Answer:
526.5 KN
Explanation:
The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.
But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.
h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg
where ρ = density of the fluid and g = acceleration due to gravity
h = ΔP/ρg
ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa
Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with
Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa
Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²
Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN
Answer:
Time period = 41654.08 s
Explanation:
Given data:
Internal volume is 210 m^3
Rate of air infiltration 
length of cracks 62 m
air density = 1.186 kg/m^3
Total rate of air infiltration 
total volume of air infiltration
Time period 
Answer:
2.379m
Explanation:
The width = 23m
The depth = 3m
The radius is denoted as R
The wetted area is = A
The perimeter perimeter = P
Hydraulic radius
R = A/P
The area of a rectangular channel
= Width multiplied by Depth
A = 23x3
A = 69m²
Perimeter = (2x3)+23
P = 6+23
P= 29
Hydraulic radius R = 69/29
= 2.379m
This answers the question
Thank you!
Answer:
a) 0.3
b) 3.6 mm
Explanation:
Given
Length of the pads, l = 200 mm = 0.2 m
Width of the pads, b = 150 mm = 0.15 m
Thickness of the pads, t = 12 mm = 0.012 m
Force on the rubber, P = 15 kN
Shear modulus on the rubber, G = 830 GPa
The average shear strain can be gotten by
τ(average) = (P/2) / bl
τ(average) = (15/2) / (0.15 * 0.2)
τ(average) = 7.5 / 0.03
τ(average) = 250 kPa
γ(average) = τ(average) / G
γ(average) = 250 kPa / 830 kPa
γ(average) = 0.3
horizontal displacement,
δ = γ(average) * t
δ = 0.3 * 12
δ = 3.6 mm
Did not engineer cables factoring wind shear