FLOW IN PIPES Applications heating cooling fluid distributions

FLOW IN PIPES Applications – heating, cooling, fluid distributions (blood and its vessels) Attention will be given on: n Laminar and Turbulent Flows n Reynolds Number n Entrance Region Flow n Pressure drops and head loss in Laminar flow in circular pipes n Minor Losses in piping system

n n n 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 NUMBER How to distinguish Turbulent and Laminar in mathematics? Osborne Reynolds manage to do that in 1880 s. Re = v. Dρ/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 NUMBER Determine 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 Blood Most 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.

n n n TURBULENT AND ITS Haemodynamic EFFECT 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 lungs In 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 REGION FLOW 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.

ENTRY LENGTH (LENGTH OF ENTRANCE REGION), Lh Laminar flow, Turbulent flow, For a pipe length over than 10 D, entrance effect is negligible and thus,

FULLY DEVELOP FLOW IN ARTERIES If we consider Re = 300, then the previous equation gives Xe=18 D. 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 n Assumptions: q q 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 n n 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.

LAMINAR FLOW IN PIPES n 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, d. P/dx is constant.

LAMINAR FLOW IN PIPES n The third assumption is that the fluid is 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 n 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(πr 2) - (P + d. P)(πr 2) – τ2πrdx = 0 n Solve Eq. (1. 1) and the result is -d. P(πr 2) = 2πrτdx

LAMINAR FLOW IN PIPES n The result from a simple force balance is shear stress as a function of pressure gradient, d. P/dx and radial position, r: and

LAMINAR FLOW IN PIPE n 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:

LAMINAR FLOW IN PIPE n 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 function of the radius, r:

LAMINAR FLOW IN PIPE n n So far, in this analysis, we have made three assumptions. First, steady flow (d. Q/dt = 0); second, fully developed tube (d. P/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 C 1.

LAMINAR FLOW IN PIPE The equation, which gives velocity as a function of radius, r, is then 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 The d. P/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.

PRESSURE DROP Pressure 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 FLOW In 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 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.

PRESSURE DROP 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 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.

HEAD LOSS In 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 h. L = 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 n n n 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 d. Q/dt=0, conservation of mass equation is applied, mass flow rate, dm/dt and mean velocity, Vm relation can be define as R X

MEAN VELOCITY AND FLOW RATE Then Vm can be expressed as For fully developed flow in pipe, the velocity is constant through the flow direction, therefore,

MEAN VELOCITY AND FLOW RATE Thus velocity can be expressed as From (a) and (b), the mean velocity then can be expressed as From (c) and (d), the relation between mean velocity and velocity at any radius, Mean velocity, Vm is at radius r=0, therefore

MEAN VELOCITY AND FLOW RATE Combining the relation between Pressure Drop and Mean velocity in laminar flow, we obtained the following relationship. The flow rate is 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°C flows 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 N < 2000, the flow is laminar. Using R 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 h. L 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.

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 2 mm 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 n n 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 n 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 noslip condition) and increases with roughness.

READING THE MOODY DIAGRAM n 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 DIAGRAM Why 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 ? q q 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. 11 x 10– 7 m 2/s. We now have

EXAMPLE 3 - MOODY DIAGRAM Thus, 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 DIAGRAM Determine 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/m 3 and μ =1. 00 x 10– 3 Pa • s. Also, for a 1. 5 in Schedule 80 pipe, D = 0. 0381 m. Then we have

EXAMPLE 4 – MOODY From. DIAGRAM 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. 0005 in 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.

n n 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 n The minor loss is calculated from the equation n where v 1 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 n Fig below shows the resistance coefficient—sudden enlargement.

SUDDEN ENLARGEMENT Table below shows the resistance coefficient— sudden enlargement

EXAMPLE 5 – SUDDEN ENLARGEMENT Determine 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 ENLARGEMENT Using 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 ENLARGEMENT To 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 ENLARGEMENT Determine 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 ENLARGEMENT If the enlargement is horizontal, z 2 – z 1 = 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 ENLARGEMENT Using γ = 9. 81 k. N/m 3 for water and h. L = 0. 40 m from Example Problem 10. 1, we have Therefore, p 2 is 1. 51 k. Pa greater than p 1.

ENERGY LOST IN GRADUAL ENLARGEMENT n n 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 ENLARGEMENT Compare gradual enlargement (left) to sudden enlargement (right).

ENERGY LOST IN GRADUAL ENLARGEMENT The energy loss for a gradual enlargement is calculated from Data for various values are given below

ENERGY LOST IN GRADUAL ENLARGEMENT The 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 ENLARGEMENT Determine 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 ENLARGEMENT From 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,

DIFFUSER If 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 n n The energy loss due to a sudden contraction, such as that sketched in Fig. 10. 6, is calculated from where v 2 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 n The following table shows the resistance coefficient— sudden contraction

EXAMPLE 8 – SUDDEN CONTRACTION Determine 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 CONTRACTION From 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 CONTRACTION The following Figure shows the data (from Reference 8) for the resistance coefficient versus the diameter ratio for several values of the cone angle. n n n 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 D 1/D 2 = 2. 0, the value of K decreases from approximately 0. 27 to 0. 10 with a radius of only 0. 05(D 2) where D 2 is the inside diameter of the smaller pipe.

CONTRACTION-ENTRANCE LOSS n A special case of a contraction occurs when a n n 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 LOSS The following figure shows four different configurations and the suggested value of K for each.

EXAMPLE 9 - CONTRACTION-ENTRANCE LOSS Determine 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 inwardprojecting tube and (b) through a well rounded inlet. Part (a): For the tube,

EXAMPLE 9 - CONTRACTION-ENTRANCE LOSS For 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 f. T 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 K is determined from the following formula For new and clean steel pipe, (Le/D) and f. T 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 f. T in the zone of complete turbulent. The velocity can be obtained from continuity principle, Q=v. A.

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 equivalentlength ratio for a fully open globe valve is 340. From Table 10. 5 we find f. T = 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 VALVE Calculate the pressure drop across a fully open globe valve placed in a 4 -in Schedule 40 steel pipe carrying 0. 0252 m 3/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 VALVE The energy loss h. L is the minor loss due to the valve only. The pressure drop is the difference between p 1 and p 2. Solving the energy equation for this difference gives But z 1 = z 2 and v 1 = v 2. Then we have

EXAMPLE 11 -PRESSURE DROP ACROSS VALVE For the pipe, From Table 10. 5 we find f. T = 0. 017 and for global valve, Le/D = 340.

EXAMPLE 11 -PRESSURE DROP ACROSS VALVE For the oil, Therefore, the pressure in the oil drops by 23. 9 k. Pa 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 m 3/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 BENDS The following figure shows that the minimum resistance for a 90° bend occurs when the ratio r/D is approximately three.

n PIPE BENDS The following figure shows a 90° bend.

PIPE BENDS n n 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. 25 in 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. 75 mm, 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 Then, we. BENDS can find f. T = 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° n 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 DEG Evaluate 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.

EXAMPLE 13 – BEND OTHER THAN 90 DEG Now 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