Answer:
The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.
Explanation:
Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as
v = (ΔPR²/4μL) [1 - (r²/R²)]
where v = fluid velocity at any point in the radial direction
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
But the maximum velocity of the fluid occurs at the middle of the pipe when r=0
Hence, maximum veloxity is
v(max) = (ΔPR²/4μL)
So, velocity at any point in the radial direction is
v = v(max) [1 - (r²/R²)]
At the point r = (R/2)
r² = (R²/4)
(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)
So,
1 - (r²/R²) = 1 - (1/4) = (3/4)
Hence, v at r = (R/2) is given as
v = v(max) × (3/4)
Hope this Helps!!!
Answer:
the length of the bridge B is L = m - 5555 ft
Explanation:
Denoting the length of the bridge B as L we get
length of bridge B = length of bridge A - 5555 ft
since length of bridge A=m
L = m - 5555 ft
Answer: The total vehicle delay is
39sec/veh
Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.
Effective red time (r) = 25sec
Arrival rate (A) = 900veh/h = 0.25veh/sec
Departure rate (D) = 1800veh/h = 0.5veh/sec
STEP1: FIND THE TRAFFIC INTENSITY (p)
p = A ÷ D
p = 0.25 ÷ 0.5 = 0.5
STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE
The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.
Dt = (A × r^2) ÷ 2(1 - p)
Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)
Dt = 156.25 ÷ 4 = 39.0625
Therefore the total vehicle delay after one cycle is;
Dt = 39
