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Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
Answer:
Ammeter
Explanation:
Instrument for measuring either direct or alternating electric current, in amperes. Ammeters vary in their operating principles and accuracies
Answer:
a) the flow under full capacity is q₂= 1.334 ft³/s
b) the velocity would be v= 3.793 ft/s
Explanation:
a) Since the pipe has 8 inches in diameter but 4 are covered with water flow ( half of a circle in area=A₁) , q₁=0.662 ft³/s then
q₁=A₁*v
then for the same velocity v but area A₂=2*A₁
flow under full capacity= q₂ = A₂*v= 2*A₁*v= 2*q₁=2*0.662 ft³/s= 1.334 ft³/s
b) when flowing at a depth of 4 inches
A₁= (1/2)*(π*D²/4) = π* (1/8)*(8 in)² = 8π in² * (1 ft²/ 144 in²) = π/18 ft² = 0.1745 ft²
then
v=q₁/A₁ = 0.662 ft³/s/0.1745 ft²= 3.793 ft/s
v= 3.793 ft/s
