Question

A \(0.083\)-in-diameter electrical wire at \(90^{\circ} \mathrm{F}\) is covered by \(0.02\)-in-thick plastic insulation $\left(k=0.075 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)$. The wire is exposed to a medium at \(50^{\circ} \mathrm{F}\), with a combined convection and radiation heat transfer coefficient of $2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}$. Determine if the plastic insulation on the wire will increase or decrease heat transfer from the wire. Answer: It helps

Short answer

Answer: The plastic insulation on the electrical wire decreases the heat transfer from the wire.

Step-by-step solution

Calculate the surface area of the wire

The first thing we need to do is calculate the surface area of the wire. The surface area of the wire cylinder can be calculated using the formula \(A = 2\pi r L\), where \(r\) is the radius and \(L\) is the length of the wire. Since we don't have the length, we can assume a length of 1 ft and the result will be in terms of heat transfer per foot. The radius of the wire can be calculated by dividing the diameter by 2. $$r_{wire} = \frac{0.083}{2} = 0.0415\, \mathrm{ft}$$ Now, plug in the value of \(r\) into the formula: $$A = 2\pi(0.0415)(1) = 0.261\, \mathrm{ft}^2$$

Calculate the heat transfer rate without insulation

Now let's calculate the heat transfer rate without insulation. We use the formula for the heat transfer rate due to combined convection and radiation: $$q = hA(T_{wire} - T_{medium})$$ Here, \(h\) is the combined convection and radiation heat transfer coefficient, \(A\) is the surface area of the wire, \(T_{wire}\) is the temperature of the wire, and \(T_{medium}\) is the temperature of the surrounding medium. Plug in the values: $$q_{no\_insulation} = 2.5(0.261)(90 - 50) = 26.1\, \mathrm{Btu/h}$$

Calculate equivalent thermal resistance for insulation

To calculate the heat transfer rate with the insulation, we first need to calculate the equivalent thermal resistance for the insulation. The formula for the thermal resistance of insulation is: $$R_{insulation} = \frac{r_{outer} - r_{inner}}{2\pi k L}$$ where \(r_{inner}\) is the radius of the wire, \(r_{outer}\) is the radius of the insulated wire, and \(k\) is the thermal conductivity of the insulation material. First, compute \(r_{outer}\): $$r_{outer} = r_{inner} + 0.02 = 0.0415+0.02 = 0.0615\, \mathrm{ft}$$ Next, plug in the values into the formula: $$R_{insulation} = \frac{0.0615 - 0.0415}{2\pi (0.075)(1)} = 0.674\, \mathrm{h \cdot ft \cdot { }^{\circ}F/ Btu }$$

Calculate the heat transfer rate with insulation

To find the heat transfer rate with insulation, we'll use the formula: $$q_{insulation} = \frac{T_{wire} - T_{medium}}{R_{insulation} + \frac{1}{hA_{insulated}}}$$ First we need the surface area of the insulated wire: $$A_{insulated} = 2\pi(0.0615)(1) = 0.387\, \mathrm{ft}^2$$ Now plug in the values to solve for \(q_{insulation}\): $$q_{insulation} = \frac{90 - 50}{0.674 + \frac{1}{2.5(0.387)}} = 14.9\, \mathrm{Btu/h}$$

Compare the heat transfer rates with and without insulation

Now, compare the heat transfer rates: $$q_{no\_insulation} = 26.1\, \mathrm{Btu/h}$$ $$q_{insulation} = 14.9\, \mathrm{Btu/h}$$ Since the heat transfer rate without insulation (\(q_{no\_insulation}\)) is higher than the heat transfer rate with insulation (\(q_{insulation}\)), we can conclude that the plastic insulation on the wire decreases the heat transfer from the wire.