/*! 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 140 Air at \(25^{\circ} \mathrm{C}\)... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 140

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, \(1.7\)-m-long pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. Air properties at the average temperature are \(k=0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735\), \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The rate of heat transfer to the pipe is (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

Short Answer

Expert verified
Question: Determine the heat transfer rate from the air to a refrigerant inside a pipe (in Watts) using the given data. Choose the correct option: (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

Step by step solution

01

1. Calculate the Reynolds number

First, we need to compute the Reynolds number (\(Re\)) of the airflow around the pipe: $$Re = \frac{Vd}{\nu}$$ where \(V = 4 \mathrm{~m} / \mathrm{s}\) is the airflow velocity, \(d = 0.05 \mathrm{~m}\) is the diameter of the pipe, and \(\nu = 1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) is the kinematic viscosity. Plugging in the values, we get: $$Re = \frac{4 \times 0.05}{1.382 \times 10^{-5}}$$
02

2. Determine the flow type

Now that we have the Reynolds number, we can determine if the airflow around the pipe is laminar or turbulent. The critical Reynolds number for pipes is \(Re_{crit} = 2300\). If \(Re < 2300\), the flow is laminar, and if \(Re > 2300\), the flow is turbulent. In our case: $$Re > 2300$$ so the flow is turbulent.
03

3. Calculate the Nusselt number

For turbulent flow over an isothermal cylinder, the appropriate correlation to determine the Nusselt number (\(Nu\)) is: $$Nu = 0.3 + \frac{0.62Re^{0.5} Pr^{1/3}}{[1 + (0.4 / Pr)^{2/3}]^{1/4}}$$ Next, plug in the given values for \(Re\) and \(Pr\), and calculate the Nusselt number: $$Nu = 0.3 + \frac{0.62Re^{0.5} Pr^{1/3}}{[1 + (0.4 / Pr)^{2/3}]^{1/4}}$$
04

4. Calculate the convective heat transfer coefficient

Now we need to calculate the convective heat transfer coefficient (\(h\)) from the Nusselt number using the formula: $$h = \frac{Nu \times k}{d}$$ where \(k = 0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity of the air and \(d\) is the diameter of the pipe. Plug in the values and solve for \(h\): $$h = \frac{Nu \times 0.0240}{0.05}$$
05

5. Calculate the heat transfer rate

Finally, we can determine the heat transfer rate from the air to the refrigerant inside the pipe using the formula: $$Q = hA(T_{air} - T_{refrigerant})$$ where \(A = \pi dL\) is the surface area of the pipe, with \(L = 1.7 \mathrm{~m}\) being the length of the pipe, and \(T_{air} = 25^{\circ}\mathrm{C}\) and \(T_{refrigerant} = -15^{\circ}\mathrm{C}\) are the temperatures of the air and refrigerant, respectively. Plug in the values and solve for \(Q\): $$Q = h(\pi \times 0.05 \times 1.7)(25 - (-15)) \mathrm{~W}$$ Compare the calculated heat transfer rate \(Q\) with the options given and choose the closest value. (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

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Key Concepts

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

Reynolds Number
Understanding the Reynolds Number is crucial in analyzing fluid flow. It helps us determine the flow regime, whether it is laminar or turbulent. It's calculated with the formula: \[ Re = \frac{Vd}{u} \] where \( V \) is the velocity of the fluid, \( d \) is the diameter of the pipe, and \( u \) is the kinematic viscosity. This gives us a dimensionless number that is useful in predicting flow patterns.
  • Laminar Flow: Occurs when \( Re < 2300 \), characterized by smooth, orderly flow layers.
  • Turbulent Flow: Happens when \( Re > 2300 \), where the flow becomes chaotic and mixed.
In the given exercise, the airflow around the pipe was found to be turbulent, indicating more complex mixing and heat transfer behavior.
Nusselt Number
The Nusselt Number is a dimensionless parameter that tells us about the convective heat transfer relative to conductive heat transfer across the boundary. It is defined as: \[ Nu = \frac{h \cdot d}{k} \] where \( h \) is the convective heat transfer coefficient, \( d \) is the characteristic length (usually diameter), and \( k \) is the thermal conductivity of the fluid. Nusselt number correlations differ for laminar and turbulent flows. For turbulent flow over an isothermal cylinder, as in the exercise, we use an empirical formula because the flow patterns are complex and unpredictable. The higher the Nusselt number, the more effective the convective heat transfer.
Convective Heat Transfer Coefficient
The Convective Heat Transfer Coefficient \( h \) measures the efficiency of heat transfer between a solid surface and a fluid. It's an essential part of designing systems involving heat exchangers. Calculated with the formula: \[ h = \frac{Nu \times k}{d} \] where \( Nu \) is the Nusselt number, \( k \) is the thermal conductivity, and \( d \) is the diameter. A higher \( h \) value indicates better convective heat flow. It鈥檚 dependent on various factors, including fluid velocity, temperature difference, and surface characteristics. Understanding \( h \) helps engineers optimize thermal systems for efficiency and performance.
Laminar vs Turbulent Flow
The nature of fluid flow is critical in predicting its behavior and heat transfer characteristics.
  • Laminar Flow: This type is smooth and orderly, with layers of fluid sliding past one another. It is predictable and less efficient in convective heat transfer compared to turbulent flow.
  • Turbulent Flow: This flow category is characterized by chaotic swirling and mixing. It enhances mixing and heat transfer but is more challenging to analyze due to its complexity.
The transition from laminar to turbulent flow is marked by the critical Reynolds number, usually around 2300 for flow inside pipes. Turbulent flows are generally desired in heat exchanger designs as they improve heat transfer rates.

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Most popular questions from this chapter

What is the difference between the upstream velocity and the free-stream velocity? For what types of flow are these two velocities equal to each other?

Air at 1 atm is flowing in parallel over a \(3-\mathrm{m}-\) long flat plate with a velocity of \(7 \mathrm{~m} / \mathrm{s}\). The air has a free stream temperature of \(120^{\circ} \mathrm{C}\) and the surface temperature of the plate is maintained at \(20^{\circ} \mathrm{C}\). Determine the distance \(x\) from the leading edge of the plate where the critical Reynolds number \(\left(\operatorname{Re}_{c r}=5 \times 10^{5}\right)\) is reached. Then, using the EES (or other) software, evaluate the local convection heat transfer coefficient along the plate. By varying the location along the plate for \(0.2 \leq x \leq 3 \mathrm{~m}\), plot the local convection heat transfer coefficient as a function of \(x\), and discuss the results.

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