Question

Hot water at an average temperature of \(90^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe $(k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( whose inner and outer diameters are \)4 \mathrm{~cm}\( and \)4.6 \mathrm{~cm}$, respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at $10^{\circ} \mathrm{C}\( in the basement, with a heat transfer coefficient of \)15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The heat transfer coefficient at the inner surface of the pipe is $120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Taking the walls of the basement to be at \)10^{\circ} \mathrm{C}$ also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.

Short answer

Question: Determine the rate of heat loss from the hot water flowing through the cast iron pipe and calculate the average velocity of the water when the temperature of the water drops by \(3^{\circ} \mathrm{C}\). Answer: To determine the rate of heat loss, we calculated the resistances for conduction, convection, and radiation. The overall heat transfer rate was found using the formula: $$Q = \frac{T_i - T_{air}}{R_{cond} + R_{conv} + R_{rad}}$$ To calculate the average velocity of the water, we first computed the volume flow rate using the formula: $$\dot{V} = \frac{\dot{Q} \Delta T}{\rho C_p}$$ Then, we determined the average water velocity using the formula: $$V_{avg} = \frac{\dot{V}}{A_{c}}$$ By substituting the obtained values, we can find the rate of heat loss and the average water velocity in the pipe.

Step-by-step solution

Calculate the Resistance of Conduction

To calculate the heat loss through conduction, we need to determine the resistance of the pipe wall to heat flow. We can do this using the following equation: $$R_{cond} = \frac{\ln \left( D_o / D_i \right)}{2 \pi k L}$$ where \(R_{cond}\) = resistance for conduction, \(D_i\) = inner diameter of the pipe, \(D_o\) = outer diameter of the pipe, \(k\) = thermal conductivity of the pipe material, \(L\) = length of the pipe section. Plug in the given values, remembering to convert the diameter values to meters: $$R_{cond} = \frac{\ln \left( 0.046 / 0.040 \right)}{2 \pi (52)(15)}$$

Calculate the Resistance of Convection

To calculate the resistance of convection between the outer surface of the pipe and the cold basement air, we use the equation: $$R_{conv} = \frac{1}{h_o A_{o}}$$ Where \(R_{conv}\) = resistance for convection, \(h_o\) = heat transfer coefficient at the outer surface, \(A_o\) = outer surface area of the pipe. Calculate the outer surface area of the pipe: $$A_{o} = \pi D_o L$$ Now we can calculate the resistance of convection: $$R_{conv} = \frac{1}{(15) \pi(0.046)(15)}$$

Calculate the Resistance of Radiation

To determine the resistance of radiation from the pipe wall to the basement walls, we use the equation: $$R_{rad} = \frac{1}{\epsilon \sigma A_{o} (T_{so} + T_{sur})(T_{so}^2 + T_{sur}^2)}$$ Where \(R_{rad}\) = resistance for radiation, \(\epsilon\) = emissivity of the pipe, \(\sigma\) = Stefan-Boltzmann constant \(= 5.670 \times 10^{-8} \, \mathrm{W/m^2 K^4}\), \(T_{so}\) = outer surface temperature of the pipe, \(T_{sur}\) = surrounding basement wall temperature. To calculate the outer surface temperature of the pipe, we will first need to determine the temperature drop caused by the conduction resistance. We can do this by calculating the heat loss due to conduction using the equation: $$Q_{cond} = \frac{T_i - T_o}{R_{cond}}$$ Where \(T_i\) = inner surface temperature, \(T_o\) = outer surface temperature. Since we know the heat transfer coefficient \(h_i\), we can also calculate the heat loss due to convection from the inner pipe wall: $$Q_{conv} = h_i A_i (T_i - T_w)$$ Where \(T_w\) = hot water temperature, \(A_i\) = inner surface area of the pipe. Solve for \(T_i\) and \(T_o\) in the two equations mentioned above and substitute the obtained values to calculate the value of \(R_{rad}\).

Determine the Rate of Heat Loss

We can find the overall heat transfer rate by adding the resistances we found earlier and using the following formula: $$Q = \frac{T_i - T_{air}}{R_{cond} + R_{conv} + R_{rad}}$$

Calculate the Average Water Velocity

To calculate the average velocity of the water in the pipe, first, we need to compute the volume flow rate. Use the formula: $$\dot{V} = \frac{\dot{Q} \Delta T}{\rho C_p}$$ Where \(\dot{V}\) = volume flow rate, \(\dot{Q}\) = rate of heat loss, \(\Delta T\) = temperature drop of the water, \(\rho\) = density of water, (assume \(\rho = 1000 \, \mathrm{kg/m^3}\)), \(C_p\) = specific heat of water, (assume \(C_p = 4186 \, \mathrm{J/kg K}\)). Now, we can determine the average water velocity with the following formula: $$V_{avg} = \frac{\dot{V}}{A_{c}}$$ Where \(V_{avg}\) = average velocity, \(A_c\) = cross-sectional area of the pipe. The cross-sectional area can be calculated as: $$A_c = \frac{\pi D_i^2}{4}$$ Substitute the values obtained, and we will get the average water velocity in the pipe.