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FLUID MECHANICS I Solution Manual 9 Question 1: Problem P5.71 z V

x

D

d

h

d ∆p = ∆p (ρ, V, ) D µ ¶ ∆p d ⇒ Cp = = C p 2 1/2ρV D model Cpm =

∆pm 5000 = 0.626 = 2 1/2 ρm Vm 0.5x998x42

geometric similarity For

dm Dm

=

dp Dp

then Cpp = Cpm

1

prototype Cpp =

∆pp = 1/2ρp Vp2

∆pp ³ 1/2 ρp

Qp π 2 4 Dp

´2 = 0.626

For best accuracy design system so that ∆pp = 15 kPa. "

8 Cpp ρp Q2p Dp = ∆pp π 2

#1/4

Dp = 0.151 m using ρp = 680 kg/m3 (Table A.3).

Question 2: Problem P5.75 Prototype Vp = 240 m/s Tp = 223 K Pp = 26.4 kP a ρp = 0.4125 kg/m3 ap = 299.5 m/s using Table A.6 (standard atmosphere). Model Lm 1 = Lp 12 Tm = 293 K Pm = unknown Vm = unknown speed of sound for model r √ cp am = Rm Tm = 1.40x287x293 = 343 m/s cv using Table A.4. Mach number for prototype M ap =

Vp = 0.801 ap

Mach number for model M am =

Vm = M ap ⇒ Vm = 275 m/s am 2

Viscosity (Assume only function of temperature, see Table A.2) µ µp = 1.71 10−5

Tm 273

µm = 1.8 10−5 N.s.m−2

¶0.7

= 1.48 10−5 N.s.m−2 (power law-Table A.2)

Reynolds number ρ p Vp L p ρ m Vm L m = Rem = µp µm Vp L p µ m = 5.25 kg.m−3 = ρp Vm Lm µp

Rep = ⇒ ρm Pressure pm

pm = ρm R Tm = 4.42x10 Pa = 4.37 atm 5

Question 3: Problem P6.79 z 1 Kv = f Le/d Le = 200 d

H= 3 m d=0.015 m 2 L = 10 m

Ke = 0.5

Bernoulli equation between 1 and 2 V12 p1 V2 p2 + + gz1 = 2 + + gz2 + ghf 1−2 2 ρ 2 ρ Losses V2 V2 LV2 + Kv +f 2 d¶ 2¸ · µ2 Le L V2 Ke + + f = 2 d d

ghf 12 = Ke ghf 1−2 Flow rate

π d2 Q˙ = V 4 ⇒ V = 1.13 m/s 3

Q = 2.010−4 m3 /s

x

with V1 ≈ 0 (large tank), p1 = p2 = patm and z2 = 0, · µ ¶ ¸ V2 Le L 1 + Ke + + f = gz1 2 d d £ 2gz1

¤ − 1 − Ke ¡ Le ¢ = 0.051 L d + d

V2

⇒f =

Using the Moody Diagram with f = 0.051 and Re =

ρV d µ

= 1.7x104 ,

² = 0.02 d ² = 0.3 mm

Question 4: Problem P6.103 z

1 H = 13.7 m 2 CV

water

x

LA = 6.10 m dA = 0.0254 m LB = 6.10 m dB = 0.508 m Assumptions: • steady flow • steady-state • 1D velocity • large reservoirs • fully developed flow Conservation of mass for CV around expansion: 0−ρ

πd2A πd2 VA + ρ B VB 4 4 d2 VB = VA 2A dB 4

Bernoulli equation from 1 to 2 V12 p1 V2 p2 + gz1 + = 2 + gz2 + + ghf 1−2 2 ρ 2 ρ using z2 =0, p1 = p2 = patm and V1 = V2 ≈ 0 · ¸ · ¸ V2 V2 LA LB ghf 1−2 = A Kent + fA + Kexp + B fB + Kexit = gH 2 dA 2 dB Kent = 0.5 sharp edged entrance (see Fig. 6.21b). Kexp = (1 − d2A /d2B )2 = 0.56 using Eq. 6.80. Kexit = 1 using Eq.6.80 with d/D → 0. ²A = ²B = 0.26 mm (Table 6.1). ²A /dA = 0.01 and ²B /dB = 0.005. v u 2gH u oi n VA = t h d4 LA Kent + Kexp + fA dA + d4A fB LdBB + Kexit B r 269 ⇒ VA = 1.12 + 240fA + 7.5fB Reynolds numbers ρVA dA = 2.5x104 VA µ ρVA dB d2A = . 2 = 1.27x104 VA µ dB ReA =

ReB =

ρVB dB µ

Iterative process to calculate the friction factors. Iteration 0 1

ReA ∝ 1.3 105

ReB ∝ 6.5 104

fA ( Moody diagram) 0.038 0.038 VA = 5.1 m/s

Q=

VA π 4

d2A

= 2.6x10−3 m3 /s

5

fB (Moody diagram) 0.03 0.031

VA (m/s 5.1 5.1

question 5: Problem 6.108 2

z

Kelbow Kexit Kv Kf 1 z=0 p1 =44.8 kPag

Kelbow d= 0.102 m (Table 6.2) ǫ = 0.046 mm (Table 6.1)

L = 24.4 m

Kv = 2.8 (Fig. 6.18b) Kelbow = 0.64 (Table 6.5) Kexit = 1.0 (exit loss Fig. 6.22) Q = 1.13x10−2 m3 /s ⇒ V = 1.39 m/s. Re =

ρV d = 1.4 105 µ

using water at 20◦ C (Table A.3). ²/d = 0.00045. Using the Moody diagram f = 0.0193 (Fig. 6.13). Bernoulli equation between 1 and 2 V12 p1 V2 p2 + gz1 + = 2 + gz2 + + ghf 1−2 2 ρ 2 ρ Losses ghf 1−2 = Kv

V2 V2 V2 V2 LV2 + Kf + 2 Kelbow + Kexit +f 2 2 2 2 d 2

using z1 = 0, V2 ≈ 0 (large reservoir) and p2 = patm , ¸ · V2 f L p1 Kv + 2Kelbow + Kexit + − 1 + Kf = − gH 2 d ρ Kf =

· ¸ 2 P1 L − gH + 1 − Kv − 2Kelbow − Kexit − f V2 ρ d Kf = 9.9 6

x

Second part Kf = 7.0 and Kv = 0.0 (see Fig. 6.18b) In this case V (and f) are unknown). From Bernoulli equation, h

 V =

2

p1 ρ

i − gH

Kv + 2Kelbow + Kf + Kexit + f Ld − 1 r V =

36 8.28 + 239f

Reynolds number Re = 1.02 105 V iterative process: Start by guessing a value for f. iteration 0 1 2

Re / 1.75x105 1.70x105

f 0.0163 0.0195 0.0195

Q=V

V (m/s) 1.72 1.67 1.67

πd2 = 1.36x10−2 m3 /s 4

7

1/2 

question 6: Problem 6.110 z

p1 =patm 1

Kv = 4.0 (Fig. 6.18b) H=40m

Kent=0.5 (Fig. 6.21) x

2 p2 = patm

Turbine Assumptions • fully developed flow • steady flow • steady state • large reservoir Bernoulli equation between 1 and 2 (multiplied by the mass flow rate) in W · 2 ¸ · 2 µ ¶¸ V1 p1 V2 p2 V2 L m ˙ + + gz1 = m ˙ + + gz2 + 2 Kent + f + Kv + P owerturbine 2 ρ 2 ρ 2 d Using z1 = H, z2 = 0, V1 ≈ 0, p1 = p2 = patm and V2 = V, · µ ¶¸ V2 L P owerturbine = m ˙ gH − 2 1 + Kent + f + Kv 2 d Use Table A.3 (20◦ C) Q˙ = 2.04 m/s πd2 /4 ρV d Re = = 1.02x105 µ ² ² = 0.005 ⇒ f (Re, ) = 0.031 d d m ˙ = ρ Q = 3.99 kg/s V =

Powerturbine = 877 W Notice that no losses are assumed in the turbine; typical efficiencies are 85%95%. 8

question 7: Problem 6.107 z p1 =patm 1 galvanised iron ǫ=0.15 mm

L=2 d = 0.05 m Kv

H=5 2 p2 = patm

∆ = 0.06 m Bernoulli equation between 1 and 2

V12 p1 V2 p2 + + gz1 = 2 + + gz2 + ghf 12 2 ρ 2 ρ Using V1 ≈ 0, p1 = p2 = patm , z2 = 0 and V = V2 , µ ¶ V2 L ghf 12 = Kent + Kv + f 2 d s V =

2gH 1 + Kent + Kv + f Ld

Some iteration are required to determine f. Use f = 0.0255 as initial guess. Summary of results Case base a b

Kent 1.0 0.5 1.0

Kv 80 80 0.25

f 0.028 0.028 0.026

Re 5.4x104 5.4x104 3.0x105

V (m/s) 1.09 1.09 5.46

% increase 0 0 +444

For Kent in base use Fig. 6.21a, in a) use Fig.6.21b. For Kv in base and a) use Fig. 6.19 (30◦ ), in b) use Fig.6.19 (90◦ ).

9

x