Chapter 2
Fundamentals of Flow in Closed Conduits 2.1. From the given data: D1 = 0.1 m, D2 = 0.15 m, V1 = 2 m/s. Using these data, the following preliminary calculations are useful: A1 =
π 2 π D = (0.1)2 = 0.007854 m2 , 4 1 4
A2 =
π 2 π D = (0.15)2 = 0.01767 m2 4 2 4
Volumetric flow rate, Q, is given by Q = A1 V1 = (0.007854)(2) = 0.0157 m3 /s According to the continuity equation, A1 V1 = A2 V2 = Q
→
V2 =
Q 0.0157 = = 0.889 m/s A2 0.01767
At 20◦ C, the density of water, ρ, is 998 kg/m3 , and the mass flow rate, ṁ, is given by ṁ = ρQ = (998)(0.0157) = 15.7 kg/s 2.2. From the given data: D1 = 0.2 m, D2 = 0.3 m, and V1 = 0.75 m/s. Using these data, the following preliminary calculations are useful: A1 =
π 2 π D = (0.2)2 = 0.03142 m2 , 4 1 4
A2 =
π 2 π D = (0.3)2 = 0.07069 m2 4 2 4
(a) According to the continuity equation, A1 V1 = A2 V2
→
V2 =
A1 V1 (0.03142)(0.75) = = 0.333 m/s A2 0.07069
(b) The volume flow rate, Q, is given by Q = A1 V1 = (0.03142)(0.75) = 0.02357 m3 /s = 23.6 L/s
2.3. From the given data: D1 = 200 mm, D2 = 100 mm, V1 = 1 m/s, and π 2 π D = (0.2)2 = 0.0314 m2 4 1 4 π 2 π A2 = D2 = (0.1)2 = 0.00785 m2 4 4 A1 =
The flow rate, Q1 , in the 200-mm pipe is given by Q1 = A1 V1 = (0.0314)(1) = 0.0314 m3 /s and hence the flow rate, Q2 , in the 100-mm pipe is Q2 =
0.0314 Q1 = = 0.0157 m3 /s 2 2
The average velocity, V2 , in the 100-mm pipe is V2 =
0.0157 Q2 = = 2 m/s A2 0.00785
2.4. The velocity distribution in the pipe is [ ( r )2 ] v(r) = v0 1 − R and the average velocity, V , is defined as ∫ 1 V = V dA, A A
A = πR2 ,
(1)
dA = 2πr dr
Combining Equations 1 and 2 yields [∫ R ] [ ] ∫ R [ ∫ R 3 ( r )2 ] r 1 2v0 2v0 R2 R4 V = v0 1 − 2πr dr = 2 r dr − dr = 2 − 2 πR2 0 R R R 2 4R2 0 0 R 2v0 R2 v0 = = 2 R 4 2 The flow rate, Q, is therefore given by Q = AV = 2.5.
πR2 V0 2
[ ] ∫ ∫ R 4 1 2r2 r4 2 2 β= v dA = V 1 − 2 + 4 2πrdr R R πR2 V02 0 0 AV̄ 2 A [∫ R [ ] ∫ R 3 ∫ R 5 ] 8 2r r 8 R2 R4 R6 = 2 rdr − dr + dr = 2 − + 2 4 R R 2 2R2 6R4 0 0 R 0 R =
4 3
(2)
2.6. D = 0.2 m, Q = 60 L/s = 0.06 m3 /s, L = 100 m, p1 = 500 kPa, p2 = 400 kPa, γ = 9.79 kN/m3 . 0.2 D = = 0.05 m 4 4 p1 p2 500 − 400 ∆h = − = = 10.2 m γ γ 9.79 γR∆h (9.79 × 103 )(0.05)(10.2) τ0 = = = 49.9 N/m2 L 100 πD2 π(0.2)2 A= = = 0.0314 m2 4 4 0.06 Q = = 1.91 m/s V = A 0.0314 8τ0 8(49.9) f= = = 0.11 2 ρV (998)(1.91)2 R=
2.7. T = 20◦ C, V = 2 m/s, D = 0.25 m, horizontal pipe, ductile iron. For ductile iron pipe, ks = 0.26 mm, and ks 0.26 = = 0.00104 D 250 (998.2)(2)(0.25) ρV D = = 4.981 × 105 Re = µ (1.002 × 10−3 ) From the Moody diagram: f = 0.0202 (flow is not fully turbulent) Using the Colebrook equation, 1 √ = −2 log f Substituting for ks /D and Re gives 1 √ = −2 log f
(
(
ks /D 2.51 √ + 3.7 Re f
)
0.00104 2.51 √ + 3.7 4.981 × 105 f
)
By trial and error leads to f = 0.0204 Using the Swamee-Jain equation, [ ] [ ] 1 ks /D 5.74 5.74 0.00104 √ = −2 log = −2 log + + 3.7 3.7 (4.981 × 105 )0.9 f Re0.9 which leads to f = 0.0205 The head loss, hf , over 100 m of pipeline is given by hf = f
L V2 100 (2)2 = 0.0204 = 1.66 m D 2g 0.25 2(9.81)
Therefore the pressure drop, ∆p, is given by ∆p = γhf = (9.79)(1.66) = 16.3 kPa If the pipe is 1 m lower at the downstream end, f would not change, but the pressure drop, ∆p, would then be given by ∆p = γ(hf − 1.0) = 9.79(1.66 − 1) = 6.46 kPa 2.8. From the given data: D = 25 mm, ks = 0.1 mm, θ = 10◦ , p1 = 550 kPa, and L = 100 m. At 20◦ C, ν = 1.00 × 10−6 m2 /s, γ = 9.79 kN/m3 , and ks 0.1 = = 0.004 D 25 π π A = D2 = (0.025)2 = 4.909 × 10−4 m2 4 4 L Q2 100 Q2 hf = f = f = 8.46 × 108 f Q2 D 2gA2 0.025 2(9.81)(4.909 × 10−4 )2 The energy equation applied over 100 m of pipe is p1 V 2 p2 V 2 + + z1 = + + z2 + hf γ 2g γ 2g which simplifies to p2 = p1 − γ(z2 − z1 ) − γhf p2 = 550 − 9.79(100 sin 10◦ ) − 9.79(8.46 × 108 f Q2 ) p2 = 380.0 − 8.28 × 109 f Q2 (a) For Q = 2 L/min = 3.333 × 10−5 m3 /s, Q 3.333 × 10−5 = = 0.06790 m/s A 4.909 × 10−4 VD (0.06790)(0.025) Re = = = 1698 ν 1 × 10−6 V =
Since Re < 2000, the flow is laminar when Q = 2 L/min. Hence, 64 64 = = 0.03770 Re 1698 p2 = 380.0 − 8.28 × 109 (0.03770)(3.333 × 10−5 )2 = 380 kPa f=
Therefore, when the flow is 2 L/min, the pressure at the downstream section is 380 kPa . For Q = 20 L/min = 3.333 × 10−4 m3 /s, V =
Q 3.333 × 10−4 = = 0.6790 m/s A 4.909 × 10−4
Re =
VD (0.6790)(0.025) = = 16980 ν 1 × 10−6
Since Re > 5000, the flow is turbulent when Q = 20 L/min. Hence, 0.25
0.25 )]2 = [ ( 0.004 )]2 = 0.0342 5.74 s /D log + 0.9 log k3.7 + 5.74 3.7 16980 Re0.9 p2 = 380.0 − 8.28 × 109 (0.0342)(3.333 × 10−4 )2 = 349 kPa f=[
(
Therefore, when the flow is 2 L/min, the pressure at the downstream section is 349 kPa . (b) Using the Colebrook equation with Q = 20 L/min, [ ] [ ] 1 ks /D 2.51 0.004 2.51 √ = −2 log √ √ + = −2 log + 3.7 3.7 f Re f 16980 f which yields f = 0.0337 . Comparing this with the Swamee-Jain result of f = 0.0342 indicates a difference of 1.5% , which is more than the 1% claimed by Swamee-Jain. 2.9. The Colebrook equation is given by 1 √ = −2 log f
(
2.51 ks /D √ + 3.7 Re f
)
Inverting and squaring this equation gives f={
[( log
0.25 ) ]}2 ks /D 2.51 √ + 3.7 Re f
This equation is “slightly more convenient” than the √ Colebrook formula since it is quasiexplicit in f , whereas the Colebrook formula gives 1/ f . 2.10. The Colebrook equation is preferable since it provides greater accuracy than interpolating from the Moody diagram. 2.11. From the given data: SG = 0.87, ν = 1.20 × 10−4 m2 /s, D = 125 mm, L = 80 m, θ = 8◦ , p2 = 150 kPa, and τ0 = 180 Pa. Quantities derived directly from the given data are: ∆z = L sin θ = 11.13 m and γ = SG · g = 8.532 kN/m3 . The shear stress, τ0 is related to the friction factor, f , as follows τ0 8τ0 f = 1 2 → fV 2 = (1) ρ 8 ρV The energy equation applied between the upstream and downstream ends of the pipe is as follows: p1 f L V 2 p2 + = + ∆z (2) γ D 2g γ Combining Equations 1 and 2, rearranging to make p1 the subject of the formula, and substituting given parameter values yields p1 = p2 + γ∆z +
4(180)(80) 4τ0 L = 150 + (8.532)(11.13) + [ ×10−3 kPa/Pa] = 706 kPa D 0.125
2.12. From the given data: Q = 500 L/s = 0.5 m3 /s, D = 750 mm, ks = 1.5 mm, L = 3 km, and ∆z = 1.5 m. For crude oil at 20◦ C, ρ = 856 kg/m3 , µ = 7.2 mPa·s, ν = µ/ρ = 8.411 × 10−6 m2 /s, and γ = ρg = 8.395 kN/m3 . The following preliminary calculations are useful: A=
πD2 π(0.5)2 = = 0.4418 m2 , 4 4
V =
VD (1.132)(0.750) = = 1.009 × 105 , ν 8.411 × 10−6
ks 1.5 = = 2.00 × 10−3 D 750
Re =
Q 0.5 = = 1.132 m/s A 0.4418
Using the Colebrook equation, [ [ ] ] 1 k /D 2.00 × 10−3 2.51 2.51 1 √ = −2 log s √ √ + + → √ = −2 log 3.7 3.7 f Re f f 1.009 × 105 f which yields f = 0.0251. Substituting into the Darcy–Weisbach equation (Equation ??) gives the head loss due to friction, hf , over the length of the pipe as hf =
(0.0251)(3000) 1.1322 fL V 2 = = 6.555 m D 2g 0.750 2(9.807)
(a) The pressure change, ∆p, can be determined using the energy equation, which gives ( ) ∆p − + ∆z = hf → ∆p = −γ(hf + ∆z) γ Substituting the given and derived data gives ∆p = −(8.395)(6.555 + 1.5) = −67.6 kPa (b) The rate, P , at which energy is being consumed is given by P = γQhf = (8.395)(0.50)(6.555) = 27.5 kW (c) If the roughness height is reduced by 70%, then ks /D = 0.3(2.00 × 10−3 ) = 6 × 10−4 . Repeating the calculations in Parts (a) and (b) for the new value of ks /D gives ∆p = −58.0 kPa,
P = 22.7 kW
Therefore, the pressure change is reduced by 14% , and the rate of energy loss is reduced by 17% . 2.13. From the given data: SG = 0.87, ν = 1.20 × 10−4 m2 /s, D = 50 mm, A = πD2 /4 = 1.963 × 10−3 m2 , ks ≈ 0, and τ0 = 150 Pa. For the given fluid, ρ = SG · 1000 = 870 kg/m3 . The shear stress, τ0 is related to the friction factor, f , as follows τ0 2 8 ρV
f= 1
→
fV 2 =
8τ0 ρ
If the flow is turbulent, the f can be evaluated using the Colebrook equation such that ( )( ( )( ) ) ks VD ks V (0.050) 8(150) 8τ0 , 0 V2 = fCE Re, ,0 V 2 = → fCE Re, D ν ρ D 1.20 × 10−4 870
→
This gives a Reynolds number (V D/ν) equal to 2241 and a flow rate (AV ) of 10.6 L/s. Since 2000 < Re < 4000 the flow is in the transition stage and the assumption of turbulent flow is not validated. Try laminar flow where f = 64/Re, which gives 64 2 8τ0 V = VD ρ ν
→
8(150) 64 V2 = V (0.050) 870 1.20 × 10−4
→
V = 8.98 m/s
This gives a Reynolds number (V D/ν) equal to 3742 and a flow rate (AV ) of 17.6 L/s. Since 2000 < Re < 4000 the flow is in the transition stage and the assumption of laminar flow is not validated. These results indicate that the flow is likely in the transitional range. The flow rate given by the turbulent-flow analysis is 10.6 L/s, and the flow rate given by the laminar-flow analysis is 17.6 L/s. Therefore the flow rate is expected to vary in the range of 10.6–17.6 L/s . 2.14. From the given data: D = 0.5 m, p1 = 600 kPa, Q = 0.50 m3 /s, z1 = 120 m, z2 = 100 m, γ = 9.79 kN/m3 , L = 1000 m, ks = 0.26 mm (ductile iron). The following preliminary calculation is useful: A=
π 2 π D = (0.5)2 = 0.1963 m2 , 4 4
V =
Q 0.50 = = 2.55 m/s A 0.1963
Using the Colebrook equation, 1 √ = −2 log f
(
ks /D 2.51 √ + 3.7 Re f
)
where ks /D = 0.26/500 = 0.00052, and at 20◦ C, Re =
(998)(2.55)(0.5) ρV D = = 1.27 × 106 µ 1.00 × 10−3
Substituting ks /D and Re into the Colebrook equation gives ( ) 0.00052 2.51 1 √ √ = −2 log + → 3.7 f 1.27 × 106 f
f = 0.0172
Applying the energy equation p1 V12 p2 V22 + + z1 = + + z2 + hf γ 2g γ 2g Since V1 = V2 , and hf is given by the Darcy-Weisbach equation, then the energy equation can be written as p1 p2 L V2 + z1 = + z2 + f γ γ D 2g
V = 5.38 m/s
Substituting known values leads to 600 p2 1000 (2.55)2 + 120 = + 100 + 0.0172 9.79 9.79 0.5 2(9.81)
→
p2 = 684 kPa
If p is the (static) pressure at the top of a 30 m high building, then p = p2 − 30γ = 684 − 30(9.79) = 390 kPa This (static) water pressure is adequate for service. 2.15. From the given data: Q = 50 L/s = 0.05 m3 /s, D = 200 mm, ks = 0.12 mm, L1 = 1 km, L2 = 3 km, pA = 500 kPa, pB = 350 kPa. For water at 20◦ C, ν = 1.00 × 10−6 m2 /s and γ = 9.789 kN/m3 . From the given data: A=
πD2 = 0.03141 m2 , 4
V =
Q = 1.592 m/s, A
Re =
VD = 3.183 × 105 , ν
ks = 6 × 10−4 D
Using the Colebrook equation with the above values of Re and ks /D gives f = 0.0187. (a) Apply the energy equation between the reservoir and Location A: ] 2 [ ] V pA 1000 1.5922 500 L1 +1 + + zA = (0.0187) +1 + + 20 = 83.27 m hp = f D 2g γ 0.2 2(9.907) 9.789 [
Apply the energy equation between the reservoir and Location B: [
] 2 [ ] L2 V 1.5922 pB 3000 350 hp = f +1 + + zB = (0.0187) +1 + − 20 = 52.08 m D 2g γ 0.2 2(9.907) 9.789 Since both of the above conditions must be met, the required pump head is hp = 74.29 m, and the required power, P , is given by P = γQhp = (9.789)(0.05)(83.27) = 40.8 kW (b) If the flow is fully turbulent, Re → ∞ and the Colebrook equation gives f = 0.0174. This friction factor differs by around 7% from the actual friction factor (= 0.0187), therefore the flow is not fully turbulent . 2.16. From the given data: D = 0.2 m, Q = 0.03 m3 /s, ks = 0.01 mm, p1 = 500 kPa, z1 = 10.00 m, z2 = 10.50 m, and L = 100 m. For water at 20◦ C, γ = 9.789 kN/m3 and ν = 10−6 m2 /s. The following preliminary calculations are useful: π 2 π D = (0.2)2 = 0.03142 m2 , 4 4 VD (0.9548)(0.2) Re = = = 190960 ν 10−6
A=
V =
Q 0.03 = = 0.9548 m/s A 0.03142
(a) Using the Jain equation to calculate the friction factor gives: ] [ ] [ 1 ks 5.74 0.01 5.74 √ = −2 log + = −2 log + 3.7D Re0.9 3.7(200) 1909600.9 f
→
f = 0.0161
Calculate the head loss using the Darcy-Weisbach equation: hf = f
L V2 100 0.95482 = 0.0161 = 0.3742 m D 2g 0.2 2(9.807)
Using the definition of the head loss: ) ( ) ( p2 p1 + z1 − + z2 = hf γ γ ( ) ( ) 500 p2 → + 10.00 − + 10.50 = 0.3742 9.807 9.789
→
p2 = 491 kPa
(b) This pressure is adequate for water supply. 2.17. From the given data: Q = 100 L/s, L = 100 m, D = 200 mm, ∆z = 0.8 m, and ∆p = −90 kPa. For water at 20◦ C, ν = 1.00 × 10−6 m2 /s and γ = 9.79 kN/m3 . The following preliminary calculations are useful, A=
πD2 π(0.2)2 = = 3.142 × 10−2 m2 , 4 4
Re =
V =
Q 0.1 = = 3.183 m/s A 3.142 × 10−2
(3.183)(0.200) VD = = 6.366 × 105 ν 1.00 × 10−6
(a) According to the energy equation, ( ) ( ) ∆p −90 hf = − + ∆z = − + 0.8 = 8.393 m γ 9.79 The corresponding friction factor can be calculated using the Darcy-Weisbach equation, which gives hf =
fL V 2 D 2g
→
f=
hf · D · 2g (8.393)(0.200)(2 × 9.807) = = 0.0325 2 L·V (100)(3.183)2
(b) Using the Colebrook equation, [ ] [ ] 1 k /D 2.51 1 ks /200 2.51 √ = −2 log s √ √ √ + → = −2 log + 3.7 3.7 f Re f 0.0325 6.366 × 105 0.0325 which yields ks = 1.2 mm . According to Table 2.1, ks for ductile iron is typically around 0.1 mm or less, so the interior surface of the pipe is likely to be in poor condition .
2.18. The head loss, hf , in the pipe is estimated by ) ( ) ( poutlet pmain hf = + zmain − + zoutlet γ γ where pmain = 400 kPa, zmain = 0 m, poutlet = 0 kPa, and zoutlet = 2.0 m. Therefore, ( ) 400 hf = + 0 − (0 + 2.0) = 38.9 m 9.79 Also, since D = 25 mm, L = 20 m, ks = 0.15 mm (from Table 2.1), ν = 1.00 × 10−6 m2 /s (at 20◦ C), the combined Darcy-Weisbach and Colebrook equation (Equation 2.36) yields, ( ) √ 1.775ν gDh k /D s f + √ log Q = −2.221D2 L 3.7 D gDhf /L [ ] √ −6 ) (9.81)(0.025)(38.9) 0.15/25 1.775(1.00 × 10 √ + = −2.221(0.025)2 log 20 3.7 (0.025) (9.81)(0.025)(38.9)/20 = 0.00265 m3 /s = 2.65 L/s The faucet can therefore be expected to deliver 2.65 L/s when fully open. 2.19. From the given data: Q = 300 L/s = 0.300 m3 /s, L = 40 m, and hf = 45 m. Assume that ν = 10−6 m2 /s (at 20◦ C) and take ks = 0.15 mm (from Table 2.1). Substituting these data into Equation 2.36 gives ( ) √ gDhf ks /D 1.775ν 2 Q = −2.221D log + √ L 3.7 D gDhf /L √ ) ( −6 ) (9.81)D(45) 0.00015 1.775(10 0.3 = −2.221D2 log + √ (40) 3.7D D (9.81)D(45)/(40) This is an implicit equation in D that can be solved numerically to yield D = 166 mm . 2.20. Since ks = 0.15 mm, L = 40 m, Q = 0.3 m3 /s, hf = 45 m, ν = 1.00 × 10−6 m2 /s, the Swamee-Jain approximation (Equation 2.37 gives [ D = 0.66 ks1.25 {
(
LQ2 ghf
)4.75
= 0.66 (0.00015)1.25
( + νQ9.4
[
(40)(0.3)2 (9.81)(45)
L ghf
]4.75
)5.2 ]0.04
+ (1.00 × 10−6 )(0.3)9.4
[
]5.2 }0.04 40 (9.81)(45)
= 0.171 m = 171 mm The calculated pipe diameter (171 mm) is about 3% higher than calculated by the Colebrook equation (166 mm).
2.21. From the given data: ks /D = 1 × 10−4 and Re = 104 − 108 . This problem can be done using MATLAB or a similar program. (a) Using the Haaland equation, the maximum percentage difference with the Colebrook equation occurs at Re = 1.526×105 where fC = 0.0172, fH = 0.0169, and the percentage error is −1.37% . (b) Using the Swamee–Jain equation, the maximum percentage difference with the Colebrook equation occurs at Re = 2.20 × 106 where fC = 0.0127, fSJ = 0.0128, and the percentage error is 0.59% . (c) Based on these results, the Swamee-Jain equation would be preferable. 2.22. The kinetic energy correction factor, α, is defined by ∫ V3 v3 ρ dA = αρ A 2 A 2 ∫
or α=
3 A v dA V 3A
(1)
Using the velocity distribution in Problem 2.4 gives [ ∫ ∫ R ( r )2 ] 2 v 3 dA = V03 1 − 2πr dr R A 0 ∫ R[ ( r )2 ( r )4 ( r )6 ] 3 = 2πV0 1−3 +3 − r dr R R R 0 ] ∫ R[ 3r3 3r5 r7 3 r − 2 + 4 − 6 dr = 2πV0 R R R 0 [ 2 ]R r 3r4 r6 r8 = 2πV03 − + − 2 4R2 2R4 8R6 0 [ ] 3 1 1 2 3 1 = 2πR V0 − + − 2 4 2 8 πR2 V03 = 4
(2)
The average velocity, V , was calculated in Problem 2.4 as V = hence
( 3
V A=
V0 2
V0 2
)3 πR2 =
πR2 V03 8
Combining Equations 1 to 3 gives α=
πR2 V03 /4 = 2 πR2 V03 /8
(3)
2.23. The kinetic energy correction factor, α, is defined by ∫ 3 v dA α= A 3 V A Using the given velocity distribution gives ∫ R ( ∫ R( ∫ r ) 37 r ) 37 3 3 3 v0 1 − v dA = 2πr dr = 2πv0 1− r dr R R 0 0 A To facilitate integration, let x=1−
r R
(1)
(2)
(3)
which gives r = R(1 − x)
(4)
dr = −R dx
(5)
Combining Equations 2 to 5 gives ∫ ∫ 1 3 3 3 v dA = 2πv0 x 7 R(1 − x)(−R)dx A 0 ∫ 1 ∫ 1 3 10 3 = 2πR2 v03 x 7 (1 − x)dx = 2πR2 v03 (x 7 − x 7 )dx 0 0 [ ]1 10 17 7 7 x7 − x7 = 2πR2 v03 10 17 0 = 0.576πR2 v03
(6)
The average velocity, V , is given by (using the same substitution as above) ∫ 1 v dA V = A A ∫ R ( ∫ 1 r ) 17 2v0 0 1 = 1 2πr dr = v − x 7 R(1 − x)(−R)dx 0 πR2 0 R R2 1 [ ] ∫ 1 1 8 7 8 7 15 1 = 2v0 (x 7 − x 7 )dx = 2v0 x 7 − x 7 8 15 0 0 = 0.817v0
(7)
Using this result, V 3 A = (0.817v0 )3 πR2 = 0.545πR2 v03
(8)
Combining Equations 1, 6, and 8 gives α=
0.576πR2 v03 = 1.06 0.545πR2 v03
The momentum correction factor, β, is defined by ∫ 2 v dA β= A 2 AV
(9)
In this case, AV 2 = πR2 (0.817v0 )2 = 0.667πR2 v02
(10)
and ∫ R
∫ 2
( r ) 27 v02 1 − 2πr dr R 0 ∫ 0 ∫ 1 2 2 9 x 7 R(1 − x)(−R)dx = 2πR2 v02 (x 7 − x 7 )dx = 2πv02 1 0 [ ]1 9 16 7 7 = 2πR2 v02 x 7 − x 7 = 0.681πR2 v02 9 16 0
v dA = A
(11)
Combining Equations 9 to 11 gives β=
0.681πR2 v02 = 1.02 0.667πR2 v02
2.24. The kinetic energy correction factor, α, is defined by ∫ 3 v dA α= A 3 V A Using the velocity distribution given by Equation ?? gives ∫ ∫ R ∫ R( ( r ) n3 r ) n3 v 3 dA = V03 1 − 2πr dr = 2πV03 1− r dr R R A 0 0
(1)
(2)
Let
r , r = R(1 − x), dr = −R dx R Combining Equations 2 to 3 gives ∫ ∫ 1 ∫ 1 3 3 3 3 2 3 v dA = 2πV0 x n R(1 − x)(−R) dx = 2πR V0 x n (1 − x) dx A 0 0 [ ]1 ∫ 1 3+n 3+n 3+2n 3 n n 2 3 2 3 = 2πR V0 x n − x n (x n − x n ) dx = 2πR V0 3+n 3 + 2n 0 0 x=1−
=
2n2 πR2 V03 (3 + n)(3 + 2n)
The average velocity, V , is given by ∫ ∫ R ( ∫ 1 1 r ) n1 2V0 0 1 V = v dA = V0 1 − 2πr dr = 2 x n R(1 − x)(−R) dx A A πR2 0 R R 1 [ ]1 ∫ 1 1+n 1+n 1+2n 1 n n = 2V0 (x n − x n ) dx = 2V0 x n − x n 1+n 1 + 2n 0 0 [ ] 2n2 = V0 (1 + n)(1 + 2n)
(3)
(4)
(5)
Using this result, [
2n2 V A= (1 + n)(1 + 2n) 3
]3 V03 πR2 =
8n6 πR2 V03 (1 + n)3 (1 + 2n)3
(6)
Combining Equations 1, 4, and 6 gives 2n2 πR2 V03 (1 + n)3 (1 + 2n)3 (3 + n)(3 + 2n) α= = 4n4 (3 + n)(3 + 2n) 8n6 2 3 πR V 0 (1 + n)3 (1 + 2n)3 Putting n = 7 gives α = 1.06 , the same result obtained in Problem 2.23. 2.25. (a) For the logarithmic velocity distribution, V0 = (1 + 1.326
√
f )V
(1)
For the power-law distribution, the average velocity can be calculated by integration, ∫ R ( 1 r ) n1 V = V0 1 − 2πr dr πR2 0 R To facilitate integration let x = 1 − r/R and hence dr = −R dx, which gives [ ] ∫ 0 1 n n − V = −2V0 x n (1 − x) dx = 2V0 n + 1 2n + 1 1
(2)
Combining Equations 1 and 2 gives the following requirement for both the average velocity (V ) and the maximum velocity (V0 ) to be equal [
] n 1 n √ − = n + 1 2n + 1 2(1 + 1.326 f )
(b) For f = 0.02, Equation 3 gives [ ] n n 1 √ − = n + 1 2n + 1 2(1 + 1.326 0.02 )
→
(3)
n = 8.32
(c) Normalizing the logarithmic velocity distribution by V0 and using the calculated value for f gives √ [ ] [ ] 2.04 f u 1 u 1 √ =1− log → = 1 − 0.243 log (4) V0 1 − (r/R) V0 1 − (r/R) 1 + 1.326 f The normalized power-law velocity distribution given by Equation 2.56 can be expressed as ( ( u r ) n1 u r )0.120 = 1− → = 1− (5) V0 R V0 R
The logarithmic and power-law velocity distributions are plotted and compared in Figure 2.1.
1.0 0.5 r R
Logarithmic law Power law
0 0.5 1.0 0.0
0.25
0.5 u V0
0.75
1.0
Figure 2.1: Comparison of turbulent velocity distributions It is apparent from Figure 2.1 that the logarithmic and power-law velocity distributions are in very close agreement 2.26. From the given data: p = 1 MPa, Dp = 65 mm, L = 300 m, ks = 0.5 mm, ks /Dp = 7.692 × 10−3 , ∆z = 2.8 m, h = 30 m and kn = 0.04. For water at 20◦ C, γ = 9.789 kN·m3 and ν = 1.004×10−6 m2 /s. The velocity of the jet, Vj , can be calculated from the energy equation, as follows √ √ Vj2 = h → Vj = 2gh = 2(9.807)(30) = 24.26 m/s 2g Equations that must be satisfied are the energy equation between the beginning and end of the hose, and also the continuity equation across the nozzle. These equations are given by, Vj2 Vj2 p Vp2 f L Vp2 + − − kn = + ∆z γ 2g Dp 2g 2g 2g Vp Ap = Vj Aj
→
Vp =
Dj2 Dp2
Vj =
Dj2 0.0652
→
[ ] Vj2 p f L Vp2 + 1− = (1 + kn ) + ∆z (1) γ Dp 2g 2g
(24.26)
→
Vp = 5742Dj2
(2)
The friction factor, f , calculated using the Colebrook equation is given by )( ) ( )( ) ( Vp (0.065) ks ks ks −3 f = fCE Re, , Rep → f = fCE Re, 7.692 × 10 , (3) D Dp D 1.004 × 10−6 Substitute Equation 3 and the given values into Equation 1, solve for Vp , and then use this value of Vp in Equation 2 to obtain Dj . The result is Vp = 2.880 m/s and Dj = 0.0224 m ≈ 22 mm . 2.27. From the given data: D = 200 mm, L = 1 km, ∆z = 7 m, p = 140 kPa, and Q = 60 L/s = 0.06 m3 /s. For water at 20◦ C, ν = 1.004 × 10−6 m2 /s and γ = 9.789 kN/m3 . The following
preliminary calculations are useful, A=
πD2 π(0.2)2 = = 3.142 × 10−2 m2 , 4 4
V =
0.06 = 1.910 m/s 3.142 × 10−2
(1.910)(0.2) VD = = 3.805 × 105 , ν 1.004 × 10−6
hv =
V2 1.9102 = = 0.1860 m 2g 2(9.807)
Re =
In order to find the pipe roughness, apply the energy equation between the pressure source and the outlet, which gives p fL V 2 V2 + ∆z − = γ D 2g 2g
→
Using the Colebrook equation, )( ) ( ks ks Re, = 0.0227 fCE Re, D D
p V2 140 + ∆z − + 7 − 0.1860 γ 2g 9.789 f= = = 0.0227 1000 L V2 0.1860 0.2 D 2g ( →
fCE
ks Re, D
)(
ks 3.805 × 10 , 200 5
) = 0.0227
→
ks = 0.324 mm
With the new connection, Q12 = 80 L/s, Q23 = 60 L/s, and the following preliminary calculations are useful, V12 =
Q12 0.08 = = 2.547 m/s, A 3.142 × 10−2
V23 =
2 V12 2.5472 = = 0.3306 m, 2g 2(9.807)
hv23 =
hv12 =
Q23 0.06 = = 1.910 m/s A 3.142 × 10−2 2 V23 1.9102 = = 0.1860 m 2g 2(9.807)
V12 D (2.547)(0.2) V23 D (1.910)(0.2) = = 5.073 × 105 , Re23 = = = 3.805 × 105 −6 ν 1.004 × 10 ν 1.004 × 10−6 ( )( ) ( )( ) ks ks ks ks f12 = fCE Re, Re12 , = 0.0226, f23 = fCE Re, Re23 , = 0.0227 D D D D Re12 =
L12 =
L 1000 = = 500 m, 2 2
L23 =
L 1000 = = 500 m 2 2
Applying the energy equation from the beginning to the end of the pipeline gives 2 2 p1 f12 L12 V12 f23 L23 V23 V2 + ∆z − − = 23 γ D 2g D 2g 2g
which yields [ ] 2 2 2 f12 L12 V12 f23 L23 V23 V23 p1 = γ −∆z + + + D 2g D 2g 2g [ ] (0.0227)(500) (0.0226)(500) = (9.789) −7 + (0.3306) + (0.1860) + 0.1860 = 219 kPa 0.2 0.2
2.28. p1 = 30 kPa, p2 = 500 kPa, therefore head, hp , added by pump is given by hp =
p2 − p1 500 − 30 = = 48.0 m γ 9.79
Power, P , added by pump is given by P = γQhp = (9.79)(Q)(48.0) = 470 kW per m3 /s 2.29. Q = 60 L/s = 0.06 m3 /s, D = 0.2 m, ks = 0.9 mm (riveted steel), ks /D = 0.9/200 = 0.00450, for 90◦ bend K = 0.3, for the entrance K = 1.0, at 20◦ C ρ = 998 kg/m3 , and µ = 1.00 × 10−3 Pa·s, therefore π 2 π D = (0.2)2 = 0.0314 m2 4 4 0.06 Q = = 1.91 m/s V = A 0.0314 ρV D (998)(1.91)(0.2) = = 3.81 × 105 Re = µ 1.00 × 10−3 A=
Substituting ks /D and Re into the Colebrook equation gives 1 √ = −2 log f
(
2.51 0.00450 √ + 3.7 3.81 × 105 f
)
which leads to f = 0.0297 Minor head loss, hm , is given by hm =
∑
K
V2 (1.91)2 = (1.0 + 0.3) = 0.242 m 2g 2(9.81)
If friction losses, hf , account for 90% of the total losses, then hf = f
L V2 = 9hm D 2g
which means that 0.0297
L (1.91)2 = 9(0.242) 0.2 2(9.81)
Solving for L gives L = 78.9 m For pipe lengths shorter than the length calculated in this problem, the word “minor” should not be used.
2.30. From the given data: p0∑ = 480 kPa, v0 = 5 m/s, z0 = 2.44 m, D = 19 mm = 0.019 m, L = 40 m, z1 = 7.62 m, and Km = 3.5. For copper tubing it can be assumed that ks = 0.0023 mm. Applying the energy and Darcy-Weisbach equations between the water main and the faucet gives p0 p1 v2 + z 0 − hf − hm = + 1 + z1 γ γ 2g 2 2 480 f (40) v v 0 v2 + 2.44 − − 3.5 = + + 7.62 9.79 0.019 2(9.81) 2(9.81) γ 2(9.81) which simplifies to v=√
6.622 107.3f − 0.2141
(1)
The Colebrook equation, with ν = 1 × 10−6 m2 /s gives [ ] ks 1 2.51 √ = −2 log √ + 3.7D Re f f [ ] 1 0.0025 2.51 √ = −2 log + v(0.019) √ 3.7(19) f f 1×10−6 ] [ 1 1.321 × 10−4 −5 √ = −2 log 3.556 × 10 + √ f v f
(2)
Combining Equations 1 and 2 gives √ [ ] 1 1.995 × 10−5 107.3f − 0.2141 −5 √ = −2 log 3.556 × 10 + √ f f which yields f = 0.0189 Substituting into Equation 1 yields v=√
6.622
= 4.92 m/s 107.3(0.0189) − 0.2141 (π ) Q = Av = 0.0192 (4.92) = 0.00139 m3 /s = 1.39 L/s (= 22 gpm) 4 This flow is very high for a faucet. The flow would be reduced if other faucets are open, this is due to increased pipe flow and frictional resistance between the water main and the faucet. 2.31. From the given data: D = 50 mm = 0.050 m, L = 30.0 m, and ks = 0. Assume water at 20◦ C, γ = 9.79 kN/m3 and ν = 1.00 × 10−6 m2 /s. The local loss coefficients are given in the following table: Location 1 2, 3, 5 4 6
Type of Fixture sharp-edged entrance 90◦ elbow tee globe valve
klocal 0.5 1.2 1.9 4.0
Hence, between the water main and the faucet the total local loss coefficient is k = 0.5 + 3(1.2) + 1.9 + 4 = 10. The energy equation between points 1 and 6 is given by [ ] 2 p1 L ∑ V V2 + z1 − f + k = + z6 γ D 2g 2g which can be conveniently expressed as [ ] 2 p1 L (∑ ) V − f + k +1 = (z6 − z1 ) γ D 2g
(1)
The friction factor, f , can be determined from the following sequence of calculations, π 2 π D = (0.050)2 = 0.001963 m2 4 4 Q Q V = = = 509.3Q A 0.001963 509.3Q(0.050) VD = = 2.547 × 107 Q Re = ν 1.00 × 10−6 0.25 0.25 0.3086 f=[ ( )]2 = [ ( )]2 = (6.563 + log Q)2 ks 5.75 5.75 log log 0 + + 3.7D Re0.9 (2.547 × 107 Q)0.9 A=
(2)
Combining Equations 1 and 2 with the given data yields [ ] 400 0.3086 30.0 (509.3Q)2 +0− + 10 + 1 = 15 9.79 (6.563 + log Q)2 0.050 2(9.81) which yields Q = 0.00991 m3 /s = 9.91 L/s = 595 L/min. ∑ 2.32. From the given data: z1 = −1.5 m, z2 = 40 m, p1 = 450 kPa, k = 10.0, Q = 20 L/s = 0.02 m3 /s, D = 150 mm (PVC), L = 60 m, T = 20◦ C, and p2 = 150 kPa. The combined energy and Darcy-Weisbach equations give [ ] 2 p1 V12 p2 V22 fL ∑ V + + z 1 + hp = + + z2 + + k (1) γ 2g γ 2g D 2g where V1 = V2 = V =
Q 0.02 = π(0.15)2 = 1.13 m/s A 4
At 20◦ C, ν = 1.00 × 10−6 m2 /s, and Re =
VD (1.13)(0.15) = = 169500 ν 1.00 × 10−6
Since PVC pipe is smooth (ks = 0), the friction factor, f , is given by ( ) ( ) 1 2.51 2.51 √ = −2 log √ √ = −2 log f Re f 169500 f
(2)
which yields f = 0.0162
(3)
Taking γ = 9.79 kN/m3 and combining Equations 1 to 3 yields [ ] 450 1.132 150 1.132 (0.0162)(60) 1.132 + + (−1.5) + hp = + + 40 + + 10 9.79 2(9.81) 9.79 2(9.81) 0.15 2(9.81) which gives hp = 11.9 m Since hp > 0, a booster pump is required . The power, P , to be supplied by the pump is given by P = γQhp = (9.79)(0.02)(11.9) = 2.3 kW 2.33. From the given data: Dp = 20 mm, De = 12 mm, ks = 0.003 mm, and Q = 50 L/min = 8.333 × 10−4 m3 /s. For water at 20◦ C, γ = 9.79 kN/m3 and ν = 1.00 × 10−6 m2 /s. The following preliminary calculations are useful, πDp2 π(0.020)2 Ap = = = 3.142 × 10−4 m2 , 4 4 V Dp (2.653)(0.020) = = 5.305 × 104 , ν 1.00 × 10−6
Re =
∆z = 2 m + 3 m = 5 m, ∑
K = 9 + 3(1.2) + 10 = 22.6,
Ve =
Vp =
Q 8.333 × 10−4 = = 2.653 m/s Ap 3.142 × 10−4
ks 0.003 = 1.50 × 10−4 = Dp 20 L = 3.5 m + 2 m + 5 m + 3 m + 4 m = 17.5 m Ae =
πDe2 π(0.012)2 = = 1.131 × 10−4 m2 4 4
Q 8.333 × 10−4 = 7.368 m/s = Ae 1.131 × 10−4
Using Re = 5.305 × 104 and ks /D = 1.50 × 10−4 in the Colebrook equation yields f = 0.0212. Applying the energy equation between the water main and the faucet exit to obtain the required pressure, pm , in the main gives [ ] 2 Vp fL ∑ V2 pm − + K = e + ∆z γ D 2g 2g [ ] pm (0.0212)(17.5) 2.6532 7.3682 − + 22.6 = + 5 → pm = 220 kPa 9.79 0.020 2(9.807) 2(9.807) 2.34. From the given data: Q = 6 L/s = 0.006 m3 /s, D1 = 75 mm, and D2 = 50 mm. For water at 20◦ C, ρ = 998 kg/m3 , and γ = 9790 N/m3 . The following preliminary calculations are useful: A2 =
πD22 π(0.050)2 = = 1.963 × 10−3 m2 , 4 4
A1 =
πD12 π(0.075)2 = = 4.418 × 10−3 m2 4 4
V2 =
Q 0.006 = = 3.056 m/s, A2 1.963 × 10−3
V1 =
Q 0.006 = = 1.358 m/s A1 4.418 × 10−3
D2 50 = = 0.667 D1 75 (a) Using the head-loss coefficient given by Equation ??, since D2 /D1 < 0.76, the head loss, hℓ , at the sudden contraction is calculated as follows, ) ( ) ( 502 D22 K = 0.42 1 − 2 = 0.42 1 − 2 = 0.2333 75 D1 hℓ = K
V22 3.0562 = (0.2333) = 0.1111 m 2g 2(9.807)
Therefore, the head loss at the sudden expansion is approximately equal to 0.11 m . (b) The pressure change, ∆p, at the expansion is calculated by applying the energy equation across the expansion, such that ( ) ( ) ∆p ∆V 2 − + = hℓ → ∆p = − γhℓ + 12 ρ∆V 2 (1) γ 2g Substituting the given and derived data into Equation 1 gives [ ] ∆p = − (9790)(0.1111) + 21 (998)(3.0562 − 1.3582 ) [×10−3 kPa/Pa] = −1.278 kPa Therefore, it is expected that the pressure will decrease by 1.28 kPa across the sudden expansion. (c) If the local head loss is not taken into account, then hℓ = 0 and Equation 1 gives [ ] ∆p = − 0 + 21 (998)(3.0562 − 1.3582 ) [×10−3 kPa/Pa] = −0.1906 kPa Therefore, if the local head loss is neglected, the pressure would be predicted to decrease by approximately 0.19 kPa . 2.35. From the given data: D = 41 mm, p = 350 kPa, De = 30 mm, ze = 1 m, h = 5 m, Kb = 1.2, Kn = 0.05, ks = 0.1 mm, and L = 15 m + 3 m = 18 m. For water at 20◦ C, γ = 9.79 kN/m3 and ν = 1.00 × 10−6 m2 /s. Applying the Bernoulli equation to the fountain gives Ve2 =h 2g
→
Ve =
√
2gh =
√
2(9.807)(5) = 9.903 m/s
Applying the continuity equation gives the velocity, V , is the pipe as V =
Ae D2 302 Ve = e2 Ve = 2 (9.903) = 5.302 m/s A D 41
Using the values of the given and derived parameters, the following preliminary calculations are useful, Re =
VD (5.302)(0.041) = = 2.174 × 105 , ν 1.00 × 10−6
ks 0.1 = = 2.439 × 10−3 D 41
V2 5.3022 = = 1.433 m, 2g 2(9.807)
Ve2 9.9032 = = 5.000 m 2g 2(9.807)
Substituting the given and derived parameters into the Colebrook equation (Equation 2.28) gives ] ] [ [ 1 ks /D 2.51 2.439 × 10−3 2.51 1 √ = −2 log √ √ → √ = −2 log + + 3.7 3.7 f Re f f 2.174 × 105 f which yields f = 0.0254. Applying the energy equation between the water main and the nozzle exit gives ( ) 2 p fL V V2 V2 − + Kv − Kn e = e + ze γ D 2g 2g 2g which gives V2 V2 p 350 − Kn e − e − z e − (1.2)(5) − 5 − 1 (0.0254)(18) fL γ 2g 2g 9.79 = − = 9.42 − Kv = 2 D 1.433 0.041 V 2g 2.36. From the given data: p0 = 400 kPa, L = 50 m, ks = 0.1 mm, K = 30, and Q = 100 L/min = 1.667 × 10−3 m3 /s. For water at 15◦ C, ν = 1.14 × 10−6 m2 /s, and γ = 9.798 kN/m3 . The following functions can be defined in terms of the unknown diameter, D, (1) A(D) =
πD2 , 4
(3) Re(D) =
V (D)D V (D)D = , ν 1.14 × 10−6
(5) hv (D) =
V (D)2 2g
Q 1.667 × 10−3 = A(D) A(D) ( ) ks (4) f (D) = fCE , Re(D) D (2) V (D) =
where fCE is the friction factor derived from the Colebrook equation. Applying the energy equation with the defined functions gives [ ] f (D)L p0 − + K hv (D) = hv (D) γ D [ ] 400 f (D)(50) − + 30 hv (D) = hv (D) → D = 0.0264 m = 26.4 mm 9.798 D 2.37. From the given data: ∆z = 42 m + 1.5 m = 43.5 m, L = 1 km, D = 600 mm, ks = 1 mm, and Kb = 0.8. For water at 20◦ C, ν = 1.00 × 10−6 m2 /s, and γ = 9.79 kN/m3 . The value of h can be calculated for any given combination of p and Q. The calculations will be illustrated for Q = 0.3 m3 /s and pB = 450 kPa. The following preliminary calculations are useful, A=
πD2 π(0.6)2 = = 0.2827 m2 , 4 4
V =
Q 1 = = 1.061 m/s A 0.2827
ks 1 = = 1.667 × 10−3 , D 600
Re =
VD (1.061)(0.6) = = 6.366 × 105 ν 1.00 × 10−6
Substituting the given and derived parameters into the Colebrook equation (Equation 2.28) gives [ ] [ ] 1 ks /D 2.51 1 1.667 × 10−3 2.51 √ = −2 log √ √ + → √ = −2 log + 3.7 3.7 f Re f f 6.366 × 105 f which yields f = 0.0227. Applying the energy equation between Points A and B gives ] 2 [ V fL pB + 5Kb = h + ∆z − D 2g γ [ ] (0.0227)(1000) 1.0612 450 h + 43.5 − + 5(0.8) = → h = 4.86 m 0.6 2(9.807) 9.79 These calculations can be repeated for the given extreme values of pB and Q leading to the following results: Q (m3 /s) 0.3 0.3 1.0 1.0
pB (kPa)
h (m)
330 450 330 450
−7.39 4.86 16.59 28.85
Since water will always be maintained in the tank, the range of h is 0 to 28.9 m . At the minimum flow rate, the minimum pressure (when h = 0 m) is approximately 403 kPa . 2.38. from the given data: L = 25 m, D = 19 mm, p0 = 200 kPa, ∆z = 0.9 m, ks = 0.1 mm, De = 10 mm, and Ke = 0.05. For water at 15◦ C, ν = 1.140×10−6 m2 /s, and γ = 9.798 kN/m3 . The following preliminary calculations are useful, A=
πD2 π(0.019)2 = = 2.835 × 10−4 m2 , 4 4
ks 0.1 = = 5.263 × 10−3 , D 19
r=
D4 = 13.03 De4
Ae =
πDe2 π(0.010)2 = = 7.853 × 10−5 m2 4 4
For any given flow rate, Q, the following functions can be defined, (1) V (Q) =
Q Q = , A 2.835 × 10−4
(3) Re(Q) =
V (Q)D V (Q)(0.019) = , ν 1.140 × 10−6
V (Q)2 2g [ ] ks (4) f (Q) = fCE , Re(Q) D (2) hv (Q) =
where fCE represents the value of f derived using the Colebrook equation.
(a) Applying the energy equation between the spigot and the hose exit (without a nozzle) gives p0 f (Q)L +hv − hv = hv γ D
→
200 f (Q)(25) − hv (Q) = 0 9.798 0.019
→
Q = 8.667×10−4 m3 /s
Therefore the flow rate is Q = 8.667 × 10−4 m3 /s = 52 L/min . The corresponding exit velocity, Ve , is given by Ve =
Q 8.667 × 10−4 = 3.06 m/s = A 2.835 × 10−4
(b) Applying the energy equation between the spigot and the hose exit (with a nozzle) gives [ ] f (Q)L 200 f (Q)(25) p0 + hv − hv − Ke rhv = rhv → + 1− − Ke r − r hv (Q) = 0 γ D 9.798 0.019 which gives Q = 7.592 × 10−4 m3 /s = 46 L/min . The corresponding exit velocity, Ve , is given by Q 7.592 × 10−4 Ve = = = 9.67 m/s Ae 7.853 × 10−5 (c) The main benefit of a nozzle is that it produces a jet that can deliver the liquid to a further location (from the exit) than is possible without the nozzle. 2.39. From the given data: L = 1 km, D = 500 mm, ∆z = 20 m, K = 45, P = 250 kW, and Q = 500 L/s = 0.5 m3 /s. For water at 20◦ C, γ = 9.79 kN/m3 , and ν = 1.00 × 10−6 m2 /s. The following preliminary calculations are useful, A=
π(0.0.5)2 πD2 = = 0.1963 m2 , 4 4
V =
Q 0.5 = = 2.547 m/s A 0.1963
hv =
V2 2.5472 = = 0.3306 m, 2g 2(9.807)
Re =
VD (2.547)(0.5) = = 1.273 × 106 ν 1.00 × 10−6
hp =
P 250 = = 51.07 m γQ (9.79)(0.5)
(a) The friction factor as a function of ks can be expressed as ( ) ( ) ks ks 6 f (ks ) = fCE , Re = fCE , 1.273 × 10 D 500 where fCE represents the value of f derived using the Colebrook equation. Applying the energy equation between the lower and upper reservoirs gives [ ] fL + K hv + hp = ∆z 0− D [ ] f (ks )(1000) + 45 (0.3306) + 51.07 = 20 → ks = 1.2 mm 0− 0.5 The value of the friction factor corresponding to ks = 1.2 mm is f = 0.0245. This friction factor will remain the same as long as the flow rate remains the same.
(b) If K is reduced by 70%, then K = 0.3(45) = 13.5. Applying the energy equation between the lower and upper reservoirs gives [ ] fL 0− + K hv + hp = ∆z D [ ] (0.0245)(1000) 0− + 13.5 (0.3306) + hp = 20 → hp = 40.66 m 0.5 The corresponding pump power, P , is given by P = γQhp = (9.79)(0.5)(40.66) = 199 kW Therefore the required pumping power is reduced by approximately 20%. 2.40. (a) Diameter of pipe, D = 0.75 m, area, A given by A=
π 2 π D = (0.75)2 = 0.442 m2 4 4
and velocity, V , in pipe Q 1 = = 2.26 m/s A 0.442 Energy equation between reservoir and A: V =
7 + hp − hf =
pA VA2 + + zA γ 2g
where pA = 350 kPa, γ = 9.79 kN/m3 , VA = 2.26 m/s, zA = 10 m, and hf =
fL V 2 D 2g
where f depends on Re and ks /D. At 20◦ C, ν = 1.00 × 10−6 m2 /s and VD (2.26)(0.75) = = 1.70 × 106 ν 1.00 × 10−6 ks 0.26 = = 3.47 × 10−4 D 750
Re =
Using the Swamee-Jain equation, ] [ [ ] ks /D 5.74 3.47 × 10−4 5.74 1 √ = −2 log + = −2 log + = 7.93 3.7 3.7 (1.70 × 106 )0.9 f Re0.9 which leads to f = 0.0159 The head loss, hf , between the reservoir and A is therefore given by hf =
fL V 2 (0.0159)(1000) (2.26)2 = = 5.52 m D 2g 0.75 2(9.81)
(1)
Substituting into Equation 1 yields 7 + hp − 5.52 =
350 2.262 + + 10 9.81 2(9.81)
which leads to hp = 44.5 m (b) Power, P , supplied by the pump is given by P = γQhp = (9.79)(1)(44.5) = 436 kW (c) Energy equation between A and B is given by V2 pA VA2 pB + + zA − hf = + B + zB γ 2g γ 2g and since VA = VB , pB = pA + γ(zA − zB − hf ) = 350 + 9.79(10 − 4 − 5.52) = 355 kPa 2.41. From the given data: Q = 0.7 m3 /s, D = 600 mm, ks = 1 mm, L0A = 1 km, LAB = 1.5 km, LBC = 1.8 km, , L0C = 4.3 km, zA = −10 m, zB = 12 m, zC = 5 m, and pC = 400 kPa. For water at 20◦ C, γ = 9.79 kN/m3 and ν = 1.00 × 10−6 m2 /s. The following preliminary calculations are useful, A=
πD2 π(0.6)2 = = 0.2827 m2 , 4 4
V =
hv =
V2 2.4762 = = 0.3125 m, 2g 2(9.807)
ks 1 = = 1.667 × 10−3 D 600
Re =
(2.476)(0.6) VD = = 1.485 × 106 ν 1.00 × 10−6
Q 0.7 = = 2.476 m/s A 0.2827
Substituting the given and derived parameters into the Colebrook equation (Equation 2.28) gives [ ] [ ] 1 ks /D 2.51 1 1.667 × 10−3 2.51 √ = −2 log √ √ + → √ = −2 log + 3.7 3.7 f Re f f 1.485 × 106 f which yields f = 0.0225. Using the energy equation, the head added by the pump, hp , is given by 0 + hp −
pC 400 f L0C (0.0225)(4300) hv = + z C → hp − (0.3125) = +5 D γ 0.6 9.79
→
hp = 96.18 m
The maximum pressure is expected at A. Applying the energy equation between the reservoir as A gives, 0+hp −
f L0A pA (0.0225)(1000) pA hv = +zA → 96.18− (0.3125) = −10 D γ 0.6 9.79
→
pA = 925 kPa
The minimum pressure could occur at B. Applying the energy equation between the reservoir as B gives, 0+hp −
f L0B pB (0.0225)(2500) pB hv = +zB → 96.18− (0.3125) = +12 D γ 0.6 9.79
→
pB = 538 kPa
(a) Based on evaluations of the pressures at key points in the pipeline, the estimated range of pressures is 400–925 kPa . (b) Expressing the maximum pressure in the pipeline as a function of the pipe diameter yields that the maximum pressure in the pipeline (at Point A) is equal to 600 kPa when D = 0.873 m = 873 mm . This was done using MATLAB. Any similar software will yield the same result. 2.42. From the given data: L = 230 m, ks = 0.2 mm, ∆z = 18 m, p = 400 kPa, and Q = 50 L/s = 0.05 m3 /s. For water at 20◦ C, γ = 9.789 kN/m3 , and ν = 1.004 × 10−6 m2 /s. Applying the energy equation between the discharge side of the pump and the reservoir, and using the Colebrook equation to calculate f , yields ( ) ( ) 4Q ks ks f Re, , L CE D 8Q2 p fL V 2 V 2 p πDν D − − = ∆z → − + 1 π 2 D4 g = ∆z γ D 2g 2g γ D Substituting the given data: ( ) ( ) 4(0.05) 0.0002 ks f Re, , (230) CE D 8(0.05)2 400 πD(1.004 × 10−6 ) D − + 1 π 2 D4 (9.807) = 18 9.789 D
→
D = 0.136 m
Therefore, the required pipe diameter is approximately 136 mm. 2.43. From the given data: ∆z = 4 m, At = 10 m2 , D = 75 mm, L = 65 m, ks = 0.5 mm, p = 550 kPa, h1 = 0.5 m, and h2 = 3 m. For water at 20◦ C, γ = 9.789 kN/m3 . The following preliminary calculations are useful, π(0.075)2 πD2 = = 4.418 × 10−3 m2 , 4 4 ( )( ) ks ks f = fCE Re, , ∞ = 0.0332, D D A=
ks 0.5 = = 6.667 × 10−3 D 75 a=
p 550 −4= − 4 = 52.19 m γ 9.789
and b = A2t
fL (0.0332)(65) = (10)2 = 7.515 × 106 s2 /m 2 2gA D 2(9.807)(4.418 × 10−3 )2 (0.075)
Applying the energy equation between the pump and the reservoir gives p f L Q2 − =4+h γ D 2gA2
(1)
and the continuity equation requires that Q = At
dh dt
Combining Equations 1 and 1 to eliminate Q gives v up u −4−h dh u γ → =u t 2 fL dt At 2gA2 D
(2)
dh = dt
√
a−h b
To integrate, use the substitution: x=
a−h , b
dh = −b dx
Therefore, the requisite integral in terms of x is ∫ x2 ∫ t2 √ √ dx √ = −b dt → (t2 − t1 ) = −2b [ x2 − x1 ] x x1 t1 The given initial and final conditions of h1 = 0.5 m and h2 = 3 m translate into x1 = 6.88 × 10−6 m2 /s2 and x2 = 6.54 × 10−6 m2 /s2 and so [√ ] √ (t2 − t1 ) = −2b 6.54 × 10−6 − 6.88 × 10−6 = 965 s = 16.1 min (3) 2.44. From the given data: L = 3 km = 3000 m, Qave = 0.0175 m3 /s, and Qpeak = 0.578 m3 /s. If the velocity, Vpeak , during peak flow conditions is 2.5 m/s, then 2.5 = which gives
0.578 Qpeak = 2 πD /4 πD2 /4
√ D=
0.578 = 0.543 m π(2.5)/4
Rounding to the nearest 25 mm gives D = 550 mm with a cross-sectional area, A, given by A=
π 2 π D = (0.550)2 = 0.238 m2 4 4
During average demand conditions, the head, have , at the suburban development is given by have =
2 pave Vave + + z0 γ 2g
(1)
where pave = 340 kPa, γ = 9.79 kN/m3 , Vave = Qave /A = 0.0175/0.238 = 0.0735 m/s, and z0 = 8.80 m. Substituting into Equation 1 gives 340 0.07352 + + 8.80 = 43.5 m 9.79 2(9.81)
have =
For ductile-iron pipe, ks = 0.26 mm, ks /D = 0.26/550 = 4.73 × 10−4 , at 20◦ C ν = 1.00 × 10−6 m2 /s, and therefore Vave D (0.0735)(0.550) = 4.04 × 104 Re = = ν 1.00 × 10−6 and the Swamee-Jain equation gives [ ] ] [ 1 ks /D 5.74 5.74 4.73 × 10−4 √ = −2 log + + = −2 log 3.7 3.7 (4.04 × 104 )0.9 fave Re0.9 and yields fave = 0.0234 The head loss between the water treatment plant and the suburban development is therefore given by L V2 3000 0.07352 hf = f = (0.0234) = 0.035 m D 2g 0.550 2(9.81) Since the head at the water treatment plant is 10.00 m, the pump head, hp , that must be added is hp = (43.5 + 0.035) − 10.00 = 33.5 m and the power requirement, P , is given by P = γQhp = (9.79)(0.0175)(33.5) = 5.74 kW During peak demand conditions, the head, hpeak , at the suburban development is given by hpeak =
2 ppeak Vpeak + + z0 γ 2g
(2)
where ppeak = 140 kPa, γ = 9.79 kN/m3 , Vpeak = Qpeak /A = 0.578/0.238 = 2.43 m/s, and z0 = 8.80 m. Substituting into Equation 2 gives hpeak =
140 2.432 + + 8.80 = 23.4 m 9.79 2(9.81)
For pipe, ks /D = 4.73 × 10−4 , and Re =
(2.43)(0.550) Vpeak D = = 1.34 × 106 ν 1.00 × 10−6
and the Swamee-Jain equation gives [ ] [ ] 1 ks /D 5.74 4.73 × 10−4 5.74 √ = −2 log + = −2 log + 3.7 3.7 (1.34 × 106 )0.9 Re0.9 fpeak
and yields fpeak = 0.0170 The head loss between the water treatment plant and the suburban development is therefore given by L V2 3000 2.432 hf = f = (0.0170) = 27.9 m D 2g 0.550 2(9.81) Since the head at the water treatment plant is 10.00 m, the pump head, hp , that must be added is hp = (23.4 + 27.9) − 10.00 = 41.3 m and the power requirement, P , is given by P = γQhp = (9.79)(0.578)(41.3) = 234 kW 2.45. From the given data: Q = 240 L/min = 4 × 10−3 m3 /s, D = 50 mm, L = 15 m, ks = 0.5 mm, and patm = 101.3 kPa. For water at 15◦ C, ρ = 999.1 kg/m3 , µ = 1.139 mPa·s, ν = µ/ρ = 1.140 × 10−6 m2 /s, γ = ρg = 9.798 kN/m3 , and pvap = 1.704 kPa. The head-loss coefficient for the reentrant intake pipe can be conservatively taken as Ke = 1.0. The following preliminary calculations are useful, Q 4 × 10−3 = = 2.037 m/s A 1.963 × 10−3
A=
πD2 π(0.05)2 = = 1.963 × 10−3 m2 , 4 4
V =
hv =
V2 2.0372 = = 0.2116 m, 2g 2(9.807)
ks 0.5 = = 0.01 D 50
Re =
VD (2.037)(0.05) = = 8.935 × 104 ν 1.140 × 10−6
Substituting the given and derived parameters into the Colebrook equation (Equation 2.28) gives [ ] [ ] 1 ks /D 2.51 1 0.01 2.51 √ = −2 log √ √ + → √ = −2 log + 3.7 3.7 f Re f f 8.935 × 104 f which yields f = 0.0386. (a) Applying the energy equation between the water in the well and the suction side of the pump gives [ ] pvap fL patm − Ke + hv = + hv + h γ D γ [ ] 101.3 (0.0386)(15) 1.704 − 1.0 + (0.2116) = + 0.2116 + h → h = 7.72 m 9.798 (0.05) 9.798 (b) Since h increases as Q decreases, the maximum value of h occurs when Q = 0. In this case, the energy equation degenerates into the hydrostatic equation as follows pvap patm = + hmax γ γ
→
101.3 1.704 = + hmax 9.798 9.798
→
hmax = 10.2 m
(c) For any pump pumping water at 15◦ C from any reservoir, the suction side of the pump should not be more that 10.2 m above the surface of the reservoir. Otherwise, the pump will not work. 2.46. From the given data: D = 0.05 m, A = πD2 /4 = 0.001963 m2 , Q = 4 L/s = 0.004 m3 /s, V = Q/A = 2.038 m/s, L = 125 m, ν = 1.00 ×10−6 m2 /s, γ = 9.79 kN/m3 , ks = 0.23 mm, Re = V D/ν = 1.02 × 105 . Using the Swamee-Jain equation, f=[
(
log
0.25 )]2 = [ ( )]2 = 0.0308 ks 0.23 5.74 5.74 log + + 3.7D Re0.9 3.7(50) (1.02 × 105 )0.9 0.25
For a sharp-edged entrance, Ke = 0.5, for an open globe valve, Kv = 10.0. The energy equation applied between the reservoir and the outlet is given by [ ] 2 [ ] 2 fL V fL V V2 + Kv − ht = → ht = 40 − Ke + + Kv + 1 (1) 40 − Ke + D 2g 2g D 2g Substituting the given and derived data into Equation 1 gives [ ] (0.0308)(125) (2.038)2 ht = 40 − 0.5 + + 10.0 + 1 = 21.27 m 0.05 2(9.81) Therefore, the power extracted by the turbine is given by P = γQht = (9.79)(0.004)(21.27) = 0.833 kW A similar problem would be encountered in calculating the power output at a hydroelectric facility . 2.47. From the given data: L = 150 m, D = 350 mm, ks = 1 mm, P = 60 kW, and ∆z = 45 m. For water at 20◦ C, γ = 9.79 kN/m3 and ν = 1.00 × 10−6 m2 /s. The following preliminary calculations are useful, A=
πD2 π0.3502 = = 0.09621 m2 , 4 4
Re =
VD QD = ν νA
P = γQht
→
→ ht =
ks 1 = = 2.857 × 10−3 D 350
Q = 2.749 × 10−7 Re 6.129 Q
Applying the energy equation between the upper and lower reservoir gives, [ ] fL Q2 ∆z − + Kin + Kout − ht = 0 D 2gA2 [ ] f (150) Q2 6.129 45 − + 0.8 + 1.0 − =0 2 0.350 2(9.807)(0.09621) Q Solving this equation simultaneously with the Colebrook equation gives Q = 0.141 m3 /s .
2.48. From the given data: L = 1 km, Q = 2 m3 /s, ks = 1.5 mm, ∆z = 23 m, and T = 20◦ C. For water at 20◦ C, ν = 1.00 × 10−6 m2 /s. From the given data, ] ] [ [ (Q/A)D 1 4Q 1 4(2) 2.547 × 106 Re = = = = ν πν D π(1.00 × 10−6 ) D D (a) Neglecting local losses, the energy equation gives ∆z =
8f LQ2 8f (1000)(2)2 f L Q2 → ∆z = → 23 = D 2gA2 gπ 2 D5 (9.807)π 2 D5
which simplifies to f = 0.0696 D5 Solving this equation simultaneously with the Colebrook equation (e.g., using MATLAB) gives D = 0.802 m = 802 mm . (b) Taking local losses into account with K = 15.3, the energy equation gives ] [ ] [ ] [ Q2 fL 8Q2 f (1000) 8(2)2 fL +K → ∆z = + K → 23 = + 15.3 ∆z = D 2gA2 D π2 D4 D π2 D4 Solving this equation simultaneously with the Colebrook equation (e.g., using MATLAB) gives D = 0.802 m = 802 mm . Therefore, local losses have no impact on the tunnel diameter that is required to meet the design objective. 2.49. The head loss is calculated using Equation 2.72. The hydraulic diameter, Dh , is given by Dh =
4(2)(1) 4A = = 1.333 m P 2(2 + 1)
and the mean velocity, V , is given by V =
Q 5 = = 2.5 m/s A (2)(1)
At 20◦ C, ρ = 998.2 kg/m3 , µ = 1.002 × 10−3 N·s/m2 , and therefore the Reynolds number, Re, is given by ρV Dh (998.2)(2.5)(1.333) Re = = = 3.32 × 106 ν 1.002 × 10−3 A median equivalent sand roughness for concrete can be taken as ks = 1.6 mm (Table 2.1), and therefore the relative roughness, ks /Dh , is given by ks 1.6 × 10−3 = = 0.00120 Dh 1.333 Substituting Re and ks /Dh into the Swamee-Jain equation (Equation 2.31) for the friction factor yields [ ] [ ] 1 ks /Dh 5.74 0.00120 5.74 √ = −2 log + = −2 log + = 6.96 3.7 3.7 (3.32 × 106 )0.9 f Re0.9
which yields f = 0.0206 The frictional head loss in the culvert, hf , is therefore given by the Darcy-Weisbach equation as fL V 2 (0.0206)(100) 2.52 hf = = = 0.493 m Dh 2g 1.333 2(9.81) 2.50. The frictional head loss is calculated using Equation 2.72. The hydraulic diameter, Dh , is given by 4A 4(2)(2) Dh = = = 2.000 m P 2(2 + 2) and the mean velocity, V , is given by V =
10 Q = = 2.5 m/s A (2)(2)
At 20◦ C, ρ = 998 kg/m3 , µ = 1.00 × 10−3 N·s/m2 , and therefore the Reynolds number, Re, is given by ρV Dh (998)(2.5)(2.000) Re = = = 4.99 × 106 µ 1.00 × 10−3 A median equivalent sand roughness for concrete can be taken as ks = 1.6 mm (Table 2.1), and therefore the relative roughness, ks /Dh , is given by ks 1.6 × 10−3 = = 0.0008 Dh 2.000 Substituting Re and ks /Dh into the Swamee-Jain equation (Equation 2.32) for the friction factor yields [ ] [ ] 1 k /D 5.74 0.0008 5.74 √ = −2 log s h + = −2 log + = 7.31 3.7 3.7 (4.99 × 106 )0.9 f Re0.9 which yields f = 0.0187 The frictional head loss in the culvert, hf , is therefore given by the Darcy-Weisbach equation as (0.0187)(500) 2.52 fL V 2 = = 1.49 m hf = Dh 2g 2.500 2(9.81) Applying the energy equation between the upstream and downstream sections (Sections 1 and 2 respectively), p1 V12 p2 V22 + + z1 = + + z2 + hf γ 2g γ 2g which gives 2.52 p2 2.52 p1 + + (0.002)(500) = + + 0 + 1.49 9.79 2(9.81) 9.79 2(9.81) Re-arranging this equation gives p1 − p2 = 4.80 kPa
2.51. The Hazen-Williams formula is given by V = 0.849CH R0.63 S0.54 , f
Sf =
hf L
Combining the above two equations to eliminate Sf , taking R = D/4, and simplifying gives ( V = 0.849CH
D 4
)0.63 (
hf L
)0.54 →
L hf = 6.82 1.17 D
(
V CH
)1.85
2.52. Comparing the Hazen–Williams and Darcy–Weisbach equations for head loss gives ( ) L V 1.85 L V2 134 1 (hf =) 6.82 1.17 =f → f = 1.85 0.17 0.15 D CH D 2g V CH D For laminar flow, Equation 2.29 gives f ∼ 1/Re ∼ 1/V , and for fully-turbulent flow Equation 2.28 gives f ∼ 1/V 0 . Since the Hazen–Williams formula requires that f ∼ 1/V 0.15 , this indicates that the flow must be in the transition regime . 2.53. From the given data: p1 = 450 kPa, p2 = 150 kPa, ks = 0.002 mm, Q = 150 L/min = 0.0025 m3 /s, ∆z1 = 27 m, ∆z2 = 10 m, and CH = 140. For water at 20◦ C, γ = 9.789 kN/m3 . (a) Applying the energy and Darcy-Weisbach equations between the water-supply pipe and the location of fixture F gives [ ] [ ] f ∆z1 2Q2 f (27) 2(0.0025)2 p2 = p1 − γ + ∆z1 → p2 = 450 − (9.789) + 27 D gπD2 D (9.807)πD2 where p2 is the pressure at fixture F. Calculating p2 for the given commercial pipe sizes yields the results shown in the following table: D (mm)
p2 (kPa)
32 38 51
111 153 177
Since the minimum allowable pressure at F is 150 kPa, a pipe diameter of 38 mm should be used for the supply line. (b) Applying the energy and Hazen–Williams equations between the water-supply pipe and the location of fixture F gives ) ] [ ( ∆z1 1.85 + ∆z1 p2 = p1 − γ 85.4 1.85 D 4.87 Q CH [ ( ) ] 27 1.85 → p2 = 450 − (9.789) 85.4 (150) + 27 1401.85 D4.87 where p2 is the pressure at fixture F, D is in cm, and Q is in L/min. Calculating p2 for the given commercial pipe sizes yields the results shown in the following table:
D (mm)
p2 (kPa)
32 38 51
97 147 176
Since the minimum allowable pressure at F is 150 kPa, a pipe diameter of 51 mm should be used for the supply line. (c) If the fixture F is supplied by both the water-supply line and the roof tank, and D = 25 mm, and the flow rate extracted at F is 150 L/min, then the following equations must hold: Q = Q1 + Q2 [ ( p2 = p1 − γ 85.4
) ] ∆z1 1.85 + ∆z1 1.85 D 4.87 Q1 CH [ ( ) ] ∆z2 1.85 p2 = 0 − γ 85.4 − ∆z2 1.85 D 4.87 Q CH Substituting known quantities yields: 150 = Q1 + Q2
) ] 27 1.85 Q1 + 27 p2 = 450 − (9.789) 85.4 1401.85 2.54.87 [ ( ) ] 10 1.85 p2 = 0 − (9.789) 85.4 Q2 − 10 1401.85 2.54.87 [
(
Solving these equations yields Q1 = 42.3 L/s, Q2 = 107.7 L/s, and p2 = 157 kPa . 2.54. The Manning equation is given by ) ( )2 ( )1 ( 1 2 12 1 D 3 hf 2 3 V = R Sf = n n 4 L
→
hf = 6.35
n2 LV 2 4
D3
2.55. Comparing the Manning and Darcy–Weisbach equations gives (hf ) = 6.35
n2 LV 2 D
4 3
=f
L V2 D 2g
→
f = 125
n2 1
D3
For laminar flow, Equation 2.29 gives f ∼ 1/Re ∼ 1/V , and for fully-turbulent flow Equation 2.28 gives f ∼ 1/V 0 . Since the Manning equation requires that f ∼ 1/V 0 , this indicates that the flow must be fully turbulent or rough . 2.56. Equating the Hazen–Williams and Manning head loss expressions ( ) ( ) L V 1.85 n2 LV 2 D0.082 1 6.82 1.17 = 6.35 → n = 1.04 0.075 4 0.93 D CH V CH D3
2.57. Choose the Darcy-Weisbach equation since this equation is applicable in all flow regimes. The Hazen-Williams and Manning equations are limited to particular flow conditions (transition and fully turbulent respectively). 2.58. (a) The Hazen-Williams roughness coefficient, CH , can be taken as 130 (Table 2.2), L = 500 m, D = 0.300 m, V = 2 m/s, and therefore the head loss, hf , is given by Equation 2.78 as ( ) ( ) L V 1.85 500 2 1.85 hf = 6.82 1.17 = 6.82 = 6.17 m D CH (0.30)1.17 130 (b) The Manning roughness coefficient, n, can be taken as 0.013 (approximation from Table 2.2), and therefore the head loss, hf , is given by Equation 2.81 as hf = 6.35
n2 LV 2 D
4 3
= 6.35
(0.013)2 (500)(2)2 4
= 10.7 m
(0.30) 3
(c) The equivalent sand roughness, ks , can be taken as 0.26 mm (Table 2.1), and the Reynolds number, Re, is given by Re =
(2)(0.30) VD = = 6.00 × 105 ν 1.00 × 10−6
where ν = 1.00 × 10−6 m2 /s at 20◦ C. Substituting ks , D, and Re into the Colebrook equation yields the friction factor, f , where [ ] [ ] 1 ks 2.51 0.26 2.51 √ = −2 log √ √ + = −2 log + → f = 0.0195 3.7D Re f 3.7(300) 6.00 × 105 f f The head loss, hf , is therefore given by the Darcy-Weisbach equation as hf = f
L V2 500 22 = 0.0195 = 6.63 m D 2g 0.30 2(9.81)
It is reasonable to assume that the Darcy-Weisbach equation yields the most accurate estimate of the head loss. In this case, the Hazen-Williams formula gives a head loss that is 7% too low, and the Manning formula yields a head loss that is 61% too high. Problem 2.52 has demonstrated that the relationship between the friction factor, f , and the Hazen-Williams coefficient, CH , is given by f=
134
1
1.85 D 0.17 V 0.15 CH
which can be re-arranged to give CH =
14.2 14.2 = = 126 f 0.54 D0.092 V 0.081 (0.0195)0.54 (0.30)0.092 (2)0.081
which is 3% lower than the assumed value of CH (130).
Problem 2.55 has demonstrated that the relationship between the friction factor, f , and the Manning coefficient, n, is given by f = 125
n2 1
D3 which can be re-arranged to give n = 0.0894f 0.5 D0.17 = 0.0894(0.0195)0.5 (0.30)0.17 = 0.010 which is 23% lower than the assumed value of n (= 0.013). 2.59. The pressure at the midpoint of the pipe as a function of time is shown in Figure 2.2.
Figure 2.2: Problem 2.59. 2.60. From the given data: L = 100 m, V = 3 m/s, and T = 20◦ C. The critical time, tc , for valve closure is 2L (1) tc = c At 20◦ C, Ev = 2.15 × 109 Pa, and ρ0 = 998 kg/m3 (Appendix B, Table B.1), therefore √ √ Ev 2.15 × 109 c= = = 1470 m/s ρ0 998 Substituting into Equation 1 gives tc =
2(100) = 0.14 s 1470
The maximum water-hammer pressure that can occur is given by ∆p = ρcV = (998)(1470)(3) = 4.40 × 106 Pa = 4400 kPa
If the water temperature is 10◦ C, Ev = 2.10 × 109 Pa, ρ0 = 999.7 kg/m3 (Appendix B, Table B.1), and √ 2.10 × 109 c= = 1450 m/s 999.7 and the maximum water-hammer pressure is given by ∆p = ρcV = (999.7)(1450)(3) = 4.35 × 106 Pa = 4350 kPa 2.61. Yes , in cases of rapid valve closure. 2.62. From the given data: T = 20◦ C, L = 150 m, D = 50 mm = 0.050 m, V = 4 m/s, e = 1.5 mm = 0.0015 m, and E = 1.655 × 105 MN/m2 = 1.655 × 1011 N/m2 . Taking ρ0 = 998 kg/m3 and Ev = 2.15 × 109 Pa, the speed of the pressure wave, c, is given by Equation 2.92 as √ √ Ev /ρ0 (2.15 × 109 )/(998) c= = = 1226 m/s 1 + (Ev D/eE) 1 + (2.15 × 109 × 0.050)/(0.0015 × 1.655 × 1011 ) Hence, the pressure increase, ∆p is given by Equation 2.86 as ∆p = ρ0 cV = (998)(1226)(4) = 4.89 × 106 Pa = 4890 kPa 2.63. From the given data: T = 20◦ C, L = 150 m, D = 50 mm = 0.050 m, V = 4 m/s, e = 2.0 mm = 0.0020 m, and E = 1.7 × 104 MN/m2 = 1.7 × 1010 N/m2 . Taking ρ0 = 998 kg/m3 and Ev = 2.15 × 109 Pa, the speed of the pressure wave, c, is given by Equation 2.92 as √ √ Ev /ρ0 (2.15 × 109 )/(998) = = 719 m/s c= 1 + (Ev D/eE) 1 + (2.15 × 109 × 0.050)/(0.0020 × 1.7 × 1010 ) Hence, the pressure increase, ∆p is given by Equation 2.86 as ∆p = ρ0 cV = (998)(719)(4) = 2.87 × 106 Pa = 2870 kPa 2.64. From the given data: Q = 0.06 m3 /s, D = 1000 mm = 1 m, L = 3.2 km = 3200 m, z1 = 10.06 m, z2 = 11.52 m, and ∆zp =1.5 m. For DIP, ks = 0.26 mm, and at 20◦ C, ν = 1.00 × 10−6 m2 /s. (a) The friction factor in the pipe is obtained using the Swamee-Jain equation from the following sequence of calculations: π 2 π 2 D = (1) = 0.7854 m2 4 4 Q 0.6 V = = = 0.764 m/s A 0.7854 VD (0.764)(1) Re = = = 7.64 × 105 ν 1.00 × 106 0.25 0.25 f=[ ( )]2 = [ ( )]2 = 0.01449 ks 5.74 0.26 5.74 log + log + 3.7D Re0.9 3.7(1000) (7.64 × 105 )0.9 A=
The energy equation gives L V2 p2 V 2 z1 + hp − f = + + z2 D 2g γ 2g ( ) 2 L V p2 z 1 + hp − f + 1 = + z2 D 2g γ ( ) 3200 0.7642 350 10.06 + hp − 0.01449 +1 = + 11.52 1 2(9.81) 9.79 which gives hp = 38.62 m. Hence the required power, P , of the pump is given by P = γQhp = (9.79)(0.6)(38.62) = 227 kW (b) At the pump, with subscripts 1 and 2 referring to just before and just after the pump, respectively: p2 V 2 + + z2 γ 2g p2 0.7642 10.06 + 38.62 = + + (10.06 − 1.5) 9.79 2(9.81) z1 + hp =
which gives p2 = 392 kPa . (c) For each of the two pipelines: D = 0.6 m and Q = 0.3 m3 /s and the friction factor in the pipe can be obtained as follows: π 2 π D = (0.6)2 = 0.283 m2 4 4 Q 0.3 V = = = 1.06 m/s A 0.283 VD (1.06)(0.6) = = 6.36 × 105 Re = ν 1.00 × 106 0.25 0.25 f=[ ( )]2 = [ ( )]2 = 0.0171 ks 5.74 0.26 5.74 log + log + 3.7D Re0.9 3.7(600) (6.36 × 105 )0.9 A=
The energy equation, with subscripts 1 and 2 referring to the source reservoir and the location 3.2 km downstream of the source, respectively, gives ( ) 2 L V p2 z 1 + hp − f + 1 = + z2 D 2g γ ( ) 3200 1.062 350 10.06 + hp − 0.0171 +1 = + 11.52 0.6 2(9.81) 9.79 which gives hp = 42.49 m. Hence the required power, P , of the pump is given by P = γQhp = (9.79)(0.6)(42.49) = 250 kW
2.65. From the given data: Q = 30 L/s, ks = 0.05 mm, D1 = 50 mm, ∆z1 = 4 m, L1 = 10 m, D2 = 75 mm, L2 = 10 m, D3 = 85 mm, and L3 = 7 m. The friction factors and pipe areas can be calculated as follows: ( ) 1 ks /D √ = −2 log , f1 = 0.01964, f2 = 0.01733, f3 = 0.01783 3.7 f A=
πD2 , 4
A1 = 1.963 × 10−3 m2 , A2 = 5.675 × 10−3 m2 , A3 = 4.418 × 10−3 m2
The energy equation between A and C gives: ∆z1 − f1
L3 Q23 L1 Q21 − f =0 3 D1 2gA21 D3 2gA23
4 − (0.01964)
Q21 7 (0.030)2 10 − (0.01783) =0 0.050 2(9.807)(1.963 × 10−3 )2 0.086 2(9.807)(4.418 × 10−3 )2
which gives Q1 = 0.00323 m3 /s = 3.23 L/s. By continuity, Q2 = Q3 − Q1 = 30 − 3.23 = 26.77 L/s = 0.02677 m3 /s. The energy equation between B and C gives: h − f2
L2 Q22 L3 Q23 − f =0 3 D2 2gA22 D3 2gA23
h − (0.01733)
(0.02677)2 7 (0.030)2 10 − (0.01783) =0 0.075 2(9.807)(5.675 × 10−3 )2 0.086 2(9.807)(4.418 × 10−3 )2
which gives h = 0.992 m . 2.66. D = 0.85 m, A = πD2 /4 = π(0.85)2 /4 = 0.567 m2 , ks = 0.26 mm (ductile iron), ks /D = 0.26/850 = 0.000306, T = 20◦ C, ρ = 998 kN/m3 , µ = 1.00 × 10−3 N·s/m2 . Assuming fully turbulent flow, then ( ) ( ) ks /D 0.000306 1 √ = −2 log = −2 log = 8.16 3.7 3.7 f which leads to f = 0.0150 for each pipe. Writing the energy equations, 100 − f
LAJ Q2AJ LBJ Q2BJ = 80 − f D 2gA2 D 2gA2
(1)
100 − f
LAJ Q2AJ LCJ Q2CJ = 60 + f D 2gA2 D 2gA2
(2)
QAJ + QBJ = QCJ
(3)
and
and the continuity equation is
Equations 1 and 3 assume that the flow is out of reservoir B. Substituting known values into Equation 1 gives 100 − 0.0150
Q2AJ Q2BJ 900 800 = 80 − 0.0150 0.85 2(9.81)(0.567)2 0.85 2(9.81)(0.567)2
which leads to Q2BJ = 1.12Q2AJ − 8.93
(4)
Substituting known values into Equation 2 gives 100 − 0.0150
Q2AJ Q2CJ 700 900 = 60 + 0.0150 0.85 2(9.81)(0.567)2 0.85 2(9.81)(0.567)2
which leads to Q2CJ = 20.4 − 1.29Q2AJ
(5)
Substituting Equations 4 and 5 into Equation 3 leads to √ √ 2 QAJ + 1.12QAJ − 8.93 − 20.4 − 1.29Q2AJ = 0 Solving for QAJ by trial and error gives QAJ = 2.84 m3 /s and Equations 4 and 5 give QBJ = 0.32 m3 /s QCJ = 3.16 m3 /s Recalculating f for each pipe (using Re and ks /D) and repeating the process until the assumed f is equal to the calculated f leads to QAJ = 2.81 m3 /s QBJ = 0.33 m3 /s QCJ = 3.14 m3 /s
(6)
These results do not differ by much from the initial result that assumes complete turbulence. 2.67. From the given data: LAB = 1.5 km, LBC = 1.5 km, LBD = 1.6 km, k1 = 1 mm, k2 = 0.5 mm, D1 = 150 mm, D2 = 200 mm, and ∆z = 8 m. For water at 20◦ C, ν = 1.00 × 10−6 m2 /s. The following preliminary calculations are useful, A1 =
π(0.15)2 πD12 = = 1.767 × 10−2 m2 , 4 4
k1 1 = = 6.667 × 10−3 , D1 150 ( ) k1 f1 = fCE = 0.0332, D1
A2 =
π(0.2)2 πD22 = = 3.142 × 10−2 m2 4 4
k2 0.5 = = 2.5 × 10−3 D2 200 ( ) k2 f2 = fCE = 0.0249 D2
αAB =
f1 LAB = 5.420 × 104 , 2gA21 D1
αBD =
f2 LBD = 1.028 × 104 2gA22 D2
αBC =
f1 LBC = 5.420 × 104 2gA21 D1
where fCE represents the friction factor given by the Colebrook equation. (a) Under the existing system, the energy equation applied between reservoirs requires that ∆z − 2
f1 LAB Q2AB =0 D1 2gA21
→
∆z − 2αAB Q2AB = 0
→
8 − 2(5.420 × 104 )Q2AB = 0
which yields QAB = 8.591 × 10−3 m3 /s = 8.59 L/s . When the new pipeline is added, the following three equations must be satisfied: QAB = QBC + QBD ∆z −
(1)
f1 LAB Q2AB f1 LBC Q2BC − =0 D1 2gA21 D1 2gA21
2 f1 LBC Q2BC f2 LBD QBD = D1 2gA21 D2 2gA22
(2)
(3)
Equations 1 to 3 can be more compactly expressed as QAB = QBC + QBD
(4)
∆z − αAB Q2AB − αBC Q2BC = 0
(5)
αBC Q2BC = αBD Q2BD
(6)
Combining Equations 4 to 6 gives √ )−1 (√ ∆z − αAB Q2AB αBD QAB − − QAB = 0 αBC αBC
(7)
which yields QAB = 11.62 × 10−3 m3 /s = 11.62 L/s . Substituting this result back into Equations 4 to 6 gives QBC = 3.52 L/s and QBD = 8.10 L/s . Therefore, the flow from the upper to the lower reservoir increases from 8.59 L/s to 11.62 L/s, an increase of approximately 35% . (b) Flow in the existing pipe will change from 8.59 L/s to 3.52 L/s, which is a change of −59% . (c) The friction factors can be determined using the calculated flows and compared to the originally assumed friction factors. The results of this analysis are as follows: Pipe
Assumed f
Recalculated f
Difference (%)
AB BC BD
0.0332 0.0332 0.0249
0.0340 0.0356 0.0275
2.4 7.2 10.4
Based on these results, the assumption of fully turbulent flow in all pipes is not validated and the determination of the flows should be repeated using the updated friction factors. This calculation loop should continue until the calculated and assumed friction factors are the same. 2.68. From the given data: LAB = 3.0 km, LAC = 2.0 km, LAD = 2.5 km, LBC = 1.8 km, DAB = 300 mm, DAC = 200 mm, DAD = 250 mm, DBC = 150 mm, zA = 3.40 m, zB = 2.80 m, zC = 3.20 m, zD = 2.60 m, pA = 450 kPa, pB = 330 kPa, pC = 300 kPa, pD = 360 kPa, and ks = 0.2 mm. For water at 20◦ C, γ = 9.79 kN/m3 , and ν = 1.00 × 10−6 m2 /s. The following preliminary calculations are useful, AAB =
2 πDAB = 0.0707 m2 , 4
AAC =
2 πDAC = 0.0314 m2 4
AAD =
2 πDAD = 0.0491 m2 , 4
ABC =
2 πDBC = 0.0177 m2 4
hA =
pA + zA = 49.37 m, γ
hB =
pB + zB = 36.51 m γ
hC =
pC + zC = 33.84 m, γ
hD =
pD + zD = 39.37 m γ
βAB =
LAB = 1.020 × 105 , 2gDAB A2AB
βAC =
LAC = 5.166 × 105 2gDAC A2AC
βAD =
LAD = 2.116 × 105 , 2gDAD A2AD
βBC =
LBC = 1.959 × 106 2gDBC A2BC
kAB = 6.667 × 10−4 , DAB
kAC = 1.000 × 10−3 DAC
kAD = 8.000 × 10−4 , DAD
kBC = 0.0013 DBC
The following functions of the flow rates can be defined, ReAB (QAB ) =
QAB DAB QAB (0.3) = , ν 1.00 × 10−6
QAD DAD QAD (0.25) = , ν 1.00 × 10−6 [ ] kAB fAB (QAB ) = fCE Re(QAB ), , DAB [ ] kAD fAD (QAD ) = fCE Re(QAD ), , DAD
ReAD (QAD ) =
ReAC (QAC ) =
QAC DAC QAC (0.2) = ν 1.00 × 10−6
QBC DBC QAB (0.15) = ν 1.00 × 10−6 [ ] kAC fAC (QAC ) = fCE Re(QAC ), DAC [ ] kBC fBC (QBC ) = fCE Re(QBC ), DBC ReBC (QBC ) =
where fCE is the value of the friction factor given by the Colebrook equation. (a) For the existing system, applying the energy equation to each of the pipe segments gives hA − βAB fAB (QAB )Q2AB = hB
→
QAB = QA = 0.0816 m3 /s = 81.6 L/s
hA − βAC fAC (QAC )Q2AC = hC
→
QAC = QB = 0.0380 m3 /s = 38.0 L/s
hA − βAD fAD (QAD )Q2AD = hD
→
QAD = QC = 0.0487 m3 /s = 48.7 L/s
(b) For the new system, hB − βBC (QBC )fBC Q2BC = hC
→
QBC = 0.0075 m3 /s = 7.5 L/s
QA = 81.6 L/s − 7.5 L/s = 74.1 L/s QB = 38.0 L/s + 7.5 L/s = 45.5 L/s QC = 48.7 L/s 2.69. For ductile-iron pipe, L Q2 hf = f = D 2gA2
(
L 1 D 2gA2
) Q2 f = rQ2
Assuming fully turbulent flow: Pipe
ks /D
f
AD BC BD AC
0.000650 0.000867 0.000743 0.00104
0.0177 0.0190 0.0183 0.0198
r 143 517 345 1173
From the given data: hA = 25 m, and hB = 20 m. Since the flow directions in the pipe network are not known in advance, flow directions can be assumed and validated. Assumed flow directions are validated when the calculated flows are real and positive. Assuming that the flow in pipe AC is from C to A and that the flow in pipe BC is from C to B (i.e. hA < hC > hB ), the Darcy–Weisbach equation gives hC = 25 + 1173Q2CA
(1)
hC = 20 + 517Q2CB
(2)
→
Q2CA − 0.441Q2CB + 0.00426 = 0
(3)
→
QCA = 0.2 − QCB
(4)
→
0.559Q2CB − 0.4QCB + 0.04426 = 0
(5)
and for flow from C to B, Combining Equations 1 and 2 gives 25 + 1173Q2CA = 20 + 517Q2CB The continuity equation requires that QCA + QCB = 0.2 Combining Equations 3 and 4 gives (0.2 − QCB )2 − 0.441Q2CB + 0.00426 = 0
Equation 5 has two real solutions that are given by QCB = 0.579 m3 /s and
QCB = 0.137 m3 /s
and substituting these results into Equation 4 gives QCA = −0.379 m3 /s
and
QCA = 0.063 m3 /s
Since only real and positive flows are acceptable, QCB = 0.137 m3 /s, QCA = 0.063 m3 /s, and the assumed flow directions are validated. [If invalid flow directions were assumed, then there would be no real and positive simultaneous solutions to the corresponding Darcy–Weisbach and continuity equations.] Assuming that the flow in pipe AD is from A to D and that the flow in pipe BD is from B to D (i.e. hA > hD < hB ), the Darcy–Weisbach equation gives 2 hD = 25 − 143QAD
(6)
hD = 20 − 345Q2BD
(7)
and for the flow from B to D, Combining Equations 6 and 7 gives 25 − 143Q2AD = 20 − 345Q2BD
→
Q2BD − 0.414Q2AD + 0.0145 = 0
(8)
→
QBD = 0.2 − QAD
(9)
→
2 0.586QAD − 0.4QAD + 0.0545 = 0
(10)
The continuity equation requires that QAD + QBD = 0.2 Combining Equations 8 and 9 gives (0.2 − QAD )2 − 0.414Q2AD + 0.0145 = 0
Equation 10 has two real solutions that are given by QAD = 0.495 m3 /s and
QAD = 0.188 m3 /s
and substituting into Equation 9 gives QBD = −0.295 m3 /s
and
QBD = 0.012 m3 /s
Since only real and positive flows are consistent with the assumed flow directions, QAD = 0.188 m3 /s, QBD = 0.012 m3 /s, and the assumed flow directions are validated. Therefore, based on these results, flow out of reservoir at A = QAD − QCA = 0.188 − 0.063 = 0.125 m3 /s flow into reservoir at B = QCB − QBD = 0.137 − 0.012 = 0.125 m3 /s The flows into and out of the reservoirs (0.125 m3 /s) are equal as expected.
2.70. From the given data: zE = zF = zG = 100 m, zA = zB = 0 m, and VAC = VBC = 2.5 m/s. Since pipes AC and BC are identical, they have the same flow. If Q is the flow through the pump, then Q π = VAC AAC = 2.5 (0.5)2 = 0.4909 m3 /s ⇒ Q = 0.9817 m3 /s 2 4 and so VCD = Q/ACD = 2.222 m/s. If hp is the head added by the pump, applying the energy equation gives 2.52 0.0055(100) 2.52 0.0050(300) (2.222)2 +0− − + hp − 2(9.81) 0.50 2(9.81) 0.75 2(9.81) Q2DE Q2DE 0.0060(500) = ]2 ]2 + 100 [ [ 0.30 2(9.81) π (0.3)2 2(9.81) π (0.3)2
0+
4
4
which simplifies to hp = 112.2Q2DE + 100.5
(1)
For lines F and G, Q2DF Q2DF 0.0060(400) + 100 + ]2 ]2 = [ [ 0.25 2(9.81) π4 (0.25)2 2(9.81) π4 (0.25)2 Q2DG Q2DG 0.0060(500) ]2 + 100 + ]2 [π [ 0.30 2(9.81) 4 (0.30)2 2(9.81) π4 (0.30)2 which simplifies to QDF = 0.7074QDG
(2)
Since E and G are identical pipes, then QDE = QDG
(3)
0.9817 = QDE + QDF + QDG
(4)
and by continuity Combining Equations 2 to 4 gives 0.9817 = QDE + 0.7074QDE + QDE which yields QDE = 0.3626 m3 /s. Substituting into Equation 1 gives hp = 112.2(0.3626)2 + 100.5 = 115.5 m Therefore, the pressure difference across the pump is 9.79(115.6) = 1128 kPa and the power consumed by the pump is γQhp /η = (9.79)(0.9817)(115.6)/0.76 = 1461 kW . 2.71. Using the Darcy-Weisbach head loss equation, r=
fL fL fL = = 0.0826 5 2 2 2 2gA D 2(9.81)(πD /4) D D
For ductile-iron, ks = 0.26 mm, and for fully turbulent flow the friction factor, f , is estimated using ) ( 1 ks /D √ = −2 log 3.7 f Using the given values of D and L in Equations 2 and 2 yields Pipe
f
AB BC CD DE EF FA BE
0.0190 0.0186 0.0210 0.0198 0.0190 0.0198 0.0183
L (m) 1000 750 800 700 900 900 950
D (mm) 300 325 200 250 300 250 350
r 646 318 4337 1172 581 1507 273
2.72. Using the Darcy-Weisbach equation to calculate frictional head losses in each pipe, the HardyCross parameters are: ( ) fL 1 r= +K , n=2 D 2gA2 Assuming the flow is fully turbulent, the friction factor, f , is given by both the Colebrook and Swamee-Jain equations as 0.25 f=[ ( )]2 ks log 3.7D Using the given pipe characteristics and the above equations for r and f yields the following results: Pipe AB BC AD CD
f 0.00965 0.01014 0.00965 0.01014
r 51.0 69.7 69.0 105.3
Applying the Hardy Cross method to any assumed flow distribution and using the derived values of r and n (= 2) yields the flow distribution shown in Figure 2.3. Taking QAB = 31.9 L/min = 0.000532 m3 /s, the pressure at B is determined using the energy equations as follows: ) ( ) ( pB VB2 pA VA2 + + zB = + + zA − rAB Q2 γ 2g γ 2g ( ) ( ) pB 340 +2 = + 1 − (51.0)(0.000532)2 γ 9.79 which yields pB = 330 kPa .
60 L/min
A
31.9 L/min
B
28.1 L/min
20 L/min 11.9 L/min
20 L/min D
8.1 L/min
C
20 L/min
Figure 2.3: Flow distribution
Figure 2.4: Final Flows
2.73. The final flows using the Hardy Cross method are shown in Figure 2.4. The results of intermediate calculations will depend on initial flow assumptions. Since the pressure at P is 500 kPa, and the network is on flat terrain, then the head at other intersections can be calculated by accounting for the frictional head losses. For each pipe, the pressure, p, is given by p = 500 ± γhf kPa
(1)
where hf is the frictional head loss, and the ± accounts for whether the flow is towards P (+) or away from P (−). The frictional head loss, hf , is given by the Darcy-Weisbach equation as
hf = f
L Q2 D 2gA2
Since D = 0.3 m and A = πD2 /4 = π(0.3)2 /4 = 0.0707 m2 , then hf = f
Q2 L = 33.99f LQ2 0.3 2(9.81)(0.0707)2
(2)
where L is given for each pipe, Q has been calculated, and f can be calculated using the Swamee-Jain equation ] [ 1 ks /D 5.74 √ = −2 log (3) + 3.7 f Re0.9 where ks /D = 0.26/300 = 0.000867, and Re is Re =
ρV D ρQD (998)(Q)(0.3) = = = 4.23 × 106 Q µ Aµ (0.0707)(1.00 × 10−3 )
(4)
Combining Equations 3 and 4 gives the following expression for f in terms of the flow rate, Q, [ ] 1 0.000867 5.74 √ = −2 log + 3.7 (4.23 × 106 Q)0.9 f which simplifies to 0.25 f={ [ ]}2 −6 log 2.34 × 10−4 + 6.22×10 Q0.9
(5)
Combining Equations 2 and 5 gives the following expression for the frictional head loss in each pipe 8.50LQ2 hf = { [ (6) ]}2 −6 log 2.34 × 10−4 + 6.22×10 0.9 Q The pressure drop, ∆p, in each pipe is equal to γhf , = 9.79hf kPa, and therefore 83.2LQ2 ∆p = { [ ]}2 −6 log 2.34 × 10−4 + 6.22×10 0.9 Q Calculating the pressure changes in each pipe, and adding/subtracting from the pressure at P (= 500 kPa) yields the following results: Pipe PB PF PH PE AB BC FI GH
L (m) 100 150 100 150 150 150 100 150
Q (m3 /s) 0.005 0.008 0.043 0.040 0.040 0.045 0.047 0.030
∆p (kPa) 0.02 0.08 1.28 1.67 1.67 2.10 1.52 0.96
Node B F H E A C I G
Pressure (kPa) 500.0 499.9 501.3 501.7 501.6 497.9 502.8 502.2
2.74. This problem can be solved using the Hardy Cross method. From the given data, ks = 0.05 mm and D = 1120 mm. Hence, 0.25 0.25 ( )]2 = [ ( )]2 = 0.01033 −5 s /D log k3.7 log 4.464×10 3.7 π 2 π A = D = (1.12)2 = 0.9852 m2 4 4 f=[
and the Hardy Cross parameters corresponding to the Darcy-Weisbach equation are n = 2 and L 1 L 1 = (0.01033) = 0.0004843L r=f 2 D 2gA 1.12 2(9.81)(0.9852)2 Using this formulation with the given pipe lengths, the r values for each of the pipes are: CD = 2.42, DE = 5.81, EF = 4.36, FC = 2.91, FG = 3.39, GH = 3.87, and HC = 4.84. The intermediate flow values will depend on the initial assumed flow distribution. The final flow values are shown in Figure 2.5.
48000 m3/d 8000 m3/d 8000 m3/d 13700
11600
C
H
D
3600
14700
5700
G
8000 m3/d
8000 m3/d
8000 m3/d
8000 m3/d
4400
F
2300
E
Figure 2.5: Flow distribution.
2.75. From the given data: ω1 = ω2 = 1200 rpm, D1 = 500 mm, D2 = 250 mm, Q1 = 250 L/s, hp1 = 63.7 m, and η1 = 81%. Applying the affinity relationships given by Equation 2.116 requires that [ ] [ ] Q1 Q2 = (ω1 )(D1 )3 (ω2 )(D2 )3 [ ] [ ] Q2 250 = (1200)(500)3 (1200)(250)3 which yields Q2 = 31 L/s. Also, [
] [ ] hp1 hp2 = (ω1 )2 (D1 )2 (ω2 )2 (D2 )2