You are stopped at a red traffic light and are first in line at the intersection. When the traffic light changes to green, you s
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Answer:
umax = 0.1259ft/s
Explanation:
Given:
•Distance between plates, B = 0.01ft
•Pressure difference decrease,
•Fluid viscosity, u = 10^-³lbf-s/ft²
Specific gravity, S = 0.80
Max velocity in the z-direction will be:
Substituting for h in the first equation, we have:
= -0.20192
Substituting dh/dz value in the first equation (umax), we have:
umax = 0.1259ft/s
Answer:
Explanation:
The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.
The answer to the above question is
The pressure at point 2 = 75.959 kPa
Explanation:
Bernoulli's equation with losses gives
hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)
Between points 1 and 2, z₁ = z₃ + 0.6 m therefore
hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
But v = Q/A
or since A = π×D²/4 we have
A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²
Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows
ε = Which gives
the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175
Substituting he above values into the equation we get
= 19.761 m
Combined head loss = 19.761 m
Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)
or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃
Where ρ = density of water, we have
260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa
Answer:
v = 21.409 m/s
Explanation:
Given data:
Total weight of the platform and weight together, W = 900 N
Diameter of the water jet = 5 cm = 0.05 m
Now,
the force exerted by the water jet is balancing the platform
also,
Force exerted by the water jet = ρAv²
where,
ρ is the density of the water = 1000 kg/m³
A is the area of the outlet of the jet =
or
A = 0.00196 m²
v is the velocity of the jet
thus,
W = ρAv²
or
900 = 1000 × 0.00196 × v²
or
v = 21.409 m/s
Hence, the velocity of the jet is 21.409 m/s