Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\) in parallel over a \(0.1-\mathrm{m}\)-long flat plate where there is an unheated starting length of \(5 \mathrm{~cm}\). The heated section of the flat plate is maintained at a constant temperature of \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at the trailing edge, \((b)\) the average convection heat transfer coefficient for the heated section, and \((c)\) the rate of heat transfer per unit width for the heated section.

Short Answer

Expert verified
Answer: The local convection heat transfer coefficient at the trailing edge is 2785.3 W/(m²·K). The average convection heat transfer coefficient for the heated section is 3543.5 W/(m²·K). The rate of heat transfer per unit width for the heated section is 885875 W/m.

Step by step solution

01

Calculate the Reynolds number

To determine whether the flow is laminar or turbulent, we will need to calculate the Reynolds number. The Reynolds number for flow over a flat plate is given by \(Re_{x}=\dfrac{\rho u x}{\mu}\), where \(\rho\) is the fluid density, \(u\) is the flow velocity, \(x\) is the distance along the plate, and \(\mu\) is the dynamic viscosity. To find these properties, consult a table of properties for saturated liquid mercury at \(250^{\circ}\mathrm{C}\): \(\rho = 13611.7~\mathrm{kg/m^3}\) \(\mu = 1.555 \times 10^{-4}~\mathrm{Pa\cdot s}\) Now, substitute the given values and mercury properties to calculate the Reynolds number at the trailing edge (\(x = 0.1\,\mathrm{m}\)): \(Re_{x}=\dfrac{13611.7~\mathrm{kg/m^3} * 0.3~\mathrm{m/s} * 0.1~\mathrm{m}}{1.555 \times 10^{-4}~\mathrm{Pa\cdot s}} = 262965\) Since \(Re_{x} < 5 \times 10^5\) (a common criterion for laminar flow), we can assume laminar flow.
02

Calculate the local convection heat transfer coefficient at the trailing edge

Now that we have determined that the flow is laminar, we can use the following equation for the local convection heat transfer coefficient: \(h_{x}=\dfrac{0.332k}{x}Re_{x}^{1/2}Pr^{1/3}\) We need to find the Prandtl number, \(Pr\), and the thermal conductivity, \(k\), from the properties table for saturated liquid mercury at \(250^{\circ}\mathrm{C}\): \(k = 29.20~\mathrm{W/(m\cdot K)}\) \(Pr = 0.025\) Now we can substitute the known values into the equation: \(h_{x}=\dfrac{0.332 * 29.20~\mathrm{W/(m\cdot K)}}{0.1~\mathrm{m}} * (262965)^{1/2} * (0.025)^{1/3} = 2785.3~\mathrm{W/(m^2\cdot K)}\) Thus, the local convection heat transfer coefficient at the trailing edge is \(2785.3~\mathrm{W/(m^2\cdot K)}\).
03

Calculate the average convection heat transfer coefficient for the heated section

For laminar flow, we can use the following equation for the average convection heat transfer coefficient: \(h_{avg}=\dfrac{0.664k}{L}Re_{L}^{1/2}Pr^{1/3}\), where \(L\) is the length of the heated section. Since the unheated starting length is \(5~\mathrm{cm}\), \(L = 0.1\,\mathrm{m} - 0.05\,\mathrm{m} = 0.05\,\mathrm{m}\). Now, substitute the known values into the equation: \(h_{avg}=\dfrac{0.664 * 29.20~\mathrm{W/(m\cdot K)}}{0.05~\mathrm{m}} * (262965)^{1/2} * (0.025)^{1/3} = 3543.5~\mathrm{W/(m^2\cdot K)}\) Thus, the average convection heat transfer coefficient for the heated section is \(3543.5~\mathrm{W/(m^2\cdot K)}\).
04

Calculate the rate of heat transfer per unit width for the heated section

Finally, we will use the following heat transfer equation to calculate the rate of heat transfer per unit width: \(q''_{x} = h_{avg}(T_{s} - T_{\infty})\) Substitute the known values into the equation: \(q''_{x} = 3543.5~\mathrm{W/(m^2\cdot K)}(50^{\circ} \,\mathrm{C} - (-200^{\circ} \,\mathrm{C})) = 3543.5~\mathrm{W/(m^2\cdot K)} * 250^{\circ} \mathrm{C} = 885875 \,\mathrm{W/m}\) Thus, the rate of heat transfer per unit width for the heated section is \(885875~\mathrm{W/m}\). In summary: (a) The local convection heat transfer coefficient at the trailing edge is \(2785.3~\mathrm{W/(m^2\cdot K)}\). (b) The average convection heat transfer coefficient for the heated section is \(3543.5~\mathrm{W/(m^2\cdot K)}\). (c) The rate of heat transfer per unit width for the heated section is \(885875~\mathrm{W/m}\).

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Key Concepts

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

Reynolds number calculation
The Reynolds number is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It compares the relative significance of viscous forces and inertial forces and is vital in determining whether a fluid flow is laminar or turbulent. In the exercise, the Reynolds number calculation involves properties of liquid mercury at a high temperature and a defined velocity flowing over a flat plate.

The formula for the Reynolds number over a flat plate is expressed as: \[ Re_{x}=\frac{\rho u x}{\mu} \] where \(\rho\) is the fluid density, \(u\) the flow velocity, \(x\) the distance along the plate, and \(\mu\) the dynamic viscosity. The solution shows that the calculated Reynolds number is well below the threshold for turbulent flow, indicating laminar flow conditions.

The calculation of the Reynolds number is critical in the subsequent steps of determining the heat transfer coefficients, as the flow type impacts the correlations used for these calculations.
Laminar flow analysis
After calculating the Reynolds number, the exercise involves analyzing the flow regime. Laminar flow, characterized by smooth and orderly fluid motion, is confirmed when the Reynolds number falls below a certain threshold, typically around 500,000 for flow over a flat plate. Laminar flow is significant for several reasons:

  • It affects the local and average convection heat transfer coefficients since different empirical correlations are used for laminar and turbulent flows.
  • It's associated with lower frictional losses compared to turbulent flow.
  • Heat transfer in laminar flow relies strongly on the temperature gradient and thermal conductivity of the fluid.

In the solution, the presence of laminar flow over the heated section of the flat plate leads to the utilization of specific formulas for heat transfer coefficient calculations tailored to laminar conditions.
Prandtl number
The Prandtl number (Pr) is a dimensionless number that signifies the ratio of momentum diffusivity to thermal diffusivity, essentially comparing the rate at which momentum and heat are transported within a fluid. In the context of the exercise, the Prandtl number is a crucial parameter for calculating the local convection heat transfer coefficient in a laminar flow situation.

The Prandtl number is calculated with fluid properties, including viscosity and thermal conductivity, and for liquid mercury at \(250^\circ C\), it is given as a very low value. This implies that thermal diffusivity is significantly higher than momentum diffusivity, a characteristic that influences the convective heat transfer in the fluid. The Prandtl number is used in the formula: \[ h_{x}=\frac{0.332k}{x}Re_{x}^{1/2}Pr^{1/3} \] illustrating its impact on heat transfer alongside the Reynolds number and thermal conductivity.
Thermal conductivity
Thermal conductivity (\(k\)) measures a material's ability to conduct heat. In the exercises' context, the thermal conductivity of liquid mercury at a specific temperature significantly influences the heat transfer process. It appears in the formulas used for calculating the local and average convection heat transfer coefficients.

High thermal conductivity indicates that the material can transfer heat efficiently. For substances like liquid mercury with high thermal conductivity, heat is rapidly transferred through the fluid, which increases the convection heat transfer coefficient. The property is thus integral to determining the effectiveness with which heat can be exchanged between the mercury and the flat plate in the given scenario.
Heat transfer rate calculation
The heat transfer rate calculation is a critical step in thermal engineering applications, as it determines the amount of heat energy transferred per unit time and area. In our exercise, we calculate the rate of heat transfer per unit width of the heated section of the flat plate, with the local and average convection heat transfer coefficients already determined.

The heat transfer rate per unit width (\(q''_{x}\)) is given by the formula: \[ q''_{x} = h_{avg}(T_{s} - T_{\infty}) \] where \(h_{avg}\) is the average convection heat transfer coefficient, \(T_{s}\) is the surface temperature, and \(T_{\infty}\) is the free-stream fluid temperature. The resulting value provides insight into the thermal performance of the heated surface. The calculation underscores the importance of understanding the mechanisms of heat exchange and being able to estimate the heat transfer rates for practical thermal management.

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Most popular questions from this chapter

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m² \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m² \(\mathrm{K}\) (g) Rate of convective heat transfer, W

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \operatorname{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

How are the average friction and heat transfer coefficients determined in flow over a flat plate?

Ambient air at \(20^{\circ} \mathrm{C}\) flows over a 30-cm-diameter hot spherical object with a velocity of \(2.5 \mathrm{~m} / \mathrm{s}\). If the average surface temperature of the object is \(200^{\circ} \mathrm{C}\), the average convection heat transfer coefficient during this process is \(\begin{array}{ll}\text { (a) } 5.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & \text { (b) } 6.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\end{array}\) (c) \(7.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(9.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(11.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (For air, use \(k=0.2514 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \mathrm{Pr}=0.7309, v=1.516 \times\) \(\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{s}=1.825 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{s}=2.577 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

A glass \((k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) spherical tank is filled with chemicals undergoing exothermic reaction. The reaction keeps the inner surface temperature of the tank at \(80^{\circ} \mathrm{C}\). The tank has an inner radius of \(0.5 \mathrm{~m}\) and its wall thickness is \(10 \mathrm{~mm}\). Situated in surroundings with an ambient temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the tank's outer surface is being cooled by air flowing across it at \(5 \mathrm{~m} / \mathrm{s}\). In order to prevent thermal burn on individuals working around the container, it is necessary to keep the tank's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine whether or not the tank's outer surface temperature is safe from thermal burn hazards.

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