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Chapter 6: Problem 9

The convection heat transfer coefficient for a clothed person standing in moving air is expressed as \(h=14.8 \mathrm{~V}^{0.69}\) for \(0.15

Short Answer

Expert verified
The rates of heat loss from the person by convection for the air velocities are: (a) For 0.5 m/s: Q鈧� = 302.67 W (b) For 1.0 m/s: Q鈧� = 476.60 W (c) For 1.5 m/s: Q鈧� = 654.88 W

Step by step solution

01

Write down the given data and formula

We are given the following data: - Convection heat transfer coefficient: \(h=14.8 V^{0.69}\) - Body surface area: \(A = 1.7 \mathrm{~m}^{2}\) - Surface temperature: \(T_{s} = 29^{\circ} \mathrm{C}\) - Surrounding air temperature: \(T_{\infty} = 10^{\circ} \mathrm{C}\) - Air velocities: \(V=(0.5, 1.0, 1.5) \mathrm{~m} / \mathrm{s}\) The formula to find the rate of heat loss: \(Q = hA(T_{s}-T_{\infty})\)
02

Calculate the convection heat transfer coefficients for each air velocity

Calculate the convection heat transfer coefficient for: (a) 0.5 m/s: \(h_{1}=14.8(0.5)^{0.69}= 9.8102\ \mathrm{W/(m^2K)}\) (b) 1.0 m/s: \(h_{2}=14.8(1.0)^{0.69}= 14.8\ \mathrm{W/(m^2K)}\) (c) 1.5 m/s: \(h_{3} = 14.8(1.5)^{0.69} = 20.150\ \mathrm{W/(m^2K)}\)
03

Calculate the rate of heat loss for each air velocity

Calculate the rate of heat loss for each convection heat transfer coefficient: (a) \(0.5 \mathrm{~m} / \mathrm{s}\): \(Q_{1} = h_{1}A (T_{s} - T_{\infty}) = 9.8102\times1.7\times(29-10) = 302.67\ \mathrm{W}\) (b) \(1.0 \mathrm{~m} / \mathrm{s}\): \(Q_{2} = h_{2}A (T_{s} - T_{\infty}) = 14.8\times1.7\times(29-10) = 476.60\ \mathrm{W}\) (c) \(1.5 \mathrm{~m} / \mathrm{s}\): \(Q_{3} = h_{3}A (T_{s} - T_{\infty}) = 20.150\times1.7\times(29-10) = 654.88\ \mathrm{W}\)
04

Present the final results

The rate of heat loss from the person by convection for each air velocity is: (a) \(0.5 \mathrm{~m} / \mathrm{s}\): \(Q_{1} = 302.67\ \mathrm{W}\) (b) \(1.0 \mathrm{~m} / \mathrm{s}\): \(Q_{2} = 476.60\ \mathrm{W}\) (c) \(1.5 \mathrm{~m} / \mathrm{s}\): \(Q_{3} = 654.88\ \mathrm{W}\)

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Key Concepts

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

Heat Loss Calculation
Heat loss calculation by convection is essential to understand how heat energy is transferred from our bodies to the surrounding air. This process can be particularly important for determining thermal comfort and energy efficiency. The formula used to calculate the heat loss rate is: \[ Q = hA(T_s - T_{\infty}) \] Where:
  • \(Q\) represents the rate of heat loss (in watts).
  • \(h\) is the convection heat transfer coefficient, which changes with air velocity.
  • \(A\) is the surface area from which heat is lost; for a clothed person, this is the body surface area.
  • \(T_s\) is the surface temperature of the body, and \(T_{\infty}\) is the surrounding air temperature.
The convection coefficient \(h\) can vary significantly based on environmental conditions like air velocity. This calculation helps in tailoring clothing and exposing duration in various climates.
Air Velocity Impact
Air velocity has a profound effect on convection heat transfer. As air velocity increases, more heat can be removed from a surface. This is because faster-moving air can carry away heat quickly, thereby increasing the heat transfer coefficient \(h\). When using the formula: \[ h = 14.8 V^{0.69} \] We see that \(h\) increases with the velocity \(V\) of air. - For low air velocity \((0.5 \text{ m/s})\), \(h\) is around \(9.81 \text{ W/(m}^2 \text{K)}\).- As air velocity reaches \(1.0 \text{ m/s}\), the coefficient raises to \(14.8 \text{ W/(m}^2 \text{K)}\).- At \(1.5 \text{ m/s}\), \(h\) peaks at \(20.150 \text{ W/(m}^2 \text{K)}\). This sequence demonstrates the influence of air motion on heat dissipation from a surface, which is crucial when determining clothing insulation or designing heating systems.
Surface Temperature
Surface temperature plays a critical role in convection heat transfer as it represents the starting point of temperature gradient essential for energy transfer. The difference between the surface temperature \(T_s\) of your body and the surrounding air temperature \(T_{\infty}\) determines the rate at which heat is lost. In our case, the surface temperature is set at \(29^{\circ} \text{C}\), while the surrounding air is \(10^{\circ} \text{C}\). This heat gradient is essential because: - The larger the temperature difference, the greater the rate of heat transfer. - With increasing air velocity, this gradient can help more rapid heat dissipation. Understanding this interplay helps in various scenarios, from clothing design to HVAC systems, ensuring environments can be adapted for thermal comfort and energy conservation.

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Most popular questions from this chapter

Air at \(5^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), is used for cooling metal plates coming out of a heat treatment oven at an initial temperature of \(300^{\circ} \mathrm{C}\). The plates \((k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) have a thickness of \(10 \mathrm{~mm}\). Using EES (or other) software, determine the effect of cooling time on the temperature gradient in the metal plates at the surface. By varying the cooling time from 0 to \(3000 \mathrm{~s}\), plot the temperature gradient in the plates at the surface as a function of cooling time. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

In turbulent flow, one can estimate the Nusselt number using the analogy between heat and momentum transfer (Colburn analogy). This analogy relates the Nusselt number to the coefficient of friction, \(C_{f}\), as (a) \(\mathrm{Nu}=0.5 C_{f} \operatorname{Re} \operatorname{Pr}^{1 / 3}\) (b) \(\mathrm{Nu}=0.5 C_{f} \operatorname{Re} \operatorname{Pr}^{2 / 3}\) (c) \(\mathrm{Nu}=C_{f} \operatorname{Re} \operatorname{Pr}^{1 / 3}\) (d) \(\mathrm{Nu}=C_{f} \operatorname{Re} \operatorname{Pr}^{2 / 3}\)

A metal plate \(\left(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\right.\), and \(\left.c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with a thickness of \(1 \mathrm{~cm}\) is being cooled by air at \(5^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the plate is \(300^{\circ} \mathrm{C}\), determine the plate temperature gradient at the surface after 2 minutes of cooling. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

What is the physical significance of the Reynolds number? How is it defined for external flow over a plate of length \(L\) ?

Most correlations for the convection heat transfer coefficient use the dimensionless Nusselt number, which is defined as (a) \(h / k\) (b) \(k / h\) (c) \(h L_{c} / k\) (d) \(k L_{c} / h\) (e) \(k / \rho c_{p}\)
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