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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: Q₁ = 302.67 W (b) For 1.0 m/s: Q₂ = 476.60 W (c) For 1.5 m/s: Q₃ = 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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Most popular questions from this chapter

A long steel strip is being conveyed through a 3 -m long furnace to be heat treated at a speed of \(0.01 \mathrm{~m} / \mathrm{s}\). The steel strip \(\left(k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8000 \mathrm{~kg} / \mathrm{m}^{3}\right.\), and \(c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) has a thickness of \(5 \mathrm{~mm}\), and it enters the furnace at an initial temperature of \(20^{\circ} \mathrm{C}\). Inside the furnace, the air temperature is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using EES (or other) software, determine the surface temperature gradient of the steel strip as a function of location inside the furnace. By varying the location in the furnace for \(0 \leq x \leq 3 \mathrm{~m}\) with increments of \(0.2 \mathrm{~m}\), plot the surface temperature gradient of the strip as a function of furnace location. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Metal plates are being cooled with air blowing in parallel over each plate. The average friction coefficient over each plate is given as \(C_{f}=1.33\left(\operatorname{Re}_{L}{ }^{-0.5}\right.\) for \(\operatorname{Re}_{L}<5 \times 10^{5}\). Each metal plate length parallel to the air flow is \(1 \mathrm{~m}\). Determine the average convection heat transfer coefficient for the plate, if the air velocity is \(5 \mathrm{~m} / \mathrm{s}\). Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Consider fluid flowing with a free stream velocity of \(1 \mathrm{ft} / \mathrm{s}\) over a flat plate, where the critical Reynolds number is \(5 \times 10^{5}\). Determine the distance from the leading edge at which the transition from laminar to turbulent flow occurs for air (at 1 atm), liquid water, isobutane, and engine oil, and mercury. Evaluate all properties at \(50^{\circ} \mathrm{F}\).

Consider steady, laminar, two-dimensional, incompressible flow with constant properties and a Prandtl number of unity. For a given geometry, is it correct to say that both the average friction and heat transfer coefficients depend on the Reynolds number only?

The coefficient of friction \(C_{f}\) for a fluid flowing across a surface in terms of the surface shear stress, \(\tau_{s}\), is given by (a) \(2 \rho V^{2} / \tau_{w}\) (b) \(2 \tau_{w} / \rho V^{2}\) (c) \(2 \tau_{w} / \rho V^{2} \Delta T\) (d) \(4 \tau_{w} / \rho V^{2}\) (e) None of them
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