- Use orifice equation: Q = K Ao (2g Δh)1/2
Assume d, find K from Fig. 5-21, and calculate Q. Repeat for another d.
First trial: d = 0.60 m
d / D = 0.60 / 1.2 = 0.5
Δh = Δp/γ = (50000 N/m2)/(9810 N/m3)
(2 g Δh)1/2 = (2 × 9.81 × 50000/9810) 1/2 = 10
For T= 15oC: ν = 1.14 × 10-6 m2/s
(2 g Δh)1/2 d/ν = 10 × 0.60 / (1.14 × 10-6) = 5.26 × 106
From Fig. 5-21: K = 0.63
Q = 0.63 (π/4) (0.60)2 × 10 = 1.78 m3/s [Discharge is too low; orifice too small]
Second trial: d = 0.72; d / D = 0.6. Repeat.
Find K = 0.65 and Q = 2.65 m3/s [Still too low]
Third trial: d = 0.84; d / D = 0.7. Repeat.
Find K = 0.70 and Q = 3.87 m3/s [Discharge is too high; orifice too large]
Fourth trial: d = 0.75; d / D = 0.625. Repeat.
Find K = 0.66 and Q = 2.91 m3/s [Discharge is short; orifice too small]
Fifth trial: d = 0.76; d / D = 0.633. Repeat.
Find K = 0.665 and Q = 3.01 m3/s [Discharge is slightly in excess of required discharge]
Thus, the required orifice diameter is: d = 0.76 m.
- Use Venturi meter equation: Q = K A2 (2g Δh)1/2
Solve for Δh: Δh = Q2 / [2g K2 A22]
Reynolds number based on throat size: Red = (4Q) / (πdν) = (4 × 0.6) / (3.1416 × 0.3 × 1.0 × 10-6)
Red = 2.54 × 106
d / D = 0.3 / 0.6 = 0.5
From Fig. 2-51; K = 1.02
Δh = (0.6)2 / [2 × 9.81 × (1.02) 2 × [(π/4) (0.3)2]2] = 3.53 m [H2O]
Specific gravity of Hg = 13.6
Δh [Hg] = Δh [H2O] / (13.6 - 1.00) = 0.28 m.
The manometer deflection is 28 cm.
- Solve for the x- component of the force:
∑Fx = ρ Q (V2x - V1x)
p1A1 - p2A2cos 30o + Fanchor, x =
ρ Q (V2x - V1x)
A1 = A2 = (π/4) D2 = 0.7854 × (2)2 = 3.1416 ft2
V2x = (Q / A2) cos 30o = (30 / 3.1416) × 0.866 = 8.27 fps
V1x = Q / A1 = 30 / 3.1416 = 9.549 fps
p1 = 10 lb/in2 × 144 in2/ft2 = 1440 lb/ft2
p2 = 8 lb/in2 × 144 in2/ft2 = 1152 lb/ft2
ρ = 62.4 lb/ft3 / 32.2 ft/s2 = 1.94 lbs⋅ s2/ft4
(1440 × 3.1416) - (1152 × 3.1416 × 0.866) + Fanchor, x =
1.94 × 30 × (8.27 - 9.549)
F anchor, x = -1464 lbs [to the left, against the flow]
F anchor, y = 0 [no horizontal force perpendicular to x direction]
Solve for the z- component of the force:
∑Fz = ρ Q (V2z - V1z)
V1z = 0
- p2A2sin 30o + Fanchor, z + Wbend + Wwater =
ρ Q V2z
Wbend = - 300 lbs
Wwater = γ [VOLUME OF BEND] = 62.4 × (π/4) D2 × [LENGTH OF BEND] = 62.4 × 3.1416 × 5 = - 980.2 lb
V2z = (Q / A2) sin 30o = (30 / 3.1416) × 0.5 = 4.775 fps
- (1152 × 3.1416 × 0.5) + Fanchor, z - 300 - 980.2 =
1.94 × 30 × (4.775)
F anchor, z = 3368 lbs [upwards, against gravity]
Force exerted by the anchor on the bend:
F = -1464 i + 0 j + 3368 k.
- ks / D = 0.004 ft / 1 ft = 0.004
Assume very high Reynolds number. Estimate f from Moody diagram: f = 0.028.
Velocity head is lost in the free jet: KE = 1
Write the energy equation between the reservoir water surface elevation and [the elevation of the] free jet.
100 - 64 = [f (L/D) + Ke + Kb + KE] [V2/(2g)]
36 = [ 0.028 × (100/1) + 0.12 + 0.35 + 1 ] [V2/(2 × 32.17)]
36 = 4.27 (V2/64.34)
V = 23.29 fps.
V2/(2g) = (23.29)2/(2 × 32.17) = 8.43 ft.
The velocity head is 8.43 ft.
A = (π/4) D2 = (3.1416/4) (1)2 = 0.7854 ft2
Q = 23.29 × 0.7854 = 18.29 cfs.
For T = 60oF : ν = 1.217 × 10-5 ft2/s
Calculate Reynolds number: Re = VD/ν = 23.29 × 1 / (1.217 × 10-5) = 1.9 × 106
From Fig. 5-4: f = 0.028. [Assumed value is OK].
The point of maximum pressure pmax is right above (upstream) of the 90o bend.
Write the energy equation between a point immediately upstream of the bend (max) and a point (2) at the exit.
The velocity heads are the same, so they cancel.
(pmax/γ) + zmax = (p2/γ) + z2 + Σ hL
p2/γ = 0
(pmax/γ) + 44 = 0 + 64 + [f (L/D) + Kb ][V2/(2g)]
(pmax/γ) + 44 = 0 + 64 + [0.028 (28/1) + 0.35](8.43)
pmax/γ = 20 + 9.56 = 29.56
pmax = γ (29.56) = 1845 lb/ft2
Or, alternatively, write the energy equation between the reservoir water surface elevation (1) and a point immediately upstream of the bend (max)
The velocity heads are the same, so they cancel.
(p1/γ) + z1 = (pmax/γ) + zmax + Σ hL
p1/γ = 0
0 + 100 = (pmax/γ) + 44 + [ f (L/D) + Ke + 1] [V2/(2g)]
56 = (pmax/γ) + [ 0.028 (72/1) + 0.12 + 1] [8.43]
56 = (pmax/γ) + 26.43
pmax/γ = 29.57 ft.
pmax = 1845 lb/ft2
Immediately upstream of the pipe entrance, the pressure head is 5 ft and the velocity head is 0.
Immediately downstream of the pipe entrance, the velocity head is 8.43 ft, and the drop in pressure head is 8.43(1 + 0.12)
Therefore, the point of minimum pressure is immediately downstream of the pipe entrance.
pmin = 62.4 × [5 - 8.43(1 + 0.12)] = -277 lb/ft2.