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A \(1 \mathrm{~m}\)-high vertical wall is maintained at \(310 \mathrm{~K}\), when the surrounding air is at 1 atm and \(290 \mathrm{~K}\). Plot the local heat transfer coefficient as a function of location up the wall. Take \(v=15.7 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) for air. Also calculate the convective heat loss per meter width of wall.

Short Answer

Expert verified
Plot the local heat transfer coefficient along the wall using the calculated Nusselt numbers. The convective heat loss is found by integrating the heat transfer along the wall.

Step by step solution

01

Identify the Problem

We need to determine the local heat transfer coefficient along a vertical wall and calculate the total convective heat loss per meter width. This involves finding the variation of heat transfer with respect to the wall's height and applying convective heat transfer formulas.
02

Calculate Grashof Number

The Grashof number \(Gr\) is calculated as: \[ Gr = \frac{g \cdot \beta \cdot (T_s - T_\infty) \cdot L^3}{v^2} \]where \(g = 9.81 \, \mathrm{m/s^2}\), \(\beta = \frac{1}{T_m} \), \(T_s = 310 \, \mathrm{K}\), \(T_\infty = 290 \, \mathrm{K}\), \(L\) is the height of the wall which is \(1 \, \mathrm{m}\), and \(v = 15.7 \times 10^{-6} \, \mathrm{m^2/s}\). First, find the mean temperature \(T_m = \frac{T_s + T_\infty}{2} = 300\, \mathrm{K}\). Calculate \(\beta = \frac{1}{T_m} = \frac{1}{300} = 3.33 \times 10^{-3} \, \mathrm{K}^{-1}\). Calculate \(Gr\) using these values.
03

Calculate Rayleigh Number

The Rayleigh number \(Ra\) is calculated as: \[ Ra = Gr \cdot Pr \]where \(Pr\) is the Prandtl number for air. For air at approximately 300 K, \(Pr \approx 0.71\). Use the \(Gr\) calculated from Step 2 to find \(Ra\).
04

Evaluate Heat Transfer Coefficient

Use the Nusselt number relation for a vertical plate to find the local heat transfer coefficient: \[ Nu_x = C \cdot \left(Gr_x \cdot Pr\right)^{1/4} \]where \(C\) is a constant, usually around 0.59 for laminar flow. Calculate \(h_x\) from \( Nu_x = \frac{h_x \cdot x}{k} \) where \(k\) for air is approximately \(0.025 \, \mathrm{W/m \, K}\). Solving gives \(h_x = \frac{0.025 \cdot Nu_x}{x}\). Repeat for various \(x\) to find the change in coefficient with height.
05

Calculate Total Convective Heat Loss

The convective heat loss per meter width of the wall \(Q\) is found by integrating the local heat transfer coefficient over the length of the wall:\[ Q = \int_0^L h_x(T_s - T_\infty) \, dx \]Substitute the average \(h\) value obtained and compute \(Q\). This represents the convective heat loss per meter width.

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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 the mechanism by which heat is transferred between a solid surface and a fluid touching that surface. In this process, the flow of the fluid carries away heat, affecting the surface temperature. For our scenario, the air flowing alongside a heated wall picks up heat from the wall (since the wall is at higher temperature). Heat transfer by convection can be categorized into natural and forced convection. For this example, we deal with natural convection, driven by the natural buoyancy forces arising from the temperature difference between the air and the wall.

Factors influencing convective heat transfer include the fluid's velocity, its properties like viscosity and thermal conductivity, and the surface's conditions. The local heat transfer coefficient, often denoted as \(h\), is a critical parameter in these calculations. It represents the capability of a section of the surface to transfer heat to the fluid. By plotting \(h\) as a function of location up the wall, we assess how efficiently different parts of the wall can dissipate heat.
Grashof Number
The Grashof number \(Gr\) is a dimensionless value in convective heat transfer that helps predict flow patterns in natural convection. It essentially compares the buoyancy forces to the viscous forces occurring in the fluid. If you imagine the heated wall, the hot air near it rises because it becomes less dense, creating a flow. Grashof number comes into play by quantifying this flow: a higher Grashof number suggests a stronger natural convection current.

To determine the Grashof number, we use the formula:\[ Gr = \frac{g \cdot \beta \cdot (T_s - T_{\infty}) \cdot L^3}{v^2} \]Where:
  • \(g\) is the acceleration due to gravity.
  • \(\beta\) is the thermal expansion coefficient (inverse of the mean temperature).
  • \(T_s\) and \(T_{\infty}\) are the surface and surrounding temperatures, respectively.
  • \(L\) is the characteristic length, the height of the wall in this case.
  • \(v\) is the kinematic viscosity of the fluid.
Understanding Grashof Number is crucial as it directly influences the calculation of other parameters like the Rayleigh and Nusselt numbers, pivotal in evaluating heat transfer.
Nusselt Number
The Nusselt number \(Nu\) is another vital dimensionless number in convective heat transfer, defining the ratio of convective to conductive heat transfer across a boundary. For an engineer or a scientist, this tells us how effective a fluid is at conducting heat away from a surface. A higher Nusselt number indicates more efficient convection, meaning the fluid can carry away more heat from the object's surface.

For vertical plates, the Nusselt number can be related to the Grashof and Prandtl numbers using empirical relations:\[ Nu_x = C \cdot (Gr_x \cdot Pr)^{1/4} \]Where:
  • \(C\) is a constant, often around 0.59 for laminar flows.
  • \(Gr_x\) and \(Pr\) are the local Grashof number and Prandtl number, respectively.
In practical applications, once the Nusselt number is found, it can be used to calculate the local heat transfer coefficient \(h\) through the equation:\[ Nu_x = \frac{h_x \cdot x}{k} \]Where \(k\) is the thermal conductivity of the fluid. The Nusselt number allows us to quantify changes in heat transfer efficiency along the height of the wall, critical in design and analysis scenarios.
Prandtl Number
The Prandtl number \(Pr\) is a dimensionless number that describes the relationship between momentum diffusivity (viscosity) and thermal diffusivity. In essence, it helps us understand how heat energy within a fluid diffuses in relation to its velocity or vice versa.

For air at approximately 300 K, the Prandtl number is around 0.71. This value illustrates that, in air, momentum diffuses faster than thermal energy. In practical terms, the Prandtl number aids in determining the thermal boundary layer's thickness when analyzing the heat transfer process for different fluids.

With reference to the heat transfer along a heated vertical wall, the Prandtl number plays a vital role in calculating both the Nusselt and Rayleigh numbers. It serves as a bridge in relations that use temperature and flow velocity to determine how convective processes evolve. In summary, the Prandtl number's importance lies in its ability to link the fluid flow characteristics with the thermal dynamics, helping engineers understand and control heat transfer rates in various applications.

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Most popular questions from this chapter

A thermocouple junction bead is modeled as a \(1 \mathrm{~mm}\)-diameter lead sphere \(\left(\rho=11,340 \mathrm{~kg} / \mathrm{m}^{3}, c=129 \mathrm{~J} / \mathrm{kg} \mathrm{K}\right)\) and is initially at a room temperature of \(20^{\circ} \mathrm{C}\). If the thermocouple is suddenly immersed in ice water to serve as a reference junction, what will be the error in indicated temperature corresponding to 1,2, and 3 times the time constant of the thermocouple? If the heat transfer coefficient is calculated to be \(2140 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), what are the corresponding times?

A natural convection heat transfer coefficient meter is intended for situations where the air temperature \(T_{e}\) is known but the surrounding surfaces are at an unknown temperature \(T_{w}\). The two sensors that make up the meter each have a surface area of \(1 \mathrm{~cm}^{2}\), one has a surface coating of emittance \(\varepsilon_{1}=0.9\), and the other has an emittance of \(\varepsilon_{2}=0.1\). The rear surface of the sensors is well insulated. When \(T_{e}=300 \mathrm{~K}\) and the test surface is at \(320 \mathrm{~K}\), the power inputs required to maintain the sensor surfaces at \(320 \mathrm{~K}\) are \(\dot{Q}_{1}=21.7 \mathrm{~mW}\) and \(\dot{Q}_{2}=8.28 \mathrm{~mW}\). Determine the heat transfer coefficient at the meter location.

An alloy cylinder \(3 \mathrm{~cm}\) in diameter and \(2 \mathrm{~m}\) high is removed from an oven at \(200^{\circ} \mathrm{C}\) and stood on its end to cool in air at \(20^{\circ} \mathrm{C}\). Give an estimate of the time for the cylinder to cool to \(100^{\circ} \mathrm{C}\) if the convective heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2}\) \(\mathrm{K}\). For the alloy, take \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=340 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), and \(k=110 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

A. \(83 \mathrm{~mm}-\mathrm{high}\) Styrofoam cup has \(1.5 \mathrm{~mm}\)-thick walls and is filled with \(180 \mathrm{ml}\) coffee at \(80^{\circ} \mathrm{C}\) and covered with a lid. The outside diameter of the cup varies from \(45 \mathrm{~mm}\) at its base to \(73 \mathrm{~mm}\) at its top. The ambient air is at \(24^{\circ} \mathrm{C}\), and the combined convective and radiative heat transfer coefficient for the outside of the cup is estimated to be \(10 \mathrm{~W} / \mathrm{m} \mathrm{K}\). (i) Determine the initial rate of heat loss through the side walls of the cup and the corresponding temperature of the outer surface. (ii) Estimate the time for the coffee to cool to \(60^{\circ} \mathrm{C}\) if the average of the heat fluxes through the lid and base are taken to be equal to the flux through the side walls. Comment on the significance of your answer to actual cooling rates experienced at the morning coffee break. Take \(k=0.035 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the Styrofoam, and \(\rho=985\) \(\mathrm{kg} / \mathrm{m}^{3}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \mathrm{K}\) for the coffee.

Two small blackened spheres of identical size-one of aluminum, the other of an unknown alloy of high conductivity-are suspended by thin wires inside a large cavity in a block of melting ice. It is found that it takes \(4.8\) minutes for the temperature of the aluminum sphere to drop from \(3^{\circ} \mathrm{C}\) to \(1^{\circ} \mathrm{C}\), and \(9.6\) minutes for the alloy sphere to undergo the same change. If the specific gravities of the aluminum and alloy are \(2.7\) and \(5.4\), respectively, and the specific heat of the aluminum is \(900 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), what is the specific heat of the alloy?

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