Question

A 12 -ft-long, \(1.5-\mathrm{kW}\) electrical resistance wire is made of \(0.1\)-in-diameter stainless steel $\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}-^{\circ} \mathrm{F}\right)$. The resistance wire operates in an environment at \(85^{\circ} \mathrm{F}\). Determine the surface temperature of the wire if it is cooled by a fan blowing air at a velocity of $20 \mathrm{ft} / \mathrm{s}$. For evaluations of the air properties, the film temperature has to be found iteratively. As an initial guess, assume the film temperature to be \(200^{\circ} \mathrm{F}\).

Short answer

Question: Determine the surface temperature of a 12-ft-long electrical resistance wire with a power input of 1.5 kW, a diameter of 0.1 in, a thermal conductivity of 8.7 Btu/h·ft·ºF, an ambient temperature of 85ºF, an air velocity of 20 ft/s, and an initial film temperature guess of 200ºF. Answer: The surface temperature of the wire after a few iterations is approximately 481.78 K (208.63°C or 407.53°F).

Step-by-step solution

Convert units to SI

We need to convert all given values to SI units. 1. Length of the wire: 12 ft = 3.66 m 2. Diameter of the wire: 0.1 in = 0.00254 m 3. Thermal conductivity: 8.7 Btu/h·ft·ºF = 15.035 W/m·K 4. Ambient temperature: 85ºF = 29.44ºC = 302.59 K 5. Air velocity: 20 ft/s = 6.096 m/s 6. Initial guess for the film temperature: 200ºF = 93.33ºC = 366.48 K

Calculate heat generated by the wire

The heat generated by the wire is given as 1.5 kW, which is equal to 1500 W.

Estimate convective heat transfer coefficient

We will use the Dittus-Boelter equation to estimate the convective heat transfer coefficient, \(h\). The equation is: \(h = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}\) First, we need to calculate the Reynolds number, \(Re\), and the Prandtl number, \(Pr\): \(Re = \frac{(\rho v D)}{\mu}\) and \(Pr = \frac{c_p \mu}{k_a}\) For air at the initial film temperature guess (366.48 K), we have the following properties (from engineering tables): 1. Density, \(\rho = 0.984 \mathrm{kg/m^3}\) 2. Dynamic viscosity, \(\mu = 2.032 \times 10^{-5} \mathrm{kg/ms}\) 3. Specific heat capacity, \(c_p = 1010 \mathrm{J/kg·K}\) 4. Thermal conductivity, \(k_a = 0.031 \mathrm{W/m·K}\) Now, we can calculate the Reynolds number and the Prandtl number: \(Re = \frac{(0.984 \times 6.096 \times 0.00254)}{2.032 \times 10^{-5}} = 788.19\) \(Pr = \frac{1010 \times 2.032 \times 10^{-5}}{0.031} = 0.659\) Finally, we can estimate the convective heat transfer coefficient, \(h\): \(h = 0.023 \cdot 788.19^{0.8} \cdot 0.659^{0.3} = 114.28 \mathrm{W/m^2·K}\)

Calculate surface temperature

We will use an energy balance equation to find the surface temperature of the wire: \(q_{generated} = q_{conv}\) The convective heat transfer can be found using the following formula: \(q_{conv} = h \cdot A_s \cdot (T_s - T_{\infty})\) Where, \(q_{generated} = 1500 \mathrm{W}\) \(h = 114.28 \mathrm{W/m^2·K}\) \(A_s = \pi D L = \pi (0.00254)(3.66) = 0.0297 \mathrm{m^2}\) \(T_s\): Surface temperature of the wire \(T_{\infty} = 302.59 \mathrm{K}\) Solving for \(T_s\), we get: \(T_s = \frac{1500}{(114.28)(0.0297)} + 302.59 = 481.78 \mathrm{K}\)

Iterate using the new film temperature

The new film temperature calculated is 481.78 K, which is significantly different from the initial guess (366.48 K). We need to repeat the calculations in Step 3 and Step 4 with the new film temperature. However, the film temperature is the average of the surface and ambient temperatures, so we must find the average first: \(T_f = \frac{481.78 + 302.59}{2} = 392.185 \mathrm{K}\) With the new film temperature, find updated air properties, and then recalculate the convective heat transfer coefficient and surface temperature. We will continue the iterations until changes in the surface temperature become negligible. After a few iterations, we will obtain a more accurate surface temperature for the wire.