Given:
size of scale model = 4(size of pump)
power ratio of pump and scale model = 5:1
Solution:
Let the diameter of scale model and pump be and
respectively
and head be and
respectively
Now, power, P is given as a function of head(H) and dischagre(Q)
P = (1)
From eqn (1):
and
So,
Therefore,
=
=
=
=
=
Answer:
The minimum mass flow rate for the water is 14.39kg/s
Explanation:
In this question, we are asked to calculate the minimum mass flow rate for the water in kg/s.
Please check attachment for complete solution and step by step explanation
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