Answer:
See attached picture.
Explanation:
Problem 2. A flat, circular hydrostatic air bearing has an outer diameter of 160 mm and a 5-mm-deep recess from the 50-mm diamet
er to the bearing center. The air pressure in the recess is 2 MPa, and the pressure around the outside of the bearing is 0.1 MPa. Determine the pressure distribution in the bearing if 0.3 kg of air is pumped through the bearing per second.
1 answer:
Answer:
Answer for the question :
"Problem 2. A flat, circular hydrostatic air bearing has an outer diameter of 160 mm and a 5-mm-deep recess from the 50-mm diameter to the bearing center. The air pressure in the recess is 2 MPa, and the pressure around the outside of the bearing is 0.1 MPa. Determine the pressure distribution in the bearing if 0.3 kg of air is pumped through the bearing per second."
is explained in the attachment.
Explanation:
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Answer:
Explanation:
By an adequate application of the Principle of Energy Conservation, the escalator need energy to elevate from to the bottom to the top. Hence:
An expression for power needed is found by deriving the equation with respect to time:
The minimum power is found by substituting known inputs:
Answer:
A. 288,030.91 cy
B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water
Explanation:
The natural material in the barrow properties are;
The mass unit weight, γ = 110.0 pcf
The water content, w = 6%
The specific gravity of the soil solids, = 2.63
The desired dry unit weight, = 122 pcf
The water content, w₁ = 5.5 %
The net section volume, = 245,000 cy = 6,615,000 ft³
A. =
/
∴ =
×
= 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs
w = (/
) × 100
∴ = (w/100) ×
= (6/100) × 807030000 lbs = 48421800 lbs
The weight of solids
W = +
= 807030000 lbs + 48421800 lbs = 855451800 lbs
V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy
V = 288,030.91 cy
The amount of cubic yards of borrow required = 288,030.91 cy
B. The volume of water in the required soil is found as follows;
= (w₁/100) ×
= (5.5/100) × 807030000 lbs = 44386650 lbs
The amount of water that must be added = -
= 44386650 lbs - 48421800 lbs = -4,035,150 lbs
Therefore, 4,035,150 lbs of water must be removed
The density of water, ρ = 8.345 lbs/gal
Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal of water must be removed from the natural material
Answer:
Plumbing using a one-way check valve to stop water flowing back on a pump when the pump shuts off.
Explanation:
Diodes are like check valves, keeping current from flowing both ways. Used to create d.c. out of a.c by rectification. Also to block flow if d.c. power like a battery is hooked up in reverse polarity.
