Question

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\) and has a uniform surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is $2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$, determine the heat transfer coefficient as a result of convection between the pipe surface and the surroundings.

Short answer

Answer: The heat transfer coefficient due to convection between the pipe surface and the surroundings is 158.7 W/m²·K.

Step-by-step solution

Calculate the heat transfer rate

To find the heat transfer rate, we need to consider the mass flow rate, the specific heat of the vapor, and the temperature drop between the inlet and exit of the pipe. We can use the formula \(Q = \dot{m} c_p \Delta T\), where \(\dot{m}\) is the mass flow rate, \(c_p\) is the specific heat of the vapor, and \(\Delta T\) is the temperature drop. In this case, we have \(\dot{m} = 0.3 \mathrm{~kg/s}\), \(c_p = 2190 \mathrm{~J/kg·K}\), and \(\Delta T = 30^{\circ} \mathrm{C}\). Calculating the heat transfer rate, \(Q\), we get: \(Q = (0.3 \mathrm{~kg/s}) (2190 \mathrm{~J/kg·K}) (30 \mathrm{~K}) = 19770 \mathrm{~W}\) So the heat transfer rate is 19770 W.

Calculate the surface area of the pipe

The surface area of the pipe is needed to determine the heat transfer coefficient. The pipe is a cylinder with a diameter of \(5 \mathrm{~cm}\) and a length of \(10 \mathrm{~m}\). To find the surface area, we can use the formula for the surface area of a cylinder, \(A = 2\pi r L\), where \(r\) is the radius of the cylinder and \(L\) is its length. In this case, we have \(r = 2.5 \mathrm{~cm}\) (5 cm diameter divided by 2) and \(L = 10 \mathrm{~m}\) (converting cm to m, we have \(r = 0.025 \mathrm{~m}\)). Calculating the surface area, \(A\), we get: \(A = 2\pi (0.025 \mathrm{~m}) (10 \mathrm{~m}) = 1.57 \mathrm{~m^2}\) So the surface area of the pipe is 1.57 m².

Calculate the temperature difference between the pipe surface and surroundings

The temperature difference between the pipe surface and surroundings is given as: \(\Delta T_s = T_{surface} - T_{surroundings} = 100^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 80^{\circ} \mathrm{C}\) So the temperature difference between the pipe surface and surroundings is 80°C.

Calculate the heat transfer coefficient

Now we can use the convective heat transfer equation to determine the heat transfer coefficient, \(h\). The equation is given by \(Q = hA\Delta T_s\), where \(Q\) is the heat transfer rate, \(A\) is the surface area of the pipe, and \(\Delta T_s\) is the temperature difference between the pipe surface and surroundings. We can rearrange this equation to solve for \(h\): \(h = \frac{Q}{A\Delta T_s}\) Substituting the values we have found, we get: \(h = \frac{19770 \mathrm{~W}}{(1.57 \mathrm{~m^2})(80 \mathrm{~K})} = 158.7 \mathrm{~W/m^2·K}\) So the heat transfer coefficient as a result of convection between the pipe surface and the surroundings is 158.7 W/m²·K.