Question

Hot water $\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( flows through an 80 -m-long PVC \)(k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( pipe whose inner diameter is \)2 \mathrm{~cm}$ and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of $1 \mathrm{~kg} / \mathrm{s}\(, entering at \)40^{\circ} \mathrm{C}$. If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(35^{\circ} \mathrm{C}\) (b) \(36^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(38^{\circ} \mathrm{C}\) (e) \(39^{\circ} \mathrm{C}\)

Short answer

Answer: 38.374°C

Step-by-step solution

Calculate the Surface Area of the Pipe

To calculate the surface area of the pipe, use the formula for the surface area of a cylinder: \(A = 2 \pi r L\) where A is the surface area, r is the radius of the pipe, and L is the length of the pipe. First, convert the diameter to radius: \(r = \dfrac{2 \mathrm{cm}}{2} = 1 \mathrm{cm} = 0.01 \mathrm{m}\) Now, plug in the values to get the surface area: \(A = 2 \pi (0.01 \mathrm{m})(80 \mathrm{m})\) \(A = 1.6 \pi \mathrm{m}^2\)

Determine the Heat Transfer Rate

To determine the overall heat transfer rate of the water through the pipe walls, we can use the following formula, considering the inner and outer surface temperatures and the thickness of the pipe walls: \(Q = U A \Delta T\) \(Q = \cfrac{k A}{\delta} (\cfrac{T_{in} - T_{out}}{R_{in} + R_{out}})\) where, Q is the heat transfer rate, A is the surface area of the pipe, 𝑘 is the pipe's thermal conductivity, 𝛿 is the thickness of the pipe walls, 𝑇_𝑖𝑛 and 𝑇_𝑜𝑢𝑡 are the temperatures of the inner and outer surface, and 𝑅_𝑖𝑛, and 𝑅_𝑜𝑢𝑡 are the inner and outer radii respectively. To find 𝛿, we subtract the inner radius from the outer radius (convert the diameter to radius first): \(\delta = \cfrac{2.5 \mathrm{cm}}{2} - \cfrac{2 \mathrm{cm}}{2} = 0.25 \mathrm{cm} = 0.0025 \mathrm{m}\) Now, plug in the values: \(Q = \cfrac{0.092 \cdot 1.6\pi \; m^2}{0.0025\; m} \cdot \cfrac{40-20}{0.01+0.01125}\) \(Q = 6796.32 \mathrm{~W}\)

Find the Outlet Temperature

To find the outlet temperature of the water, we can use the heat transfer equation along with the mass flow rate and specific heat of the water: \(\Delta T = \cfrac{Q}{m \cdot c_p}\) where \(\Delta T\) is the temperature difference between the inlet and outlet, Q is the heat transfer rate, m is the mass flow rate, and \(c_p\) is the specific heat of the water. Using the given mass flow rate and specific heat, plug in the values: \(\Delta T = \cfrac{6796.32 \mathrm{~W}}{1\frac{\mathrm{kg}}{\mathrm{s}}\cdot 4.179 \frac{\mathrm{kJ}}{\mathrm{kg}\cdot \mathrm{K}} \cdot 1000 \frac{\mathrm{W}}{\mathrm{kJ}}}\) \(\Delta T = 1.6263\, \mathrm{K}\) Since the initial temperature of the water is \(40^{\circ} \mathrm{C}\) and the interior surface temperature is \(35^{\circ} \mathrm{C}\), the outlet temperature of the water will be lower than the initial temperature. Therefore, the outlet temperature will be: \(T_{out} = T_{in} - \Delta T\) \(T_{out} = 40^{\circ} \mathrm{C}- 1.6263^{\circ} \mathrm{C} = 38.374^{\circ} \mathrm{C}\) Since the result is closer to \(38 ^{\circ} \mathrm{C}\), the correct answer is (d) \(38^{\circ} \mathrm{C}\).