Question

A long 12-cm-diameter steam pipe whose external surface temperature is \(90^{\circ} \mathrm{C}\) passes through some open area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit of its length when the air is at \(1 \mathrm{~atm}\) pressure and $7^{\circ} \mathrm{C}\( and the wind is blowing across the pipe at a velocity of \)65 \mathrm{~km} / \mathrm{h}$.

Short answer

Question: Determine the rate of heat loss per unit length from a steam pipe with a 12 cm diameter and a surface temperature of 90°C. The surrounding air is at 7°C and 1 atm pressure, with a wind velocity of 65 km/h. Answer: The rate of heat loss from the steam pipe per unit length is 110.77 W/m.

Step-by-step solution

Convert units

First, let's convert the given units into the SI units. The wind velocity should be converted from km/h to m/s and the diameter from cm to m. Wind velocity: \(65 \mathrm{~km} / \mathrm{h} = 65 * (1000 \mathrm{~m/} 1 \mathrm{~km}) * (1 \mathrm{~h} / 3600 \mathrm{~s}) = 18.06 \mathrm{~m} / \mathrm{s}\) Diameter: \(12 \mathrm{~cm} = 12 / 100 = 0.12 \mathrm{~m}\)

Calculate the Reynolds Number

We'll need to find the Reynolds Number (Re) for external flow over the cylindrical pipe. The Reynolds Number can be found using the following formula: Re = \( \dfrac{VD\rho}{\mu} \), where V is the wind velocity, D is the diameter, \(\rho\) is the density of air at the given temperature, and \(\mu\) is the dynamic viscosity of air. At \(7^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure, we can find \(\rho\) and \(\mu\) from standard air property tables: \(\rho = 1.25 \mathrm{~kg} / \mathrm{m^3}\) \(\mu = 1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\) Now, calculate Re: Re = \( \dfrac{(18.06 \mathrm{~m} / \mathrm{s})(0.12 \mathrm{~m})(1.25 \mathrm{~kg} / \mathrm{m^3})}{1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}} = 1.805 \times 10^5 \)

Calculate the convective heat transfer coefficient

Since we have the Reynolds Number, we can find the Nusselt Number (Nu) for the flow over the pipe. For turbulent flow over a cylinder, we'll use the Churchill-Bernstein equation: Nu = \(0.3 + \dfrac{0.62 * Re^{1/2} * Pr^{1/3}}{[1 + (0.4 / Pr)^{2/3}]^{1/4}} * (1 + (Re/282000)^{5/8})^{4/5}\) where Pr is the Prandtl Number of air. At \(7^{\circ} \mathrm{C}\), we have Pr = 0.71: Nu = \(0.3 + \dfrac{0.62 * (1.805 \times 10^5)^{1/2} * (0.71)^{1/3}}{[1 + (0.4 / 0.71)^{2/3}]^{1/4}} * (1 + ((1.805 \times 10^5)/282000)^{5/8})^{4/5} = 181.9\) Now, we can calculate the convective heat transfer coefficient (h) using the relation: h = \(\dfrac{Nu * k}{D}\) where k is the thermal conductivity of air. At \(7^{\circ} \mathrm{C}\), we have k = \(0.0241 \mathrm{~W} / (\mathrm{m} \cdot{} K)\): h = \(\dfrac{(181.9)(0.0241 \mathrm{~W/} (\mathrm{m} \cdot{} K))}{0.12 \mathrm{~m}} = 3.641 \mathrm{~W} / (\mathrm{m}^2 \cdot{} K)\)

Calculate the rate of heat loss per unit length

Finally, we can determine the rate of heat loss per unit length (q) using the convective heat transfer coefficient: q = \(h * A * \Delta T\) where A is the surface area per unit length of the pipe and \(\Delta T\) is the temperature difference between the pipe surface and the air. A = \(\pi D = \pi (0.12 \mathrm{~m}) = 0.377 \mathrm{~m}^2 / \mathrm{m}\) \(\Delta T = (90 - 7)^\circ \mathrm{C} = 83^\circ \mathrm{C}\) Now, find q: q = \((3.641 \mathrm{~W} / \mathrm{m}^2 \cdot K) (0.377 \mathrm{~m}^2 / \mathrm{m}) (83 K) = 110.77 \mathrm{~W} / \mathrm{m}\) The rate of heat loss from the pipe per unit of its length is \(110.77 \mathrm{~W} / \mathrm{m}\).