Reynolds, an engineering professor in early 1880 demonstrated two different types of flow through an experiment:

7 STEADY FLOW IN PIPES 7.1 Reynolds Number Reynolds, an engineering professor in early 1880 demonstrated two different types of flow through an experiment: Laminar flow Turbulent flow Reynolds apparatus dye dye filament water outlet Laminar flow at low velocity Dye injection dye filament remained intact throughout the length of the tube Dye filament fluid particles move in a straight line considered as moving in layers P.7-1

Turbulent flow at high velocity Dye injection dye diffused over the whole cross-section fluid particles do not move in a straight line velocity in average sense Dye filament The transition of flow is due to change of velocity, size of pipe and properties of fluid. Reynolds explained the phenomena by considering the forces acting on the fluid particle. When the motion of a fluid particle in a stream is disturbed, its inertia will tend to carry it on in the new direction, but the viscous forces due to the surrounding fluid will tend to make it conform to the motion of the rest of the stream. The criterion that determines whether flow will be viscous or turbulent is therefore the ratio of the inertial force to the viscous force acting on the particle. Hence, for a particular flow, i.e. inertial force viscous force = constant (7.1) By using dimensional analysis, Reynolds derived a criterion to distinguish between laminar and turbulent flow. Reynolds Number, Re = inertial force viscous force = ρvd µ (7.) where ρ = density of the fluid, kg/m 3 v = velocity of the flow, m/s d = diameter of the pipe, m µ = dynamic viscosity, Ns/m P.7-

Since ν = ρ µ (7.3) where ν = kinematic viscosity, m /s Hence Reynolds Number, Re = ν vd (7.4) Reynolds Number is a dimensionless number In general, the flow is laminar when Re - small transitional when Re - intermediate turbulent when Re - large The flow in pipe can be treated as laminar when Re < 000 transitional when 000 < Re < 4000 turbulent when Re > 4000 P.7-3

Worked example: What is the critical velocity of a water flow through a circular pipe of diameter cm so that the flow is laminar? (dynamic viscosity of water is 0.89*10-3 Ns/m, density of water = 1000 kg/m 3 ) Answer As the Reynolds number is ρvd Re = µ 1000* v*0.0 = 3 0.89*10 < 000 i.e. v 3 000*089*10 = 1000*0.0 = 8.9 cm/s Hence the critical velocity is 8.9 cm/s. P.7-4

7. Laminar Flow in Pipes 7..1 Hagen - Poiseuille Equation It was discovered independently by: G.H.L. Hagen - a German engineer in 1893 J.L.M. Poiseuille - a French physician in 1840 It states that the head loss experienced by the water when it flows through a pipe is directly proportional to the rate of flow, (Q), and inversely proportional to the fourth power of the diameter of the pipe (d 4 ). Q i.e. h f = k 4 d where k is a constant (7.5) For a laminar flow, the shear stress on the cylindrical surface is given by r dp τ = dx For Newtonian fluid, dv τ = µ dr Equating the two equations, dv r dp µ = dr dx When integrating the above equation with respect to r with the boundary condition v = 0 when r = R (i.e. no slip condition), the result is 1 dp v = - (R r ) (7.6) 4µ dx From (7.10), we can see that the velocity distribution is in parabolic form with maximum velocity at r = 0. P.7-5

umax The maximum velocity at centre (r = 0) 1 dp v max = R 4µ dx dp by putting dx = - p L v max = R p 4µ L The corresponding discharge, Q is R Q = πrvdr 0 R 1 dp = πr[ ( 0 4µ dx R r )] dr = - π 4 R dp 8µ dx (7.7) by putting dp dx = - p L Q = π 4 d p 18µ L & R = d/ (7.8) - known as Hagen-Poiseuille equation flow rate, Q The mean velocity, v = x sec tional area, A = d p 3µ L (7.9) When compare with v max, 1 v = vmax (7.10) From Hagen-Poiseuille law, P.7-6

Q h f = k * 4 d By substituting the expression of Q in (7.8) and consider the change of pressure head as head loss, i.e. p = ρgh f 18µ L then k = ρgπ Therefore, k depends on the properties of fluid, ρ and µ and pipe length, L. Hagen-Poiseuille equation can be expressed in terms of head loss, h f and average velocity, µ L v h f = 64* * * ρvd d g ρvd Since Re = µ Therefore head loss in Hagen-Poiseuille equation 64 L v h f = * * Re d g (7.11) L v If h f can be expressed as h f = f * * d g (7.1) where f - friction factor Then the friction factor in laminar flow is 64 f = Re (7.13) Noted that in UK publication, h f is often written as L v h f = 4f * * d g 16 and hence f = Re P.7-7

Worked examples: 1. For a laminar flow, d = 50 mm, P 1 - P = 0 kn/m L = 00 m, µ = 10-3 Ns/m Determine the flow rate, Q in m 3 /s. Answer From Hagen-Poiseuille equation, 4 πd (P1 P ) Q = 18µ L 4 3 π*(0.5) *(0*10 ) = 3 18*10 *00 = 9.59 m 3 /s m 3 /s P.7-8

. Oil of viscosity 0.101 Ns/m and specific gravity of 0.85 flows through 3000 m of 300 mm-dia. pipe at the rate of 0.0444 m 3 /s. What is the lost in head in the pipe? Answer Since 4 πd (P1 P ) Q = 18µ L Now, Q = 0.0444 m 3 /s, µ = 0.101 Ns/m, L = 3000 m, d = 0.3 m then 4 π(0.3) (P1 P ) 0.0444 = 18*0.101*3000 Hence P 1 - P 0.0444*18* 0.101*3000 = 4 π*(0.3) = 67.67 kn/m = lost in pressure P1 P Lost in head = ρg 67.67 = m of oil 850*9.81 = 8.1 m of oil P.7-9

7.3 Turbulent Flow in Pipes 7.3.1 Darcy - Weisbach Formula Darcy, Weisbach and others found that a formula for pipe friction loss could be expressed as L v h f = f d g where f is friction factor (7.14) The above equation can be applied in both laminar (refer 7..) and turbulent flows and is known as Darcy - Weisbach formula. It is found that friction factor depends on density of the fluid, ρ velocity of the flow, v diameter of the pipe, d viscosity of the fluid, µ wall roughness, ε ε wall roughness i.e. f = f(ρ, v, d, µ, ε) By dimensional analysis, f = f ρ vd ε, µ d = f ( Reynolds number, relative roughness) ρvd where Reynolds number = and µ relative roughness = ε d (7.15) P.7-10

Once the Reynolds number and relative roughness have been determined, the corresponding value of the friction factor can be obtained from a graphical relationship known as the Moody diagram. Typical values of surface roughness New pipe surface Roughness, ε (m) Glass, brass, copper and lead smooth Wrought iron, steel 0.46*10-4 Cast iron.6*10-4 Concrete 3*10-4 to 30*10-4 7.3. Moody Diagram Moody diagram has been used extensively in solving pipe flow problems. Two equations are related to the Moody diagram for laminar flow, the friction factor is 64 f = Re This is the straight portion of the diagram when Re < 000. for a turbulent flow, friction factor is 1 = ε d.51.0 * log + f 3.7 Re f (7.16) This is known as Colebrook-White formula. P.7-11

P.7-13

Worked examples: 1. Determine the head loss for flow of 140 L/s of oil, ν = 0.00001 m /s, through 400 m of a 00 mm diameter cast iron pipe. Answer Given Q = 140 L/s = 0.14 m 3 /s d = 00 mm = 0. m ν = 0.00001 m /s = 10-5 m /s L = 400 m ε = 0.6 mm (cast iron) v = Q πd 4 = 014. π * 0. 4 m/s = 4.456 m/s vd Re = = 0. * 4. 456 = 8.91*10 4 ν 10 5 4 ε.6*10 = = 0.0013 d 0. From the Moody diagram, f(8910, 0.0013) = f = 0.038 i.e. h f 400 4.456 = 0.038* * 0. *9. 81 = 48.17 m of oil m of oil P.7-14

. Solving the previous example by using Colebrook-White formula. Answer As 1 f ε d = 0. log + 37.. Re 51 f where i.e. or ε = 0.0013; Re = 8910 d 1 0. 0013 51. = 0. log + f 37. 8910 f 0 0801 + log(1 + 1f. ) 6.908 = 0 f Since this is a non-linear equation, it has to be solved by trial & error or iterations. f = 0.03365 Hence h f = 0.03365* 400 * 4. 456 0. * 981. = 47.9 m of oil m of oil P.7-15

7.4 Minor Losses In section 7.3, the head loss in long, straight sections of pipe can be calculated by use of the friction factor obtained from Moody diagram or the Colebrook White equation. This is called friction loss or major loss. Most pipe systems consist of considerably more than straight pipes. These pipe fittings add to the overall head loss of the system. These losses are called minor losses. In some cases, the minor losses may be greater than the friction loss. Since the flow pattern in fittings and valves is quite complex, the theory is very week. The losses are commonly measured experimentally and correlated with the pipe flow patterns. 7.4.1 Different Types of Minor Losses Minor losses are losses due to the inclusion of a pipe fittings in a pipeline. Some examples are entrances or exit of a pipe expansions or contractions of a pipe bends, elbow and tees valves of open or partially closed gradual expansions or contractions Minor losses is given by h L = K* v g where K is a constant (7.17) P.7-16

Component K a. Elbows Regular 90, flanged 0.3 Regular 90, threaded 1.5 Long radius 90, flanged 0. Long radius 90, threaded 0.7 Long radius 45, flanged 0. Regular 45, threaded 0.4 b. 180 return bends 180 return bends, flanged 0. 180 return bends, threaded 1.5 c. Tees Line flow, flanged 0. Line flow, threaded 0.9 Branch flow, flanged 1.0 Branch flow, threaded.0 d. Valves Globe, fully open 10 Gate, fully open 0.15 Ball valve, fully open 0.05 e. Others Entrance loss 0.5 Exit loss 1.0 7.4. Modified Bernoulli s Equation The original Bernoulli s equation should be extended to include the friction loss and minor losses. i.e. total energy at 1 = total energy at + energy loss on the way p γ 1 v1 p v + + z1 = + + z g γ g + Σ fl i d i i v g + ΣK v i g (7.18) P.7-17

Worked examples: 1. Find the discharge through the pipe in the figure below. The minor loss coefficient for entrance is 0.5. The pipe diameter is 15 mm and the pipe roughness produces a friction factor of 0.05. 1 Answer 15m 150m Applying Bernoulli s equation between pt.1 and p γ 1 v1 p v + + z1 = + + z g γ g + fl d v g + K v g or v 15 = (1 + K + fl d )* v g = (1+0.5+0.05*150/0.015) * v g = 51.5 v g = 1.08 m/s Hence Q = A *v = 1.08*π*0.015 /4 m 3 /s = 0.191 L/s P.7-18

. Find the discharge through the pipe in the figure below for H = 0 m. The minor loss coefficients for entrance, elbows and globe valve are 0.5, 0.8 (each) and 10 respectively. The kinematic viscosity of water is 1.0*10-6 m /s. 1 globe valve 40m elbows dia. = 150mm ε = 0.0003 0m 30m 40m Answer p1 v1 p v Using + + z1 = + + z + Σ fl i γ g γ g d i i v g + ΣK v i g Σ fl i d i i v g = f v L i d g Σ f v = ( 30 + 0 + 40) 015. * 981. = 30.58 f v ΣK v i g = v g Σ v K i = (. 05+ *. 08+ 10) * 981. = 0.617 v As P 1 -P = 0, v 1 = 0, z 1 -z = 0 m, i.e. 0 = v g + Σ fl i d i v i g + ΣK v i g = (0.051+30.58f + 0.617)v = (0.668+30.58f) v P.7-19

or v = 547. 1+ 4578. f Since Re = 1.47*10 5 v ; ε/d = 0.00 f v (m/s) Re f cal. 0.030 3.551 5.*10 5 0.030 0.030 3.818 5.6*10 5 0.035 0.035 3.797 5.6*10 5 0.035 (ok) v = 3.797 m/s Since Q = A *v = π*0.15 /4 * 3.797 m 3 /s = 0.067 m 3 /s P.7-0

7.5 Pipe Systems In many pipe systems, there is more than one pipe involved. The governing mechanisms for the flow in multiple pipe systems are the same as for the single pipe system discussed previously. 7.5.1 Resistance Coefficients for Pipelines in Series and Parallel In general, the equation of head loss can be expressed as h f = k*q (7.19) Pipes are in series if they are connected end to end so the fluid flow in a continuous line is a constant. Q h1 h h3 Q Q Q hn Q By continuity of flow, Q is same for each pipe. The total loss of the system is given as h f = h 1 + h + h 3 + + h n = k 1 *Q + k *Q + k 3 *Q + + k n *Q = (k 1 + k + k 3 + + k n )*Q (7.0) The effective resistance coefficient is k eff = k 1 + k + k 3 + + k n (7.1) i.e. the total head loss is the summation of the individual pipe. P.7-1

For pipes connected in parallel, the fluid can flow from one to the other by a number of alternative routines. Q Q1 Q Q3 Q Qn hf The head loss for individual pipe is the same as the total head loss. The total flow rate is the summation of the individual pipe. Since h f = k i *Q i or Q i = h k f i Hence Total Q = Q 1 + Q + Q 3 + + Q n (7.) = hf + h f + h + + k1 k kf f 3 k n = ( 1 k + 1 1 k + 1 k + + 1 ) 3 k n h f = hf keff 1 1 = k eff k + 1 1 k + 1 k + + 1 3 k n (7.3) P.7-

Worked examples: 1. Two reservoirs are connected by a pipeline which is 150 mm in diameter for the first 6 m and 5 mm in diameter for the remaining 15 m. The water surface in the upper reservoir is 6 m above that in the lower. By neglecting any minor losses, calculate the rate of flow in m3/s. Friction coefficient f is 0.04 for both pipes. Answer The velocities v 1 and v are related by the continuity equation. i.e. A 1 v 1 = A v v 1 A = v d = v A d 1 5 = v =.5 v 150 Friction in the 150 mm pipe h f1 = f1 L1 v1 = 004. * 6 v d g 015. g 1 = 1.6 v 1 = 1.6*.5 * v g g Similarly, friction in the 5 mm pipe h f = = f L d v g 1 1 004. * 15v 0. 5 g = 8.1 v g =.67 v g Hence, total head loss = h f1 + h f = 10.77 v g Applying Bernoulli s equation between the two top water surfaces, p 1 = p = 0 (P atm ) v 1 = v = 0 (water surfaces) z 1 = 6 m; z = 0 p1 v1 p v + + z1 = + + z +h L γ g γ g P.7-3

or z 1 z = h L 6 = 10.77 v g or v = 6* * 981. 10. 77 = 3.31 m/s Hence Q = A v = π *. 0 5 * 331. m 3 /s 4 = 0.13 m 3 /s P.7-4

. Two reservoirs have a difference of level of 6 m and are connected by two pipes laid in parallel. The first pipe is 600 mm diameter of 3000 m long and the second one is 300 mm diameter of 000 m long. By neglecting all the minor losses, calculate the total discharge if f = 0.04 for both pipes. Answer For parallel pipes, h f = 1 1 = f L d 1 v1 g f L d v g Apply Bernoulli s equation to the points on the free surfaces and from the result of the previous worked example, level difference = head loss f1 L1 v1 f L H = = d g d 6 = 1 004. * 3000 v 06. * 981. 1 v g = 0. 04 * 000 v 03. * 981. Therefore, v 1 = 0.767 m/s v = 0.664 m/s Hence Q 1 = A 1 v 1 = π * 06. * 0. 767 4 Q = A v = π *. 03 * 0664. 4 = 0.17 m 3 /s = 0.047 m 3 /s Total discharge, Q = Q 1 + Q = 0.17 + 0.047 m 3 /s = 0.64 m 3 /s P.7-5

7.5. Branched-pipe Problem h1 reservoir 1 pipe1, k1 h3 Q1 Q3 reservoir 3 pipe 3, k3 J Q pipe, k reservoir h Assume h 1 > h > h 3 and the 3 pipes intersect at junction J. As h 1 is the highest head, the flow in pipe 1 must be toward J. As h 3 is the lowest head, Q 3 is flowing from J to the reservoir 3. The flow Q s direction is unknown because it depends on the head at junction J. If h J be the head at junction J. There are two possible cases (i) h 1 > h J > h, or (ii) h > h J > h 3 For case (i), h 1 > h J > h, Q is from J to reservoir. Q 1 - Q - Q 3 = 0 h 1 h J = k 1 *Q 1 h J - h = k *Q h J - h 3 = k 3 *Q 3 (7.4) For case (ii), h > h J > h 3, Q is from reservoir to J. Q 1 + Q - Q 3 = 0 h 1 h J = k 1 *Q 1 h h J = k *Q h J - h 3 = k 3 *Q 3 (7.5) P.7-6

Both sets of equations have 4 unknowns Q 1, Q, Q 3 and h J. We have to determine which case controls the problem. It is determined by assuming h J = h, i.e. no flow from J to reservoir. Therefore Q 1 = Q 3 = h h h k 1 1 h k 3 3 (7.6) (7.7) If Q 1 > Q 3, Q is from J to reservoir - case (i). If Q 1 < Q 3, Q is from reservoir to J - case (ii). P.7-7

Worked example: Three reservoirs are connected as the figure below. Determine the flow, Q 1, Q and Q 3 with k 1 = 3.058, k = 8.860 and k 3 = 0.403 s /m 5 and h f = k i *Q i. reservoir 1 00m Q1 J Q reservoir 140m reservoir 3 Q3 180m Answer Step 1 Pipe h i (m) k i (s /m 5 ) 1 00 3.058 180 8.860 3 140 0.403 Step - calculate Q 1 and Q 3 Q 1 = h 1 h k 1 = 00 180 3058. =.557 m 3 /s Q 3 = h h k 3 3 = 180 140 0403. = 9.963 m 3 /s Q 1 < Q 3 case (ii) i.e. h > h > h 3 P.7-8

Step 3 - set up the equations For case (ii), we have 00 - h = 3.058Q 1 00 h or Q 1 = 3. 058 m 3 /s 180 - h = 8.860Q 180 h or Q = 8. 860 m 3 /s h - 140 = 0.403Q 3 h 140 or Q 3 = 0. 403 m 3 /s Since Q 1 + Q - Q 3 = 0 therefore 00 h 180 h + 3. 058 8. 860 - h 140 0. 403 = 0 Step 4 - solve for h (180 < h < 140) by iterations h (m) Q 1 + Q + Q 3 (m 3 /s) Error 180.557 + 0 9.963-7.406 160 3.617 + 1.50 7.045-1.96 150 4.044 + 1.840-4.981 +0.903 153 3.90 + 1.746-5.680-0.014 15.9 3.95 + 1.749-5.658 +0.016 15.95 3.9 + 1.747-5.669 +0.000 Therefore, the head h at the junction is 15.95 m and Q 1 = 3.9 m 3 /s (towards J) Q = 1.747 m 3 /s (towards J) Q 3 = 5.669 m 3 /s (towards reservoir C) P.7-9

7.5.3 Hardy-Cross Method (Reference only) water in loop 1 loop loop 3 water out loop 4 loop 5 loop 6 The supply of water system to a city is a complicated network of pipelines. The commonly used technique is the Hardy-Cross method. The two assumptions made by Hardy-Cross method are: The algebraic sum of head loss around each loop must be zero. 1 4 +ve 3 h 14 + h 43 - h 3 - h 1 = 0 (7.8) The net flow out of each junction must be equal to zero. 1 7 6 5 3 4 Q 17 + Q 47 - Q 76 - Q 75 - Q 7 - Q 73 = 0 (7.9) P.7-30

Method of Analysis Assume h i = k i *Q i The sign convention is clockwise positive for the discharge and head loss. Initially a flow rate, Q is assumed. A correction for discharge, Q is then evaluated and the new flow rate is Q + Q. the head loss for each member can be approximated as h i = k i *(Q i + Q) = k i *[Q i + Q i Q + ( Q) ] k i *Q i + k i *Q i * Q Summation around the loop Σ h i = Σ[k i *Q i + k i *Q i * Q] = 0 (by assumption 1) i.e. Σh i + Q*Σ(h i / Q i ) = 0 hence Q = 1 Σhi h Σ( i Q ) i (7.30) Σh i is the sum of head loss around a loop which can be +ve or -ve. Σ(h i / Q i ) is the sum of the ratio (head loss/flow) for each member of the loop. The ratio is a magnitude and therefore is +ve only. Q is the correction flow for a loop. Each pipe within the loop will have this correction. Any pipe belonging to or more loops, the correction for that particular pipe will contribute from every loop. P.7-31

Worked examples: 1. Determine the flow in each branch of the loop as shown below by using (i) equivalent pipe method, and (ii) Hardy-Cross method. 3 m /s pipe 1, k 3 m /s Q1 Q pipe, k Answer (i) As a parallel system h f1 = h f k*q 1 = k*q or Q 1 = Q As Q = m 3 /s, = Q 1 + Q = Q + Q (1+ )Q = Q = 0.884 m 3 /s Q 1 = Q = *0.884 = 1.1716 m 3 /s P.7-3

(ii) Assume Q 1 = 1.5 m 3 /s Q = 0.5 m 3 /s 3 m /s 1.5m 3 /s +ve 3 m /s -0.5m 3 /s 1 st iteration Pipe k i Q i ±h i abs(h i /Q i ) 1 1 1.5.5 1.5-0.5-0.5 1.0 Σ = 1.75.5 nd iteration Q = 1 Σhi h Σ( i Q ) i 175. = - m 3 /s * 5. = -0.35 m 3 /s Pipe k i Q i ±h i abs(h i /Q i ) 1 1 1.15 1.35 1.15-0.85-1.445 1.70 Σ = -0.15.85 Q = - 015. m 3 /s * 85. = 0.015 m 3 /s P.7-33

3 rd iteration Pipe k i Q i ±h i abs(h i /Q i ) 1 1 1.1715 1.374 1.1715-0.885-1.378 1.655 Σ = -0.0004.865 0. 0004 Q = - m 3 /s *. 865 = 0.000071 m 3 /s Q 1 = 1.1715 m 3 /s Q = 0.885 m 3 /s (after 3 iterations) P.7-34

. Find the flow in the pipeline using Hardy Cross Method. K for vertical members are 3 and horizontals are 5. Answer 3 0.5m /s 300m 00mm Assume the flow rates are A 500m 00mm B 500m 00mm 300m C 00mm 300m 00mm D 500m E 500m F 00mm 00mm 0.1m 3 /s 0. m 3 /s 3 0.m /s 3 0.5m /s 3 3 0. m /s 0.05m /s A B C 0.3 m 3 /s 0.15 3 m /s 3 3 D 0. m /s E 0.15m /s F 0.05 3 m /s 1 st Iteration 0.1m 3 /s 0. m 3 /s 3 0.m /s Pipe k i Q i ±h i Abs(h i /Q i ) AB 5 0.0 0.000 1.00 BE 3 0.15 0.0675 0.45 DE 5-0.0-0.000 1.00 AD 3-0.30-0.700 0.90-0.05 3.35 BC 5 0.05 0.015 0.5 CF 3 0.05 0.0075 0.15 EF 5-0.15-0.115 0.75 BE 3-0.15-0.0675 0.45-0.16 1.6 0. 05 Q 1 = * 335. Q = 016. *. 16 = 0.03 m 3 /s = 0.05 m 3 /s P.7-35

nd Iteration Pipe k i Q i ±h i Abs(h i /Q i ) AB 5 0.30 0.650 1.15 BE 3 0.130 0.0509 0.39 DE 5-0.170-0.1441 0.85 AD 3-0.70-0.183 0.81-0.04657 3. BC 5 0.100 0.05 0.5 CF 3 0.100 0.03 0.3 EF 5-0.100-0.05 0.5 BE 3-0.130-0.05087 0.39067-0.0087 1.69067 0. 4657 Q 1 = * 3. 0. 087 Q = *. 16907 = 0.0071 m 3 /s = 0.006 m 3 /s 3rd Iteration Pipe k i Q i ±h i Abs(h i /Q i ) AB 5 0.38 0.80 1.19 BE 3 0.131 0.0517 0.39 DE 5-0.163-0.130 0.81 AD 3-0.63-0.067 0.79-0.00498 3.181479 BC 5 0.106 0.056364 0.530868 CF 3 0.106 0.033818 0.31851 EF 5-0.094-0.0440 0.46913 BE 3-0.131-0.05174 0.393979-0.00557 1.715 Q 1 0. 498 = * 31815. Q 0. 0557 = *. 1715 = 0.00078 m 3 /s = 0.00163 m 3 /s P.7-36

4th Iteration Pipe k i Q i ±h i Abs(h i /Q i ) AB 5 0.38 0.839 1.19 BE 3 0.130 0.0511 0.39 DE 5-0.16-0.1308 0.81 AD 3-0.6-0.055 0.79-0.0018 3.176597 BC 5 0.108 0.058105 0.539005 CF 3 0.108 0.034863 0.33403 EF 5-0.09-0.045 0.460995 BE 3-0.130-0.05108 0.391444-0.00061 1.714847 0. 018 Q 1 = * 31766. 0. 0061 Q = *. 17148 = 0.0000 m 3 /s = 0.00018 m 3 /s 5th Iteration Pipe k i Q i ±h i Abs(h i /Q i ) AB 5 0.38 0.844 1.19 BE 3 0.131 0.0511 0.39 DE 5-0.16-0.1304 0.81 AD 3-0.6-0.05 0.78-0.00014 3.17606 BC 5 0.108 0.05898 0.539896 CF 3 0.108 0.034979 0.33938 EF 5-0.09-0.0434 0.460104 BE 3-0.131-0.05109 0.391515-0.00016 1.715453 0. 00014 Q 1 = * 31761. 0. 00016 Q = *. 17155 =.*10-5 m 3 /s = 4.6*10-5 m 3 /s Therefore OK P.7-37

Therefore, the flow rates after five iterations will be given by 3 0.5m /s 3 3 0.38m /s 0.108 m /s A B C 0.6 m 3 /s 0.131 3 m /s 3 3 D 0.16 m /s E 0.09m /s F 0.108 3 m /s 0.1m 3 /s 0. m 3 /s 3 0.m /s P.7-38

Class Exercise 7.1: A steady push on the piston causes a flow rate of 0.4 cm 3 /s through the needle. The fluid has S.G. = 0.9 and µ = 0.00 Ns/m. Determine the head loss at the needle and hence the force F required to maintain the flow. Neglect the head loss in the piston only. (F = 0.01 kn) D=1cm Q D=0.5mm F 1.5cm 3cm P.7-39

Class Exercise 7.: Oil, with density = 900 kg/m 3 and ν = 1*10-5 m /s, flows at 0. m 3 /s through a 0-cm diameter pipe 500 m long cast-iron pipe. The roughness of iron is 0.6 mm. Determine the head loss in the pipe. (h f = 117 m) P.7-40

Class Exercise 7.3: Water flows at a velocity of 1 m/s in a 150 mm new ductile iron pipe. Estimate the head loss over 500 m using Darcy-Weisbach equation. (ε = 0.6 mm and µ = 10-3 Ns/m ). (h f = 4.04 m) P.7-41

Class Exercise 7.4: The distance between two sections A and B of a 300 mm diameter pipe is 300 m. The elevations of A and B are 90 m and 75 m and the pressures are 80 kpa and 350 kpa respectively. Find the direction of flow of water and calculate the head loss due to friction and the value of the friction factor for the pipe if the flow is 14 L/s. (A to B, h = 7.87 m, f = 0.0386) P.7-4

Class Exercise 7.5: Three pipes A, B & C are interconnected and discharge water from a reservoir as shown below. With the provided pipe characteristics, determine the flow rate in each pipe. Neglect all the minor loss. Pipe Diameter, Length, f mm m A 150 600 0.0 B 100 500 0.03 C 00 150 0.04 (Q A = 0.075 m 3 /s, Q B = 0.036 m 3 /s, Q c = 0.0988 m 3 /s) P.7-43

Tutorial: Steady Pipe Flow 1. Glycerine of viscosity 0.9 Ns/m and density 160 kg/m 3 is pumped along a horizontal pipe 6.5 m long of diameter d = 0.01 m at a flow rate of Q = 1.8 L/min. Determine the flow Reynolds number and verify whether the flow is laminar or turbulent. Calculate the pressure loss in the pipe due to frictional effects.. If oil (ν = 4*10-5 m /s, S.G. = 0.9) flows from the upper to the lower reservoir at a rate of 0.08 m 3 /s in the 15-cm smooth pipe, determine the elevation of the oil surface in the upper reservoir. (K for entrance, exit and bend are 0.5, 1 and 0.19 respectively) 60m 130mPD 7m 130m 3. Two reservoirs having a difference of surface level of 4 m are connected by two parallel pipes each 1600 m long and of diameters 450 mm and 300 mm. To repair a length of 10 m of the 450 mm diameter pipe midway between the reservoirs, the total flow is diverted over this length to the 300 mm pipe. Calculate the percentage reduction in discharge resulting from the diversion. Consider only the friction losses and take f = 0.04. 300mm 450mm 1600m 4m 300mm 450mm 10m 300mm 450mm 4m 300mm P.7-44