CHAPTER 3 FLUID DYNAMICS 45 QP = discharge through cross-section P-P AP = cross-sectional area through P-P vp = fluid mean velocity through P-P QR = discharge through cross-section R-R AR = cross-sectional area through R-R vR = fluid mean velocity through R-R Discharge at section P = Discharge at section R QP = QR AP vP = AR vR 3.3.1 APPLICATION OF CONTINUITY EQUATION FOR PIPE a) Principle of continuity equation in single pipe A liquid is flowing from left to right and the pipe is in the same direction. By the continuity principle, the discharge must be the same at each section. The mass going into the pipe is equal to the mass going out of the pipe. Discharge at section 1 = Discharge at section 2 = = QP =QR
CHAPTER 3 FLUID DYNAMICS 46 Example 3.3 Referring to the Figure 3.3, If the area A1 = 3 10-3 m2 and A2 = 10 10-3 m2 and the upstream mean velocity, 1 = 2.3 m/s, calculate the downstream mean velocity. Solution: , 1 = 3 10−3 2 2 = 10 10−3 2 1 = 2.3 / 1 = 2 11 = 22 2 = 1 2 1 2 = 3 10−3 10 10−3 (2.3) = . / Figure 3.3 b) Principle of continuity equation in branch pipe Total discharge into the junction = Total discharge out of the junction = + = + Example 3.4 Water flows through a pipe with a diameter of 50 mm. Then the pipe split into two, one of the pipes has a diameter 25 mm with the velocity of flow 0.4 m/s and the other one has a diameter 15 mm with the velocity 0.6 m/s. Calculate the velocity in the main pipe. Solution:
CHAPTER 3 FLUID DYNAMICS 47 Given; d1 = 50 mm = 0.05 m d2 = 25 mm = 0.025 m v2 = 0.4 m/s d3 = 15 mm = 0.015 m v3 = 0.6 m/s 1 = 1 2 4 = (0.05) 2 4 = . − 2 = 2 2 4 = (0.025) 2 4 = . − 3 = 3 2 4 = (0.015) 2 4 = . − 1 = 2 + 3 11 = 22 + 33 (1.963 10−3 )1 = (4.909 10−4 )(0.4) + (1.767 10−4 )(0.6) 1 = (4.909 10−4 )(0.4) + (1.767 10−4 )(0.6) (1.963 10−3) = . / 3.4 BERNOULLI’S THEOREM Bernoulli’s Theorem states that the total energy of each particle of a body of fluid is the same provided that no energy enters or leaves the system at any point. The division of this energy between potential, pressure and kinetic energy may vary, but the total remains constant. In symbols: = + 2 2 + = By Bernoulli’s Theorem, Total energy per unit weight at section 1 = Total energy per unit weight at section 2 1 + 1 2 2 + 1 = 2 + 2 2 2 + 2
CHAPTER 3 FLUID DYNAMICS 48 z = potential head p = pressure head g v 2 2 = velocity head H = Total head Bernoulli’s Equation has some restrictions in its applicability, they are: i. the flow is steady ii. the density is constant (which also means the fluid is compressible) iii. friction losses are negligible iv. The equation relates the state at two points along a single streamline (not conditions on two different streamlines). 3.4.1 APPLICATION OF BERNOULLI’S EQUATION
CHAPTER 3 FLUID DYNAMICS 49 a) Pipe Lines i. Horizontal Pipe = + + Example 3.5 Water flows through a pipe 36 m from the sea level as shown in figure 3.5. Pressure in the pipe is 410 kN/m2 and the velocity is 4.8 m/s. Calculate total energy of every weight of unit water above the sea level. Solution: Total energy per unit weight: = + 2 2 + = 410 103 1000 9.81 + (4.8) 2 2 9.81 + 36 = . Figure 3.5 = 1000 3 = 1000 9.81 3
CHAPTER 3 FLUID DYNAMICS 50 Example 3.6 A very large pipe carries water with a very slow velocity and empties into a small pipe with a high velocity. If pressure at P2 is 7000 N/m2 lower than pressure at P1, what is the velocity of the water at small pipe? Solution: , 1 − 2 = 7000 / 2 1 = 2 = 0 1 = 0 1 + 1 2 2 + 1 = 2 + 2 2 2 + 2 1 − 2 = 2 2 2 1 − 2 = 2 2 2 2 = √ 2(1 − 2) 2 = √ 2(7000) 1000 = . / = 1000 3 ii. Inclined Pipe
CHAPTER 3 FLUID DYNAMICS 51 1 + 1 2 2 + 1 = 2 + 2 2 2 + 2 Example 3.7 A bent pipe labeled AB measures 3 m and 5 m respectively above the datum line. The diameter A and B are both 5 cm and 20 cm. The water pressure at inlet is 350 kN/m2 . If the velocity at B is 1 m/s, determine the pressure at B in kN/m2 . Solution: , 1 = 350 / 2 = 350 103 / 2 1 = 5 = 0.05 2 = 20 = 0.2 1 = 3 2 = 5 2 = 1 / 1 + 1 2 2 + 1 = 2 + 2 2 2 + 2 2 = [ 1 + 1 2 − 2 2 2 + (1 − 2)] 2 = [ 350 103 1000 9.81 + 162 − 1 2 2 9.81 + (3 − 5)] 1000 9.81 2 = 457880 /2 = . / = 1000 9.81 /3 1 = 2 11 = 22 1 = 2 1 2 1 = 2 2 1 2 2 1 = 0.2 2 0.052 (1) = /
CHAPTER 3 FLUID DYNAMICS 52 b) Venturi Meter Venturi meter is a device used for measuring the rate of flow of a non-viscous, incompressible fluid in non-rotational and steady-stream lined flow. Although venturi meters can be applied to the measurement of gas, they are most commonly used for liquids. The following treatment is limited to incompressible fluids. Figure 3.6: Venturi meter i. Horizontal Venturi Meter Theory Discharge = √ ( − ) A1 = Area at entrance (m2 ) g = Gravity (9.81 m2 /s)
CHAPTER 3 FLUID DYNAMICS 53 m = Area ratio (value of m > 1) H = Pressure difference expressed as a head of the liquid flowing in venturi meter (m) Actual Discharge = = √ ( − ) Cd = Coefficient of discharge Area ratio = @ = A1 = Area of entrance (m2 ) A2 = Area of throat (m2 ) Pressure Difference = − @ = ( − ) P1 – P2 = Pressure difference between entrance and throat (N/m2 ) x = Difference of level in U-tube manometer (m) ωsub = specific weight of fluid in pipe (N/m3 ) ωHg = Specific weight of mercury (N/m3 )
CHAPTER 3 FLUID DYNAMICS 54 Basic Step to Calculate the Actual Discharge of Horizontal Venturi Meter Example 3.8 A horizontal venturi meter is used to measures the flowrate of oil (Soil = 0.8). Diameter of the entering is 420 mm and the throat is 280 mm. A differential mercury U-tube manometer (Smercury = 13.6) is connected to the entering and throat indicated difference level of 70 mm. Calculate the actual discharge if Cd = 0.98. Solution: Given; Soil = 0.8 SHg = Smercury = 13.6 d1 = 420 mm = 0.42 m d2 = 280 mm = 0.28 m Cd = 0.98 Difference level (u-tube), x = 70 mm = 0.07 m 1 • Find value of A1 & A2 : • = 2 •Area ratio : • = @ = 3 • Pressure Difference : • = − @ = − 4 •Actual Discharge : • = = ( −)
CHAPTER 3 FLUID DYNAMICS 55 1 = 1 2 4 = (0.42) 2 4 = . 2 = 2 2 4 = (0.28) 2 4 = . = 1 2 = 0.1385 0.0616 = . = ( − 1) = 0.07 ( (13.6 9810) (0.8 9810) − 1) = . = 1√ 2 (2 − 1) = (0.98)(0.1385)√ 2(9.81)(1.12) (2.252 − 1) = . / ii. Inclined Venturi Meter Theory Discharge = √ ( − ) A1 = Area at entrance (m2 )
CHAPTER 3 FLUID DYNAMICS 56 g = Gravity (9.81 m2 /s) m = Area ratio (value of m > 1) H = Pressure difference expressed as a head of the liquid flowing in venturi meter (m) Actual Discharge = = √ ( − ) Cd = Coefficient of discharge Area ratio = @ = A1 = Area of entrance (m2 ) A2 = Area of throat (m2 ) Pressure Difference = − + ( − ) @ = ( − ) P1 – P2 = Pressure difference between entrance and throat (N/m2 ) Z1 – Z2 = height difference between entrance and throat from datum line x = Difference of level in U-tube manometer (m) ωsub = specific weight of fluid in pipe (N/m3 ) ωHg = Specific weight of mercury (N/m3 )
CHAPTER 3 FLUID DYNAMICS 57 Basic Step to Calculate the Actual Discharge of Incline Venturi Meter Example 3.9 A vertical venturi meter measures the flow of oil of specific gravity 0.82 and has an entrance of 125 mm diameter and throat of 50 mm diameter. There are pressure gauges at the entrance and at the throat, which is 300 mm above the entrance. If the coefficient for the meter is 0.97 and pressure difference is 27.5 kN/m2 , calculate the actual discharge in m3 /s. Solution: Given; P1 –P2 = 27.5 kN/m2 = 27.5 x 103 N/m2 Soil = 0.82 d1 = 125 mm = 0.125 m d2 = 50 mm = 0.05 m 1 • Find value of A1 & A2 : • = 2 •Area ratio : • = @ = 3 • Pressure Difference : • = − + − @ = − 4 •Actual Discharge : • = = ( −) ωsub = Soil x ωwater ωsub = 0.82 x 9810 ωsub = 8044.2 N/m3
CHAPTER 3 FLUID DYNAMICS 58 Cd = 0.97 Difference level (u-tube), (Z1 – Z2) = - 300 mm = - 0.3 m 1 = 1 2 4 = (0.125) 2 4 = . = 1 2 2 2 = 0.1252 0.052 = . = (1 − 2) + (1 − 2 ) = (27.5 103 ) (8044.2) + (− 0.3) == . = 1√ 2 (2 − 1) = (0.97)(0.0123)√ 2(9.81)(3.1186) (6.252 − 1) = . / c) Orifice Meter The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it. There is a pressure tap upstream from the orifice plate and another just downstream. There are three recognized methods of placing the taps and the coefficient of the meter will depend upon the position of the taps. The principle of the orifice meter is identical with that of the venturi meter. The reduction at the cross section of the flowing stream in passing through the orifice increases the velocity head at the expense of the pressure head, and the reduction in pressure between the taps is measured by a manometer. Bernoulli's equation provides a basis for correlating the increase in velocity head with the decrease in pressure head.
CHAPTER 3 FLUID DYNAMICS 59 Figure 3.7: Part of orifice meter Example 3.10 A meter orifice has a 100 mm diameter rectangular hole in the pipe. Diameter of the pipe is 250 mm. Coefficient of discharge, Cd = 0.65 and specific gravity of oil in the pipe is 0.9. The pressure difference that is measured by the manometer is 750 mm. Calculate the flow rate of the oil through the pipe. Solution: Given, d1 = 250 mm = 0.25 m d2 = 100 mm = 0.10 m Cd = 0.65 oil = 0.9 N/m3 x = 750 mm = 0.75 m i. Find A1; 1 = 1 2 4 = (0.25) 2 4 = . ii. Area ratio, m; = 1 2 2 2 = 0.252 0.1 2 = . iii. Pressure difference, H = ( − 1) = 0.75 ( (13.6 9810) (0.9 9810) − 1) = .
CHAPTER 3 FLUID DYNAMICS 60 iv. Actual discharge, Qactual; = 1√ 2 (2 − 1) = (0.65)(0.0491)√ 2(9.81)(10.58) (6.252 − 1) = . / d) Simple Pitot Tube The Pitot tube is a device used to measure the local velocity along a streamline (Figure 5.10). The Pitot tube has two tubes: one is a static tube (a), and another is an impact/stagnation tube (b). The opening of the impact tube is perpendicular to the flow direction. The opening of the static tube is parallel to the direction of flow. The two legs are connected to the legs of a manometer or an equivalent device for measuring small pressure differences. The static tube measures the static pressure, since there is no velocity component perpendicular to its opening. The impact tube measures both the static pressure and impact pressure (due to kinetic energy). In terms of heads, the impact tube measures the static pressure head plus the velocity head. Figure 3.8: Pitot tube
CHAPTER 3 FLUID DYNAMICS 61 Advantages of Pitot tube 1. It is easy and economical to install and remove. 2. It has no moving parts which helps in minimizing frictional losses. 3. It is small in size. 4. It is low in cost. 5. It causes very less pressure loss. Example 3.11 A Pitot Tube is used to measure air velocity in a pipe attached to a mercury manometer. Head difference of that manometer is 6 mm water. The weight density of air is 1.25 kg/m3 . Calculate the air velocity if coefficient of the pitot tube, C = 0.94. Solution: = √ Given; ℎ = 0.006 = 1.25 /2 = 0.94 = ℎ = ℎ ℎ = ℎ ℎ = 0.006 1000 1.25 ℎ = . = 0.94√2(9.81)(4.8) = . /
CHAPTER 3 FLUID DYNAMICS 62 PROBLEMS 1. On a circular conduit there are different diameters: diameter D1 = 2 m changes into D2 = 3 m. The velocity in the entrance profile was measured: v1 = 3 m/s. Calculate the discharge and mean velocity at the outlet profile (see fig. 1). 2. Raw oil flow a pipe of 40 mm diameter and entered a pipe 25 mm diameter. The volume flow rate is 3.75 liter/s. Calculate the flow velocity for both pipes and the density of raw oil if the mass flow rate is 3.23 kg/s. 3. Oil flows through a pipe RS and split into two pipes which are ST and SU. The following information is given; - Diameter pipe RS = 250 mm - Diameter pipe ST = 200 mm - Specific gravity, soil = 0.92. Calculate; i. Discharge and mass flowrate of oil at pipe RS if velocity is 2.5 m/s ii. Diameter pipe SU if velocity at pipe ST is 1.5 m/s and at pipe SU is 3 m/s. 4. A horizontal venturi meter is used to measure fluid flow from the tank with the inlet and throat diameter of 105 mm and 45 mm respectively. A total of 2000 liters of water collected in 3 minutes at different readings of mercury levels in the U tube of 256 mm. Calculate the coefficient of discharge. Given the relative density of mercury is 13.6. 5. Incline venturi meter at 30⁰ measured the flow of oil in a pipe with diameter of d1=150 mm and the diameter of throat is, d2=80 mm. The difference between the throat and the entrance of the meter is measured by the U-tube containing mercury which is being in contact with the oil. Calculate the pressure difference and difference in the level of mercury in the U-tube if the distance at the entrance and the throat is, L1=250 cm and the oil flowing in the pipe is, v1=3.7 m/s. assume specific gravity of oil is, soil=0.9. Assume of specific gravity mercury is 13.6.
CHAPTER 1 INRODUCTION OF FLUID 63 ENERGY LOSS IN PIPELINES 4 2
CHAPTER 4 ENERGY LOSS IN PIPELINES 64 A pipe is defined as a closed conduit of circular section through which the fluid flows, filling the complete cross-section. The fluid in the pipe has no free surface. It will be at a pressure which may vary along the pipe. Losses of energy in a pipeline cannot be ignored. When the shock losses and friction loss have been determined, they are inserted in Bernoulli’s equation in the usual way. 1 + 1 2 2 + 1 = 2 + 2 2 2 + 2 + 1 = Pressure energy / head 1 2 2 = Kinetic energy / head 1 = Potential energy / head H = Total energy / head losses 4.1 VELOCITY PROFILE IN THE ROUND PIPE SYSTEM Velocity profile shows the difference in intensity of resistance of fluid particles across the flow, due to cohesive and adhesive forces. The velocity profile in turbulent flow is flatter in the central part of the pipe (i.e., in the turbulent core) than in laminar flow. The flow velocity drops rapidly, extremely close to the walls. This is due to the diffusivity of the turbulent flow.
CHAPTER 4 ENERGY LOSS IN PIPELINES 65 Figure 4.1: Velocity profile in pipelines system 4.2TYPES OF HEAD LOSS a) Shock Loss at Sudden Enlargement Sudden enlargement occurs when liquid flows from a smaller pipe to a larger pipe abruptly as shown in Figure 4.5. This causes the velocity to decrease abruptly, causing turbulence and, hence, energy loss. The head loss due to sudden enlargement can be represented by the head loss coefficient K, which depends on the ratio of the two pipe diameters. It can be calculated using the following equation: Figure 4.2: Sudden enlargement in pipelines
CHAPTER 4 ENERGY LOSS IN PIPELINES 66 Loss of head at enlargement, hL g v v 2 2 1 2 V1 = velocity liquid at smaller pipe (m/s) V2 = velocity liquid at larger pipe (m/s) Example 4.1 A pipe carrying 1800 l/min of water increases suddenly from 10 cm to 15 cm diameter. Find the head loss due to the sudden enlargement Solution: , 1 = 1800 / = 1800 0.001 60 = 0.03 3⁄ 1 = 10 = 0.1 2 = 15 = 0.15 the head loss due to the sudden enlargement: ℎ = (1 − 2) 2 2 ℎ = (3.892 − 1.698) 2 19.62 = . 1 = 11 1 = 1 1 1 = 0.03 (4) (0.1) 2 = . / 2 = 22 2 = 2 2 2 = 0.03 (4) (0.15) 2 = . / b) Shock Loss at Sudden Contraction In a sudden contraction, the flow converges to form a vena contracta at section (C) in the smaller pipe. The loss of energy in the convergence from sections (1) to (C) is small and the
CHAPTER 4 ENERGY LOSS IN PIPELINES 67 Loss of head at sudden contraction, hC g v Cc 2 1 1 2 2 2 main loss occurs in the enlargement from sections (C) to (2). It is usual to ignore the loss from sections (1) to (C) and treat the loss from (C) to (2) as if it was due to a sudden enlargement from the area of the vena contracta Ac to the area A2 of the smaller pipe. (Figure 4.3) Figure 4.3: Sudden contraction in pipelines Cc = coefficient of contraction V2 = velocity at smaller pipe (m/s) Example 4.2 A pipe carrying 1800 l/min of water decreases suddenly from 35 cm to 15 cm diameter. Find the head loss due to the sudden contraction if coefficient contraction, CC is 0.65. Solution:
CHAPTER 4 ENERGY LOSS IN PIPELINES 68 , 1 = 1800 / = 1800 0.001 60 = 0.03 3⁄ 1 = 35 = 0.35 2 = 15 = 0.15 = 0.65 the head loss due to the sudden contraction: ℎ = ( 1 − 1) 2 2 2 2 ℎ = ( 1 0.65 − 1) 2 1.6982 19.62 = . 1 = 11 1 = 1 1 1 = 0.03 (4) (0.35) 2 = . / 2 = 22 2 = 2 2 2 = 0.03 (4) (0.15) 2 = . / c) Frictional Resistance to Flow When there is a loss of head due to friction in a pipeline in terms of the velocity head; we assume that the frictional resistance per unit area of the pipe wall is proportional to the square of the mean velocity of flow. Consider a cylinder of fluid of length L completely filling the pipe of cross sectional area A and moving with a mean velocity v (Figure 4.4). The force acting on the cylinder is the force due to pressure difference and the force due to frictional resistance. Since the velocity is constant and there is no acceleration, the resultant of these two forces in the direction of motion must be zero. Figure 4.4: Friction loss at pipelines
CHAPTER 4 ENERGY LOSS IN PIPELINES 69 Loss of head due to friction, hf g v d fL 2 4 2 f = friction factor L = length of pipeline (m) d = diameter pipe (m) v = velocity liquid in pipe (m/s) Example 4.3 Determine the loss of head due to friction in a pipe 14 m long and 2 m diameter which carries 1.5 m/s oil. Take into consideration f = 0.05. Solution : , = 14 1 = 2 1 = 1.5 / = 0.05 the head loss due to the friction loss: ℎ = 4 2 2 ℎ = 4 (0.05)(14) (1.5) 2 (2)(19.62) = . d) Loss at Entry The entrance loss is the head loss that occurs when a liquid flows from a large tank into a pipe. At the entrance to the pipe, the liquid must accelerate from zero velocity at the liquid surface in the tank to the velocity corresponding to the flow rate through the pipe.
CHAPTER 4 ENERGY LOSS IN PIPELINES 70 Loss of head at sharp entrance, hi g v 2 2 1 2 Figure 4.5: loss at entry v = velocity liquid at the pipe (m/s) e) Loss at Rounded Exit The exit loss is associated with liquid flow from a pipe into a large tank as shown in Figure 4.6. As the liquid enters the tank, its velocity is decreased to very nearly zero. Similar to entrance loss, the exit loss can be calculated as equation below. Figure 4.6: Loss at rounded exit ν1 ≈ 0 ν2 ν2 ≈ 0 ν1
CHAPTER 4 ENERGY LOSS IN PIPELINES 71 v = velocity liquid at the pipe before discharge into large reservoir (m/s) 4.3 HEAD LOSS IN PIPELINE SYSTEMS All pipeline problems should be solved by applying Bernoulli’s theorem between points for which the total energy is known and including expressions for any loss of energy due to shock or to friction, thus 1 + 1 2 2 + 1 = 2 + 2 2 2 + 2 + ℎ + 4.3.1 DISCHARGE TO ATMOSPHERE Bernoulli equation; + 2 2 + = + 2 2 + + Assume that; PA = PB = Patm, vA = 0 (large reservoir) So that; − = 2 2 + 1
CHAPTER 4 ENERGY LOSS IN PIPELINES 72 Head losses; = (ℎ ) + (ℎ) = 0.51 2 2 + 4 1 2 2 Substitute 2 in 1; − = 2 2 + 0.51 2 2 + 4 1 2 2 Example 4.6 Water is discharged from a reservoir into the atmosphere through a pipe 100 m long. There is a sharp entrance to the pipe and the diameter is 30 mm for the first 70 m long pipe. The pipe then suddenly contracts to the half diameter of the first pipe for the remaining of its length. Taken Cc = 0.7 and f = 0.01 for both pipes, calculate the total loss of head if the discharge is 0.2 dm3 /s Solution: Given; Q = 0.2 dm3 /s x 1/103 = 0.0002 m3 /s d1 = 30 mm = 0.03 m L1 = 70 m d2 = (½)d1 = (½)0.03 = 0.015 m L2 = 100 – 70 = 30 m f = 0.01 Cc = 0.7 1 = 1 2 4 = (0.3) 2 4 = . − 2 = 2 2 4 = (0.015) 2 4 = . − 1 = 11 1 = 1 1 = 0.0002 7.0686 10−4 = . / 2 = 22 2 = 2 2 = 0.0002 1.7671 10−4 = . / 2
CHAPTER 4 ENERGY LOSS IN PIPELINES 73 Total head losses which are occurred; = (ℎ ) + (ℎ1) + (ℎ) + (ℎ2) = 0.51 2 2 + 41 1 2 1 2 + ( 1 − 1) 2 ( 2 2 2 ) + 42 2 2 2 2 = 0.5(0.2829) 2 19.62 + 4(0.01)(70)(0.2829) 2 (0.3)(19.62) + ( 1 0.7 − 1) 2 ( (1.1318) 2 19.62 ) + 4(0.01)(30)(1.1318) 2 (0.015)(19.62) = 0.002 + 0.3827 + 0.012 + 5.224 = . 4.3.2 HYDRAULIC GRADIENTS Bernoulli equation; + 2 2 + = + 2 2 + + Assume that; PA = PB = Patm, vA = vB = 0 (large reservoir) So that; − = 1
CHAPTER 4 ENERGY LOSS IN PIPELINES 74 Head losses; = (ℎ ) + (ℎ1) + (ℎ) + (ℎ2) + (ℎ) = 0.51 2 2 + 41 1 2 1 2 + (1 − 2) 2 2 + 42 2 2 2 2 + 2 2 2 Substitute 2 in 1; − = 0.51 2 2 + 41 1 2 1 2 + (1 − 2) 2 2 + 42 2 2 2 2 + 2 2 2 Example 4.7 Two reservoirs are connected by a pipeline which is 150 mm in diameter for the first 6 m and 225 mm in diameter for the remaining 15 m. The entrance and exit are sharp and the change of section is sudden. The water surface in the upper reservoir is 6 m above that in the lower. Tabulate the losses of head which occur and calculate the rate of flow in m3 /s. Friction coefficient, f is 0.01 for both pipes. Solution: Given; ZA – ZB = 6 m f = 0.01 d1 = 150 mm = 0.15 m L1 = 6 m d2 = 225 mm = 0.225 m L2 = 15 m Solution; Bernoulli equation; + 2 2 + = + 2 2 + + Assume that; PA = PB = Patm, vA = vB = 0 (large reservoir) So that; − = Head losses; = (ℎ ) + (ℎ1) + (ℎ) + (ℎ2) + (ℎ) = 0.51 2 2 + 41 1 2 1 2 + (1 − 2) 2 2 + 42 2 2 2 2 + 2 2 2 Substitute 2 in 1; − = 0.51 2 2 + 41 1 2 1 2 + (1 − 2) 2 2 + 42 2 2 2 2 + 2 2 2 2
CHAPTER 4 ENERGY LOSS IN PIPELINES 75 1 = 1 2 4 = (0.15) 2 4 = . 2 = 2 2 4 = (0.225) 2 4 = . 1 = 2 11 = 22 1 = (0.0398) (0.0177) 2 1 = (. ) Total head losses; ℎ = 0.51 2 2 = 0.5(2.252) 2 19.62 = . ℎ1 = 41 1 2 1 2 = 4(0.01)(6)(2.252) 2 (0.15)(19.62) = . ℎ = (1 − 2) 2 2 = (2.252 − 2) 2 19.62 = . ℎ1 = 42 2 2 2 2 = 4(0.01)(15)2 2 (0.225)(19.62) = . ℎ = 2 2 2 = 2 2 19.62 = . − = (ℎ ) + (ℎ1) + (ℎ) + (ℎ2) + (ℎ) 6 = 0.1292 2 + 0.4132 2 + 0.082 2 + 0.1362 2 + 0.0512 2 6 = (0.809)2 2 2 = √ 6 0.809 2 = . / = 22 = (0.0398)(2.72) = . /
CHAPTER 4 ENERGY LOSS IN PIPELINES 76 PROBLEMS 1. The horizontal pipe carrying water with the rate of flow of 3.8 m3 /min. the pipe is 300 mm in diameter suddenly decreased to 200 mm. Given coefficient of contraction is 0.55. Calculate the head loss due to the sudden contraction. 2. A 100 mm diameter pipe drain the water at a rate of 1.8 m3 /minit, grows rapidly to a diameter of 150 mm. Calculate; i. The head loss due to shock ii. The pressure difference between points before and after enlargement. 3. Water from a large reservoir is discharged to atmosphere through a pipe of 100 m long. The diameter is 30 mm for the first 70 m long pipe. The water then enters a second pipe which is half the diameter of the first pipe. Taking friction coefficient, f = 0.01 for both pipes and Cc = 0.7, calculate the total loss of head if the discharge is 0.2 dm3 /s. 4. A 65 m long horizontal pipeline is connected to a reservior at one end discharges freely into the atmosphere at the other end. For the first 35 m of its length from the reservior, the pipe is 15 cm in diameter. Then the diameter of the pipe suddenly enlarges to 30 cm. The entrance is sharp and the flow rate of water entering the pipe is 0.048 m3 /s. Calculate the difference of level from the surface of the reservior to the outlet if the friction coefficient, f = 0.01 for both section of the pipe. 5. Two reservoirs are connected by a pipeline which is 150 mm in diameter for the first 9 m and 300 mm in diameter for the remaining 16 m. The entrance and exit are sharp and the change of section is sudden. The water surface in the upper reservoir is 9 m above that in the lower. Tabulate the losses of head which occur and calculate the rate of flow in m3 /s. Friction coefficient, f is 0.01 for both pipes.
77 REFERENCES Cengel, Y. A., and Cimbala, J. M., (2017). Fluid Mechanics: Fundamental and Application Fourth Edition. McGraw-Hill Education, Hibbler R.C (2017). Fluid Mechanics (2nd Edition). Pearson Yahaya Ramli., (2017). Mekanik Bendalir Teori dan Penggunan. Edisi Pertama. UTM Press Norhaslinda Mustafar (2016). Fluid Mechanics, First Edition. Politeknik Seberang Perai. Janna W.S, (2015). Introduction to Fluid Mechanics Fifth Edition. CRC Press Taylor & Francia Group. Douglas, J.F., Gasiorek J.M., Swaffield, J, A. and Jack L (2011). Fluid Mechanics, 6th Edition. Pearson Education Canada.
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