flow in pipes applications – heating, cooling, fluid distributions (blood and its vessels)...
TRANSCRIPT
FLOW IN PIPES
Applications – heating, cooling, fluid distributions (blood and its vessels)
Attention will be given on: Laminar and Turbulent Flows Reynolds Number Entrance Region Flow Pressure drops and head loss in Laminar
flow in circular pipes Minor Losses in piping system
Define and explain Laminar and turbulent flows in pipes, Reynolds number, and Critical Reynolds Number.
Explain friction loss in Laminar flow. Derive and Explain Hagen - Poiseuille equation.
Explain and describe the important of Poiseuille’s law in Biomedical application.
Explain the important of laminar and turbulent flow in biomedical engineering.
Solve problems about laminar and turbulent flow.
FLOW IN PIPES
Fluid flow can be classified as external or internal.
We focus on internal for flow in pipes.
LAMINAR AND TURBULENT FLOWS The flow appears to be
smooth and steady. The stream has a fairly uniform diameter and there is little or no evidence of mixing of the various parts of the stream.The flow has very low velocity – highly ordered motion
The flow has a rather high velocity – highly disordered motion. The elements of fluid appear to be mixing chaotically within the stream.
REYNOLDS NUMBERHow to distinguish Turbulent and Laminar in
mathematics?Osborne Reynolds manage to do that in 1880s.
Re = vDρ/µ where fluid density ρ, fluid viscosity µ , pipe
diameter D, and average velocity of flow, v.Re is the ratio of the inertial forces to viscous forces
in the fluid. Dimensionless - no UNITS.
How to calculate average velocity,v in laminar flow?The mean velocity is half of the maximum velocity.
Vmean = Vmax/2
REYNOLDS NUMBER How to use it ?
Critical Reynolds Number – flow become turbulent.
Larger Re – Inertial is bigger than viscous, viscous cannot prevent the random and rapid fluctuation of the fluids and vice versa.
The Re value is different for different geometries and flow conditions.
For circular pipe, Re cr = 2300.
EXAMPLE 1–REYNOLDS NUMBERDetermine whether the flow is laminar or turbulent if
glycerine at 25°C flows in a pipe with a 150-mm inside diameter. The average velocity of flow is 3.6 m/s.
We must first evaluate the Reynolds number
Laminar and Turbulent in Human BloodMost of human blood flow is laminar, having Re
of 300 or less.
However, it is possible for turbulent to occur at very high flow rates in the descending aorta, for example in highly conditioned athletes. Sometimes it is also common in pathological conditions (narrowed (stenotic) arteries and across stenotic heart valves.
TURBULENT AND ITS EFFECT Haemodynamic studies have shown that diseased
cardiac valves, whether stenosed or incompetent, create regions of increased turbulence and shear stresses that are large enough to damage the vascular endothelium leading to endothelial dysfunction.
Endothelial cells line the entire circulatory system, from the heart to the smallest capillary. These cells reduce friction of the flow of blood allowing the fluid to be pumped further and control of blood pressure.
A key feature of endothelial dysfunction is the inability of arteries and arterioles to dilate fully in response to an appropriate stimulis.
Laminar and Turbulent in the lungsIn the lungs, fully developed laminar flow
probably occurs only in very small airways with low Re.
When air flow at higher rates in larger diameter tubes, like Trachea, the flow is often turbulent.
Much of the flow in intermediate sized airways will be transitional flow.
ENTRANCE REGIONFLOW ENTERING CIRCULAR
PIPE
Boundary layer region - Viscous effects and the velocity changes are significance
The inviscid flow region – Frictional effects are negligible and velocity in radial directions is constant.
The region from the pipe inlet to the point at which the boundary layer merges at the centreline is called the Hydrodynamic Entrance Region, and the length is Hydrodynamic Entry Length, Lh.
1/4
Dh 1.359ReL
ENTRY LENGTH (LENGTH OF ENTRANCE REGION), Lh
Laminar flow,
D0.05ReL Dh
Turbulent flow,
For a pipe length over than 10D, entrance effect is negligible and thus,
D10Lh
FULLY DEVELOP FLOW IN ARTERIES
If we consider Re = 300, then the previous equation gives Xe=18D. This means that, an entrance length equal to 18 pipe diameters is required for fully developed flow in human system.
In human cardiovascular system, it is not common to see fully developed flow in arteries. The vessels continually branch, with the distance between branches not often being greater than 18 diameters.
LAMINAR FLOW IN PIPES Assumptions:
Steady, laminar flow of incompressible liquid with constant properties in the fully developed region of a straight circular pipe
No acceleration since it is steady and fully develop. No motion in the radial direction, velocity in radial direction is
zero We try to obtain the velocity profile and also a relation to the
friction factor.
LAMINAR FLOW IN PIPES
Consider a free-body diagram of the cylinder of fluid and sum the forces acting on that cylinder.
First, we will need to make some assumptions. Assume that the flow is steady. This means that the flow is not changing with time; that the derivative of flow rate with respect to time is equal to zero.
0dt
dQ
LAMINAR FLOW IN PIPES
Second, assume that the flow is through a long tube with a constant cross-section. This flow condition is known as uniform flow. For steady flows in long tubes with a constant cross-section, the flow is fully developed and therefore, the pressure gradient, dP/dx is constant.
LAMINAR FLOW IN PIPES The third assumption is that the fluid is
Newtonian. Newtonian flow is flow in which the shearing stress, t, in the fluid is constant. In other words, the viscosity is constant with respect to the shear rate, y, and the whole process is carried out at a constant temperature.
LAMINAR FLOW IN PIPES
Now let the x direction be the axial direction of the pipe with the downstream (to the right) being positive.If the flow is unchanging with time, then the sum of forces in the x dire<
P(πr2) - (P + dP)(πr2) – τ2πrdx = 0
Solve Eq. (1.1) and the result is
-dP(πr2) = 2πrτdx
LAMINAR FLOW IN PIPES
The result from a simple force balance is shear stress as a function of pressure gradient, dP/dx and radial position, r:
and
dx
dPr
2
dx
dPRtubewall
2
LAMINAR FLOW IN PIPE From the definition of viscosity that the shear stress
is also related to the shear rate:
From the relation of shear stress, pressure drop and velocity gradient
Then producing differential equation with the variables velocity,V and radius,r:
dr
dV
dr
dV
dx
dPr 2
rdrdx
dPdV
21
LAMINAR FLOW IN PIPE
The next step in the analysis is to solve the previous differential equation, which gives the velocity of each point in the tube as a functionof the radius, r:
1
2
2
1C
r
r
dx
dPV
LAMINAR FLOW IN PIPE So far, in this analysis, we have made three
assumptions. First, steady flow (dQ/dt = 0); second, fully developed tube (dP/dx is constant); and third, viscosity is constant.
Now we make assumption four, which is the no slip condition. This means V at the wall is zero when r equals the radius of the tube. Therefore, set r = Rtube = R and V=0 to solve for C1.
LAMINAR FLOW IN PIPE
22
1
22
10
2
1
1
2
R
dx
dPC
Cr
dx
dP
The equation, which gives velocity as a function of radius, r, is then
][4
1 22
Rrdx
dPV
The final assumption is that the flow is laminar whereby this parabolic velocity profile represents the velocity profile across a constant cross section.
LAMINAR FLOW IN PIPES; VELOCITY PROFILE
mVu 2max
The dP/dx must have –ve value for pressure drops that cause a positive velocity (pressure must decrease in the flow direction due to the viscous effect). The maximum velocity will occur at the centreline, where r=0.
In fluid flow, it is convenient to work with average velocity which remains constant in incompressible flow when the cross sectional area of pipe is constant.
][4
1 22
Rrdx
dPV
PRESSURE DROPPressure drop occurs as the fluid flows along straight lengths of
pipe and tubing. It causes pressure to decrease along the pipe and they increase the amount of power that a pump must deliver the fluid. It is caused by friction, changes in kinetic energy, etc.
Friction may occur between the fluid & the pipe work, but friction also occurs within the fluid as sliding between adjacent layers of fluid takes place. The friction within the fluid is due to the fluid’s viscosity.
When fluids have a high viscosity, the speed of flow tends to be low, and resistance to flow becomes almost totally dependant on the viscosity of the fluid, this condition is known as ‘Laminar flow’.
PRESSURE DROP IN LAMINAR FLOWIn laminar flow, the fluid seems to flow in several layers, one on another as described in below figures.
Because of the viscosity, shear stress is created between the layers of fluid.
WHY ENERGY IS LOST ???Overcoming the frictional forces produced by the shear stress (act opposite direction to flow).
PRESSURE DROP IN LAMINAR FLOW
2
32
2
821
D
LVm
R
LVmPPP
The pressure drop in laminar flow can be expressed as below.
Pressure drop will be 0 if the viscosity is 0, when there is no frictions. It also mean that the pressure drop depends entirely on the viscous effects.
The above equation is used to determine the pressure drop in laminar flow. However, the following equation can be used to determine the pressure drop for all cases of fully develop internal flow (Laminar or Turbulent, Circular or non circular pipes, Smooth or rough surfaces, Horizontal or inclines) and known as Darcy’s equation.
Where friction factors, f can be defined as
PRESSURE DROP
)1(328
2221 D
LV
R
LVPPP mm
)2(2
2
mV
D
LfP
m
w
Vf
2
8
PRESSURE DROP IN LAMINAR
Both (1) and (2) equation can be used to determine the pressure drop for circular pipe in laminar flow, equating both, we find,
The equation shows that for laminar flow the friction factors is a function of Reynolds number only and independent of surface roughness.
eRf
64
HEAD LOSSIn piping system analysis, ΔP = ρgh express
loss in terms of pressure.
The pressure loss can also be expressed in terms of length of water (m) which is
hL = ΔPL/pg
It represents the additional height that the fluid need to be raised by a pump to overcome the frictional losses in the pipe.
MEAN VELOCITY AND FLOW RATE Consider the steady,laminar flow in
pipe as shown in the above figure. The velocity of the fluid is the function
of radius,r and distance x and can be shown as u(r,x).
Since dQ/dt=0, conservation of mass equation is applied, mass flow rate,dm/dt and mean velocity,Vm relation can be define as
R
X
c
A
cm
.
dAx,ruAVmc
MEAN VELOCITY AND FLOW RATEThen Vm can be expressed as
)a(rdr)x,r(uR
2
R
rdr2)x,r(u
A
dA)x,r(uV
R
022
A
c
A c
mcc
For fully developed flow in pipe, the velocity is constant through the flow direction, therefore,
)r(uu0x
)x,r(u
MEAN VELOCITY AND FLOW RATE
)R
r1(V2)r(u
2
2
m
Thus velocity can be expressed as
From (a) and (b), the mean velocity then can be expressed as
)b(]Rr[dx
dP
4
1)r(u 22
)c(dx
dP
8
RV
2
m
From (c) and (d), the relation between mean velocity and velocity at any radius,
Mean velocity, Vm is at radius r=0, therefore
mmax V2u
MEAN VELOCITY AND FLOW RATE
L32
PD
L32
D)PP(
L8
R)PP(V
2221
221
m
Combining the relation between Pressure Drop and Mean velocity in laminar flow, we obtained the following relationship.
The flow rate is
L128
4DP
L128
4D)PP(2R
L8
2R)PP(AVV
2121cm
This relationship is known as the Hagen–Poiseuille equation.
The equation is valid only for laminar flow (NR < 2000).
EXAMPLE 2 - ENERGY LOSS
Determine the energy loss if glycerine at 25°Cflows 30 m through a 150-mm-diameter pipe with an average velocity of 4.0 m/s.First, we must determine whether the flow is laminar or turbulent by evaluating the Reynolds number:
From Appendix B, we find that for glycerin at 25°C
EXAMPLE 2 - ENERGY LOSS
Because NR < 2000, the flow is laminar. Using Darcy’s equation, we get
Notice that each term in each equation is expressed in the units of the SI unit system. Therefore, the resulting units for hL are m or Nm/N. This means that 13.2 Nm of energy is lost by each newton of the glycerine as it flows along the 30 m of pipe.
POISEUILLE’S LAW AND AIR RESISTANCE IN PULMONARY
When we breathe air flows through trachea to the lungs. Although the air is not very viscous, there is a noticeable resistance to the air flow. It causes the pressure drop along the airway and it decrease in the direction of the flow.
When air flows through relatively small diameters tubes as in the terminal brochioles, the flow is laminar.
When it flows at higher rates in larger diameter tubes, like the trachea, the flow is often turbulent.
Much of the flow in the intermediate sized airways can be transitional flow which is difficult to predict either laminar or turbulent.
POISEUILLE’S LAW AND AIR RESISTANCE IN PULMONARY
We learn about Poiseuille’s law that relevant to laminar flow. This law applies to air flow and also blood flow. The relation can be expressed as below
Analogous to V=IR in electrical, Voltage drop,V similar to Pressure gradient, ΔP/L; electric current similar to flow rate and resistance to flow can be expressed as
Resistance is inversely related to the fourth power of the diameter, the resistance in the airways is not predominantly in the smallest diameter airways. Why? See next slide.
L128
DPV
4
4D
128cetansisRe
POISEUILLE’S LAW AND AIR RESISTANCE IN PULMONARY
Airways branches and become narrower, but they are also numerous. Therefore the smaller bronchioles contribute relatively little resistance because of their increased numbers.
The major site of airway resistance is the medium-sized bronchi.
Airways with less than 2mm diameter only contribute about 20% of air resistance.
25 -40% is contributed by upper airways including the mouth, nose, pharynx, larynx and trachea.
FRICTION LOSS IN TURBULENT FLOW
Using Darcy equations we can calculate the friction losses in turbulent flow. It depends on the surface roughness of the pipe as well as Reynolds number (IN LAMINAR, LOSSES ONLY DEPEND ON THE REYNOLD NUMBER)
The є , the average wall roughness can be obtained from tables (experiment has been conducted to determine the value). The average value is for new and clean pipe.
FRICTION LOSS IN TURBULENT FLOW
Roughness value, є for new and clean pipe
MOODY DIAGRAM FOR TURBULENT FLOW
One of the most widely used methods for evaluating the friction factor employs the Moody diagram shown below.
MOODY DIAGRAM –IMPORTANT OBSERVATION
For a given Reynolds number of flow, as the relative roughness is increased, the friction factor f decreases.
For a given relative roughness , the friction factor f decreases with increasing Reynolds number until the zone of complete turbulence is reached.
MOODY DIAGRAM
The transition region is shown in the shaded area ( 2300<Re<4000)
The friction factors alternate between laminar and turbulent flow.
MOODY DIAGRAM
Within the zone of complete turbulence, the Reynolds number has no effect on the friction factor.
As the relative roughness increases, the value of the Reynolds number at which the zone of complete turbulence begins also increases.
The friction factor is a minimum for a smooth pipe (but still not zero because of the no-slip condition) and increases with roughness.
READING THE MOODY DIAGRAM Check your ability to read the Moody diagram correctly by verifying the following values for friction factors for the given values of Reynolds number and relative roughness, using Fig. 8.6:
USE OF THE MOODY DIAGRAMWhy do we need the Moody diagram?
The Moody diagram is used to help determine the value of the friction factor, f for turbulent flow.
How ? First determine the value of the Reynolds number
(calculations), Then determine the relative roughness (dividing Diameter
of the pipe to the pipe roughness).
Therefore, the basic data required are:1. The pipe inside diameter,
2. The pipe material,
3. The flow velocity, and the kind of fluid and its temperature, from which the viscosity can be found.
EXAMPLE 3- MOODY DIAGRAM
Determine the friction factor f if water at 70°C is flowing at 9.14 m/s in an uncoated ductile iron pipe having an inside diameter of 25 mm.
The Reynolds number must first be evaluated to determine whether the flow is laminar or turbulent:
Here D=0.025 m and kinematic viscosity=4.11x10–7 m2/s. We now have
EXAMPLE 3- MOODY DIAGRAMThus, the flow is turbulent. Now the relative roughness must be evaluated. From Table 8.2 we find ε = 2.4 x 10–4 m. Then, the relative roughness is
The steps are as follows:
1. Locate the Reynolds number on the abscissa of the Moody diagram:
EXAMPLE 3- MOODY DIAGRAM
2. Project vertically until the curve for D/ε = 104 is reached. Because 104 is so close to 100, that curve can be used.
3. Project horizontally to the left, and read f = 0.038.
EXAMPLE 4 – MOODY DIAGRAMDetermine the friction factor f if ethyl alcohol at 25°C is flowing at 5.3 m/s in a standard 1.5-in Schedule 80 steel pipe.
Evaluating the Reynolds number, we use the equation
Given, ρ = 787 kg/m3 and μ =1.00 x 10–3 Pa•s. Also, for a 1.5in Schedule 80 pipe, D = 0.0381 m. Then we have
EXAMPLE 4 – MOODY DIAGRAMFrom Fig. 8.6, f = 0.0225. You must interpolate on both
NR and D/ε to determine this value, and you should expect some variation. However, you should be able to read the value of the friction factor f within 0.0005in this portion of the graph.
MINOR LOSSES IN PIPES
The amount of energy losses that occurs as fluid flow through devices as enlargements and contractions in the size of paths.
It is called minor losses since the energy losses is small in comparison with the energy losses due to friction in long, straight section pipes.
Sudden Enlargement
As a fluid flows from a smaller pipe into a larger pipe through a sudden enlargement, its velocity abruptly decreases, causing turbulence, which generates an energy loss.
Figure below shows the sudden enlargement.
SUDDEN ENLARGEMENT
The minor loss is calculated from the equation
where v1 is the average velocity of flow in the smaller pipe ahead of the enlargement.
By making some simplifying assumptions about the character of the flow stream as it expands through the sudden enlargement, it is possible to analytically predict the value of K from the following equation:
SUDDEN ENLARGEMENT Fig below shows the resistance coefficient—sudden
enlargement.
SUDDEN ENLARGEMENTTable below shows the resistance coefficient—
sudden enlargement
EXAMPLE 5 – SUDDEN ENLARGEMENTDetermine the energy loss that will occur as 100 L/min of water flows through a sudden enlargement from a 1-in copper tube (Type K) to a 3-in tube (Type K). See Appendix H for tube dimensions.
EXAMPLE 5 – SUDDEN ENLARGEMENTUsing the subscript 1 for the section just ahead of the enlargement and 2 for the section downstream from the enlargement, we get
EXAMPLE 5 – SUDDEN ENLARGEMENTTo find a value for K, the diameter ratio is needed. We find that
Try to obtained from graph( Resistance coefficient – Sudden enlargement ), K = 10.2. Then we have
This result indicates that 0.40 Nm of energy is dissipated from each Newton of water that flows through the sudden enlargement.
EXAMPLE 6 – SUDDEN ENLARGEMENTDetermine the difference between the pressure ahead of a sudden enlargement and the pressure downstream from the enlargement. Use the data from Example 5.
First, we write the energy equation:
EXAMPLE 6 – SUDDEN ENLARGEMENTIf the enlargement is horizontal, z2 – z1 = 0. Even if it were vertical, the distance between points 1 and 2 is typically so small that it is considered negligible. Now, calculating the velocity in the larger pipe, we get
EXAMPLE 6 – SUDDEN ENLARGEMENTUsing γ = 9.81 kN/m3 for water and hL = 0.40m from Example Problem 10.1, we have
Therefore, p2 is 1.51 kPa greater than p1.
ENERGY LOST IN GRADUAL ENLARGEMENT
If the transition from a smaller to a larger pipe can be made less abrupt than the square-edged sudden enlargement, the energy loss is reduced.
This is normally done by placing a conical section between the two pipes as shown in the below figure.
ENERGY LOST IN GRADUAL ENLARGEMENTCompare gradual enlargement (left) to sudden
enlargement (right).
The energy loss for a gradual enlargement is calculated from
Data for various values are given below
ENERGY LOST IN GRADUAL ENLARGEMENT
ENERGY LOST IN GRADUAL ENLARGEMENTThe energy loss calculated from previous
does not include the loss due to friction at the walls of the transition.
For relatively steep cone angles, the length of the transition is short and therefore the wall friction loss is negligible.
EXAMPLE 7 - GRADUAL ENLARGEMENTDetermine the energy loss that will occur as 100 L/min of water flows from a 1-in copper tube (Type K) into a 3-in copper tube (Type K) through a gradual enlargement having an included cone angle of 30 degrees.
EXAMPLE 7 - GRADUAL ENLARGEMENTFrom Graph (Resistance coefficient – gradual enlargement), we find that K = 0.48. Then we have
Compared with the sudden enlargement described in Example 5, the energy loss decreases by 33 % when 30 degrees the gradual enlargement is used.
DIFFUSER
Another term for an enlargement is a diffuser.
The function of a diffuser is to convert kinetic energy (represented by velocity head) to pressure energy (represented by the pressure head) by decelerating the fluid as it flows from the smaller to the larger pipe.
The theoretical maximum pressure after the expansion could be computed from Bernoulli’s equation,
DIFFUSERIf the diffuser is in a horizontal plane, the elevation
terms can be cancelled out. Then the pressure increase across the ideal diffuser
is
This is often called pressure recovery.In a real diffuser, energy losses do occur and the
general energy equation must be used:
SUDDEN CONTRACTION
The energy loss due to a sudden contraction, such as that sketched in Fig. 10.6, is calculated from
where v2 is the velocity in the small pipe downstream from the contraction.
Figure 10.8 illustrates what happens as the flow stream converges. The lines in the figure represent the paths of various parts of the flow stream called streamlines.
RESISTANCE COEFFICIENT - SUDDEN CONTRACTION
SUDDEN CONTRACTION
SUDDEN CONTRACTION
The following table shows the resistance coefficient—sudden contraction
EXAMPLE 8 – SUDDEN CONTRACTIONDetermine the energy loss that will occur as 100 L/min of water flows from a 3-in copper tube (Type K) into a 1-in copper tube (Type K) through a sudden contraction.
Head lost is
For the copper tube,
EXAMPLE 8 – SUDDEN CONTRACTIONFrom graph, K = 0.42. Then we have
GRADUAL CONTRACTION
The energy loss in a contraction can be decreased substantially by making the contraction more gradual.
The following figure shows such a gradual contraction, formed by a conical section between the two diameters with sharp breaks at the junctions.
GRADUAL CONTRACTIONThe following Figure shows the data (from Reference 8) for the
resistance coefficient versus the diameter ratio for several values of the cone angle.
As the cone angle of the contraction decreases, the resistance coefficient actually increases
The reason is that the data include the effects of both the local turbulence caused by flow separation and pipe friction.
For the smaller cone angles, the transition between the two diameters is very long, which increases the friction losses.
GRADUAL CONTRACTION
GRADUAL CONTRACTION
The following figure shows a contraction with a 120° included angle and D1/D2 = 2.0, the value of K decreases from approximately 0.27 to 0.10 with a radius of only 0.05(D2) where D2 is the inside diameter of the smaller pipe.
CONTRACTION-ENTRANCE LOSS A special case of a contraction occurs when a
fluid flows from a relatively large reservoir or tank into a pipe.
The fluid must accelerate from a negligible velocity to the flow velocity in the pipe.
The ease with which the acceleration is accomplished determines the amount of energy loss, and therefore the value of the entrance resistance coefficient is dependent on the geometry of the entrance.
CONTRACTION - ENTRANCE LOSSThe following figure shows four different configurations and the suggested
value of K for each.
EXAMPLE 9 - CONTRACTION-ENTRANCE LOSSDetermine the energy loss that will occur as 100 L /min of water flows from a reservoir into a 1-in copper tube (Type K) (a) through an inward-projecting tube and (b) through a well rounded inlet.
Part (a): For the tube,
EXAMPLE 9 - CONTRACTION-ENTRANCE LOSSFor an inward-projecting entrance, K = 1.0. Then we have
For well rounded entrance, K = 0.04. Then we have
RESISTANCE COEFFICIENTS FOR VALVES AND FITTINGS
Valves are used to control the amount of flow and may be globe valves, angle valves, gate valves, butterfly valves, any of several types of check valves, and many more.
RESISTANCE COEFFICIENTS FOR VALVES AND FITTINGS
However, the method of determining the resistance coefficient K is different. The value of K is reported in the form
The term fT is the friction factor in the pipe to which the valve or fitting is connected, taken to be in the zone of complete turbulence.
Le is the length of straight pipe of the same nominal diameter as the valve that would have the same resistant as the valve, called the equivalent length.
D is the actual inside diameter of the pipe.
RESISTANCE COEFFICIENTS FOR VALVES AND FITTINGS
Some system designers prefer to compute the equivalent length of pipe for a valve and combine that value with the actual length of pipe.
Equation (10–8) can be solved for Le
Table 10.4 shows the resistance in valves and fittings expressed as equivalent length in pipe diameters, Le>D.
RESISTANCE COEFFICIENTS FOR VALVES AND FITTINGS
RESISTANCE COEFFICIENTS FOR VALVES AND FITTINGS
The following table shows the friction factor in zone of complete turbulence for new, clean, commercial steel pipe
HOW TO CALCULATE ENERGY LOSS FOR VALVE AND FITTINGS?
Energy loss is determined by using the following equation
)g2
v(Kh
2
L
K is determined from the following formula
Te f)D
L(K
For new and clean steel pipe, (Le/D) and fT are obtained from their respective table, then calculate K value.
For material other than steel, the pipe wall roughness,є is determined from table Pipe Roughness Design value (see slide 40). Then, compute D/є and use the Moody diagram to determine fT in the zone of complete turbulent.
The velocity can be obtained from continuity principle, Q=vA.
EXAMPLE 10 – CALCULATE THE EQUIVALENT LENGTH
Determine the resistance coefficient K for a fully open globe valve placed in a 6-in Schedule 40 steel pipe.
From Table 10.4 we find that the equivalent-length ratio for a fully open globe valve is 340. From Table 10.5 we find fT = 0.016 for a 6-in pipe. Then,
EXAMPLE 10 – CALCULATE THE EQUIVALENT LENGTH
Using D=0.154 m for the pipe, we find the equivalent length
EXAMPLE 11-PRESSURE DROP ACROSS VALVECalculate the pressure drop across a fully open globe valve placed in a 4-in Schedule 40 steel pipe carrying 0.0252 m3/s of oil (sg = 0.87)
A sketch of the installation is shown in Fig. 10.24. To determine the pressure drop, the energy equation should be written for the flow between points 1 and 2:
EXAMPLE 11-PRESSURE DROP ACROSS VALVEThe energy loss hL is the minor loss due to the valve only. The pressure drop is the difference between p1 and p2. Solving the energy equation for this difference gives
But z1 = z2 and v1 = v2. Then we have
EXAMPLE 11-PRESSURE DROP ACROSS VALVEFor the pipe,
From Table 10.5 we find fT = 0.017 and for global valve,Le/D = 340.
EXAMPLE 11-PRESSURE DROP ACROSS VALVEFor the oil,
Therefore, the pressure in the oil drops by 23.9 kPa as it flows through the valve. Also, an energy loss of 2.802 m is dissipated as heat from each pound of oil that flows through the valve.
EXAMPLE 11-PRESSURE DROP ACROSS VALVE (NON STEEL)
Calculate the energy loss for the flow of 500 m3/h of water through a standard tee connected to a 6-in uncoated ductile iron pipe. The flow is through the branch.
Use the Procedure for Computing the Energy Loss.
PIPE BENDSThe following figure shows that the minimum resistance for a 90°
bend occurs when the ratio r/D is approximately three.
PIPE BENDS
The following figure shows a 90° bend.
PIPE BENDS The following figure shows a 90° bend pipe.
If Ro is the radius to the outside of the bend, Ri is the radius to the inside of the bend and Do is the outside diameter of the pipe or tube. The radius to the centerline of the pipe or tube called mean radius, r can be expressed as
EXAMPLE 12 - PIPE BENDS
A distribution system for liquid propane is made from 1.25in drawn steel tubing with a wall thickness of 0.083 in. Several 90° bends are required to fit the tubes to the other equipment in the system. The specifications call for the radius to the inside of each bend to be 200 mm.When the system carries 160 L /min of propane at 25°C, compute the energy loss to each bend.
EXAMPLE 12- PIPE BENDS
The radius r must be computed from
where Do = 31.75mm, the outside diameter of the tube as found from Appendix G.Completion of the calculation gives
EXAMPLE 12- PIPE BENDS
We now must compute the velocity to complete the evaluation of the energy loss from Darcy’s equation:
The relative roughness is
EXAMPLE 12- PIPE BENDSThen, we can find fT = 0.0108 from the Moody diagram
(Fig. 8.6) in the zone of complete turbulence. Then
Now the energy loss can be computed:
BEND AT ANGLES OTHER THAN 90°
Reference 2 recommends the following formula for computing the resistance factor K for bends at angles other than 90°
where K is the resistance for one 90° bend found from right table.
EXAMPLE 13 – BEND OTHER THAN 90 DEGEvaluate the energy loss that would occur if the drawn steel tubing described in Example Problem 10.10 is coiled for 4.5 revolutions to make a heat exchanger. The inside radius of the bend is the same 200 mm used earlier and the other conditions are the same.
Let’s start by bringing some data from Example Problem 10.10.
EXAMPLE 13 – BEND OTHER THAN 90 DEGNow we can compute the value of KB for the complete coil using Eq. (10–10). Note that each revolution in the coil contains four 90° bends. Then,
The total bend resistance is
EXAMPLE 13 – BEND OTHER THAN 90 DEG
Then the energy loss is found from