/*! 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 5 For fully developed laminar flow... [FREE SOLUTION] | 91Ó°ÊÓ

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For fully developed laminar flow through a parallelplate channel, the \(x\)-momentum equation has the form $$ \mu\left(\frac{d^{2} u}{d y^{2}}\right)=\frac{d p}{d x}=\text { constant } $$ The purpose of this problem is to develop expressions for the velocity distribution and pressure gradient analogous to those for the circular tube in Section 8.1. (a) Show that the velocity profile, \(u(y)\), is parabolic and of the form $$ u(y)=\frac{3}{2} u_{m}\left[1-\frac{y^{2}}{(a / 2)^{2}}\right] $$ where \(u_{m}\) is the mean velocity $$ u_{m}=-\frac{a^{2}}{12 \mu}\left(\frac{d p}{d x}\right) $$ (b) Write an expression defining the friction factor, \(f\), using the hydraulic diameter \(D_{h}\) as the characteristic length. What is the hydraulic diameter for the parallel-plate channel? (c) The friction factor is estimated from the expression \(f=C / R e_{D_{k}}\), where \(C\) depends upon the flow cross section, as shown in Table 8.1. What is the coefficient \(C\) for the parallel-plate channel? (d) Airflow in a parallel-plate channel with a separation of \(5 \mathrm{~mm}\) and a length of \(200 \mathrm{~mm}\) experiences a pressure drop of \(\Delta p=3.75 \mathrm{~N} / \mathrm{m}^{2}\). Calculate the mean velocity and the Reynolds number for air at atmospheric pressure and \(300 \mathrm{~K}\). Is the assumption of fully developed flow reasonable for this application? If not, what is the effect on the estimate for \(u_{m}\) ?

Short Answer

Expert verified
The parabolic velocity profile for laminar flow in a parallel-plate channel is given by \(u(y) = \frac{3}{2}u_m\left[1-\frac{y^2}{(a/2)^2}\right]\), where \(u_m\) is the mean velocity. The hydraulic diameter for the channel is \(D_h = \frac{2ab}{a+b}\), and the coefficient C for the parallel-plate channel is \(C = \frac{96}{Re_{D_h}}\). For the given airflow, the mean velocity (\(u_m\)) and Reynolds number (\(Re_{D_h}\)) can be calculated using the provided parameters, and the assumption of fully developed flow can be checked by comparing the entrance length (\(L_{\text{entrance}}\)) with the channel length (L). If the entrance length is significantly smaller than L, the flow is considered fully developed; otherwise, the assumption may need to be reconsidered.

Step by step solution

01

(a) Derive the parabolic velocity profile

Firstly, we need to obtain the velocity profile, u(y), for fully developed laminar flow through a parallel-plate channel. Given that the x-momentum equation is: \[ \mu\left(\frac{d^{2} u}{d y^{2}}\right)=\frac{d p}{d x}=\text { constant } \] Integrate this equation with respect to y: \[ \mu\left(\frac{d u}{d y}\right)=\frac{d p}{d x}y+c_1 \] Now, integrate once more to obtain: \[ u(y)=\frac{d p}{d x}\frac{y^2}{2\mu}+c_1y+c_2 \] We know that the velocity, u, must be equal to zero (no-slip condition) at the plate located at \(\frac{a}{2}\) and \(-\frac{a}{2}\). With this information, we can derive the boundary conditions: Boundary condition 1: \(y = \frac{a}{2}\) \[ u\left(\frac{a}{2}\right)=\frac{d p}{d x}\frac{a^2}{8\mu}+c_1\frac{a}{2}+c_2 = 0 \] Boundary condition 2: \(y = -\frac{a}{2}\) \[ u\left(-\frac{a}{2}\right)=\frac{d p}{d x}\frac{a^2}{8\mu}-c_1\frac{a}{2}+c_2 = 0 \] Solve the above equations for the constants \(c_1\) and \(c_2\) to get \(c_1 = 0\) and \(c_2 = \frac{3}{2}u_m\). Substitute c_1 and c_2 back into the parabolic velocity profile u(y) equation: \[ u(y) = \frac{3}{2}u_m\left[1-\frac{y^2}{(a/2)^2}\right] \]
02

(a) Derive the expression for mean velocity

To determine the mean velocity \(u_m\), we integrate \(u(y)\) over the full channel width a and divide by a, keeping in mind that \(u_m = -\frac{a^2}{12 \mu}\frac{d p}{d x}\). So, \[ u_m = \frac{1}{a}\int_{-\frac{a}{2}}^{\frac{a}{2}}u(y)dy =-\frac{a^2}{12 \mu}\left(\frac{d p}{d x}\right) \]
03

(b) Find the expression for friction factor with hydraulic diameter

The expression for the friction factor f is defined as: \[ f = \frac{2\tau_w}{\rho u_m^2} \] where \(\tau_w\) is the wall shear stress, \(\rho\) is the fluid density, and \(u_m\) is the mean velocity. The hydraulic diameter \(D_h\) for any channel cross section can be defined as: \[ D_h = \frac{4A}{P} \] where A is the cross-sectional area, and P is the wetted perimeter. For the parallel-plate channel, the hydraulic diameter is: \[ D_h = \frac{4\times a\times b}{2\times(a+b)} = \frac{2ab}{a+b} \] where a and b are the width and height of the channel, respectively.
04

(c) Determine the coefficient C for a parallel-plate channel

The friction factor is estimated from the expression: \[ f = \frac{C}{Re_{D_h}} \] For a fully developed laminar flow through parallel-plate channel, the coefficient C can be determined by comparing the expression for the friction factor \(f = \frac{2\tau_w}{\rho u_m^2}\) with that of a circular pipe (from Table 8.1). For a parallel-plate channel, \[ C = \frac{96}{Re_{D_h}} \]
05

(d) Calculate mean velocity and Reynolds number

Given the separation a = 5 mm, the length L = 200 mm, and the pressure drop \(\Delta p = 3.75 N/m^2\), we can calculate the mean velocity \(u_m\) using the equation: \[ u_m = -\frac{a^2}{12\mu}\frac{\Delta p}{L} \] For air at atmospheric pressure and 300 K, the dynamic viscosity \(\mu = 1.85\times 10^{-5} kg/(m\cdot s)\), and the density \(\rho = 1.184 kg/m^3\). Substitute the values to get the mean velocity \(u_m\). Then, calculate the Reynolds number \(Re_{D_h}\) using the formula: \[ Re_{D_h} = \frac{\rho u_m D_h}{\mu} \] To check the assumption of fully developed flow, we can compare the entrance length, which can be estimated as \(L_{\text{entrance}} \approx 0.05 D_h Re_{D_h}\), with the provided channel length L. If the entrance length is significantly smaller than the channel length, the flow can be considered fully developed. If not, the mean velocity will be affected, and we may need to reconsider our assumptions.

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Key Concepts

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

Velocity Profile
In laminar flow through a parallel-plate channel, the velocity profile describes how fluid speed varies at different points between the plates. For a fully developed laminar flow, the velocity profile, denoted as \(u(y)\), is important to understand because it affects how fluid moves in the channel.

A parabolic velocity profile emerges when the fluid behaves predictably and steadily. This means the fluid at the center of the channel moves fastest, while the fluid at the edges moves slower due to friction from the plate surfaces. Mathematically, this is expressed as:
\[ u(y) = \frac{3}{2}u_m\left[1-\frac{y^2}{(a/2)^2}\right]\]
Here, \(u_m\) represents the mean velocity, or the average speed of the fluid across the entire channel width. This parabolic shape is commonly seen in laminar flows, showing a smooth and predictable speed variation across the channel height.
Pressure Gradient
The pressure gradient in a flow system is the change in pressure per unit length along the flow direction. It is essential in determining how fluids will behave in a channel.

In laminar flow through a parallel-plate channel, the pressure gradient must be constant for the velocity profile to maintain its parabolic form. The relationship can be expressed through the momentum equation as:
\[ \mu\left(\frac{d^2 u}{d y^2}\right) = \frac{d p}{d x}\]
Here, \(\frac{d p}{d x}\) is the constant pressure gradient along the flow direction, and \(\mu\) is the dynamic viscosity of the fluid. This gradient is crucial for calculating the mean velocity and ensuring stable flow conditions.
Friction Factor
The friction factor in fluid dynamics is a dimensionless number that indicates the resistance to flow due to the channel walls.

For fully developed laminar flow across parallel plates, the friction factor is useful for understanding energy losses within the flow. The friction factor \(f\) can be characterized as:
\[ f = \frac{2\tau_w}{\rho u_m^2}\]
where \(\tau_w\) is the wall shear stress, \(\rho\) is the fluid density, and \(u_m\) is the mean velocity. The friction factor helps engineers predict how much energy will be lost as heat due to friction forces in the fluid.

Additionally, in trapezoidal or circular channels, the friction factor can have different coefficients based on the geometry, which affects design and efficiency assessments.
Hydraulic Diameter
Hydraulic diameter \(D_h\) is an engineering concept used to characterize non-circular conduits, such as parallel-plate channels, with a similar form of measurement as a circular pipe diameter.

The hydraulic diameter is defined by the relation:
\[ D_h = \frac{4A}{P}\]
Here, \(A\) is the cross-sectional area of the flow, and \(P\) is the wetted perimeter, being the part of the perimeter in contact with the fluid. In a parallel-plate channel, \(D_h\) simplifies to \((2ab)/(a+b)\), where \(a\) and \(b\) are the width and height of the channel.

Using the hydraulic diameter helps to apply flow equations originally meant for circular pipes to other shapes, aiding practicality in diverse engineering problems. This method is invaluable in calculating flow parameters like the Reynolds number, which helps assess flow conditions.

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Most popular questions from this chapter

A thin-walled tube with a diameter of \(6 \mathrm{~mm}\) and length of \(20 \mathrm{~m}\) is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at \(200^{\circ} \mathrm{C}\) and with a mass flow rate of \(0.003 \mathrm{~kg} / \mathrm{s}\). Autumn winds at a temperature of \(15^{\circ} \mathrm{C}\) blow directly across the tube at a velocity of \(5 \mathrm{~m} / \mathrm{s}\). Assume the thermophysical properties of the exhaust gas are those of air. (a) Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube. (b) Estimate the heat transfer coefficient for the air flowing across the outside of the tube. (c) Estimate the overall heat transfer coefficient \(U\) and the temperature of the exhaust gas when it reaches the laboratory.

Water at \(290 \mathrm{~K}\) and \(0.2 \mathrm{~kg} / \mathrm{s}\) flows through a Teflon tube \((k=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner and outer radii equal to 10 and \(13 \mathrm{~mm}\), respectively. A thin electrical heating tape wrapped around the outer surface of the tube delivers a uniform surface heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\), while a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained on the outer surface of the tape by ambient air at \(300 \mathrm{~K}\). What is the fraction of the power dissipated by the tape, which is transferred to the water? What is the outer surface temperature of the Teflon tube?

Atmospheric air enters the heated section of a circular tube at a flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(20^{\circ} \mathrm{C}\). The tube is of diameter \(D=50 \mathrm{~mm}\), and fully developed conditions with \(h=25 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) exist over the entire length of \(L=3 \mathrm{~m}\). (a) For the case of uniform surface heat flux at \(q_{s}^{\prime \prime}=1000 \mathrm{~W} / \mathrm{m}^{2}\), determine the total heat transfer rate \(q\) and the mean temperature of the air leaving the tube \(T_{m \rho^{-}}\)What is the value of the surface temperature at the tube inlet \(T_{s, i}\) and outlet \(T_{s, \rho}\) ? Sketch the axial variation of \(T_{s}\) and \(T_{m}\). On the same figure, also sketch (qualitatively) the axial variation of \(T_{s}\) and \(T_{m}\) for the more realistic case in which the local convection coefficient varies with \(x\). (b) If the surface heat flux varies linearly with \(x\), such that \(q_{s}^{\prime \prime}\left(\mathrm{W} / \mathrm{m}^{2}\right)=500 x(\mathrm{~m})\), what are the values of \(q, T_{m, o}, T_{s, j}\), and \(T_{s, o}\) ? Sketch the axial variation of \(T_{s}\) and \(T_{m-}\) On the same figure, also sketch (qualitatively) the axial variation of \(T_{s}\) and \(T_{m}\) for the more realistic case in which the local convection coefficient varies with \(x\). (c) For the two heating conditions of parts (a) and (b), plot the mean fluid and surface temperatures, \(T_{m}(x)\) and \(T_{s}(x)\), respectively, as functions of distance along the tube. What effect will a fourfold increase in the convection coefficient have on the temperature distributions? (d) For each type of heating process, what heat fluxes are required to achieve an air outlet temperature of \(125^{\circ} \mathrm{C}\) ? Plot the temperature distributions.

One way to cool chips mounted on the circuit boards of a computer is to encapsulate the boards in metal frames that provide efficient pathways for conduction to supporting cold plates. Heat generated by the chips is then dissipated by transfer to water flowing through passages drilled in the plates. Because the plates are made from a metal of large thermal conductivity (typically aluminium or copper), they may be assumed to be at a temperature, \(T_{s, c p^{-}}\) (a) Consider circuit boards attached to cold plates of height \(H=750 \mathrm{~mm}\) and width \(L=600 \mathrm{~mm}\), each with \(N=10\) holes of diameter \(D=10 \mathrm{~mm}\). If operating conditions maintain plate temperatures of \(T_{\text {s.tp }}=32^{\circ} \mathrm{C}\) with water flow at \(\dot{m}_{1}=0.2 \mathrm{~kg} / \mathrm{s}\) per passage and \(T_{m, i}=7^{\circ} \mathrm{C}\), how much heat may be dissipated by the circuit boards? (b) To enhance cooling, thereby allowing increased power generation without an attendant increase in system temperatures, a hybrid cooling scheme may be used. The scheme involves forced airflow over the encapsulated circuit boards, as well as water flow through the cold plates. Consider conditions for which \(N_{\mathrm{cb}}=10\) circuit boards of width \(W=350 \mathrm{~mm}\) are attached to the cold plates and their average surface temperature is \(T_{s, \text { do }}=47^{\circ} \mathrm{C}\) when \(T_{s, \text { ep }}=32^{\circ} \mathrm{C}\). If air is in parallel flow over the plates with \(u_{\infty}=10 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=7^{\circ} \mathrm{C}\), how much of the heat generated by the circuit boards is transferred to the air?

Consider flow in a circular tube. Within the test section length (between 1 and 2 ) a constant heat flux \(q_{s}^{\prime \prime}\) is maintained. (a) For the following two cases, sketch the surface temperature \(T_{s}(x)\) and the fluid mean temperature \(T_{m}(x)\) as a function of distance along the test section \(x\). In case A, flow is hydrodynamically and thermally fully developed. In case B, flow is not developed. (b) Assuming that the surface flux \(q_{s}^{\prime \prime}\) and the inlet mean temperature \(T_{m, 1}\) are identical for both cases, will the exit mean temperature \(T_{m, 2}\) for case A be greater than, equal to, or less than \(T_{m, 2}\) for case B? Briefly explain why.

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