Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a 12 -cm-diameter and \(8.5-\mathrm{m}\)-long tube at \(75^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\), and is cooled by a refrigerant evaporating outside at \(-10^{\circ} \mathrm{C}\). If the average heat transfer coefficient on the inner surface is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the exit temperature of water is (a) \(18.4^{\circ} \mathrm{C}\) (b) \(25.0^{\circ} \mathrm{C}\) (c) \(33.8^{\circ} \mathrm{C}\) (d) \(46.5^{\circ} \mathrm{C}\) (e) \(60.2^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: (d) 46.5°C

Step by step solution

01

Calculate the surface area of the tube

The surface area A of a tube is given by \(A = 2\pi rL\), where r is the radius and L is the length of the tube. Given the diameter of \(12\text{ cm}\), we have a radius \(r=6\text{ cm}=0.06\text{ m}\). And the length of the tube \(L=8.5\text{ m}\). Now we can calculate the surface area of the tube: \(A = 2\pi(0.06)(8.5) = 3.198\mathrm{~m}^2\).
02

Calculate the rate of heat transfer

The rate of heat transfer \(q\) is given by \(q= hA \Delta T\), where \(h = 500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the average heat transfer coefficient, and \(\Delta T = (75-(-10)) = 85\mathrm{~K}\) is the temperature difference. Now we can calculate the rate of heat transfer: \(q = (500)(3.198)(85) = 135950 \mathrm{~W}\).
03

Calculate the heat loss by the water to find the exit temperature

The heat loss by the water equals the rate of heat transfer and is given by the formula: \(q=mc_p\Delta T_{exit}\), where \(m = 0.35 \mathrm{~kg} / \mathrm{s}\) is the mass flow rate, \(c_p=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) is the specific heat of water, and \(\Delta T_{exit}\) is the temperature difference between the entrance and exit of the water. Now we can solve for \(\Delta T_{exit}\): \(\Delta T_{exit} = \frac{q}{mc_p} = \frac{135950}{(0.35)(4180)}= 29.2\)°C$.
04

Determine the exit temperature of water

To find the exit temperature of the water, subtract the temperature difference, \(\Delta T_{exit}\), from the entrance temperature: \(T_{exit}=T_{entrance} - \Delta T_{exit} = 75°\text{C} - 29.2°\text{C} = 45.8°\text{C}\). Now, we need to find which option from the given choices matches with our calculated exit temperature. Comparing our answer to the given choices, the closest option is: (d) \(46.5^{\circ} \mathrm{C}\). Therefore, the exit temperature of the water is approximately \(46.5^{\circ} \mathrm{C}\).

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Key Concepts

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

Convective Heat Transfer
Convective heat transfer is an essential concept when analyzing how heat moves through fluids such as water or air. This process is driven by the movement of the fluid itself—when warmer parts of a fluid flow move to colder regions, heat is transferred. This can occur naturally, as in the case of buoyancy-driven movement, or be assisted by external forces like pumps or fans, which is known as forced convection.

In the exercise, water flows through a pipe and is subject to cooling by a refrigerant. The heat transfer coefficient, represented by the symbol 'h', reflects the efficiency of heat being transferred from the water inside the tube to the surrounding refrigerant. A higher 'h' value means that the fluid is effectively losing heat. The specific formula used here, \(q= hA \Delta T\), encapsulates the core idea of convective heat transfer by relating the heat transfer rate (q), the heat transfer coefficient (h), the contact area (A), and the temperature difference (\(\Delta T\)) between the tube's water and the refrigerant.
Thermal Analysis
In the realm of physics and engineering, thermal analysis is a branch that deals with how heat moves within different materials and across interfaces. It incorporates assessing temperature variations, heat transfer rates, and energy change within a system to predict how temperature changes over time.

To carry out a thermal analysis, one must understand the properties of the materials involved, such as specific heat capacity, which is a measure of how much energy is needed to raise the temperature of a unit mass of a substance by one degree. In the given exercise, the water's specific heat capacity \(c_p\) plays a pivotal role in determining how much heat is lost, which in turn affects the water exit temperature. After calculating the rate of heat transfer using convective heat transfer principles, we apply the concept to quantify the water's heat loss through the formula \(q = mc_p \Delta T_{exit}\). This approach is a classic example of applying thermal analysis to solve real-world problems.
Heat Exchanger Design
Heat exchangers are devices designed to efficiently transfer heat from one medium to another. They are widely used in various industries such as power generation, chemical processing, and HVAC systems. Design considerations include the type of heat exchanger, the materials used, surface area for heat transfer, and the overall thermal performance.

The exercise demonstrates a simple heat exchanger where water transfers its heat to a refrigerant through the pipe's wall. The effectiveness of this process depends on the pipe's dimensions (which influence the surface area available for heat transfer), the properties of the flowing fluid, and operational conditions such as mass flow rate and inlet temperature. The solution process involves determining the exit temperature, it implicitly touches upon the critical aspects of heat exchanger design—calculating the necessary surface area and understanding how the flow rate and temperature difference drive the thermal performance of the system.

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Most popular questions from this chapter

A tube with a square-edged inlet configuration is subjected to uniform wall heat flux of \(8 \mathrm{~kW} / \mathrm{m}^{2}\). The tube has an inside diameter of \(0.622\) in and a flow rate of \(2.16 \mathrm{gpm}\). The liquid flowing inside the tube is ethylene glycol-distilled water mixture with a mass fraction of \(2.27\). Determine the friction coefficient at a location along the tube where the Grashof number is \(\mathrm{Gr}=35,450\). The physical properties of the ethylene glycol-distilled water mixture at the location of interest are \(\operatorname{Pr}=13.8, v=18.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\), and \(\mu_{b} / \mu_{s}=1.12\). Then recalculate the fully developed friction coefficient if the volume flow rate is increased by 50 percent while the rest of the parameters remain unchanged.

What fluid property is responsible for the development of the velocity boundary layer? For what kinds of fluids will there be no velocity boundary layer in a pipe?

In a gas-fired boiler, water is being boiled at \(120^{\circ} \mathrm{C}\) by hot air flowing through a 5 -m-long, 5 -cm-diameter tube submerged in water. Hot air enters the tube at 1 atm and \(300^{\circ} \mathrm{C}\) at a mean velocity of \(7 \mathrm{~m} / \mathrm{s}\), and leaves at \(150^{\circ} \mathrm{C}\). If the surface temperature of the tube is \(120^{\circ} \mathrm{C}\), determine the average convection heat transfer coefficient of the air and the rate of water evaporation, in \(\mathrm{kg} / \mathrm{h}\).

What is the generally accepted value of the Reynolds number above which the flow in smooth pipes is turbulent?

In a manufacturing plant that produces cosmetic products, glycerin is being heated by flowing through a \(25-\mathrm{mm}\)-diameter and \(10-\mathrm{m}\)-long tube. With a mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\), the flow of glycerin enters the tube at \(25^{\circ} \mathrm{C}\). The tube surface is maintained at a constant surface temperature of \(140^{\circ} \mathrm{C}\). Determine the outlet mean temperature and the total rate of heat transfer for the tube. Evaluate the properties for glycerin at \(30^{\circ} \mathrm{C}\).

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