Water is boiled at \(150^{\circ} \mathrm{C}\) in a boiler by hot exhaust gases \(\left(c_{p}=1.05 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) that enter the boiler at \(400^{\circ} \mathrm{C}\) at a rate of \(0.4 \mathrm{~kg} / \mathrm{s}\) and leaves at \(200^{\circ} \mathrm{C}\). The surface area of the heat exchanger is \(0.64 \mathrm{~m}^{2}\). The overall heat transfer coefficient of this heat exchanger is (a) \(940 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(1056 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1145 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(1230 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(1393 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short Answer

Expert verified
What is the overall heat transfer coefficient of a heat exchanger with the following conditions: - A hot exhaust gas enters the heat exchanger at 400°C and leaves at 200°C. - The exhaust gas has a specific heat capacity of 1.05 kJ/kg·K. - The mass flow rate of the exhaust gas is 0.40 kg/s. - The surface area of the heat exchanger is 0.64 m². a) 800 W/m²·K b) 1056 W/m²·K c) 1200 W/m²·K d) 1500 W/m²·K Answer: b) 1056 W/m²·K

Step by step solution

01

Calculate the temperature difference

To calculate the temperature difference (\(\Delta T\)), we subtract the outlet temperature of the exhaust gases from their inlet temperature: \(\Delta T = T_{\text{inlet}} - T_{\text{outlet}} = 400^{\circ}C - 200^{\circ}C = 200^{\circ}C\)
02

Find the heat transfer rate

We can find the heat transfer rate (q) using the formula \(q = mc\Delta T\), where \(m\) is the mass flow rate, \(c\) is the specific heat capacity of the exhaust gases, and \(\Delta T\) is the temperature difference. Plugging in the provided values, we get: \(q = (0.4 \,\text{kg/s})(1.05 \, \text{kJ/kg} \cdot^{\circ}\text{C})(200^{\circ}C) = 84\,\text{kJ/s}\)
03

Convert heat transfer rate from kJ/s to W.

Since we need the overall heat transfer coefficient in \(\text{W}/\text{m}^2\cdot\text{K}\). we need to convert the heat transfer rate from kJ/s to W. We can do this by multiplying it with 1000, as 1 kJ/s equals 1000 W. \(q = 84\,\text{kJ/s} \times 1000 = 84000\,\text{W}\)
04

Calculate the overall heat transfer coefficient

Now, we can calculate the overall heat transfer coefficient, \(U\), using the formula \(q = UA\Delta T\). We have the heat transfer rate (q), the surface area (A), and the temperature difference (\(\Delta T\)), so we can rearrange the formula to solve for \(U\): \(U = \frac{q}{A\Delta T} = \frac{84000 \, \text{W}}{(0.64 \, \text{m}^2)(200^{\circ}\text{C})} = 1056 \, \text{W/m}^2\cdot\text{K}\) The overall heat transfer coefficient of the heat exchanger is \(1056 \, \text{W/m}^2\cdot\text{K}\), which corresponds to choice (b).

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 Vaia!

Key Concepts

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

Understanding Heat Exchangers
Heat exchangers are pivotal devices in thermal systems, as they allow for the transfer of heat between two or more fluids at different temperatures without mixing them. In our problem, the heat exchanger facilitates the transfer of energy from hot exhaust gases to water to boil it. An effective heat exchanger maximizes the surface area available for heat transfer while minimizing resistance to heat flow. Factors like design, materials used, and flow configuration influence the heat exchanger's performance.

Although not explicitly detailed in the exercise, the type of heat exchanger (whether it's a shell and tube, plate, or finned-tube design, for example) and the fluid flow arrangement (counterflow, parallel flow, etc.) can significantly impact its effectiveness and the overall heat transfer coefficient. These are important considerations in the real-world application of the concepts being learned.
Calculating Heat Transfer Rate
The heat transfer rate is a measure of the thermal energy transferred per unit time. In the given textbook example, we need to calculate this rate to progress towards finding the overall heat transfer coefficient. The solution provided uses the formula q = mcΔT, which factors in the mass flow rate (m), the specific heat capacity (c), and the temperature change (ΔT).

Understanding how to accurately determine each variable in this formula is crucial. The specific heat capacity, which represents the amount of heat required to change a substance's temperature by a certain amount, is a key term in this equation. Small errors made in calculating the heat transfer rate can magnify when applied to subsequent calculations, making precision essential in solving problems involving thermal systems. For the heat transfer rate to be useful in further calculations (like finding the overall heat transfer coefficient), it has to be expressed in consistent units, hence the conversion from kJ/s to W in the solution.
Specific Heat Capacity and Its Role
Specific heat capacity is a material-specific thermal property that plays a pivotal role in determining how a substance will react to the addition or removal of heat. It is defined as the amount of heat energy required to raise the temperature of one kilogram of the substance by one degree Celsius (or one Kelvin).

Different materials have different specific heat capacities. For instance, water has a high specific heat capacity, which means it can absorb a lot of heat before it starts to get hot. In our example, the hot exhaust gases have a known specific heat capacity (c_p = 1.05 kJ/kg·°C), which is used to calculate the heat transfer rate to the water. Understanding specific heat capacity is essential not only for problems like this but also has practical applications such as designing thermal systems, heating and cooling processes, and even in culinary practices where controlling temperature is crucial.

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

Can the temperature of the hot fluid drop below the inlet temperature of the cold fluid at any location in a heat exchanger? Explain.

A cross-flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine (a) the exit temperature of the hot fluid and \((b)\) the rate of heat transfer in the heat exchanger.

A shell-and-tube heat exchanger is used for cooling \(47 \mathrm{~kg} / \mathrm{s}\) of a process stream flowing through the tubes from \(160^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). This heat exchanger has a total of 100 identical tubes, each with an inside diameter of \(2.5 \mathrm{~cm}\) and negligible wall thickness. The average properties of the process stream are: \(\rho=950 \mathrm{~kg} / \mathrm{m}^{3}, k=0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=3.5 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and \(\mu=2.0 \mathrm{mPa} \cdot \mathrm{s}\). The coolant stream is water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a flow rate of \(66 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(10^{\circ} \mathrm{C}\), which yields an average shell-side heat transfer coefficient of \(4.0 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the tube length if the heat exchanger has \((a)\) a 1 -shell pass and a 1 -tube pass and (b) a 1-shell pass and 4-tube passes.

The cardiovascular counter-current heat exchanger has an overall heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Arterial blood enters at \(37^{\circ} \mathrm{C}\) and exits at \(27^{\circ} \mathrm{C}\). Venous blood enters at \(25^{\circ} \mathrm{C}\) and exits at \(34^{\circ} \mathrm{C}\). Determine the mass flow rates of the arterial blood and venous blood in \(\mathrm{g} / \mathrm{s}\) if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and the surface area of the heat transfer to occur is \(0.15 \mathrm{~cm}^{2}\).

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 \(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

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.

Sign-up for free