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Chapter 7: Problem 15

Explain under what conditions the total rate of heat transfer from an isothermal flat plate of dimensions \(L \times 2 L\) would be the same, independent of whether parallel flow over the plate is directed along the side of length \(L\) or \(2 L\). With a critical Reynolds number of \(5 \times 10^{5}\), for what values of \(R e_{L}\) would the total heat transfer be independent of orientation?

Short Answer

Expert verified
The total rate of heat transfer from an isothermal flat plate of dimensions \(L \times 2L\) will be the same, independent of the orientation, when its Nusselt numbers in both orientations are equal. For turbulent flow, the Nusselt number is proportional to the Reynolds number raised to the power of \(\frac{4}{5}\). The heat transfer will be independent of orientation for any value of Reynolds number (\(Re_L\)) that satisfies the equation \((Re_L)^\frac{4}{5} = \text{Constant}\). For the given critical Reynolds number of \(5 \times 10^5\), this condition is met when \(Re_L\) is greater than or equal to the critical value.

Step by step solution

01

Write the equation for heat transfer rate for both orientations

We can express the total heat transfer rate in the two orientations as: \[Q_1 = h_1A_1\] \[Q_2 = h_2A_2\] where \(Q_1\) and \(Q_2\) are the heat transfer rates for flow along the side of length \(L\) and \(2L\), respectively, \(h_1\) and \(h_2\) are the convective heat transfer coefficients, and \(A_1\) and \(A_2\) are the surface areas of the plate in the two orientations.
02

Use Nusselt number relationship for both orientations

We will use the relationship between Nusselt number, heat transfer coefficient, and thermal conductivity to write the heat transfer coefficients for both orientations: \[h_1 = \frac{Nu_1k}{L}\] \[h_2 = \frac{Nu_2k}{2L}\]
03

Substitute heat transfer coefficients into heat transfer rate equations

Now, we can substitute the expressions we found in Step 2 into the heat transfer rate equation: \[Q_1 = (\frac{Nu_1k}{L})(L\times L)\] \[Q_2 = (\frac{Nu_2k}{2L})(2L \times L)\]
04

Equate heat transfer rates and simplify the equation

For the total heat transfer to be independent of the orientation, the heat transfer rates in both orientations must be equal, that is: \[Q_1 = Q_2\] Substitute the expressions for \(Q_1\) and \(Q_2\) from Step 3 into the equation above and simplify: \[\frac{Nu_1}{L^2} = \frac{Nu_2}{L^2}\] Since both sides share the same denominator, we can cancel it out, leaving us with: \[Nu_1 = Nu_2\]
05

Determine the relationship between Nusselt numbers and Reynolds numbers

For this step, we will assume that the flow is turbulent. Therefore, we can use the relationship between Nusselt number and Reynolds number for turbulent flow, which is given by: \[Nu_L \propto Re_L^\frac{4}{5}\] Applying this relationship to both orientations, we find: \[Nu_1 = C_1(Re_{L1})^\frac{4}{5}\] \[Nu_2 = C_2(Re_{L2})^\frac{4}{5}\] Remember that, for the flow along the side of length \(L\), the Reynolds number is denoted as \(Re_{L1}\), and for the flow along the side of length \(2L\), the Reynolds number is denoted as \(Re_{L2}\).
06

Solve for the values of Reynolds numbers for which the total heat transfer is independent of orientation

Now that we have the relationship between Nusselt numbers and Reynolds numbers for both orientations, we can substitute these expressions into the equation from Step 4: \[C_1(Re_{L1})^\frac{4}{5} = C_2(Re_{L2})^\frac{4}{5}\] Since we want the total heat transfer to be independent of the orientation, the values of the Reynolds numbers must be equal: \[Re_{L1} = Re_{L2} = Re_L\] Now, we can solve for the values of Reynolds numbers which satisfy the condition: \[C_1(Re_L)^\frac{4}{5} = C_2(Re_L)^\frac{4}{5}\] Since \(C_1\) and \(C_2\) are constants, we can divide both sides by the same constant, resulting in: \[(Re_L)^\frac{4}{5} = \text{Constant}\] The heat transfer will be independent of orientation for any value of Reynolds number that satisfies this equation. For the given critical Reynolds number of \(5 \times 10^5\), it is clear that this condition will be met when \(Re_L\) is greater than or equal to the critical value.

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Key Concepts

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

Reynolds Number
The Reynolds Number is a fundamental concept in fluid dynamics. It provides insight into the flow type over a surface, distinguishing between laminar and turbulent flow. Calculating the Reynolds Number helps determine the nature of flow which affects heat transfer rates. Reynolds Number Calculation
To find the Reynolds Number, use the formula: \[ Re = \frac{\rho v L}{\mu} \]
- \( \rho \) is the fluid density.
- \( v \) is the velocity of the fluid flow.
- \( L \) is the characteristic length.
- \( \mu \) is the dynamic viscosity of the fluid.
Critical Reynolds Number
The flow transitions from laminar to turbulent at the critical Reynolds Number. For many scenarios involving flat plates, this critical value is around \(5 \times 10^5\). If \(Re > 5 \times 10^5\), expect a turbulent flow, influencing the convective heat transfer characteristics.
Nusselt Number
The Nusselt Number is a dimensionless number representing the ratio of convective to conductive heat transfer across a boundary. In simpler terms, it provides a measure of how effective convection is in transferring heat in comparison to conduction.Understanding Nusselt Number
The Nusselt Number \( Nu \) can be related to the heat transfer coefficient \( h \), characteristic length \( L \), and thermal conductivity \( k \) using the formula:\[ Nu = \frac{hL}{k} \]
- \( h \) is the convective heat transfer coefficient.
- \( L \) is the characteristic length (like the side of a plate).
- \( k \) is the thermal conductivity of the fluid.
Relationship with Reynolds Number
In turbulent flow, the Nusselt Number is proportional to the Reynolds Number raised to the power of \( \frac{4}{5} \) and it is influenced by both flow velocity and fluid properties, which reflects on the efficiency of heat transfer.
Convective Heat Transfer
Convective Heat Transfer is a mode of heat transfer between a surface and a fluid flowing over it. The effectiveness depends on the temperature difference and the flow of the fluid.How It Works
Convective Heat Transfer occurs when- A fluid flows over a surface like a plate.
- Heat energy exchanges between the surface and the fluid.
- The rate of heat transfer is influenced by the fluid's velocity and temperature gradient.
Heat Transfer Coefficient
In convection, the heat transfer coefficient \( h \) is crucial. It determines how much heat a fluid can absorb or release when in contact with a surface. Calculating this involves both Nusselt Number and other properties, like fluid thermal conductivity.ApplicationsThis type of heat transfer is common in weather systems, engineering, and HVAC systems as it effectively regulates temperature and facilitates energy efficiency.
Turbulent Flow
Turbulent Flow, characterized by chaotic changes in pressure and flow velocity, leads to higher energy mixing and increased heat transfer rates. In contrast to laminar flow, turbulent flow does not occur in smooth layers. Characteristics of Turbulent Flow
- Unpredictable and irregular paths.
- Increased mixing of fluid layers.
- Enhanced momentum and heat diffusion.
Importance in Heat Transfer In turbulent conditions, the heat transfer rate is typically higher due to the increased interaction between fluid layers, promoting effective energy exchange. The presence of eddies and swirls helps in better mixing and consequently more uniform heat distribution. When studying turbulent flow, always check if the Reynolds Number exceeds the critical value, indicating a shift to turbulence. This understanding is vital in designing systems for efficient thermal management in industries, appliances, and climate infrastructure.

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Most popular questions from this chapter

Consider the velocity boundary layer profile for flow over a flat plate to be of the form \(u=C_{1}+C_{2} y\). Applying appropriate boundary conditions, obtain an expression for the velocity profile in terms of the boundary layer thickness \(\delta\) and the free stream velocity \(u_{\infty}\). Using the integral form of the boundary layer momentum equation (Appendix G), obtain expressions for the boundary layer thickness and the local friction coefficient, expressing your result in terms of the local Reynolds number. Compare your results with those obtained from the exact solution (Section 7.2.1) and the integral solution with a cubic profile (Appendix \(G\) ).

Consider atmospheric air at \(u_{\infty}=2 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=300 \mathrm{~K}\) in parallel flow over an isothermal flat plate of length \(L=1 \mathrm{~m}\) and temperature \(T_{s}=350 \mathrm{~K}\). (a) Compute the local convection coefficient at the leading and trailing edges of the heated plate with and without an unheated starting length of \(\xi=1 \mathrm{~m}\). (b) Compute the average convection coefficient for the plate for the same conditions as part (a). (c) Plot the variation of the local convection coefficient over the plate with and without an unheated starting length.

Air at a pressure of 1 atm and a temperature of \(50^{\circ} \mathrm{C}\) is in parallel flow over the top surface of a flat plate that is heated to a uniform temperature of \(100^{\circ} \mathrm{C}\). The plate has a length of \(0.20 \mathrm{~m}\) (in the flow direction) and a width of \(0.10 \mathrm{~m}\). The Reynolds number based on the plate length is 40,000 . What is the rate of heat transfer from the plate to the air? If the free stream velocity of the air is doubled and the pressure is increased to \(10 \mathrm{~atm}\), what is the rate of heat transfer?

Fluid velocities can be measured using hot-film sensors, and a common design is one for which the sensing element forms a thin film about the circumference of a quartz rod. The film is typically comprised of a thin \((\sim 100 \mathrm{~nm})\) layer of platinum, whose electrical resistance is proportional to its temperature. Hence, when submerged in a fluid stream, an electric current may be passed through the film to maintain its temperature above that of the fluid. The temperature of the film is controlled by monitoring its electric resistance, and with concurrent measurement of the electric current, the power dissipated in the film may be determined. Proper operation is assured only if the heat generated in the film is transferred to the fluid, rather than conducted from the film into the quartz rod. Thermally, the film should therefore be strongly coupled to the fluid and weakly coupled to the quartz rod. This condition is satisfied if the Biot number is very large, \(B i=\bar{h} D / 2 k \geqslant 1\), where \(\bar{h}\) is the convection coefficient between the fluid and the film and \(k\) is the thermal conductivity of the rod. (a) For the following fluids and velocities, calculate and plot the convection coefficient as a function of velocity: (i) water, \(0.5 \leq V \leq 5 \mathrm{~m} / \mathrm{s}\); (ii) air, \(1 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\). (b) Comment on the suitability of using this hot-film sensor for the foregoing conditions.

An \(L=1\)-m-long vertical copper tube of inner diameter \(D_{i}=20 \mathrm{~mm}\) and wall thickness \(t=2 \mathrm{~mm}\) contains liquid water at \(T_{w}=0^{\circ} \mathrm{C}\). On a winter day, air at \(V=3 \mathrm{~m} / \mathrm{s}, T_{\infty}=-20^{\circ} \mathrm{C}\) is in cross flow over the tube. (a) Determine the heat loss per unit mass from the water (W/kg) when the tube is full of water. (b) Determine the heat loss from the water (W/kg) when the tube is half full.
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