Assume that the voltage applied to a load is V = 208-30° V and the current flowing through the load is I = 515° A. (a) Calcula
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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