Answer:
im trying to do a challenge because this kid deleted all my answers
Explanation:
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
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
![Q = A \times V](https://tex.z-dn.net/?f=Q%20%3D%20A%20%5Ctimes%20V)
where A is Area
![A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2](https://tex.z-dn.net/?f=A%20%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20%2840%5Ctimes%2010%5E%7B-3%7D%29%5E2%20%3D%201.256%5Ctimes%2010%5E%7B-3%7D%20m%5E2)
![Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s](https://tex.z-dn.net/?f=Q%20%3D%201.256%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%2020%20%3D%200.02513%20m%5E3%2Fs)
minimum flow rate provided by pump is 0.02513 m^3/s
It is auxillary sorry i couldn’t help it happens to the best of us