Ghosh - 550Page 110/05/2018

Engineering Bernoulli Equation

We introduce here the most general form of the Bernoulli equation for steady, incompressible flows. Unlike the inviscid Bernoulli equation, this equation may be applied for viscous flows. We derive the general form of head-loss using the control volume approach as given in the system diagram below.

We chose a general system showing pipe area changes, valves or fittings, as well as, power devices and heating devices in the system. The Engineering Bernoulli equation may be written as:

In the above equation, is the rate of heat input to the system, is the rate of shaft work on the system, is the rate of shear work on the system, is the rate of non-mechanical work on the system, and e is the specific energy, or energy per unit mass. We know, is the sum of specific internal energy, specific kinetic energy, and specific potential energy. In most of our mechanical engineering applications, and by choosing the control surface perpendicular to the flow direction. However, in general cases, andmay not be zero. If heat is added to the system, . On the other hand, if heat is generated and given off by the system, . Similarly, if we have pumps, compressors, fans, etc. doing work on the system . On the other hand, turbines that extract work out of the system have . We write the simplified form of the above equation for steady flows as:

[Assuming]

Since most of the control volume has no flow, except through the areas (1) and (2) we may be able to simplify the above R.H.S. by holding fluid properties constant over (1) and (2). In this process, the only term that creates trouble is the term, because weknow V cannot be constant over the pipe cross-section in a viscous flow. Thus, we write:

where, the flow’s average velocity is utilized as constant and we define a new variable  as the Kinetic energy flux coefficientfrom the above expression. The other terms such as the internal energy, pressure density, and “z” are reasonably constant over small pipe cross-sections. Thus,

If , we may interpret the above result in a perfect energy conversion equation (where there are no losses through the system boundaries):

The left-hand side of this equation shows the loss of mechanical energy term per unit mass. Recall the flow energy/mass, K.E./mass, and P.E./mass as the basic mechanical energy terms in a fluid flow system. We notice that the drop of these terms on the left-hand side results in the gain of the internal energy/mass and given out as heat generated. Thus the drop of mechanical energy per unit mass, defined as the “head loss”, which can also be interpreted as the gain of non-mechanical energy per unit mass. We represent the head loss, hl, in two categories, hlmajor and hlminor: . Major head loss is a result of pipe friction as mentioned before, whereas minor head loss is a result of localized losses, due to valves and fittings, sudden expansions and contractions, pipe bends, etc. Furthermore, minor losses are expressed in two different ways:

(i) or,

(ii)

Tables and charts are available in various references listing the minor loss coefficients, K, or the equivalent length, Le, for various pipe flow components. Whenever you use values from references to solve problems, make sure that you check the range of applicability for these constants.

Pipe flow problems can be categorized into two specific categories: (a) the problems that can be solved directly by the use of tables and charts, and (b) problems that require iterations to solve. Remember the Moody diagrams can be used to look up the friction factor only if one knows the flow Reynolds number. Since , any problem that does not supply the , (or Q), pipe diameter, or fluid type (making  unknown), would require the second iterative approach. In these problems, both Moody diagram [f vs. ReD for various ] and the Darcy-Weisbach formula: are used to solve a problem iteratively.

We shall illustrate the use of head loss calculation in a design example given below. Since the problem demonstrates the use of a power device () in a pipe system, we shall write the Engineering Bernoulli equation (A) as:

………… (C)

where, the definition of hl has been introduced using the right hand side of equation (B).Remember that you must use this above form across power devices in problems using them. For simpler problems not recognizing the use of power devices, the head loss calculations may be performed using the equation (setting in above).

Example 1. (Head-Loss Calculations using equation (C))

Air is supplied to a steel-making process through a 6-in. diameter, smooth circular duct that terminates abruptly into a large plenum chamber. A new young engineer suggests that appreciable power can be saved by replacing the existing duct, which includes two 90 bends (centerline radius = 2 ft.) with a combination duct-diffuser. There is room for a diffuser with area ratio, AR = 1.35. The proposed new duct would eliminate the elbows, 8 feet of duct, and an additional piece of duct as long as the diffuser. The air speed required in the smooth duct is 150 ft/sec, and the outlet pressure is essentially atmospheric. How much power would be saved by making the changes suggested by the young engineer? The blower efficiency is 80 percent.

Given

Find

Solution

First, find out head loss for each system using first law of thermodynamics (or, the Engineering Bernoulli equation) and then the power required to drive each system.

Old System:

We need to determine the major and minor loss coefficients from charts and tables as follows.

First, for the 90 bend, r = 2’. Thus we compute the r/D and use it in the figure shown here for 90 pipe bends (selection shown in red) to obtain the equivalent pipe length for minor head-loss calculations.

Also, Kexit = 1.0 for most types of pipe exits. Therefore,

New System:

With AR = 1.35, the bestperformancerecoverycoefficient obtainable from

the figure (selection shown in red) above is for Cp = .4 and for that

Now, (from Fox & McDonald)

Exit loss at diffuser exit is given by (once again using Kexit = 1.0),

Now, apply the Engineering Bernoulli Equation between “0” and “4”(equation(C)).

Since the conditions are such that p4 = patm and can be neglected, assuming , and po are the same in old and new configurations,

or,

Flow

Furthermore, for smooth pipe, the Moody diagram

(selection shown in red in the figure below)


Also,

(by dividing above by 550, since 1 hp = 550 ft lb/sec)

This effort is required if the blower is ideal. However, for a blower efficiency of .8, it needs more power input from electrical system.

Critical Assessment: This problem illustrates how to use charts and tables in a pipe design problem. Note how some values are used for optimum performance.

Example 2. (Head Loss Calculations using equation (B)):

A hydraulic press is powered by a remote high-pressure pump. The gage pressure at the pump outlet is 20 MPa, whereas the pressure required for the press is 19 MPa (gage), at a flow rate of 0.032 m3/min. The press and pump are connected by 50 m of smooth, drawn steel tubing. The fluid is SAE 10W at 40 C. Determine the minimum tubing diameter that may be used.

  1. Statement of the Problem

a)Given

  • P1 = 20 MPa (gage) @ the pump
  • P2 = 19 MPa (gage) @ the press
  • Q = 0.032 m3/min = 5.33  10-4 m3/s
  • L = 50 m
  • Fluid is SAE 10W at 40 C  = 0.038 Ns/m2 = 3.75  10-5 m2/s
  • Tube is made of smooth, drawn steel e = 0.0015  10-3 m

b)Find

  • Minimum tubing diameter
  1. System Diagram
  1. Assumptions
  • Steady state condition
  • Constant property fluid (,,)
  • No elevation (z1 = z2)
  • 12 = 2.0 (for laminar flow in a pipe)
  • hlm = 0

(because there is no contraction, expansion, bend, elbow, valve, or fitting between the pump and press.)

  1. Governing Equations
  • Energy Equation (Engineering Bernoulli Equation)
  • Major Loss
  • Friction Factor for Laminar Flow
  1. Detailed Solution

First of all, assume that the flow in the pipe is laminar (if it is not laminar, the friction factor equation changes).

Considering the assumptions listed, the energy equation becomes:

Therefore,

The average velocity, , can be expressed as . Thus,

… (1)

Finally, the diameter is

After plugging in values,

D = 0.0142543 m = 14.2543 mm

Considering the equation (1), it can be said that the pressure drop decreases as the pipe diameter increases. We want the pressure drop to be just a bit less than 1 MPa, so the diameter has to be just a bit bigger than what we have calculated.

Answer:

The minimum tubing diameter is D = 0.0143 m = 14.3 mm.

  1. Critical Assessment

We assumed the flow in the pipe to be laminar. Let us check if this assumption is reasonable or not by calculating Reynolds number and comparing it with the critical Reynolds number for the pipe flow.

= 1266 < 2300 = ReCr Laminar flow

The assumption was reasonable.

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