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On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given by
U = Usf[1 − k/kj] and q = Uk
Where U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal to
- a)2499
- b)2501
Correct answer is between '2499,2501'. Can you explain this answer?
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Verified Answer
On an urban road, the free mean speed was measured as 70 kmph and the...
Kj = 1000s = 1000/7 = 142.86veh/km
View all questions of this test
qmax = vsfkj/4 = 70 × 142.864 = 2500veh/hr
Most Upvoted Answer
On an urban road, the free mean speed was measured as 70 kmph and the...
Solution:
Given data:
Usf = 70 kmph
Average spacing between vehicles under jam condition = 7.0 m
We know that the maximum flow (qmax) is obtained when the density (k) is equal to the jam density (kj).
Maximum flow (qmax) = U(kj) = Usf(kj)[1 - kj/kj] = 0
So, the maximum flow (qmax) for this condition is zero.
Now, to find the maximum flow rate (q) per lane, we need to use the speed-flow-density equation.
U = Usf[1 - k/kj] ... (1)
q = Uk ... (2)
From equation (1), we can write:
k = kj[1 - U/Usf]
Substituting this in equation (2), we get:
q = U.kj[1 - U/Usf]
To find the maximum flow rate (q) per lane, we need to find the value of U that maximizes q.
Differentiating q w.r.t U, we get:
dq/dU = kj(Usf - 2U)
Setting dq/dU = 0, we get:
Usf - 2U = 0
U = Usf/2 = 35 kmph
Substituting this value of U in equation (1), we get:
k = kj[1 - U/Usf] = kj/2
Substituting these values of U and k in equation (2), we get:
q = U.kj/2 = 35 x kj/2
Substituting the given value of average spacing between vehicles under jam condition in the equation:
kj = 3600/7 veh/km
Substituting this value of kj in the above equation, we get:
q = 35 x 3600/14 veh/hr
q = 9000 veh/hr
So, the maximum flow rate (q) per lane for this condition is 9000 veh/hr, which lies between the given options of 2499 and 2501.
Given data:
Usf = 70 kmph
Average spacing between vehicles under jam condition = 7.0 m
We know that the maximum flow (qmax) is obtained when the density (k) is equal to the jam density (kj).
Maximum flow (qmax) = U(kj) = Usf(kj)[1 - kj/kj] = 0
So, the maximum flow (qmax) for this condition is zero.
Now, to find the maximum flow rate (q) per lane, we need to use the speed-flow-density equation.
U = Usf[1 - k/kj] ... (1)
q = Uk ... (2)
From equation (1), we can write:
k = kj[1 - U/Usf]
Substituting this in equation (2), we get:
q = U.kj[1 - U/Usf]
To find the maximum flow rate (q) per lane, we need to find the value of U that maximizes q.
Differentiating q w.r.t U, we get:
dq/dU = kj(Usf - 2U)
Setting dq/dU = 0, we get:
Usf - 2U = 0
U = Usf/2 = 35 kmph
Substituting this value of U in equation (1), we get:
k = kj[1 - U/Usf] = kj/2
Substituting these values of U and k in equation (2), we get:
q = U.kj/2 = 35 x kj/2
Substituting the given value of average spacing between vehicles under jam condition in the equation:
kj = 3600/7 veh/km
Substituting this value of kj in the above equation, we get:
q = 35 x 3600/14 veh/hr
q = 9000 veh/hr
So, the maximum flow rate (q) per lane for this condition is 9000 veh/hr, which lies between the given options of 2499 and 2501.
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On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer?
Question Description
On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer?.
On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow -density equation is given byU = Usf[1 − k/kj] and q = UkWhere U = space-mean speed (kmph); Usf = free mean speed (kmph); k = density (veh/km); kj = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal toa)2499b)2501Correct answer is between '2499,2501'. Can you explain this answer?.
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