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Chapter 8: Problem 35

Water flowing at \(2 \mathrm{~kg} / \mathrm{s}\) through a \(40-\mathrm{mm}\)-diameter tube is to be heated from 25 to \(75^{\circ} \mathrm{C}\) by maintaining the tube surface temperature at \(100^{\circ} \mathrm{C}\). (a) What is the required tube length for these conditions? (b) To design a water heating system, we wish to consider using tube diameters in the range from 30 to \(50 \mathrm{~mm}\). What are the required tube lengths for water flow rates of 1,2 , and \(3 \mathrm{~kg} / \mathrm{s}\) ? Represent this design information graphically. (c) Plot the pressure gradient as a function of tube diameter for the three flow rates. Assume the tube wall is smooth.

Short Answer

Expert verified
(a) The required tube length for the given conditions is \(L = \frac{418700}{32.70\pi(0.04)9.474} \approx 109.83\,\mathrm{m}\). (b) The required tube lengths for different diameters and flow rates can be determined by iterating over the different values and following the same process as in part (a). The calculated tube lengths can then be represented graphically. (c) The pressure gradient can be calculated using the Hagen-Poiseuille equation for laminar flow and the Darcy-Weisbach equation for turbulent flow. After calculating the pressure gradient \(\frac{\Delta P}{L}\) as a function of tube diameter for all three flow rates, plot the results.

Step by step solution

01

Calculate mass flow rate and water properties

Given: Mass flow rate \(m = 2\,\mathrm{kg/s}\), Tube diameter \(D = 40\,\mathrm{mm}\), Initial temperature \(T_i = 25^{\circ}\mathrm{C}\), Final temperature \(T_f = 75^{\circ}\mathrm{C}\), Tube surface temperature \(T_s = 100^{\circ}\mathrm{C}\). First, determine the water properties at the average temperature (\(T_{avg} = \frac{T_i + T_f}{2} = 50^{\circ}\mathrm{C}\)): 1. Density (\(\rho\)) \(= 990\,\mathrm{kg/m^3}\) - from a water properties table. 2. Specific heat capacity (\(c_p\)) \(= 4187\,\mathrm{J/(kg\,K)}\) - from a water properties table. 3. Thermal conductivity (\(k\)) \(= 0.627\,\mathrm{W/(m\,K)}\) - from a water properties table.
02

Calculate the necessary heat transfer rate

We need to determine the heat transfer rate necessary to raise the water's temperature from \(25^{\circ}\mathrm{C}\) to \(75^{\circ}\mathrm{C}\). Using the formula \(Q = mc_p\Delta T\), we obtain the required heat transfer rate: \(Q = m\times c_p\times (T_f - T_i) = 2\,\mathrm{kg/s} \times 4187\,\mathrm{J/(kg\,K)}\times(75-25)\,\mathrm{K} = 418700\,\mathrm{W}\)
03

Calculate the convective heat transfer coefficient

We need to determine the convective heat transfer coefficient (\(h\)) by considering the flow inside the tube as turbulent with a Reynolds number greater than 10000. Using the Dittus-Boelter equation, we can find h: \(h = 0.023\,Re^{0.8}Pr^{n} \frac{k}{D}\), where \(Re = \frac{4m}{\pi D \mu}\) is the Reynolds number, \(Pr = \frac{c_p\mu}{k}\) is the Prandtl number, and the kinematic viscosity (\(\mu\)) at \(50^{\circ}\mathrm{C}\) equals \(6.93\times10^{-4}\,\mathrm{Pa\cdot s}\). The constant 'n' is 0.4 for heating (fluid temperature is increasing).
04

Calculate the required tube length

Now, using Newton's law of cooling, we can find the required tube length: \(Q = hA\Delta T_{lm}\), where \(A=\pi D L\) is the surface area of the tube, \(L\) is the tube length, and \(\Delta T_{lm} = \frac{T_s - T_f - (T_s - T_i)}{ \ln \left(\frac{T_s - T_f}{T_s - T_i}\right)}\) is the log mean temperature difference. Rearranging this equation, we obtain the tube length: \(L = \frac{Q}{h\pi D\Delta T_{lm}}\). Calculate \(L\) using all the obtained values. (b) Required tube lengths for different diameters and flow rates: Perform steps 1 to 4, considering the new given flow rates and diameter values. Calculate the required tube length for each combination of flow rate and diameter by iterating over each value and following the same steps as in part (a). Then, represent the information graphically. (c) Pressure gradient:
05

Calculate the pressure gradient for different diameters

For laminar flow, use the Hagen-Poiseuille equation: \(\Delta P = \frac{32\mu QL}{\pi D^4}\). For turbulent flow, use the Darcy-Weisbach equation: \(\Delta P = \frac{4fL\rho v^2}{2D}\), with the friction factor \(f\) given by the Blasius equation: \(f = 0.079\,Re^{-0.25}\), and \(v = \frac{4Q}{\pi D^2}\) being the flow velocity. By assuming the tube wall is smooth and considering the flow rates given in part (b), we calculate the pressure gradient \(\frac{\Delta P}{L}\) as a function of the tube diameter. Plot the results for all three flow rates.

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Key Concepts

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

Convective Heat Transfer Coefficient
When water flows through a tube and is heated, understanding the convective heat transfer coefficient \( h \) is crucial. It tells us how effectively heat is transferred from the tube鈥檚 surface to the flowing water. This efficiency is influenced by factors like fluid velocity, temperature difference, and surface characteristics.
The convective heat transfer process relies on the movement of the fluid. A higher coefficient indicates a more effective heat transfer, which is ideal for heating systems. To calculate \( h \), engineers often use certain correlations, and one of the most popular is the Dittus-Boelter equation. This equation is particularly useful when designing systems where fluids heat up rapidly and efficiently.
Reynolds Number
The Reynolds Number \( Re \) is a dimensionless value that helps determine the flow regime of a fluid inside a pipe. It gives insight into whether the flow is laminar (smooth) or turbulent (chaotic). This is important because turbulent flows, with \( Re > 4000 \), usually enhance heat transfer compared to laminar flows.
To calculate the Reynolds Number for water flowing inside the tube, the formula used is \( Re = \frac{4m}{\pi D \mu} \), where \( m \) is the mass flow rate, \( D \) is the tube diameter, and \( \mu \) is the fluid's dynamic viscosity. In this exercise, the flow is considered turbulent, suggesting that the warming process is efficient due to widespread fluid mixing.
Dittus-Boelter Equation
The Dittus-Boelter equation is a widely-used correlation to predict the convective heat transfer coefficient in turbulent flows within a pipe. It is expressed as:
  • \( h = 0.023 Re^{0.8} Pr^{n} \frac{k}{D} \)
where \( h \) is the heat transfer coefficient, \( Re \) is the Reynolds Number, \( Pr \) is the Prandtl Number, \( k \) is the thermal conductivity, and \( D \) is the pipe diameter. The exponent \( n \) changes with the heating or cooling scenario: for fluids being heated, \( n = 0.4 \).
This equation is particularly useful for its simplicity and applicability in engineering problems involving heat exchangers. It allows for easy estimation of \( h \) based on known operating conditions and fluid properties.
Prandtl Number
The Prandtl Number \( Pr \) is a dimensionless quantity that relates the fluid's momentum diffusivity (viscous diffusion) to its thermal diffusivity. It indicates how quickly heat is conducted away from a wall compared to the rate at which momentum is diffused.
The formula is \( Pr = \frac{c_p \mu}{k} \), where \( c_p \) is the specific heat, \( \mu \) is the dynamic viscosity, and \( k \) is the thermal conductivity. For water at moderate temperatures, \( Pr \) typically falls in a range that supports effective heat transfer, making it an important factor in calculating \( h \) using the Dittus-Boelter equation. Understanding \( Pr \) helps in assessing whether a fluid will efficiently transfer heat, which is crucial for designing heating and cooling systems.

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Most popular questions from this chapter

The problem of heat losses from a fluid moving through a buried pipeline has received considerable attention. Practical applications include the trans- Alaska pipeline, as well as power plant steam and water distribution lines. Consider a steel pipe of diameter \(D\) that is used to transport oil flowing at a rate \(\dot{m}_{o}\) through a cold region. The pipe is covered with a layer of insulation of thickness \(t\) and thermal conductivity \(k_{i}\) and is buried in soil to a depth \(z\) (distance from the soil surface to the pipe centerline). Each section of pipe is of length \(L\) and extends between pumping stations in which the oil is heated to ensure low viscosity and hence low pump power requirements. The temperature of the oil entering the pipe from a pumping station and the temperature of the ground above the pipe are designated as \(T_{m, i}\) and \(T_{s}\), respectively, and are known. Consider conditions for which the oil (o) properties may be approximated as \(\rho_{o}=900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p, o}=2000\) \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}, \quad \nu_{o}=8.5 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}, \quad k_{o}=0.140 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(P r_{o}=10^{4}\); the oil flow rate is \(\dot{m}_{o}=500 \mathrm{~kg} / \mathrm{s}\); and the pipe diameter is \(1.2 \mathrm{~m}\). (a) Expressing your results in terms of \(D, L, z, t, \dot{m}_{o}\), \(T_{m, i}\) and \(T_{s}\), as well as the appropriate oil \((o)\), insulation ( \(i\) ), and soil \((s)\) properties, obtain all the expressions needed to estimate the temperature \(T_{m \rho o}\) of the oil leaving the pipe. (b) If \(T_{s}=-40^{\circ} \mathrm{C}, T_{m, i}=120^{\circ} \mathrm{C}, t=0.15 \mathrm{~m}, k_{i}=0.05\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}, k_{s}=0.5 \mathrm{~W} / \mathrm{m}+\mathrm{K}, z=3 \mathrm{~m}\), and \(L=100 \mathrm{~km}\), what is the value of \(T_{m \rho}\) ? What is the total rate of heat transfer \(q\) from a section of the pipeline? (c) The operations manager wants to know the tradeoff between the burial depth of the pipe and insulation thickness on the heat loss from the pipe. Develop a graphical representation of this design information.

In Chapter 1, it was stated that for incompressible liquids, flow work could usually be neglected in the steady-flow energy equation (Equation 1.12d). In the trans-Alaska pipeline, the high viscosity of the oil and long distances cause significant pressure drops, and it is reasonable to question whether flow work would be significant. Consider an \(L=100 \mathrm{~km}\) length of pipe of diameter \(D=1.2 \mathrm{~m}\), with oil flow rate \(\dot{m}=500 \mathrm{~kg} / \mathrm{s}\). The oil properties are \(\rho=900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=2000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=0.765\) \(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Calculate the pressure drop, the flow work, and the temperature rise caused by the flow work.

The surface of a 50 -mm-diameter, thin-walled tube is maintained at \(100^{\circ} \mathrm{C}\). In one case air is in cross flow over the tube with a temperature of \(25^{\circ} \mathrm{C}\) and a velocity of \(30 \mathrm{~m} / \mathrm{s}\). In another case air is in fully developed flow through the tube with a temperature of \(25^{\circ} \mathrm{C}\) and a mean velocity of \(30 \mathrm{~m} / \mathrm{s}\). Compare the heat flux from the tube to the air for the two cases.

A thick-walled, stainless steel (AISI 316) pipe of inside and outside diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{o}=40 \mathrm{~mm}\) is heated electrically to provide a uniform heat generation rate of \(\dot{q}=10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The outer surface of the pipe is insulated, while water flows through the pipe at a rate of \(\dot{m}=0.1 \mathrm{~kg} / \mathrm{s}\).

Engine oil is heated by flowing through a circular tube of diameter \(D=50 \mathrm{~mm}\) and length \(L=25 \mathrm{~m}\) and whose surface is maintained at \(150^{\circ} \mathrm{C}\). (a) If the flow rate and inlet temperature of the oil are \(0.5 \mathrm{~kg} / \mathrm{s}\) and \(20^{\circ} \mathrm{C}\), what is the outlet temperature \(T_{m, o}\) ? What is the total heat transfer rate \(q\) for the tube? (b) For flow rates in the range \(0.5 \leq \dot{m} \leq 2.0 \mathrm{~kg} / \mathrm{s}\), compute and plot the variations of \(T_{m, o}\) and \(q\) with \(\dot{m}\). For what flow rate(s) are \(q\) and \(T_{m, \rho}\) maximized? Explain your results.
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