/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Engine oil flows at a rate of \(... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 29

Engine oil flows at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) through a \(5-\mathrm{mm}-\) diameter straight tube. The oil has an inlet temperature of \(45^{\circ} \mathrm{C}\) and it is desired to heat the oil to a mean temperature of \(80^{\circ} \mathrm{C}\) at the exit of the tube. The surface of the tube is maintained at \(150^{\circ} \mathrm{C}\). Determine the required length of the tube. Hint: Calculate the Reynolds numbers at the entrance and exit of the tube before proceeding with your analysis.

Short Answer

Expert verified
The required length of the tube, \(L\), can be calculated using the equation: \(L = \frac{\dot{m} c_p \Delta T}{h \pi D \Delta T_{lm}}\) After calculating the Reynolds numbers at the entrance and exit, determining the heat transfer coefficient, \(h\), and finding the log mean temperature difference, \(\Delta T_{lm}\), we can plug in the given parameters and the specific heat capacity, \(c_p\), of the engine oil to find the tube length.

Step by step solution

01

1 - Identifying given parameters

From the exercise, we have the following given parameters: - Mass flow rate of engine oil, \(\dot{m} = 1 \mathrm{~kg/s}\) - Diameter of the tube, \(D = 5 \mathrm{~mm}\) - Inlet temperature of engine oil, \(T_{in} = 45^{\circ} \mathrm{C}\) - Desired mean exit temperature, \(T_{out}^{mean} = 80^{\circ} \mathrm{C}\) - Surface temperature of the tube, \(T_s = 150^{\circ} \mathrm{C}\)
02

2 - Calculate the Reynolds numbers at the entrance and exit of the tube

To calculate the Reynolds numbers, we need to know the velocity of the oil in the tube. To find this, we can use the equation \(\dot{m} = \rho A v\), where \(\dot{m}\) is the mass flow rate, \(\rho\) is the density of the oil, \(A\) is the cross-sectional area of the tube, and \(v\) is the velocity of the oil. Rearranging, we get: \(v = \frac{\dot{m}}{\rho A}\) For simplicity, let's assume that the density and velocity don't change significantly with temperature. So, we can find the Reynolds numbers at entrance and exit using the entrance temperature: \(Re = \frac{\rho v D}{\mu}\) Where \(\rho\) is the density of the oil, \(v\) is the velocity, \(D\) is the diameter of the tube, and \(\mu\) is the dynamic viscosity. Calculate the density, velocity, and Reynolds numbers using the appropriate values for the engine oil and given parameters.
03

3 - Determine the heat transfer coefficient

After calculating the Reynolds numbers at the entrance and exit, we need to find the heat transfer coefficient, \(h\), of the engine oil in the tube. One of the correlations used to estimate the heat transfer coefficient is the Dittus-Boelter equation, given by: \(Nu = \frac{h D}{k} = C Re^m Pr^n\) Where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, \(C\), \(m\), and \(n\) are constants. For this exercise, we'll assume: - The constants \(C\), \(m\), and \(n\) are appropriate for the engine oil and flow conditions. - The Reynolds number is constant within the range calculated in Step 2. With these assumptions, determine the heat transfer coefficient, \(h\), for the engine oil in the tube.
04

4 - Calculate the required tube length

Now that we have the heat transfer coefficient \(h\), we can calculate the required tube length, \(L\), to achieve the desired temperature change. The rate of heat transfer between the tube and the oil is given by: \(\dot{Q} = h A_s \Delta T_{lm}\) Where \(A_s\) is the surface area of the tube, and \(\Delta T_{lm}\) is the log mean temperature difference between the oil and the tube surface, given by: \(\Delta T_{lm} = \frac{(T_s - T_{out}) - (T_s - T_{in})}{\ln((T_s - T_{out}) / (T_s - T_{in}))}\) Moreover, the rate of heat transfer can also be expressed as: \(\dot{Q} = \dot{m} c_p \Delta T\) Where \(c_p\) is the specific heat capacity of the oil, and \(\Delta T = T_{out} - T_{in}\). Equating the two expressions for \(\dot{Q}\) and solving for the tube length, \(L\), we get: \(L = \frac{\dot{m} c_p \Delta T}{h \pi D \Delta T_{lm}}\) Calculate the required tube length, \(L\), using the parameters from Steps 1-3, and the expression above.

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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 is a vital concept in fluid dynamics that helps us determine the type of flow within a pipe or around an obstacle. It's calculated using the formula:\[Re = \frac{\rho v D}{\mu}\]where:
  • \( \rho \) - Density of the fluid
  • \( v \) - Velocity of the fluid
  • \( D \) - Diameter of the pipe
  • \( \mu \) - Dynamic viscosity of the fluid
The Reynolds Number tells us whether the flow is laminar or turbulent. A low Reynolds Number (typically below 2000) indicates a laminar flow, where the fluid flows in parallel layers with no disruption between them.
Conversely, a high Reynolds Number (above 4000) indicates turbulent flow, characterized by chaotic fluid movement.
In between these two ranges lies the transitional flow region. Understanding this concept helps engineers and scientists design systems to optimize heat transfer conditions.
In this exercise, calculating the Reynolds Number at the entrance of the tube indicates how the oil behaves when it begins its journey through the pipe.
Understanding this behavior is crucial for determining the appropriate equations and methods to describe and facilitate heat transfer efficiently.
Nusselt Number
The Nusselt Number is a dimensionless value that characterizes the heat transfer at the fluid-surface boundary. It's used to determine the convective heat transfer coefficient, \( h \), and can be expressed by the equation:\[ Nu = \frac{h D}{k} \]where:
  • \( h \) - Convective heat transfer coefficient
  • \( D \) - Characteristic length (often diameter for pipes)
  • \( k \) - Thermal conductivity of the fluid
An important understanding of the Nusselt Number is that it helps quantify how efficient a surface is in transferring heat to or from a fluid in motion.
A higher Nusselt Number indicates a more effective convective heat transfer compared to conductive heat transfer alone.
This number is closely related to both the Reynolds Number and the Prandtl number.
The Nusselt Number changes depending on whether the system is in a laminar, turbulent, or transitional state.
This means that knowing the Reynolds Number can help us determine an appropriate Nusselt Number, which in turn reflects the effectiveness of the heat transfer process in the system.
For this exercise, evaluating the Nusselt Number allows students to connect the fluid flow characteristics to the rate of heat transfer, moving one step closer to figuring out the tube's required length.
Dittus-Boelter Equation
The Dittus-Boelter equation is a practical correlation for estimating the heat transfer coefficient in turbulent flow through pipes. It offers a simplified way to calculate the Nusselt Number under the assumption that the flow regime and fluid properties fit the criteria defined by this equation. It is given as:\[ Nu = C \times Re^m \times Pr^n \]where:
  • \( Nu \) - Nusselt Number
  • \( C \), \( m \), \( n \) - Empirically determined constants for specific conditions
  • \( Re \) - Reynolds Number
  • \( Pr \) - Prandtl Number, a measure of momentum diffusivity over thermal diffusivity
This equation assumes turbulent flow conditions and is typically applied where more detailed calculations are unnecessary.
For many engineering applications, the constants are set as \( C = 0.023 \), \( m = 0.8 \), and \( n = 0.3 \), when the fluid is being heated.
With these constants, the Dittus-Boelter equation can provide a practical approximation for the Nusselt Number, facilitating easier computation of the heat transfer coefficient.
In the context of this exercise, using this equation allows us to derive the heat transfer coefficient between the oil and the pipe walls.
Once the heat transfer coefficient is known, it's possible to calculate the length of the tube needed to reach the desired oil exit temperature.

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Most popular questions from this chapter

A circular tube of diameter \(D=0.2 \mathrm{~mm}\) and length \(L=\) \(100 \mathrm{~mm}\) imposes a constant heat flux of \(q^{\prime \prime}=20 \times 10^{3}\) \(\mathrm{W} / \mathrm{m}^{2}\) on a fluid with a mass flow rate of \(\dot{m}=0.1 \mathrm{~g} / \mathrm{s}\). For an inlet temperature of \(T_{m, i}=29^{\circ} \mathrm{C}\), determine the tube wall temperature at \(x=L\) for pure water. Evaluate fluid properties at \(\bar{T}=300 \mathrm{~K}\). For the same conditions, determine the tube wall temperature at \(x=L\) for the nanofluid of Example \(2.2\).

Fluid enters a thin-walled tube of \(5-\mathrm{mm}\) diameter and \(2-\mathrm{m}\) length with a flow rate of \(0.04 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(T_{m, i}=85^{\circ} \mathrm{C}\). The tube surface is maintained at a temperature of \(T_{s}=25^{\circ} \mathrm{C}\), and for this operating condition, the outlet temperature is \(T_{m, o}=31.1^{\circ} \mathrm{C}\). What is the outlet temperature if the flow rate is doubled? Fully developed, turbulent flow may be assumed to exist in both cases, and the fluid properties may be assumed to be independent of temperature.

8.106 Consider the pharmaceutical product of Problem 8.27. Prior to finalizing the manufacturing process, test trials are performed to experimentally determine the dependence of the shelf life of the drug as a function of the sterilization temperature. Hence, the sterilization temperature must be carefully controlled in the trials. To promote good mixing of the pharmaceutical and, in turn, relatively uniform outlet temperatures across the exit tube area, experiments are performed using a device that is constructed of two interwoven coiled tubes, each of 10 -mm diameter. The thin-walled tubing is welded to a solid high thermal conductivity rod of diameter \(D_{r}=40 \mathrm{~mm}\). One tube carries the pharmaceutical product at a mean velocity of \(u_{p}=0.1 \mathrm{~m} / \mathrm{s}\) and inlet temperature of \(25^{\circ} \mathrm{C}\), while the second tube carries pressurized liquid water at \(u_{w}=0.12 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(127^{\circ} \mathrm{C}\). The tubes do not contact each other but are each welded to the solid metal rod, with each tube making 20 turns around the rod. The exterior of the apparatus is well insulated. (a) Determine the outlet temperature of the pharmaceutical product. Evaluate the liquid water properties at \(380 \mathrm{~K}\). (b) Investigate the sensitivity of the pharmaceutical's outlet temperature to the velocity of the pressurized water over the range \(0.10

A mass transfer operation is preceded by laminar flow of a gaseous species B through a circular tube that is sufficiently long to achieve a fully developed velocity profile. Once the fully developed condition is reached, the gas enters a section of the tube that is wetted with a liquid film (A). The film maintains a uniform vapor density \(\rho_{\Lambda_{S}}\) along the tube surface. (a) Write the differential equation and boundary conditions that govern the species A mass density distribution, \(\rho_{A}(x, r)\), for \(x>0\). (b) What is the heat transfer analog to this problem? From this analog, write an expression for the average Sherwood number associated with mass exchange over the region \(0 \leq x \leq L\). (c) Beginning with application of conservation of species to a differential control volume of extent \(\pi r_{o}^{2} d x\), derive an expression (Equation 8.86) that may be used to determine the mean vapor density \(\rho_{\mathrm{A}, \mathrm{m}, \mathrm{Q}}\) at \(x=L\). (d) Consider conditions for which species \(\mathrm{B}\) is air at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) and the liquid film consists of water, also at \(25^{\circ} \mathrm{C}\). The flow rate is \(\dot{m}=2.5 \times 10^{-4} \mathrm{~kg} / \mathrm{s}\), and the tube diameter is \(D=10 \mathrm{~mm}\). What is the mean vapor density at the tube outlet if \(L=1 \mathrm{~m}\) ?

To slow down large prime movers like locomotives, a process termed dynamic electric braking is used to switch the traction motor to a generator mode in which mechanical power from the drive wheels is absorbed and used to generate electrical current. As shown in the schematic, the electric power is passed through a resistor grid \((a)\), which consists of an array of metallic blades electrically connected in series (b). The blade material is a high- temperature, high electrical resistivity alloy, and the electrical power is dissipated as heat by internal volumetric generation. To cool the blades, a motor-fan moves high-velocity air through the grid. (a) Treating the space between the blades as a rectangular channel of \(220-\mathrm{mm} \times 4-\mathrm{mm}\) cross section and \(70-\mathrm{mm}\) length, estimate the heat removal rate per blade if the airstream has an inlet temperature and velocity of \(25^{\circ} \mathrm{C}\) and \(50 \mathrm{~m} / \mathrm{s}\), respectively, while the blade has an operating temperature of \(600^{\circ} \mathrm{C}\). (b) On a locomotive pulling a 10 -car train, there may be 2000 of these blades. Based on your result from part (a), how long will it take to slow a train whose total mass is \(10^{6} \mathrm{~kg}\) from a speed of \(120 \mathrm{~km} / \mathrm{h}\) to \(50 \mathrm{~km} / \mathrm{h}\) using dynamic electric braking?
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