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Chapter 8: Problem 58

Oil at \(150^{\circ} \mathrm{C}\) flows slowly through a long, thin-walled pipe of \(30-\mathrm{mm}\) inner diameter. The pipe is suspended in a room for which the air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient at the outer tube surface is \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the heat loss per unit length of tube. 8.59 Exhaust gases from a wire processing oven are discharged into a tall stack, and the gas and stack surface temperatures at the outlet of the stack must be estimated. Knowledge of the outlet gas temperature \(T_{m, o}\) is useful

Short Answer

Expert verified
The estimated heat loss per unit length of the tube is approximately 135.08 W/m.

Step by step solution

01

Identify given information

We know the following information from the problem statement: - Inner oil temperature: \(T_{inner} = 150^{\circ} \mathrm{C}\) - Inner diameter of the tube: \(d_{inner} = 30 \mathrm{~mm}\) - Room temperature: \(T_{room} = 20^{\circ} \mathrm{C}\) - Convection coefficient: \(h = 11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
02

Calculate the temperature difference

First, we need to find the temperature difference between the inner and outer surfaces of the tube. This can be calculated using the given temperatures: $$ \Delta T = T_{inner} - T_{room} $$ Substituting the values: $$ \Delta T = 150^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 130^{\circ} \mathrm{C} $$
03

Calculate the heat transfer per unit area

Now, we'll use the convection coefficient given to find the heat transfer per unit area. The heat transfer per unit area is given as: $$ q'' = h \cdot \Delta T $$ Substituting the given convection coefficient and calculated temperature difference: $$ q'' = 11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 130^{\circ} \mathrm{C} = 1430 \mathrm{~W} / \mathrm{m}^2 $$
04

Calculate the heat loss per unit length

Finally, we need to find the heat loss per unit length. We can do this by multiplying the heat transfer per unit area by the outer surface area per unit length of the tube: $$ q' = q'' \times \pi d_{outer} $$ Assuming that the tube walls are thin, we can approximate the outer diameter as equal to the inner diameter. Thus, $$ q' = 1430 \mathrm{~W} / \mathrm{m}^2 \times \pi (30 \times 10^{-3} \mathrm{~m}) = 135.08 \mathrm{~W} / \mathrm{m} $$
05

State the final answer

The estimated heat loss per unit length of the tube is approximately 135.08 W/m.

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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 one of the primary mechanisms of thermal energy movement between a surface and a fluid in motion. It's an essential concept when estimating how much heat is lost or gained in systems involving fluids, like the flow of oil through a pipe.

Convective heat transfer occurs because of the relative movement between the fluid (such as air or oil) and a solid boundary (like the inside of the pipe). This movement transports heat energy from the warmer to the cooler region. In our exercise, the oil inside the long, thin-walled pipe is at a higher temperature compared to the surrounding air. The difference in temperature results in heat transfer from the oil through the pipe wall and finally to the air via convection.

In practice, the rate of convective heat transfer is quantified by the convection coefficient, represented by the symbol 'h' in units of \(W/m^2鈰匥\). The higher the convection coefficient, the more efficient the heat transfer. The convection coefficient depends on properties such as the type of fluid, fluid velocity, and temperature, as well as the nature of the surface.
Temperature Difference Calculation
The temperature difference calculation is a fundamental step when dealing with heat transfer exercises. It involves finding the difference in temperature between two points, typically between a surface and the surrounding fluid. In the concerned exercise, we are looking at the temperature difference between the inner surface of the tube holding hot oil and the air outside the tube.

This calculation is simple but crucial because it's the driving force for heat transfer. No matter the method of heat transfer 鈥� conduction, convection, or radiation 鈥� a difference in temperature is necessary for heat to flow. For our problem, we calculate the temperature difference \(\Delta T\) by subtracting the ambient air temperature \(T_{room}\) from the oil temperature inside the pipe \(T_{inner}\).

This resultant \(\Delta T\) is then used in conjunction with the convection coefficient to determine the rate of heat loss, as heat naturally flows from the hot oil to the colder room air.
Heat Transfer Per Unit Area
Heat transfer per unit area, often denoted as \(q''\), is a crucial measurement that quantifies the amount of heat passing through a specific area in a given time. It's a vital parameter in many engineering applications, from HVAC system design to thermal insulation in buildings.

When examining heat loss or heat gain, knowing the heat transfer per unit area helps in predicting the thermal performance of materials and systems. Using the previously determined temperature difference and the convection coefficient, you can calculate the heat transfer per unit area using the formula \(q'' = h \cdot \Delta T\).

The value of \(q''\) obtained reflects the intensity of heat transfer across a defined area of the surface. This information can then be used to estimate the total heat loss or gain by considering the overall area involved in the heat transfer process. For the pipe in our exercise, once \(q''\) is calculated, we can estimate the heat lost per unit length of the tube by considering the outer surface area exposed to the convective environment.

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Most popular questions from this chapter

8.106 Consider the pharmaceutical product of Problem 8.27. Prior to finalizing the manufacturing process, test trials are performed to experimentally determine the dependence of the shelf life of the drug as a function of the sterilization temperature. Hence, the sterilization temperature must be carefully controlled in the trials. To promote good mixing of the pharmaceutical and, in turn, relatively uniform outlet temperatures across the exit tube area, experiments are performed using a device that is constructed of two interwoven coiled tubes, each of 10 -mm diameter. The thin-walled tubing is welded to a solid high thermal conductivity rod of diameter \(D_{r}=40 \mathrm{~mm}\). One tube carries the pharmaceutical product at a mean velocity of \(u_{p}=0.1 \mathrm{~m} / \mathrm{s}\) and inlet temperature of \(25^{\circ} \mathrm{C}\), while the second tube carries pressurized liquid water at \(u_{w}=0.12 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(127^{\circ} \mathrm{C}\). The tubes do not contact each other but are each welded to the solid metal rod, with each tube making 20 turns around the rod. The exterior of the apparatus is well insulated. (a) Determine the outlet temperature of the pharmaceutical product. Evaluate the liquid water properties at \(380 \mathrm{~K}\). (b) Investigate the sensitivity of the pharmaceutical's outlet temperature to the velocity of the pressurized water over the range \(0.10

Consider a thin-walled, metallic tube of length \(L=1 \mathrm{~m}\) and inside diameter \(D_{i}=3 \mathrm{~mm}\). Water enters the tube at \(\dot{m}=0.015 \mathrm{~kg} / \mathrm{s}\) and \(T_{m, i}=97^{\circ} \mathrm{C}\). (a) What is the outlet temperature of the water if the tube surface temperature is maintained at \(27^{\circ} \mathrm{C}\) ? (b) If a \(0.5-\mathrm{mm}\)-thick layer of insulation of \(k=0.05\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) is applied to the tube and its outer surface is maintained at \(27^{\circ} \mathrm{C}\), what is the outlet temperature of the water? (c) If the outer surface of the insulation is no longer maintained at \(27^{\circ} \mathrm{C}\) but is allowed to exchange heat by free convection with ambient air at \(27^{\circ} \mathrm{C}\), what is the outlet temperature of the water? The free convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\).

Engine oil is heated by flowing through a circular tube of diameter \(D=50 \mathrm{~mm}\) and length \(L=25 \mathrm{~m}\) and whose surface is maintained at \(150^{\circ} \mathrm{C}\). (a) If the flow rate and inlet temperature of the oil are \(0.5 \mathrm{~kg} / \mathrm{s}\) and \(20^{\circ} \mathrm{C}\), what is the outlet temperature \(T_{m, o}\) ? What is the total heat transfer rate \(q\) for the tube? (b) For flow rates in the range \(0.5 \leq \dot{m} \leq 2.0 \mathrm{~kg} / \mathrm{s}\), compute and plot the variations of \(T_{m, o}\) and \(q\) with \(\dot{m}\). For what flow rate(s) are \(q\) and \(T_{m, \rho}\) maximized? Explain your results.

A cold plate is an active cooling device that is attached to a heat-generating system in order to dissipate the heat while maintaining the system at an acceptable temperature. It is typically fabricated from a material of high thermal conductivity, \(k_{\text {cp, }}\), within which channels are machined and a coolant is passed. Consider a copper cold plate of height \(H\) and width \(W\) on a side, within which water passes through square channels of width \(w=h\). The transverse spacing between channels \(\delta\) is twice the spacing between the sidewall of an outer channel and the sidewall of the cold plate. Consider conditions for which equivalent heat-generating systems are attached to the top and bottom of the cold plate, maintaining the corresponding surfaces at the same temperature \(T_{s}\). The mean velocity and inlet temperature of the coolant are \(u_{m}\) and \(T_{m i}\), respectively. (a) Assuming fully developed turbulent flow throughout each channel, obtain a system of equations that may be used to evaluate the total rate of heat transfer to the cold plate, \(q\), and the outlet temperature of the water, \(T_{m, o}\), in terms of the specified parameters. (b) Consider a cold plate of width \(W=100 \mathrm{~mm}\) and height \(H=10 \mathrm{~mm}\), with 10 square channels of width \(w=6 \mathrm{~mm}\) and a spacing of \(\delta=4 \mathrm{~mm}\) between channels. Water enters the channels at a temperature of \(T_{m, i}=300 \mathrm{~K}\) and a velocity of \(u_{m}=2 \mathrm{~m} / \mathrm{s}\). If the top and bottom cold plate surfaces are at \(T_{s}=360 \mathrm{~K}\), what is the outlet water temperature and the total rate of heat transfer to the cold plate? The thermal conductivity of the copper is \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while average properties of the water may be taken to be \(\rho=984 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=489 \times\) \(10^{-6} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=3.15\). Is this a good cold plate design? How could its performance be improved?

In a particular application involving fluid flow at a rate \(\dot{m}\) through a circular tube of length \(L\) and diameter \(D\), the surface heat flux is known to have a sinusoidal variation with \(x\), which is of the form \(q_{s}^{\prime \prime}(x)=q_{s, m}^{\prime \prime} \sin (\pi x / L)\). The maximum flux, \(q_{s, m}^{n}\), is a known constant, and the fluid enters the tube at a known temperature, \(T_{m, i}\) Assuming the convection coefficient to be constant, how do the mean temperature of the fluid and the surface temperature vary with \(x\) ?
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