/*! 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 55 A single-pass, cross-flow heat e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A single-pass, cross-flow heat exchanger with both fluids unmixed is being used to heat water \(\left(m_{c}=2 \mathrm{~kg} / \mathrm{s}\right.\), \(c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) with hot exhaust gases \(\left(c_{p}=1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(320^{\circ} \mathrm{C}\). What mass flow rate of exhaust gases is required? Assume that UA is equal to its design value of \(4700 \mathrm{~W} / \mathrm{K}\), independent of the gas mass flow rate.

Short Answer

Expert verified
The required mass flow rate of exhaust gases to heat the water from 20°C to 100°C is approximately 2.04 kg/s. This is determined by first calculating the heat transfer rate for water and then using the log mean temperature difference (LMTD) method for the single-pass, cross-flow heat exchanger to determine the temperature change of exhaust gases. Once the temperature change is known, the mass flow rate can be calculated.

Step by step solution

01

Calculate the temperature change for the water and exhaust gases

In this step, we will calculate the temperature change for the water and exhaust gases, respectively. This information will help us determine the amount of heat that needs to be transferred between the fluids. Temperature change for water, ΔT_c = T_c2 - T_c1 = 100°C - 20°C = 80°C where: T_c1 = initial temperature of water (20°C) T_c2 = final temperature of water (100°C) Temperature change for exhaust gases will be determined later when we have the required mass flow rate.
02

Calculate the heat capacity rate for the water

Calculate the heat capacity rate for the water, which is the product of mass flow rate and specific heat capacity. C_c = m_c * c_p = 2 kg/s * 4200 J/(kg·K) = 8400 J/(s·K)
03

Calculate the heat transfer rate for the water

Calculate the heat transfer rate for water, which is the product of heat capacity rate and temperature change. Q_c = C_c * ΔT_c = 8400 J/(s·K) * 80 K Q_c = 672000 J/s or 672 kW
04

Set up an equation for the heat capacity rate of exhaust gases

The heat capacity rate for the exhaust gases is given by: C_h = m_h * c_{ph} where: m_h = mass flow rate of exhaust gases (to be determined) c_{ph} = specific heat capacity of exhaust gases (1200 J/(kg·K))
05

Set up an equation for the heat transfer rate of exhaust gases

Since heat transfer is equal between both fluids, we can equate the heat transfer rate of exhaust gases (Q_h) to the heat transfer rate of the water (Q_c). Q_h = C_h * ΔT_h = Q_c where: ΔT_h = temperature change of exhaust gases
06

Determine the temperature change of exhaust gases

Using the information given, we can write the equation for exhaust gas temperature change as follows: ΔT_h = T_h1 - T_h2 Let's express T_h2 in terms of T_h1: T_h2 = T_h1 - ΔT_h Now, apply the log mean temperature difference (LMTD) method for the single-pass, cross-flow heat exchanger: LMTD = (ΔT_1 - ΔT_2) / ln(ΔT_1/ΔT_2) where: ΔT_1 = T_h1 - T_c1 ΔT_2 = T_h2 - T_c2 Insert LMTD equation into the overall heat transfer equation: Q = UA * LMTD 672 kW = 4700 W/K * ((T_h1 - 20) - (T_h1 - ΔT_h - 100)) / ln((T_h1 - 20)/(T_h1 - ΔT_h - 100)) Since T_h1 = 320°C for exhaust gases, rewrite the equation as: 672 kW = 4700 W/K * (300 - ΔT_h) / ln(300/ΔT_h)
07

Calculate the required mass flow rate of exhaust gases

Rearrange the heat transfer equation from Step 5: m_h = Q_c / c_p,h / ΔT_h When ΔT_h is determined, the mass flow rate of exhaust gases m_h can be calculated. Solve the equation from Step 6 numerically to obtain a value for ΔT_h and then calculate m_h accordingly. After solving the equation numerically, we find that ΔT_h ≈ 273 K. Now, we can calculate the mass flow rate m_h: m_h ≈ 672 kW / (1200 J/(kg·K) * 273 K) ≈ 2.04 kg/s Therefore, the required mass flow rate of exhaust gases to heat the water from 20°C to 100°C is approximately 2.04 kg/s.

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.

Cross-Flow Heat Exchanger
A cross-flow heat exchanger is a type of device used to transfer heat between two fluids flowing perpendicularly to each other. In these setups, one fluid flows over tubes or plates that carry the other fluid, facilitating efficient heat transfer. Cross-flow heat exchangers are widely used in various industries such as automotive, aerospace, and HVAC due to their compact design and effectiveness.
A specific characteristic of cross-flow heat exchangers is whether the fluids are mixed or unmixed. In this exercise, both fluids are described as unmixed, which means each fluid remains in its own stream without intermingling with the other. This design affects the heat transfer efficiency and the approach to calculating temperature changes and other thermal properties. By understanding the arrangement of fluid paths, we get a clearer perspective of how the heat transfer process is optimized in this configuration.
Heat Transfer Rate
The heat transfer rate is an essential factor that defines how much heat energy is exchanged between the two fluids in the heat exchanger per unit time. This rate is generally denoted by the symbol Q and is measured in watts (W) or kilowatts (kW).
In a heat exchanger, the heat transfer rate can be calculated using the formula: \[ Q = C \times \Delta T \]where \( C \) is the heat capacity rate (which is the product of mass flow and specific heat capacity), and \( \Delta T \) is the temperature difference.
For the given exercise, the heat transfer rate for the water is calculated to be 672 kW. This value reflects the amount of energy transferred from the hot exhaust gases to the cooler water, ensuring the water's temperature rises from 20°C to 100°C as desired. Understanding the heat transfer rate helps in designing or selecting a heat exchanger that meets specific thermal requirements.
Log Mean Temperature Difference
The Log Mean Temperature Difference (LMTD) is a critical parameter when analyzing heat exchangers. It represents the effective average temperature difference between the hot and cold fluids across the heat exchanger. This value helps to simplify complex temperature variations over the exchanger’s surface into a single, manageable figure.
The LMTD is calculated using the formula:\[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \]where \( \Delta T_1 \) and \( \Delta T_2 \) are the temperature differences at each end of the heat exchanger.
In the problem, the LMTD is employed to relate the overall heat transfer to the heat exchanger’s thermal design coefficient, \( UA \). It enables us to integrate the uneven temperature gradients of the fluids into a succinct mathematical model, which is necessary for calculating other thermal properties of the system.
Mass Flow Rate Calculation
Calculating the mass flow rate of the exhaust gases is a step that demands careful balancing of thermal equations. The mass flow rate is crucial since it determines how much fluid mass passes through the heat exchanger per unit time. With a known heat transfer rate, the mass flow rate can be calculated by rearranging the heat transfer formula to isolate the mass flow variable, \( m \).
The exercise provided a step-by-step solution that led to finding \( \Delta T_h \), the temperature change of the exhaust gases, which is required for this calculation. By knowing \( Q_c \), the heat transfer for water, and \( c_{ph} \), the specific heat of the gases, the relationship can be expressed as:\[ m_h = \frac{Q_c}{c_{ph} \times \Delta T_h} \]With \( \Delta T_h \) calculated as 273 K, the result is a mass flow rate \( m_h \) of approximately 2.04 kg/s. This value is essential for ensuring the thermal process achieves the desired heating, demonstrating practical application of mass flow calculations in engineering solutions.

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

The condenser of a steam power plant contains \(N=1000\) brass tubes \(\left(k_{\mathrm{t}}=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), each of inner and outer diameters, \(D_{i}=25 \mathrm{~mm}\) and \(D_{o}=\) \(28 \mathrm{~mm}\), respectively. Steam condensation on the outer surfaces of the tubes is characterized by a convection coefficient of \(h_{o}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If cooling water from a large lake is pumped through the condenser tubes at \(m_{c}=400 \mathrm{~kg} / \mathrm{s}\), what is the overall heat transfer coefficient \(U_{o}\) based on the outer surface area of a tube? Properties of the water may be approximated as \(\mu=9.60 \times\) \(10^{-4} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\mathrm{Pr}=6.6 .\) (b) If, after extended operation, fouling provides a resistance of \(R_{f, i}^{\prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), at the inner surface, what is the value of \(U_{o}\) ? (c) If water is extracted from the lake at \(15^{\circ} \mathrm{C}\) and \(10 \mathrm{~kg} / \mathrm{s}\) of steam at \(0.0622\) bars are to be condensed, what is the corresponding temperature of the water leaving the condenser? The specific heat of the water is \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

As part of a senior project, a student was given the assignment to design a heat exchanger that meets the following specifications: \begin{tabular}{lccc} \hline & \(\dot{m}(\mathrm{~kg} / \mathrm{s})\) & \(T_{m, i}\left({ }^{\circ} \mathrm{C}\right)\) & \(T_{m, \theta}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline Hot water & 28 & 90 & \(-\) \\ Cold water & 27 & 34 & 60 \\ \hline \end{tabular} Like many real-world situations, the customer hasn't revealed, or doesn't know, additional requirements that would allow you to proceed directly to a final configuration. At the outset, it is helpful to make a first-cut design based upon simplifying assumptions, which can be evaluated to determine what additional requirements and trade-offs should be considered by the customer. (a) Design a heat exchanger to meet the foregoing specifications. List and explain your assumptions. Hint: Begin by finding the required value for \(U A\) and using representative values of \(U\) to determine \(A\). (b) Evaluate your design by identifying what features and configurations could be explored with your customer in order to develop more complete specifications.

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.

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.

Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate of \(15 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(1100 \mathrm{~K}\), passes through a bundle of tubes, while the air, which has a flow rate of \(10 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(300 \mathrm{~K}\), is in cross flow over the tubes. The tubes are unfinned, and the overall heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the total tube surface area required to achieve an air outlet temperature of \(850 \mathrm{~K}\). The exhaust gas and the air may each be assumed to have a specific heat of \(1075 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

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.