Explanation:
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If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum h
1 answer:
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Answer:
The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²
The shear stress at a distance 0.5-in away from the pipe wall is 0
Explanation:
Given;
pressure drop per unit length of pipe = 0.6 psi/ft
length of the pipe = 12 feet
diameter of the pipe = 1 -in
Pressure drop per unit length in a circular pipe is given as;
make shear stress (τ) the subject of the formula
Where;
τ is the shear stress on the pipe wall.
ΔP is the pressure drop
L is the length of the pipe
r is the distance from the pipe wall
Part (a) shear stress at a distance of 0.3-in away from the pipe wall
Radius of the pipe = 0.5 -in
r = 0.5 - 0.3 = 0.2-in = 0.0167 ft
ΔP = 0.6 psi/ft
ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²
Part (b) shear stress at a distance of 0.5-in away from the pipe wall
r = 0.5 - 0.5 = 0
Answer:
z = 3,737 10⁵ m
Explanation:
a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation
P V = n R T
P = (n r / V) T
We replace
P = (n R / V) T₀
b) Let's apply this equation in the points
Lower
.z = 0
P₀ = 610 Pa
P₀ = (nR / V) T₀
Higher.
P = 10 Pa
P = (n R / V) T₀ e^{- C z}
We replace
P = P₀ e^{- C z}
e^{- C z} = P / P₀
C z = ln P₀ / P
z = 1 / C ln P₀ / P
Let's calculate
z = 1 / 1.1 10⁻⁵ ln (610/10)
z = 3,737 10⁵ m