No clue sorry man I would help but I need help too
Answer:
Statement 1: All balls hit the ground at the same time
Explanation:
When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.
Answer:
a) -8 lb / ft^3
b) -70.4 lb / ft^3
c) 54.4 lb / ft^3
Explanation:
Given:
- Diameter of pipe D = 12 in
- Shear stress t = 2.0 lb/ft^2
- y = 62.4 lb / ft^3
Find pressure gradient dP / dx when:
a) x is in horizontal flow direction
b) Vertical flow up
c) vertical flow down
Solution:
- dP / dx as function of shear stress and radial distance r:
(dP - y*L*sin(Q))/ L = 2*t / r
dP / L - y*sin(Q) = 2*t / r
Where dP / L = - dP/dx,
dP / dx = -2*t / r - y*sin(Q)
Where r = D /2 ,
dP / dx = -4*t / D - y*sin(Q)
a) Horizontal Pipe Q = 0
Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)
dP / dx = -8 + 0
dP/dx = -8 lb / ft^3
b) Vertical pipe flow up Q = pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)
dP / dx = 8 - 62.4
dP/dx = -70.4 lb / ft^3
c) Vertical flow down Q = -pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)
dP / dx = -8 + 62.4
dP/dx = 54.4 lb / ft^3
Answer:
a) Friction factor for this duct = 0.0239
b) ε = 0.006 ft
Explanation:
Given data :
Flow rate = 11000 ft^3 /min
Pressure drop = 1.2 in per 1500 ft of duct
<u>a) Determine the value of the friction factor for this duct</u>
Friction factor for this duct = 0.0239
<u>b) Determine the approximate size of the equivalent roughness of the surface of the duct</u>
ε = 0.006 ft
attached below is the detailed solution to the given problem
