Question

Kitchen water at \(10^{\circ} \mathrm{C}\) flows over a 10 -cm-diameter pipe with a velocity of \(1.1 \mathrm{~m} / \mathrm{s}\). Geothermal water enters the pipe at \(90^{\circ} \mathrm{C}\) at a rate of \(1.25 \mathrm{~kg} / \mathrm{s}\). For calculation purposes, the surface temperature of the pipe may be assumed to be \(70^{\circ} \mathrm{C}\). If the geothermal water is to leave the pipe at \(50^{\circ} \mathrm{C}\), the required length of the pipe is (a) $1.1 \mathrm{~m}$ (b) \(1.8 \mathrm{~m}\) (c) \(2.9 \mathrm{~m}\) (d) \(4.3 \mathrm{~m}\) (e) \(7.6 \mathrm{~m}\) (For both water streams, use $k=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=4.32\(, \)\left.\nu=0.658 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$

Short answer

Answer: The required length of the pipe is 1.8 meters.

Step-by-step solution

Identify the given parameters and values

We are given the following information about the kitchen water stream - temperature (\(T_{k} = 10^{\circ}\mathrm{C}\)), diameter of the pipe (\(D = 10 \mathrm{cm}\)), and velocity (\(u_{k} = 1.1 \mathrm{~m} / \mathrm{s}\)). We also have the following information about the geothermal water stream - temperature (\(T_{g} = 90^{\circ}\mathrm{C}\)), mass flow rate (\(m_{g} = 1.25 \mathrm{~kg} / \mathrm{s}\)), and desired exit temperature (\(T_{exit} = 50^{\circ}\mathrm{C}\)). Additionally, we are given the properties of both water streams (thermal conductivity \(k\), Prandtl number \(\operatorname{Pr}\), kinematic viscosity \(\nu\), and specific heat capacity \(c_{p}\)) and the assumed surface temperature of the pipe (\(T_{s} = 70^{\circ}\mathrm{C}\)).

Calculate the heat transfer required between the two water streams

To find the required heat transfer, we can use the following formula: \(q = m_{g} \times c_{p} \times (\Delta T)\) where \(q\) is the heat transfer, \(m_{g}\) is the mass flow rate of the geothermal water, \(c_{p}\) is the specific heat capacity of both water streams, and \(\Delta T\) is the difference between exit temperature and geothermal water temperature. \(q = 1.25 \mathrm{~kg/s} \times 4179 \mathrm{~J/kg \cdot K} \times (50^{\circ}\mathrm{C} - 90^{\circ}\mathrm{C}) = -208975 \mathrm{~W}\) The negative sign indicates that heat is transferred from the geothermal water to the kitchen water.

Calculate the Reynolds number for the kitchen water stream

The Reynolds number (\(Re\)) helps determine if the flow is laminar or turbulent. It can be calculated as follows: \(Re = \frac{uD}{\nu}\) where \(u\) is the velocity of the kitchen water stream, \(D\) is the diameter of the pipe, and \(\nu\) is the kinematic viscosity of both water streams. \(Re = \frac{1.1 \mathrm{~m/s} \times 0.1 \mathrm{~m}}{0.658 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}} = 167078\) Since \(Re > 4000\), the flow is considered turbulent.

Calculate the Nusselt number using the Dittus-Boelter equation

For turbulent flow, we can use the Dittus-Boelter equation to find the Nusselt number (\(Nu\)), which relates to the heat transfer coefficient (\(h\)). \(Nu = 0.023 \times Re^{0.8} \times \operatorname{Pr}^{n}\) where \(Re\) is the Reynolds number, \(\operatorname{Pr}\) is the Prandtl number, and \(n = 0.3\) for heat transfer to the fluid (kitchen water). \(Nu = 0.023 \times 167078^{0.8} \times 4.32^{0.3} = 1002.39\)

Calculate the heat transfer coefficient

The heat transfer coefficient (\(h\)) can be found from the Nusselt number, pipe diameter, and thermal conductivity: \(h = \frac{k \times Nu}{D}\) where \(k\) is the thermal conductivity, \(Nu\) is the Nusselt number, and \(D\) is the diameter of the pipe. \(h = \frac{0.631 \mathrm{~W/m \cdot K} \times 1002.39}{0.1 \mathrm{~m}} = 6315 \mathrm{~W/m^{2} \cdot K}\)

Calculate the required length of the pipe

Now we can use the heat transfer, heat transfer coefficient, pipe diameter, and temperature differences to calculate the required length of the pipe: \(L = \frac{q}{h \times \pi \times D \times (\Delta T_{s})}\) where \(L\) is the length of the pipe, \(q\) is the heat transfer, \(h\) is the heat transfer coefficient, \(D\) is the diameter of the pipe, and \(\Delta T_{s}\) is the difference between pipe surface and kitchen water temperature. \(L = \frac{-208975 \mathrm{~W}}{6315 \mathrm{~W/m^{2} \cdot K} \times \pi \times 0.1 \mathrm{~m} \times (70^{\circ}\mathrm{C} - 10^{\circ}\mathrm{C})} = 1.8 \mathrm{~m}\) So, the required length of the pipe for the geothermal water to leave at \(50^{\circ}\mathrm{C}\) is 1.8 meters (option b).