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Chapter 9: Problem 62

A hot fluid \(\left(k_{\text {fluid }}=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) is flowing as a laminar fully-developed flow inside a pipe with an inner diameter of \(35 \mathrm{~mm}\) and a wall thickness of \(5 \mathrm{~mm}\). The pipe is \(10 \mathrm{~m}\) long and the outer surface is exposed to air at \(10^{\circ} \mathrm{C}\). The average temperature difference between the hot fluid and the pipe inner surface is \(\Delta T_{\text {avg }}=10^{\circ} \mathrm{C}\), and the inner and outer surface temperatures are constant. Determine the outer surface temperature of the pipe. Evaluate the air properties at \(50^{\circ} \mathrm{C}\). Is this a good assumption?

Short Answer

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Question: Calculate the outer surface temperature of a pipe with fluid flowing inside, given the following information: Inner diameter of the pipe: 35 mm Wall thickness: 5 mm Length of the pipe: 10 m Thermal conductivity of the pipe material: 0.72 W/mK Fluid temperature: 65 掳C Average temperature difference between the fluid and the outer surface: 10 掳C Heat transfer coefficient for air at 50 掳C: h_air (To be evaluated) Answer: To find the outer surface temperature of the pipe, follow these steps: 1. Calculate the thermal resistance of the pipe using the given dimensions and thermal conductivity. 2. Determine the heat transfer through the pipe wall using the thermal resistance and the given temperature difference. 3. Calculate the temperature difference between the outer and inner surfaces of the pipe using the heat transfer and the heat transfer coefficient for air (h_air). 4. Determine the outer surface temperature using the calculated temperature difference and given inner surface temperature. 5. Check if the assumption about air properties at 50 掳C is valid by comparing the calculated outer surface temperature to the assumed air temperature. Following these steps, the outer surface temperature of the pipe can be calculated once the heat transfer coefficient (h_air) is evaluated.

Step by step solution

01

Find the heat transfer through the pipe wall

To find the heat transfer through the pipe wall, we will need to use the thermal resistance concept. The formula for the thermal resistance through a cylindrical wall is: \(R_{\text {pipe }}=\frac{L \ln \frac{r_{o}}{r_{i}}}{2 \pi k_{\text {pipe}}}\) where \(L\) is the length of the pipe, \(r_{i}\) and \(r_{o}\) are the inner and outer radii of the pipe, and \(k_{\text {pipe}}\) is the thermal conductivity of the pipe material. We are given that \(L=10\,\text m\) and \(k_{\text {fluid}}=0.72\,\text{W/mK}\). We can find \(r_{i}\) and \(r_{o}\) using the inner diameter and wall thickness of the pipe: \(r_{i}=\frac{35}{2} \times 10^{-3}\,\text m\) \(r_{o}=\left(\frac{35}{2}+5\right) \times 10^{-3}\,\text m\) Now we can calculate the thermal resistance \(R_{\text {pipe}}\).
02

Calculate the heat transfer from the fluid through the pipe wall

The formula for the heat transfer through the pipe wall is: \(q=\frac{\Delta T_{\text {avg}}}{R_{\text {pipe}}}\) We are given that \(\Delta T_{\text {avg}}=10^{\circ}\text C\). Calculate the heat transfer using the previously calculated thermal resistance.
03

Find the temperature difference between the outer and inner surface of the pipe

Now that we have the heat transfer through the pipe wall, we can use the following formula to find the temperature difference between the outer and inner surface: \(\Delta T_{\text {outer-inner}}=\frac{q}{h_{\text {air}}A_{\text {outer}}}\) We need to find the heat transfer coefficient \(h_{\text {air}}\) and the outer surface area \(A_{\text {outer}}\). The heat transfer coefficient can be evaluated once we know the air properties at \(50^{\circ} \mathrm{C}\). The outer surface area can be calculated as: \(A_{\text {outer}}=2 \pi r_{o} L\)
04

Calculate the outer surface temperature of the pipe

Once we have found the temperature difference between the outer and inner surface of the pipe, we can calculate the outer surface temperature using following formula: \(T_{\text {outer}}=T_{\text {inner}}-\Delta T_{\text {outer-inner}}\) We are given that the pipe inner surface temperature can be found as: \(T_{\text {inner}}=T_{\text {fluid}}-\Delta T_{\text {avg}}\) Now, we can calculate the outer surface temperature of the pipe.
05

Check if the assumption about air properties at \(50^{\circ} \mathrm{C}\) is correct

Finally, we need to check whether the assumption about air properties at \(50^{\circ} \mathrm{C}\) is good. To do this, we can compare the calculated outer surface temperature of the pipe and the assumption temperature. If the difference is small, the assumption is valid. Otherwise, we might need to re-evaluate the air properties at a different temperature and repeat the previous steps to recalculate the outer surface temperature of the pipe.

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Key Concepts

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

Thermal Resistance
Thermal resistance is a key concept when analyzing heat transfer through materials. It quantifies how strongly a material opposes the flow of heat. Imagine you have a cylindrical pipe with heat flowing from the inside to the outside. This pipe acts like a barrier to heat flow. We calculate the thermal resistance using the formula:\[R_{\text {pipe }}=\frac{L \ln \frac{r_{o}}{r_{i}}}{2 \pi k_{\text {pipe}}}\]Where:
  • \(L\) is the length of the pipe,
  • \(r_{i}\) and \(r_{o}\) are the inner and outer radii,
  • \(k_{\text {pipe}}\) is the thermal conductivity of the pipe material.
This formula helps us understand that a longer pipe or a material with lower thermal conductivity increases resistance to heat flow. Think of thermal resistance like wearing a thicker sweater鈥攖he thicker it is, the harder it is for heat to escape, keeping you warmer. In the exercise, it's shown how this resistance is used to determine the ease with which heat transfers through the pipe wall.
Laminar Flow
Laminar flow refers to a type of flow where a fluid moves smoothly in parallel layers, with minimal disruption between them. It is crucial in this exercise because it ensures the uniformity of the heat transfer across the surface of the pipe. This type of flow occurs at lower velocities and allows us to make certain simplifications in calculations. It鈥檚 akin to cars moving smoothly on a highway: each one follows the same direction without overtaking or disturbing the flow of others. In laminar flow, the speed and direction of fluid particles at each point remain constant. This makes it easier to predict how heat moves within the fluid. When dealing with laminar flow in a pipe, engineers often use the Reynolds number to determine the flow regime. If this number is below a certain threshold (typically 2,300 for pipe flow), the flow is considered laminar. Understanding whether the flow is laminar helps us apply the correct heat transfer equations and expect steady, predictable outcomes in thermal resistance calculations.
Thermal Conductivity
Thermal conductivity is a property that represents a material's ability to transfer heat. It is denoted by the symbol \(k\) and measured in Watts per meter-Kelvin (W/m路K). In simpler terms, it's a measure of how well heat is conducted through a material.Materials with high thermal conductivity, such as metals, allow heat to flow quickly. On the other hand, materials with low thermal conductivity, like foam or fiberglass, resist heat flow. This property is vital in determining how much heat is lost or gained through materials.In the context of a pipe, the thermal conductivity coefficient, \(k_{\text{pipe}}\), tells us how efficiently heat moves across the pipe wall. If the pipe's material has high thermal conductivity, the heat flows effortlessly from the fluid inside to the outside environment. It's like having a fast-track lane for heat transfer, compared to a slow lane in poor conductors.Understanding thermal conductivity helps engineers select appropriate materials for specific applications, ensuring that heat transfer occurs effectively and efficiently.
Temperature Difference
Temperature difference, often symbolized as \(\Delta T\), is crucial in heat transfer as it drives the flow of heat from a hot region to a cold one. In our exercise, the average temperature difference \(\Delta T_{\text {avg}}\) between the fluid's surface and the pipe's surface is given as 10掳C. This difference creates a thermal "pressure" that pushes heat through the pipe wall. If there's a large temperature difference, heat moves faster because it wants to equalize the temperatures. This principle is observed in simple daily life. Think of opening a window on a cold day; heat rushes out quickly due to the larger temperature contrast between the inside and outside air.This concept also underpins calculations in the exercise: determining how much heat moves through and the consequent outer surface temperature of the pipe relies on understanding \(\Delta T\). It says, the more significant the temperature "push," the more rapid the heat transfer process.

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Most popular questions from this chapter

Under what conditions can the outer surface of a vertical cylinder be treated as a vertical plate in natural convection calculations?

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A group of 25 power transistors, dissipating \(1.5 \mathrm{~W}\) each, are to be cooled by attaching them to a black-anodized square aluminum plate and mounting the plate on the wall of a room at \(30^{\circ} \mathrm{C}\). The emissivity of the transistor and the plate surfaces is \(0.9\). Assuming the heat transfer from the back side of the plate to be negligible and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the size of the plate if the average surface temperature of the plate is not to exceed \(50^{\circ} \mathrm{C}\). Answer: \(43 \mathrm{~cm} \times 43 \mathrm{~cm}\)

A solar collector consists of a horizontal aluminum tube of outer diameter \(5 \mathrm{~cm}\) enclosed in a concentric thin glass tube of \(7 \mathrm{~cm}\) diameter. Water is heated as it flows through the aluminum tube, and the annular space between the aluminum and glass tubes is filled with air at \(1 \mathrm{~atm}\) pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of \(20 \mathrm{~W}\) per meter length, and the temperature of the ambient air outside is \(30^{\circ} \mathrm{C}\). Approximating the surfaces of the tube and the glass cover as being black (emissivity \(\varepsilon=1\) ) in radiation calculations and taking the effective sky temperature to be \(20^{\circ} \mathrm{C}\), determine the temperature of the aluminum tube when equilibrium is established (i.e., when the net heat loss from the tube by convection and radiation equals the amount of solar energy absorbed by the tube). For evaluation of air properties at \(1 \mathrm{~atm}\) pressure, assume \(33^{\circ} \mathrm{C}\) for the surface temperature of the glass cover and \(45^{\circ} \mathrm{C}\) for the aluminum tube temperature. Are these good assumptions?

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at \(25^{\circ} \mathrm{C}\). Each chip dissipates \(0.18 \mathrm{~W}\) of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?
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