Mecanica De Fluidos
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Publicado: 24 de abril de 2012
CHAPTER EIGHT REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION
8.1
υ=
Q 0.20 ft 3 /s 144 in 2 = × = 2.29 ft/s; D = 4.0 in(1 ft/12 in) = 0.333 ft A π (4.0 in ) 2 / 4 ft 2
NR =
υ Dρ (2.29)(0.333)(1.26)(1.94) = = 249 Laminar η 7.5 × 10−3
from App. D
8.2
Let NR = 4000 = υD/v: v = 4.38 × 10−6 ft2/s—App. A; D = (2/12)ft N ν 4000(4.38 × 10−6 ) ft0.3048 m m = 0.105 × = 0.03204 υ= R = D 2 /12 s ft s Let NR = 2000 = υDρ/η N η 2000(4.0 × 10 −2 ) υmax = R = = 0.894 m/s Dρ (0.10)(0.895)(1000) Q = Aυ =
8.3
π (0.10 m ) 2
4
× 0.894 m/s = 7.02 × 10−3 m3/s
8.4
υ = Q/A =
0.25 ft 3 /s = 10.72 ft/s; D = 0.1723 ft 0.02333 ft 2
a) b) c) d)
υ D (10.72)(0.1723) = = 1.53 × 105 (v from App. A) −5 ν 1.21 × 10 υ Dρ (10.72)(0.1723)(1.53) NR= = = 4.28 × 105 (ρ, η from App. B) η 6.60 × 10 −6 υ Dρ (10.72)(0.1723)(1.86) NR = = 253 (ρ, η from App. B) = η 1.36 × 10 −2 υ Dρ (10.72)(0.1723)(0.87)(1.94) NR = = = 3.28 × 104 (η from App. D) η 9.5 × 10 −5
NR =
8.5
NR =
4 1 m3 /s = 6.667 × 10−5 m3/s: Let NR = 2000 Q = 4.0 L/min × 60000 L/min
υ Dρ QDρ QDρ 4Qρ 4Qρ 4Q = = = = : D min = 2 πD η πηD πηN R π N Rν Aη η
94
Chapter 8 a)
Dmin =
4(6.667 × 10−5 ) 4.244 × 10−8 4.244 × 10−8 = 0.0647 m = 64.7 mm = = π (2000)(ν ) ν 6.56 × 10−7 3-in Type K copper tube—D = 73.8 mm
b)
Dmin =
4.244 × 10 −8(680) = 0.101 m; 5-in tube, D = 122 mm 2.87 × 10−4 4.244 × 10 −8(790) 3 = 0.0186 m; -in tube , D = 18.9 mm −3 1.8 × 10 4 4.244 × 10−8 (906) 1 = 3.59 × 10−4 m; -in tube , D = 4.57 mm −1 8 1.07 × 10 Smallest listed
c)Dmin =
d)
Dmin =
8.6
NR =
υ Dρ υ Dρ (2.97)(0.0779)(890) = 4.12 × 10−3 Pa⋅s : η= = 4 η NR 5 × 10
Q 8.50 L/min 1 m3 /s × = 2.97 m/s = A 4.768 × 10−3 m 2 60000 L/min From App. D, oil must be heated to 100°C for SAE 10 oil.
υ=
8.7 At 212°F At 104°F
Auto. Hydraulic Oil Medium Hydraulic Oil υD (10)(0.4011) 10(0.4011) = = 5.11 × 104 turb. NR = = 5.11 × 104 turb. NR = −5 ν7.85 × 10 7.85 × 10−5 (10)(0.4011) 10(0.4011) NR = = 9328 turb. NR = = 5563 turb. −4 4.30 × 10 7.21 × 10−4
8.8
NR =
υ D (3.06)(0.0475) = = 1.12 × 105 Turbulent −6 ν 1.30 × 10
υ=
Q 325 L/min 1 m3 /s × = 3.06 m/s = A 1.772 × 10−3 m 2 60000 L/min
8.9
NR =
υ Dρ (0.899)(0.0243)(860) = = 4.76 × 104 Turbulent η 3.95 × 10 −4
υ=
Q 25 L/min 1 m3 /s = × = 0.899 m/s A 4.636 × 10−4m 2 60000 L/min
8.10
NR =
υ D (1.78)(0.0134) = = 6.62 × 104 Turbulent ν 3.60 × 10−7
υ=
Q 15.0 L/min 1 m3 /s = × = 1.78 m/s A 1.407 × 10−4 m 2 60000 L/min
REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION
95
8.11
NR =
υ D (8.59)(1.563) = = 9.59 × 105 ν 1.40 × 10−5
υ= υ=
Q 16.5 ft 3 /s = = 8.59 ft/s A 1.920 ft 2
8.12
3 Q 0.40gal 1 hr 1 ft = × × × = 0.732 ft/s A hr 7.48 gal 3600 s 2.029 × 10−5 ft 2 υ Dρ (0.732)(0.00508)(0.88)(1.94) = = 1.02 Laminar NR = η 6.2 × 10 −3
8.13
NR =
υ Dρ (0.732)(0.00508)(0.88)(1.94) = = 33.4 Laminar η 1.90 × 10 −4 υ Dρ N η (4000)(4.01 × 10−5 ) = 0.424 ft/s : υ= R = η Dρ (0.2423)(1.56)
Note: sg of oil may be slightly lower at 160°F. 8.14 NR =
Q = Aυ = 4.609 × 10−2 ft2 × 0.424ft/s = 1.96 × 10−2 ft3/s 8.15
Q 45 L/min 1 m3 /s = × = 2.67 m/s A 2.812 × 10−4 m 2 60000 L/min υ Dρ (2.67)(0.01892)(0.89)(1000) NR = = = 5.61 × 103 Turbulent −3 η 8 × 10 Note: η from App. D.
υ=
8.16
NR =
υ Dρ (2.67)(0.01892)(890) = = 15.0 very low—Laminar η 3.0
8.17
Q 45 L/min 1 m3 /s = × = 0.423 m/s A 1.772 × 10−3 m 2 60000 L/min υ Dρ (0.423)(0.0475)(890) NR = = = 2237 CriticalZone η 8 × 10−3
υ=
8.18
NR =
υ Dρ (0.423)(0.0475)(890) = = 5.97 very low—Laminar η 3.0
8.19
υ=
Q 1.65 gal/min 1 ft 3 /s = × = 14.65 ft/s A 2.509 × 10−4 ft 2 449 gal/min υ D (14.65)(0.01788) = = 1105 Laminar NR = ν 2.37 × 10−4
8.20
NR =
υ D (14.65)(0.01788) = = 6237 Turbulent ν 4.20 × 10−5
96
Chapter 8
Changing from laminar flow, through critical zone, into...
8.1
υ=
Q 0.20 ft 3 /s 144 in 2 = × = 2.29 ft/s; D = 4.0 in(1 ft/12 in) = 0.333 ft A π (4.0 in ) 2 / 4 ft 2
NR =
υ Dρ (2.29)(0.333)(1.26)(1.94) = = 249 Laminar η 7.5 × 10−3
from App. D
8.2
Let NR = 4000 = υD/v: v = 4.38 × 10−6 ft2/s—App. A; D = (2/12)ft N ν 4000(4.38 × 10−6 ) ft0.3048 m m = 0.105 × = 0.03204 υ= R = D 2 /12 s ft s Let NR = 2000 = υDρ/η N η 2000(4.0 × 10 −2 ) υmax = R = = 0.894 m/s Dρ (0.10)(0.895)(1000) Q = Aυ =
8.3
π (0.10 m ) 2
4
× 0.894 m/s = 7.02 × 10−3 m3/s
8.4
υ = Q/A =
0.25 ft 3 /s = 10.72 ft/s; D = 0.1723 ft 0.02333 ft 2
a) b) c) d)
υ D (10.72)(0.1723) = = 1.53 × 105 (v from App. A) −5 ν 1.21 × 10 υ Dρ (10.72)(0.1723)(1.53) NR= = = 4.28 × 105 (ρ, η from App. B) η 6.60 × 10 −6 υ Dρ (10.72)(0.1723)(1.86) NR = = 253 (ρ, η from App. B) = η 1.36 × 10 −2 υ Dρ (10.72)(0.1723)(0.87)(1.94) NR = = = 3.28 × 104 (η from App. D) η 9.5 × 10 −5
NR =
8.5
NR =
4 1 m3 /s = 6.667 × 10−5 m3/s: Let NR = 2000 Q = 4.0 L/min × 60000 L/min
υ Dρ QDρ QDρ 4Qρ 4Qρ 4Q = = = = : D min = 2 πD η πηD πηN R π N Rν Aη η
94
Chapter 8 a)
Dmin =
4(6.667 × 10−5 ) 4.244 × 10−8 4.244 × 10−8 = 0.0647 m = 64.7 mm = = π (2000)(ν ) ν 6.56 × 10−7 3-in Type K copper tube—D = 73.8 mm
b)
Dmin =
4.244 × 10 −8(680) = 0.101 m; 5-in tube, D = 122 mm 2.87 × 10−4 4.244 × 10 −8(790) 3 = 0.0186 m; -in tube , D = 18.9 mm −3 1.8 × 10 4 4.244 × 10−8 (906) 1 = 3.59 × 10−4 m; -in tube , D = 4.57 mm −1 8 1.07 × 10 Smallest listed
c)Dmin =
d)
Dmin =
8.6
NR =
υ Dρ υ Dρ (2.97)(0.0779)(890) = 4.12 × 10−3 Pa⋅s : η= = 4 η NR 5 × 10
Q 8.50 L/min 1 m3 /s × = 2.97 m/s = A 4.768 × 10−3 m 2 60000 L/min From App. D, oil must be heated to 100°C for SAE 10 oil.
υ=
8.7 At 212°F At 104°F
Auto. Hydraulic Oil Medium Hydraulic Oil υD (10)(0.4011) 10(0.4011) = = 5.11 × 104 turb. NR = = 5.11 × 104 turb. NR = −5 ν7.85 × 10 7.85 × 10−5 (10)(0.4011) 10(0.4011) NR = = 9328 turb. NR = = 5563 turb. −4 4.30 × 10 7.21 × 10−4
8.8
NR =
υ D (3.06)(0.0475) = = 1.12 × 105 Turbulent −6 ν 1.30 × 10
υ=
Q 325 L/min 1 m3 /s × = 3.06 m/s = A 1.772 × 10−3 m 2 60000 L/min
8.9
NR =
υ Dρ (0.899)(0.0243)(860) = = 4.76 × 104 Turbulent η 3.95 × 10 −4
υ=
Q 25 L/min 1 m3 /s = × = 0.899 m/s A 4.636 × 10−4m 2 60000 L/min
8.10
NR =
υ D (1.78)(0.0134) = = 6.62 × 104 Turbulent ν 3.60 × 10−7
υ=
Q 15.0 L/min 1 m3 /s = × = 1.78 m/s A 1.407 × 10−4 m 2 60000 L/min
REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION
95
8.11
NR =
υ D (8.59)(1.563) = = 9.59 × 105 ν 1.40 × 10−5
υ= υ=
Q 16.5 ft 3 /s = = 8.59 ft/s A 1.920 ft 2
8.12
3 Q 0.40gal 1 hr 1 ft = × × × = 0.732 ft/s A hr 7.48 gal 3600 s 2.029 × 10−5 ft 2 υ Dρ (0.732)(0.00508)(0.88)(1.94) = = 1.02 Laminar NR = η 6.2 × 10 −3
8.13
NR =
υ Dρ (0.732)(0.00508)(0.88)(1.94) = = 33.4 Laminar η 1.90 × 10 −4 υ Dρ N η (4000)(4.01 × 10−5 ) = 0.424 ft/s : υ= R = η Dρ (0.2423)(1.56)
Note: sg of oil may be slightly lower at 160°F. 8.14 NR =
Q = Aυ = 4.609 × 10−2 ft2 × 0.424ft/s = 1.96 × 10−2 ft3/s 8.15
Q 45 L/min 1 m3 /s = × = 2.67 m/s A 2.812 × 10−4 m 2 60000 L/min υ Dρ (2.67)(0.01892)(0.89)(1000) NR = = = 5.61 × 103 Turbulent −3 η 8 × 10 Note: η from App. D.
υ=
8.16
NR =
υ Dρ (2.67)(0.01892)(890) = = 15.0 very low—Laminar η 3.0
8.17
Q 45 L/min 1 m3 /s = × = 0.423 m/s A 1.772 × 10−3 m 2 60000 L/min υ Dρ (0.423)(0.0475)(890) NR = = = 2237 CriticalZone η 8 × 10−3
υ=
8.18
NR =
υ Dρ (0.423)(0.0475)(890) = = 5.97 very low—Laminar η 3.0
8.19
υ=
Q 1.65 gal/min 1 ft 3 /s = × = 14.65 ft/s A 2.509 × 10−4 ft 2 449 gal/min υ D (14.65)(0.01788) = = 1105 Laminar NR = ν 2.37 × 10−4
8.20
NR =
υ D (14.65)(0.01788) = = 6237 Turbulent ν 4.20 × 10−5
96
Chapter 8
Changing from laminar flow, through critical zone, into...
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