Question

Hot water at \(90^{\circ} \mathrm{C}\) enters 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 and \(4.6 \mathrm{~cm}\), respectively, at an average velocity of \(1.2 \mathrm{~m} / \mathrm{s}\). The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at $10^{\circ} \mathrm{C}\( in a basement, with a convection heat transfer coefficient of \)12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine \((a)\) the rate of heat loss from the water and (b) the temperature at which the water leaves the basement.

Short answer

Answer: (a) The rate of heat loss from the water is 1342.3 W. (b) The temperature at which water leaves the basement is approximately 87.31°C.

Step-by-step solution

Calculate the surface area of the exposed pipe

We need to find the surface area of the cast iron pipe to calculate the heat transfer between the pipe and the surrounding air. The surface area of a cylindrical pipe is given by \(A=2 \pi r L\), where \(r\) is the outer radius and \(L\) is the length of the pipe. Since the diameter of the outer surface is 4.6 cm, the outer radius is 2.3 cm or 0.023 m. The pipe length is given as 15 m. Let's calculate the surface area: \(A=2 \pi (0.023\,\text{m})(15\,\text{m})\)

Calculate the heat transfer through the pipe wall

To calculate the heat transfer through the pipe wall, we'll use Fourier's Law for conduction: \(q = kA\frac{T_i - T_o}{x}\), where: \(q\) - Heat transfer rate through the pipe wall \(k\) - Thermal conductivity of the pipe material (52 W/mK) \(A\) - Surface area calculated in step 1 \(T_i\) - Inner surface temperature of the pipe (90°C) \(T_o\) - Outer surface temperature of the pipe (unknown) \(x\) - Thickness of the pipe wall, which is the difference between the outer and inner radius First, let's calculate the thickness of the pipe wall: \(x = r_o - r_i = 0.023\,\text{m} - 0.020\,\text{m} = 0.003\,\text{m}\)

Calculate the heat transfer from the outer surface of the pipe to the surrounding air

To calculate the heat transfer from the outer surface of the pipe to the surrounding air, we'll use Newton's Law of Cooling: \(q = hA(T_o - T_\infty)\), where: \(q\) - Heat transfer rate from the outer surface of the pipe \(h\) - Convection heat transfer coefficient (12 W/m²K) \(A\) - Surface area calculated in step 1 \(T_o\) - Outer surface temperature of the pipe (unknown) \(T_\infty\) - Temperature of the surrounding air (10°C) Since the heat transfer through the pipe wall is equal to the heat transfer from the outer surface to the surrounding air, we can equate equations from Step 2 and Step 3: \(kA\frac{T_i - T_o}{x} = hA(T_o - T_\infty)\) Note that the surface area \(A\) cancels out from both sides of the equation. Now we can solve for the unknown outer surface temperature of the pipe (\(T_o\)).

Solve for the outer surface temperature of the pipe and heat transfer rate

Solve for \(T_o\) from the equation derived in Step 3, and then calculate the heat transfer rate using the Fourier's law formula from Step 2: \(T_o = \frac{k(T_i - T_\infty) + hT_ix}{k + hx} \approx 88.14\,\text{°C}\) \(q = k \frac{T_i - T_o}{x} \approx 1342.3\,\text{W}\) So, (a) the rate of heat loss from the water is 1342.3 W.

Calculate the temperature at which water leaves the basement

To find the temperature at which the water leaves the basement, we can use the energy balance equation: \(q = m\dot{c_p}(T_i - T_\text{out})\) where \(m\dot{}\) is the mass flow rate of the water, \(c_p\) is the specific heat capacity of water (approximated to 4190 J/kgK), \(T_i\) is the initial temperature of the water (90°C), and \(T_\text{out}\) is the outlet temperature of the water. First, we need to find the mass flow rate of the water. We can do this using the formula: \(m\dot{} = \rho VA\) where \(\rho\) is the density of water (approximated to 1000 kg/m³), \(V\) is the average velocity of the water (1.2 m/s), and \(A\) is the cross-sectional area of the pipe. Since the inner diameter of the pipe is 4 cm, the inner radius is 2 cm or 0.02 m. The cross-sectional area of the pipe is given by: \(A = \pi r_i^2 \approx 0.001256\,\text{m}^2\) Now we can calculate the mass flow rate: \(m\dot{} \approx (1000\,\text{kg/m³})(1.2\,\text{m/s})(0.001256\,\text{m}^2) \approx 1.507\,\text{kg/s}\) We can now use the energy balance equation to find the temperature at which water leaves the basement: \(T_\text{out} = T_i - \frac{q}{m\dot{c_p}}\approx 90\,\text{°C} - \frac{1342.3\,\text{W}}{(1.507\,\text{kg/s})(4190\,\text{J/kgK})} \approx 87.31\,\text{°C}\) So, (b) the temperature at which water leaves the basement is approximately 87.31°C.