Bernoulli and Pipe Flow

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1 Civil Engineering Hydraulics Mechanics of Fluids Head Loss Calculations Bernoulli and The Bernoulli equation that we worked with was a bit simplistic in the way it looked at a fluid system All real systems that are in motion suffer from some type of loss due to friction It takes something to move over a rough surface 2 1

2 Bernoulli and 3 Consider flow in a constant-diameter pipe Bernoulli and If we look at the energy at points 1 and 2. p1 v12 p2 v z1 g = + + z2 g ρ 2 ρ 2 2

3 Bernoulli and Since this system is horizontal, z1 = z2 so p1 v12 p2 v 22 + = + ρ 2 ρ 2 5 Bernoulli and The pipe has a constant diameter and the flow is constant at both sections so the velocity at each point is the same. v1 = v2 p1 v12 p2 v22 + = + ρ 2 ρ 2 6 p1 p2 = ρ ρ 3

4 Bernoulli and Since we are dealing with an uncompressible fluid, the pressures at points 1 and 2 should be the same. p1 p2 = p1 = p2 ρ ρ 7 Bernoulli and If we ran the system experimentally and measured the two pressures, they would not be the same. p1 p2 = p1 = p2 ρ ρ 8

5 Bernoulli and The pressure at point 2 would be lower than the pressure at point 1. p1 p2 = p1 = p2 ρ ρ 9 Bernoulli and The pressure is being lost (actually the pressure energy) due to friction as the flow moves along the pipe. p1 p2 = p1 = p2 ρ ρ 10 5

6 11 To be able to quantify what is happening we need to look an element of flow in the pipe. 12 The grey element on the right is an element of the flow, a slice like a quarter. 6

7 The thickness of the quarter is dz and it has a perimeter length of P (πd) This makes the area in contact with the sides of the pipe equal to Pdz = πddz At the wall there is a shearing stress, τw, which is the stress between the wall and the outer layer of the fluid 13 Then the force from the wall on the section of fluid (opposing the fluid flow) will be equal to Fretarding = τ w Pdz = τ w π Ddz 1 7

8 The pressure on the upstream (left face) of the section will produce a force accelerating the section F = τ Pdz = τ π Ddz retarding Faccelerating 15 w w πd2 = pa = p As we move from the left face to the right face (dz) the change in pressure will equal dp F = τ Pdz = τ π Ddz retarding Faccelerating 16 w w πd2 = pa = p 8

9 So this will add an additional retarding force πd2 Fretarding = τ w π Ddz + ( p + dp ) 2 πd Faccelerating = pa = p 17 Since both the velocity and mass flow rates as the same on both sides of the section, there is no net force on the section (conservation of momentum) Fretarding = τ w π Ddz + ( p + dp ) Faccelerating = pa = p Faccelerating Fretarding 18 πd2 πd2 =0 9

10 Fretarding = τ w π Ddz + ( p + dp ) Therefore Faccelerating = pa = p Faccelerating Fretarding πd2 πd2 =0 πd2 πd2 τ w π Ddz ( p + dp ) =0 πd2 τ w π Ddz dp = 0 p Rearranging πd2 dp = 0 πd2 τ w π Ddz = dp τ π D dp w 2 = dz πd τ w π Ddz 10

11 So we have an expression for the rate at which the pressure changes as the flow moves downstream πd2 dp = 0 πd2 τ w π Ddz = dp τ π D dp w 2 = dz πd τ w π Ddz 21 In the simplest form τ w π D dp = π D 2 dz τ dp w = D dz 22 11

12 Generalizing to any shape pipe we substitute the hydraulic diameter for the pipe diameter 23 τ w dp = Dh dz At this point, we can introduce another dimensionless ratio, the friction factor, f 2 τ w dp = Dh dz 12

13 The friction factor, f, is the ratio of the friction forces to the inertia forces. τ dp w = Dh dz 25 f = τ w 1 2 ρv 2 Combining the two expressions. 26 f 1 2 ρv dp 2 = Dh dz 13

14 Isolating the differentials f ρv 2 dz = dp 2 Dh 27 Now back to the pipe we originally considered L f ρv 2 dz = dp 2 D 0 h 28 1

15 Integrating L f ρv 2 dz = dp 2 D 0 h 29 flρv 2 = p2 p1 2 Dh Usually we break this expression into two terms flρv 2 ρv 2 fl = = p2 p1 2 Dh 2 Dh 30 15

16 So we have a modified form of the Bernoulli equation that takes into account the friction losses in the system ρv 2 fl 2 Dh p1 v + + z1 g ρ ρ = p2 + v 2 + z g 2 ρ 2 The pressure change term takes that form because we are using energy terms in this expression. Reducing p1 v12 v 2 fl p2 v z1 g = + + z2 g ρ 2 2 Dh ρ

17 And rewriting the expression in terms of head p1 v12 p v2 v 2 fl + + z1 = z2 ρg 2 g 2 g Dh ρg 2 g 33 So the downstream energy (at point 2) is lower than the energy at point 1 p1 v12 p v2 v 2 fl + + z1 = z2 ρg 2 g 2 g Dh ρg 2 g 3 17

18 A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be The 2-nominal refers to the pipe size. At one time, it would have been the inside diameter of the pipe in inches but that is no longer the case. Like the 2 by in lumber, it is now just a reference. Table C.1 in the back of the book gives the critical dimensions for the pipe. If no schedule (has to do with wall thickness and strength of the pipe) you may assume that it is the most common, Schedule

19 A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be We can draw a sketch of the system and use the modified Bernoulli expression to answer the pressure drop. 37 A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be We have the volumetric flow rate so we can calculate the average velocity at points 1 and 2. Since they have the same cross sectional area, their average velocities will be the same

20 A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be If we set the reference plane through the left end of the system at point 1, then we can calculate the elevation of point 2 from the reference plane from the information given in the problem. 39 A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be Using the loss term we developed we can determine the pressure drop (in units of head) due to the friction in the pipe. 0 20

21 A 2-nominal pipe is inclined at an angle of 30 with the horizontal and conveys m3/s of water uphill. Determine the pressure drop in the pipe if it is 7 m long. Take the friction factor f to be The substitutions and computation for the problem ate on pages 22 and 225 of the text. 1 2 A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. Head Loss 21

22 A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. 3 Head Loss A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. Head Loss 22

23 A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. We have the length of the pipe, the hydraulic diameter, the mass density of the fluid, and the velocity. What we don t have is the friction factor for this flow. There is no simple way to calculate this. Under some flow regimes, there is a good approximation but to make the calculation we first need to calculate the Reynolds number of the flow and the relative roughness of the pipe. 5 Head Loss A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. For most common piping materials, we can find what is known as the characteristic dimension of wall roughness which is given the symbol, ε (epsilon). Table 5.2 in the chapter gives some examples of values for ε. In this example, the conduit is galvanized so ε in feet is shown. 6 Head Loss 23

24 A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. We now calculate a new dimensionless ratio known as the relative roughness which is the ratio of the characteristic roughness to the hydraulic diameter. 7 Head Loss A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. We also need to calculate the Reynolds number of the flow. 8 Head Loss 2

25 A rectangular galvanized duct in by 2 ft conveys heated air (T=98 F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.0 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct. If the flow had been laminar, we could have used 6/Re to calculate f. Since this flow is turbulent, we have to use the Moody Diagram. Figure 5.15 page Head Loss Moody Diagram On the right side of the diagram we locate a curve representing the relative roughness. 50 Head Loss 25

26 Moody Diagram On the right side of the diagram we locate a curve representing the relative roughness. On the bottom of the diagram, we locate the Reynolds number. 51 Head Loss Moody Diagram On the right side of the diagram we locate a curve representing the relative roughness. On the bottom of the diagram, we locate the Reynolds number. 52 Move straight up from the Reynolds Number until you intersect with the curve representing the relative roughness. Head Loss 26

27 Moody Diagram Move from the intersection point to the left hand axis, that value is the friction factor. On the bottom of the diagram, we locate the Reynolds number. 53 On the right side of the diagram we locate a curve representing the relative roughness. Move straight up from the Reynolds Number until you intersect with the curve representing the relative roughness. Head Loss Homework A -nominal riveted steel pipe 35 m long is laid horizontally and is to convey castor oil at a rate of 0.1 m3/s. Determine the pressure drop in the pipe. 27

28 Homework A 12-standard, type K copper tube conveys water at a rate of gpm. If the tube is 15 ft long, determine the pressure drop. Homework A 15-ft-long annulus made of -nominal, schedule 0 pipe and 2-nominal, schedule 0 pipe conveys carbon disulfide at a volume flow rate of 0.3 ft3/s. Determine the pressure drop over the 15-ft length. Both pipes are made of cast iron. 28

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