both Fluid Streams

Now we will calculate the heat transfer, \(Q\), on both fluid streams using the effectiveness, NTU, and the minimum heat capacity rate: \(Q = eff \cdot C_{min} \cdot (T_{h,i} - T_{c,i}) \)
05

Calculate the Cold Water Exit Temperature

We can calculate the exit temperature of the cold water, \(T_{c,o}\), using the formula: \(T_{c,o} = T_{c,i} + \frac{Q}{C_{c}} \) This will give us the cold water exit temperature, solving the problem.

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.

Heat Capacity Rate
The heat capacity rate is a crucial concept in thermal engineering that defines the ability of a fluid to absorb or release heat as it flows through a heat exchanger. It is calculated using the relation C = mc_p, where m is the mass flow rate (kg/h) and c_p represents the fluid's specific heat capacity (in kJ/(kgcdot K)). In this context, specific heat capacity is the amount of heat required to raise the temperature of the unit mass of the fluid by one degree Celsius. The heat capacity rate is pivotal in determining how effectively a heat exchanger can alter a fluid's temperature.
For water with a mass flow rate of 10,000 kg/h and a specific heat capacity of 4.18 kJ/(kgcdot K), the heat capacity rate can be calculated, enabling us to further assess the performance of the heat exchanger.
Capacity Ratio CR
The capacity ratio, CR, in heat exchanger design, is a dimensionless number that compares the heat capacity rates of the two fluids involved. It is calculated by dividing the minimum heat capacity rate (C_{min}) by the maximum heat capacity rate (C_{max}), expressed as CR = C_{min} / C_{max}. This ratio is central to understanding which fluid dictates the heat exchange process's capacity, and it influences the design and analysis of a heat exchanger. A lower capacity ratio generally suggests that there is a wide difference between the heat capacity rates, potentially leading to a limited heat transfer effectiveness.
NTU (Number of Transfer Units)
The Number of Transfer Units (NTU) is a dimensionless parameter that measures the size and effectiveness of a heat exchanger relative to the flow rates of the fluids passing through it. It considers the heat exchanger's capacity ratio and effectiveness (which is the ratio of actual heat transfer to the maximum possible heat transfer). The NTU is obtained from the relation that accounts for these factors, particularly in complex scenarios such as a single-pass, cross-flow heat exchanger with unmixed fluids. It is crucial for predicting the heat exchanger's performance and can be used with effectiveness to understand and design thermal systems.
Mass Flow Rate
Mass flow rate (m) is a fundamental concept reflecting the quantity of mass passing through a cross-section of a pipe or channel per unit time. Typically measured in kilograms per hour (kg/h), it is key in calculating the heat capacity rate of a fluid in the heat exchanger. The greater the mass flow rate, the more thermal energy the fluid can carry, affecting the heat exchanger's design and determining the potential for temperature change in the fluid being heated or cooled.
Specific Heat Capacity
Specific heat capacity (c_p) reflects the amount of heat that must be added to or removed from a unit mass of a substance to change its temperature by one degree Celsius. Its unit is kJ/(kgcdot K), indicating energy per mass per temperature change. In the context of our water-based heat exchanger, a high specific heat capacity means the water can absorb or release significant amounts of heat with a comparatively small change in temperature, highlighting the importance of this property in thermal system design and analysis.
Cross-flow Heat Exchanger
A cross-flow heat exchanger is a type of heat exchanger where the two fluids pass perpendicular to each other. In single-pass, unmixed cross-flow heat exchangers, one fluid flows through tubes while the other fluid passes around the tubes without any mixing. This setup greatly affects the heat transfer coefficients and the overall effectiveness of the heat exchanger. Understanding this geometry is essential for accurate heat transfer calculations and for anticipating the thermal performance in industrial applications.
Heat Transfer Calculation
Heat transfer calculation involves determining the amount of thermal energy (Q) exchanged between hot and cold fluids in a heat exchanger. The formula used is typically dependent on the effectiveness of the exchanger, the mass flow rates, specific heat capacities of each fluid, and their inlet temperatures. The calculated heat transfer helps in evaluating the energy balance and is used to find out the exit temperatures of the fluids, ensuring that the heat exchanger meets its intended design specifications.
Fluid Exit Temperature
Fluid exit temperature is the temperature of a fluid as it exits the heat exchanger. In the context of the cold water in our example, knowing the inlet temperature, mass flow rate, specific heat capacity, and the amount of heat absorbed or lost allows us to calculate the exit temperature. This is crucial in applications where specific exit temperatures must be achieved for process control or safety reasons, such as in HVAC systems, chemical plants, and power generation facilities.

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

A single-pass, cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30 to \(80^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225 and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the required surface area.

An energy storage system is proposed to absorb thermal energy collected during the day with a solar collector and release thermal energy at night to heat a building. The key component of the system is a shelland-tube heat exchanger with the shell side filled with \(n\)-octadecane (see Problem 8.47). (a) Warm water from the solar collector is delivered to the heat exchanger at \(T_{h, i}=40^{\circ} \mathrm{C}\) and \(\dot{m}=2 \mathrm{~kg} / \mathrm{s}\) through the tube bundle consisting of 50 tubes, two tube passes, and a tube length per pass of \(L_{l}=2 \mathrm{~m}\). The thin-walled, metal tubes are of diameter \(D=25 \mathrm{~mm}\). Free convection exists within the molten \(n\)-octadecane, providing an average heat transfer coefficient of \(h_{o}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside of each tube. Determine the volume of \(n\) octadecane that is melted over a 12 -h period. If the total volume of \(n\)-octadecane is to be \(50 \%\) greater than the volume melted over \(12 \mathrm{~h}\), determine the diameter of the \(L_{j}=2.2\)-m-long shell. (b) At night, water at \(T_{c, i}=15^{\circ} \mathrm{C}\) is supplied to the heat exchanger, increasing the water temperature and solidifying the \(n\)-octadecane. Do you expect the heat transfer rate to be the same, greater than, or less than the heat transfer rate in part (a)? Explain your reasoning.

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 \(U_{c}\) ?

A boiler used to generate saturated steam is in the form of an unfinned, cross-flow heat exchanger, with water flowing through the tubes and a high- temperature gas in cross flow over the tubes. The gas, which has a specific heat of \(1120 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\), enters the heat exchanger at \(1400 \mathrm{~K}\). The water, which has a flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), enters as saturated liquid at \(450 \mathrm{~K}\) and leaves as saturated vapor at the same temperature. If the overall heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and there are 500 tubes, each of \(0.025-\mathrm{m}\) diameter, what is the required tube length?

The human brain is especially sensitive to elevated temperatures. The cool blood in the veins leaving the face and neck and returning to the heart may contribute to thermal regulation of the brain by cooling the arterial blood flowing to the brain. Consider a vein and artery running between the chest and the base of the skull for a distance \(L=250 \mathrm{~mm}\), with mass flow rates of \(3 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\) in opposite directions in the two vessels. The vessels are of diameter \(D=5 \mathrm{~mm}\) and are separated by a distance \(w=7 \mathrm{~mm}\). The thermal conductivity of the surrounding tissue is \(k_{\mathrm{r}}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the arterial blood enters at \(37^{\circ} \mathrm{C}\) and the venous blood enters at \(27^{\circ} \mathrm{C}\), at what temperature will the arterial blood exit? If the arterial blood becomes overheated, and the body responds by halving the blood flow rate, how much hotter can the entering arterial blood be and still maintain its exit temperature below \(37^{\circ} \mathrm{C}\) ? Hint: If we assume that all the heat leaving the artery enters the vein, then heat transfer between the two vessels can be modeled using a relationship found in Table 4.1. Approximate the blood properties as those of water.

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.