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Fundamentals of Momentum, Heat, and Mass Transfer

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Riska Listiarini Iskandar

Academic year: 2023

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Those familiar with previous editions will note that only one of the original writers (JW) continues as an active member of the writing team. Conservation of energy: Control-volume approach 65 6.1 Integral relation to conservation of energy 65 6.2 Uses of the integral expression 71.

Introduction to Momentum Transfer

This important simplification – the reduction from stress, a tensor, to pressure, a scalar – can also be demonstrated for the case of zero shear stress in flowing fluid. Of specific importance in momentum transfer is the description of the point-to-point variation in pressure.

Figure 1.1 Density at a point.
Figure 1.1 Density at a point.

Fluid Statics

The fluid force on the curved plate is the negative of this or WFCO. Calculate (a) the pressure at the butyl alcohol/benzene interface, (b) the pressure at the benzene/water interface, and (c) the pressure at the bottom of the tank.

Figure 2.1 Pressure forces on a static fluid element.
Figure 2.1 Pressure forces on a static fluid element.

Description of a Fluid in Motion

The straight line is useful in relating the fluid velocity components to the flow field geometry. This expression represents the velocity at a radial distance, r, from the center of the flow section.

Figure 3.1 Unsteady fl ow with respect to a fixed coordinate system.
Figure 3.1 Unsteady fl ow with respect to a fixed coordinate system.

Newton ’ s Second Law of Motion

Control-Volume Approach

Considering the directional component equations, (5-5a) and (5-5b), of the overall momentum balance, the external forces acting on the fluid in the control volume are The outer boundary of the control volume is at radiusr1, and the inner boundary is atr2.

Figure 5.2 Fluid fl ow through a control volume.
Figure 5.2 Fluid fl ow through a control volume.

Conservation of Energy

Such a pressure is called the stagnation pressure, which is greater than the static pressure by an amount equal to the change in the kinetic energy of the flow. In evaluating the surface integral, the choice of control volume coincided with the location of the pressure taps in sections (1) and (2). The selected control volume consists of a unit length of fluid surrounding the axis as shown in Figure 6.7.

When the velocity at the surface of the control volume is zero, the viscous term is zero. The upper limit of the control volume is just below the liquid surface and can therefore be considered to be at the same height as the liquid. The resulting expression, Equation (6-10), is, along with Equations (4-1) and (5-4), one of the basic expressions for the control volume analysis of fluid-flow problems.

A special case of the integral expression for the conservation of energy is the Bernoulli equation, equation (6-11). Develop an expression for the fluid velocity, v(r), where is the distance from the center of the shaft. Under these circumstances, please calculate the pressure change between the inlet and the outlet of the pumping station.

Figure 6.1 Reversible and irreversible thermodynamic cycles.
Figure 6.1 Reversible and irreversible thermodynamic cycles.

Shear Stress in Laminar Flow

The velocity profile in this case is parabolic; since the shear stress is proportional to the derivative of the velocity, the shear stress varies in a linear fashion. If the boundary or wall moves, the fluid layer moves at the velocity of the boundary, hence the name no-slip (boundary) condition. An understanding of the existence of viscosity requires an investigation of the motion of fluid on a molecular basis.

For multicomponent gas mixtures at low density, Wilke4 proposed this empirical formula for the mixture viscosity: In Fig. 7.6(a), for example, the shear stress tyx at the top of the element acts on the area DxDz. The distance between the plates is constant at 0.03 in. and the surface area of ​​the top plate in contact with the liquid is 0.95 ft2.

We first want to determine the viscosity of mayonnaise and then the speed of the moving plate. If the average velocity is 2 fps, determine the magnitude of the shear stress on the tube wall. The distance between the plates is 0.03 in., and the area of ​​the top plate in contact with the liquid is 0.1 ft2.

Figure 7.2 Velocity and shear stress pro fi les for fl ow between two parallel plates.
Figure 7.2 Velocity and shear stress pro fi les for fl ow between two parallel plates.

Analysis of a Differential Fluid Element in Laminar Flow

This expression can be integrated over a given length of tube to find the pressure drop and associated drag force on the tube resulting from the flow of a viscous fluid. -11) Considering a Newtonian fluid in laminar flow makes it possible to make the substitution of m(dvx/dy) forty, which yields. Thus, for our present case, flow is not the result of a pressure gradient, but rather the manifestation of the gravitational acceleration on a fluid.

In Chapter 9, the methods introduced in this chapter will be used to derive fluid flow differential equations for a general control volume. If the pipe diameter is doubled at a constant pressure loss, what percentage change will occur in the flow rate? Assume that the flow is a well-developed laminar flow with a zero pressure gradient, and that the atmosphere causes no shear at the outer surface of the film.

The width of the annulus is very small compared to the diameter of the drum, so that the flow in the annulus is equivalent to the flow between two flat plates. Assuming fully developed, laminar, incompressible flow, calculate the volumetric flow rate and average velocity of the fluid in the pipe. Assuming that the flow is laminar and fully developed with a pressure drop of 256 lbf/ft2, calculate the density of this fluid if the no-slip boundary condition applies.

Figure 8.1 Control volume for fl ow in a circular conduit.
Figure 8.1 Control volume for fl ow in a circular conduit.

Differential Equations of Fluid Flow

The net mass flow out of the control volume in the x direction is rvxjxDx rvxjxDyDz. Equation (9-2) can be arranged in a slightly different form to illustrate the use of the essential derivative. The reader can check that the terms on the left-hand side of equations (9-15) are all of the form.

Because no assumptions are made regarding the compressibility of the fluid, these equations are valid for both compressible and incompressible flows. Assuming there is no slippage between the fluid and the shaft at the surface of the shaft, this is the case. So vv(s, n, t) andPP(s, n, t). The substantial derivatives of the velocity and pressure gradients in equation (9-20) must be expressed in terms of streamline coordinates so that equation (9-20) can be integrated.

Examples 1 and 2 gave examples of using the Navier–Stokes equations in rectangular and cylindrical coordinates, respectively. Use the appropriate form of the Navier–Stokes equations to derive an expression for the velocity of the liquid film as it is drawn up the belt. Derive an expression for the shear stress using the appropriate form of the Navier–Stokes equations.

Figure 9.2 illustrates the various forces acting on the control volume. Summing the forces in the x direction, we obtain
Figure 9.2 illustrates the various forces acting on the control volume. Summing the forces in the x direction, we obtain

Inviscid Fluid Flow

Ñf ˆ v

Assume the core diameter is 200 feet and the static pressure at the center of the core is 38 psf below ambient pressure. Euler's equation can be used to relate the pressure gradient in the core to the fluid acceleration. Plot the streamline pattern and find the velocity that each vortex induces on the other vortex.

Maximum surface velocity and its position (x,y) 10.21 When a double is added to a uniform flow, so that the original part of the double faces the flow, a cylindrical flow is formed. Draw the flow lines when the duplicate is turned so that the sink faces the flow. Using potential theory for external pressure, calculate the tensile force in each bolt if the free-stream fluid is air at sea level and the free-stream wind speed is 25 m/s.

Determine the velocity components for this flow, and determine whether the flow is rotational or irrotational.

Dimensional Analysis and Similitude

This parameter can be interpreted as a measure of the ratio between inertial and gravitational forces. The number of dimensionless groups used to describe a situation involving variables is equal to r, which indicates the rank of the dimensional matrix of the variables. Below is an example of Randi's evaluation, as well as the application of the Buckingham method.

An important use and application of the dimensionless parameters listed in Table 11.2 is to use experimental results obtained using models to predict the performance of full-scale prototype systems. The relationship between the forces on the model and the forces experienced by the prototype can be determined by comparing values ​​of Eu between the model and the prototype. Find the airspeed required to test the model and find the ratio of model drag to full-scale drag.

Neglecting friction, what should be the size and speed of the waves in the model. What will be the time scale of the flow around the model relative to the full scale vehicle. Furthermore, it is assumed that the ratio between the forward speed and the propeller rotational speed must be constant (V/Nd ratio, where Ni is the number of propeller revolutions).

Table 11.2 lists several dimensionless groups that pertain to fl uid fl ow. Similar tables will be include in later chapters that list dimensionless parameters common to heat transfer and to mass transfer.
Table 11.2 lists several dimensionless groups that pertain to fl uid fl ow. Similar tables will be include in later chapters that list dimensionless parameters common to heat transfer and to mass transfer.

Viscous Flow

This is again due to the change from laminar to turbulent flow in the boundary layer. The pressure gradient plays a major role in inflow separation, as can be seen using the boundary layer equation (12-7). While the average value of the turbulent fluctuations is zero, these fluctuations contribute to the average value of certain flow quantities.

Using an approach similar to that of Section 7.3, let us consider momentum transfer in the turbulent flow illustrated in Figure 12.12. In rough tubes, the degree of roughness is found to affect the flow in the turbulent core but not in the laminar substrate. Flow in smooth circular pipes, it is found that at most of the cross section, the velocity profile can be related to.

The change in boundary layer thickness for turbulent flow over a smooth flat plate can be obtained from the von Kármán momentum integral. The boundary layer is known to be initially laminar and to transition to turbulent flow at a Rexof value of about 2105. If each plate area is 500 m2, determine the total drag force if both sides are exposed to the flow.

Figure 12.2 Drag coefficient for circular cylinders as a function of Reynolds number. Shaded regions indicate areas influenced by shear stress.
Figure 12.2 Drag coefficient for circular cylinders as a function of Reynolds number. Shaded regions indicate areas influenced by shear stress.

Flow in Closed Conduits

By performing the procedure described in Chapter 11 for solving for the unknown exponents in each group, we see that the dimensionless parameters become Since the pressure drop is due to fluid friction, this parameter is often written as DP/replaced by where "head loss" is; so, p1. The third group p, the ratio of pipe roughness to diameter, is the so-called relative roughness.

-1) Experimental data have shown that the pressure loss at fully developed flow is directly proportional to the L/D ratio. -2) The function f2, which varies with relative roughness and Reynolds number, is denoted by the friction factor f. With a factor of 2 inserted on the right-hand side, Equation (13-3) is the defining relation forff, the Fanning friction factor. Another friction factor commonly used is the Darcy friction factor, fD, defined by equation (13-4 ).

The student should be careful to note which friction factor he is using to properly calculate the friction head loss from equation (13-3) or (13-4). The student can easily verify that the Fanning friction factor is the same as the skin friction coefficientCf. Our task now becomes that of determining appropriate relationships for that theory and experimental data.

Gambar

Figure 1.1 Density at a point.
Figure 1.4 Element in a static fluid.
Figure 1.9 Capillary effects with a tube inserted in (a) water and (b) mercury.
Figure 2.4 Uniformly accelerated fuel tank.
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