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FLUID MECHANICS I Solution Manual 7 Question 1: Problem P3.158 z

11

1

x

2 CO2

D2 = 6 cm

D1 = 10 cm

p1 = 170 kPa T1 = 20 Deg. C

H=8 cm

Meriam red oil

Assumptions: • ρ is constant ? -needs to be checked. • 1D flow at 11, 1 and 3 • steady-state • Steady flow • neglect losses Conservation of mass for a CV: Z Z ∂ρ ~. n dV ol + ρ(V ˆ ) dA = 0 CV ∂t CS 0 = −ρ

π D22 π D12 V11 + ρ V2 4 4 0 = −m ˙ 1+m ˙2 1

Using the perfect gas law ρ=

p1 = 3.07 kg/m3 RT1

using RCO2 = 189 m2 /s2 K (see Table A.4). Manometer p1 = p2 + ρwater SG gH ⇒ p2 = 169.4kPa Bernoulli equation from 1 to 2 V12 p1 V2 p2 + gz1 + = 2 + gz3 + + ghf 2 ρ 2 ρ V1 ≈ 0 this is the stagnation point z1 = z2 and there is no loss, therefore s 2(p1 − p2 ) V2 = ρ V2 = 20.5m/s ¿ Speed of sound, incompressibility OK. flow rate πD22 Q˙ = V2 = 0.058m3 /s 4

Question 2: Problem P3.165 Assumptions: • steady flow • steady state • incompressible • no losses • 1D flow at Points 1 and 2 Conservation of mass: Z CV

∂ρ dV ol + ∂t

Z ~. n ρ(V ˆ ) dA = 0 CS

0 − m˙ 1 + m˙ 2 = 0 m˙ 1 = m˙ 2 D2 ⇒ V1 = V2 22 D1 2

z

1

2

D2

D1

h

flow rate Q = V2

πD22 = V2 A2 4

Manometer p1 = p2 + (ρm − ρ)gh p1 − p2 = (ρm − ρ)gh Bernoulli equation from 1 to 2 V12 p1 V2 p2 + gz1 + = 2 + gz2 + + ghf 2 ρ 2 ρ z1 = z2 (horizontal) and no losses V22 2

µ

D2 D1

¶4

p1 V2 p2 = 2 + ρ 2 ρ v u 2(p1 −p2 ) u ρ V2 = t 2 4 (1 − (D D1 ) ) +

v u u Q = A2 t

2(ρm −ρ)gh ρ (D2 4 (1 − D1 ) )

3

x

Question 3: Problem P3.172 s D2 = 0.075 m

H= 1.83 cm D1 = 0.025 m D3 = ?

1

2

3

x

patm = 1.01 105 Pa

water at 35 Deg. C sketch of p variation

p

patm pv x Assumptions • pseudo-steady state in tank • steady flow • cavitation occurs at pv • pv = 5.809 103 Pa (linear interpolation in Table A.5) Conservation of mass: ρ

πD12 πD32 πD22 V1 = ρ V3 = ρ V2 4 4 4 D12 V1 = D32 V3 = D22 V2 µ ¶2 D3 ⇒ V1 = V3 D1 4

Bernoulli equation from s to 3 Vs2 ps V2 p3 + gzs + = 3 + gz3 + + ghf s−3 2 ρ 2 ρ Vs ≈ 0, zs = H, ps = patm = p3 , z3 = 0 and no losses p ⇒ V3 = 2gH = 6.0m/s Bernoulli equation from 1 to 3 V12 p1 V2 p3 + gz1 + = 3 + gz3 + + ghf 1−3 2 ρ 2 ρ Using z1 = z3 , p3 = patm , p1 = pv and no losses, µ 2 ¶2 2 V3 pv V2 patm D3 + = 3 + 2 D1 2 ρ 2 ρ Solve for D3 D34

(patm − pv )/ρ +

=

V32 2

V32 2D14

D3 = 0.040m To avoid cavitation we need to increase p1 (pv in above relationship) which means that D3 must be decreased, i.e. D3 < 0.040m

Question 4: Problem P3.161 Assumptions: • steady flow • steady-state • 1D exit flow at 1 and 2 • az ≈ 0 for fluid in tube at point of motion • no losses (but once flow starts up in tube there will be mixing losses. Conservation of mass: 0−ρ

πD12 πD22 V1 + ρ V2 4 4 D2 V2 = V1 12 D2 5

z 2 1 water

V2

V1

D2 p2 = patm

h

water tube patm = p1 + ρgh Bernoulli equation from 1 to 2 V12 p1 V2 p2 + gz1 + = 3 + gz2 + + ghf 2−3 2 ρ 2 ρ sing z1 = z2 and no losses, µ ¶4 V12 patm − ρgh V12 D1 patm + = + 2 ρ 2 D2 ρ Ã ! µ ¶ 4 D1 V12 1− = gh 2 D2 s V1 =

2gh (1 − D41 /D42 )

6

x

Question 4: P3.173 y

Q2 p2 D2

Q1

D1

Fx

x

p1 Fy

D3 p3 Q3 Assumptions: • steady flow • steady-state • 1D exit flow at 1, 2 and 3 • no losses • no shear forces on CV Mass flow rates: Vi =

˙

Qi πDi2 4

m ˙ i = ρQ˙ i

7

i 1 2 3

Q˙ i (m3 /s) 0.142 0.071 0.071

Vi (m/s) 7.8 15.2 8.7

m ˙ i (kg/s) 142 71 71

Ai (m2 ) 0.0181 0.0045 0.0082

conservation of momentum x momentum 0−m ˙ 1 V1 + m ˙ 2 V2 sinθ2 + m ˙ 3 V3 sinθ3 = +p1

πD12 πD22 πD32 − p2 sinθ2 − p3 sinθ3 + Fx 4 4 4

Bernoulli equation from 1 to 2 V12 p1 V2 p2 + gz1 + = 3 + gz2 + + ghf 2−3 2 ρ 2 ρ Using z1 = z2 and no losses, ρ p2 = p1 + (V12 − V22 ) 2 p2 = 87.1 103 Pa(gage) Bernoulli equation from 1 to 3 (similar to previous equations) ρ p3 = p1 + (V12 − V32 ) 2 p2 = 165.0 103 Pa(gage) back to x momentum πD12 πD32 πD12 sinθ2 + p3 sinθ3 − p1 4 4 4 Fx = 540 + 473 − 1108 + 196 + 1036 − 3113

Fx = m ˙ 2 V2 sinθ2 + m ˙ 3 V3 sinθ3 − m ˙ 1 V1 + p2

Fx = −1976 N Fx acts to the left. y momentum πD32 πD22 cosθ2 + p3 cosθ3 + Fy 4 4 πD22 πD32 Fy = m ˙ 2 V2 cosθ2 − m ˙ 3 V3 cosθ3 + p2 cosθ2 − p3 cosθ3 4 4 Fy = 935 − 397 + 339 − 870

0+m ˙ 1 (0) + m ˙ 2 V2 cosθ2 − m ˙ 3 V3 cosθ3 = 0 − p2

Fy = 7.3 N Fy acts to the top. 8

patm = 101 kPa z

4

CVA

CVB

x alcohol SG = 0.79

F=425 N

1

2

D1 = 0.05 m

D2 = 0.02 m

Question 6: P3.149 Assumptions: • steady flow • steady-state • 1D exit flow at 1-4 • no losses • no shear forces on CVA conservation of mass for CVA 0−m ˙ 2+m ˙ 3+m ˙4 x momentum for CVA 0−m ˙ 2 V2 = −Fx µ ¶ m ˙2 Fx = m ˙ 2 V2 = m ˙2 ρA2 ρ = rhowater SG = 788kg/m3 πD22 A2 = = 3.1 10−4 m2 4 m ˙2=

p ρA2 Fx = 10.3 kg/s 9

3

⇒ V2 = 42 m/s conservation of mass for CVB 0−m ˙ 1+m ˙2 ⇒m ˙1=m ˙2 V1 =

m ˙1 = 6.7 m/s ρA1

Bernoulli equation from 1 to 2 V12 p1 V2 p2 + gz1 + = 2 + gz2 + + ghf 1−2 2 ρ 2 ρ Using z1 = z2 , p2 = patm and no losses, ρ p1 = patm + (V22 − V12 ) 2 p1 = 785 kPa

10