Question

The resistance wire of a \(1000-\mathrm{W}\) iron is 15 in long and has a diameter of \(D=0.08\) in. Determine the rate of heat generation in the wire per unit volume in \(\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{3}\) and the heat flux on the outer surface of the wire in $\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}$ as a result of this heat generation.

Short answer

Answer: The rate of heat generation per unit volume is 78,215,435.6 Btu/h⋅ft³, and the heat flux on the outer surface of the wire is 130,383.37 Btu/h⋅ft².

Step-by-step solution

Convert the power to Btu/h

To convert the power from Watts (W) to British thermal units per hour (Btu/h), we can use the conversion factor 1W = 3.412142 Btu/h. Therefore, the power P in Btu/h is: P (Btu/h) = 1000 W * 3.412142 Btu/h/W = 3412.142 Btu/h

Find the volume of the wire

The wire has a cylindrical shape with a diameter D = 0.08 in and a length L = 15 in. First, we need to find the radius r of the cylinder, which is half of the diameter: r = D/2 = 0.08 in / 2 = 0.04 in Now, we can find the volume V of the cylinder using the formula: V = π * r^2 * L So, V = π * (0.04 in)^2 * 15 in = 0.0754 in³ Let's convert the volume from in³ to ft³ using the conversion factor 1 in³ = 0.000578704 ft³: V = 0.0754 in³ * 0.000578704 ft³/in³ = 0.0000436 ft³

Calculate the rate of heat generation per unit volume

Now that we have the power in Btu/h and the volume in ft³, we can find the rate of heat generation per unit volume Q_v by dividing the power by the volume: Q_v = P / V Q_v = 3412.142 Btu/h / 0.0000436 ft³ = 78215435.6 Btu/h⋅ft³

Find the surface area of the wire

Now, to find the heat flux on the outer surface of the wire, we need to find the surface area A of the wire. Since the wire has a cylindrical shape, we can use the formula for the surface area of a cylinder without the top and bottom: A = 2 * π * r * L So, A = 2 * π * 0.04 in * 15 in = 3.7699 in² Let's convert the surface area from in² to ft² using the conversion factor 1 in² = 0.00694444 ft²: A = 3.7699 in² * 0.00694444 ft²/in² = 0.02617 ft²

Calculate the heat flux on the outer surface of the wire

Finally, to find the heat flux Q_f on the outer surface of the wire, we need to divide the power (P) by the surface area (A): Q_f = P / A Q_f = 3412.142 Btu/h / 0.02617 ft² = 130383.37 Btu/h⋅ft² So the rate of heat generation in the wire per unit volume is 78215435.6 Btu/h⋅ft³, and the heat flux on the outer surface of the wire is 130383.37 Btu/h⋅ft².