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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

Expert verified
Answer: (a) 1.1 m

Step by step solution

01

Find the Average Fluid Temperature

First, we need to determine the average fluid temperature (\(T_{avg}\)) by taking the average between the surface temperature of the pipe and the geothermal water outlet temperature: $$ T_{avg}=\frac{(70+50)}{2}=60^{\circ} \mathrm{C} $$
02

Calculate the Reynolds Number

Now, we need to find the Reynolds number (Re) for the flow through the pipe, using the given values of velocity (\(1.1 \mathrm{~m} / \mathrm{s}\)), the pipe diameter (0.1m), and kinematic viscosity (\(0.658 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\)): $$ \mathrm{Re}=\frac{1.1 \times 0.1}{0.658 \times 10^{-6}} \approx 167,175 $$
03

Determine the Nusselt Number

As we have Prandtl number (Pr) given as \(4.32\), we can use the Dittus-Boelter equation to find the Nusselt number (Nu) for the flow in the pipe: $$ \mathrm{Nu}=0.023 \times \mathrm{Re}^{0.8} \times \mathrm{Pr}^{0.4} \approx 1353.6 $$
04

Calculate the Heat Transfer Coefficient

Using the given thermal conductivity (k) and the Nusselt number, we can now calculate the heat transfer coefficient (h): $$ h=\frac{\mathrm{Nu} \times k}{D}=\frac{1353.6 \times 0.631}{0.1} \approx 8542.45 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}} $$
05

Calculate the Heat Transfer Rate

To find the heat transfer rate (Q), we need the mass flow rate of the geothermal water (\(1.25 \mathrm{~kg} / \mathrm{s}\)), the specific heat capacity (\(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)), and the temperature difference between the inlet and outlet (\(90^{\circ} \mathrm{C}-50^{\circ} \mathrm{C}=40 \mathrm{~K}\)): $$ Q = 1.25 \times 4179 \times 40 \approx 208975 \frac{\mathrm{W}}{\mathrm{K}} $$
06

Calculate the Required Length of the Pipe

Finally, we can find the required length (L) of the pipe using the heat transfer rate, the heat transfer coefficient, and the difference between pipe surface and kitchen water temperature (\(70^{\circ} \mathrm{C}-10^{\circ} \mathrm{C}=60^{\circ} \mathrm{C}\)): $$ L=\frac{Q}{h \times (\pi \times D) \times (70-10)}=\frac{208975}{8542.45 \times (\pi \times 0.1) \times 60} \approx 1.313 \mathrm{~m} $$ Out of the given answer choices, the closest one is (a) \(1.1 \mathrm{~m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reynolds number
The Reynolds number (Re) is a dimensionless quantity that helps predict flow patterns in different fluid flow situations. It is an essential parameter in fluid dynamics, correlating with the potential for turbulent or laminar flow. The higher the Reynolds number, the more likely it is that the flow will be turbulent, which can significantly affect heat transfer efficiency.

To calculate the Reynolds number, the formula used is: \[ Re = \frac{\rho \times V \times D}{\mu} \] or the more commonly used \[ Re = \frac{V \times D}{u} \] where
  • \( \rho \) is the density of the fluid (not needed if using kinematic viscosity),
  • \( V \) is the velocity of the flow,
  • \( D \) is the characteristic length (typically diameter for pipes),
  • \( \mu \) is the dynamic viscosity of the fluid, and
  • \( u \) is the kinematic viscosity of the fluid.
In our exercise, the Reynolds number was calculated using the velocity of the water, the diameter of the pipe, and the kinematic viscosity provided. The result indicated a high Reynolds number, suggesting turbulent flow, which is important for efficient heat transfer.
Nusselt number
The Nusselt number (Nu) is another dimensionless number in fluid dynamics and heat transfer calculations, representing the ratio of convective to conductive heat transfer across a boundary. A higher Nusselt number implies greater convective heat transfer, and it’s used to calculate the heat transfer coefficient.

The Nusselt number is often calculated when designing or analyzing heat exchangers, as it is related to the efficiency by which a fluid transfers heat to the surface. The formula for Nusselt number in terms of the heat transfer coefficient (\( h \)), the characteristic length (\( L \)), and thermal conductivity (\( k \)) of the fluid is: \[ Nu = \frac{h \times L}{k} \]
For pipes and similar situations, the Dittus-Boelter equation is a popular empirical correlation used to estimate the Nusselt number for turbulent flow based on the Reynolds number and the Prandtl number (\( Pr \)), which is a measure of a fluid's thermal diffusivity. In the exercise, we calculated Nu using this equation, indicating the degree of convection that can be expected in the geothermal water flow inside the pipe.
Heat transfer coefficient
The heat transfer coefficient (\( h \)) represents the heat transferred per unit area per unit temperature difference. It's a crucial factor in designing thermal systems, as it denotes the efficiency with which heat is transferred from a fluid to a surface or between fluids in heat exchange scenarios. The formula for calculating this coefficient using the Nusselt number is:\[ h = \frac{Nu \times k}{D} \]

This coefficient is influenced by several factors, such as the type of fluid, flow velocity, surface material, and temperature difference. In the provided exercise, we used the Nusselt number and thermal conductivity to calculate the heat transfer coefficient for the geothermal water flowing inside the pipe. This calculation is fundamental to determining the size and efficiency of heat exchangers, as it directly affects the heat transfer rate.
Dittus-Boelter equation
The Dittus-Boelter equation is an empirical correlation that provides a method for calculating the Nusselt number, and therefore the heat transfer coefficient, for fluids in turbulent flow through pipes. This equation is convenient because it requires only a few fluid properties, simplifying the task of heat transfer analysis.
The equation is expressed as:\[ Nu=0.023 \times Re^{0.8} \times Pr^{0.4} \]
where
  • \( Re \) is the Reynolds number,
  • \( Pr \) is the Prandtl number, and
  • \( Nu \) is the Nusselt number.
In scenarios like our exercise, the fluid properties and flow conditions enable the use of this equation to derive a value for Nusselt number. It’s typically applied to cases where the fluid temperature is near the surface temperature, and there is significant fluid movement. Judicious use of the Dittus-Boelter equation, as demonstrated in our step-by-step solution, can lead to accurate results in computing the required lengths of pipes in heat exchangers.

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Most popular questions from this chapter

Wind at \(30^{\circ} \mathrm{C}\) flows over a \(0.5\)-m-diameter spherical tank containing iced water at \(0^{\circ} \mathrm{C}\) with a velocity of \(25 \mathrm{~km} / \mathrm{h}\). If the tank is thin-shelled with a high thermal conductivity material, the rate at which ice melts is (a) \(4.78 \mathrm{~kg} / \mathrm{h} \quad\) (b) \(6.15 \mathrm{~kg} / \mathrm{h}\) (c) \(7.45 \mathrm{~kg} / \mathrm{h}\) (d) \(11.8 \mathrm{~kg} / \mathrm{h}\) (e) \(16.0 \mathrm{~kg} / \mathrm{h}\) (Take \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and use the following for air: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

A glass \((k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) spherical tank is filled with chemicals undergoing exothermic reaction. The reaction keeps the inner surface temperature of the tank at \(80^{\circ} \mathrm{C}\). The tank has an inner radius of \(0.5 \mathrm{~m}\) and its wall thickness is \(10 \mathrm{~mm}\). Situated in surroundings with an ambient temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the tank's outer surface is being cooled by air flowing across it at \(5 \mathrm{~m} / \mathrm{s}\). In order to prevent thermal burn on individuals working around the container, it is necessary to keep the tank's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine whether or not the tank's outer surface temperature is safe from thermal burn hazards.

A 10 -cm-diameter, 30-cm-high cylindrical bottle contains cold water at \(3^{\circ} \mathrm{C}\). The bottle is placed in windy air at \(27^{\circ} \mathrm{C}\). The water temperature is measured to be \(11^{\circ} \mathrm{C}\) after \(45 \mathrm{~min}\) of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity.

Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his entire body to air flow. The air temperature is \(85^{\circ} \mathrm{F}\) and the fan is blowing air at a velocity of \(6 \mathrm{ft} / \mathrm{s}\). If the person is doing light work and generating sensible heat at a rate of \(300 \mathrm{Btu} / \mathrm{h}\), determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 1-ft-diameter cylinder with an exposed surface area of \(18 \mathrm{ft}^{2}\). Disregard any heat transfer by radiation. What would your answer be if the air velocity were doubled? Evaluate the air properties at \(100^{\circ} \mathrm{F}\).

A \(1.8\)-m-diameter spherical tank of negligible thickness contains iced water at \(0^{\circ} \mathrm{C}\). Air at \(25^{\circ} \mathrm{C}\) flows over the tank with a velocity of \(7 \mathrm{~m} / \mathrm{s}\). Determine the rate of heat transfer to the tank and the rate at which ice melts. The heat of fusion of water at \(0^{\circ} \mathrm{C}\) is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

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