Transcript CE 2710 Tutorial Traffic Flow Shock Wave March 27

Introduction to
Transportation Engineering
Instructor Dr. Norman Garrick
Hamed Ahangari
27th March 2014
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Traffic Stream Analysis
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Key equations
Flow
headway (h)
- (s/veh)
Flow rate (q)
- (veh/h)
q=3600/(h) (1)
T=3sec
T=0
sec
h1-2=3sec
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Key equations
Density
Spacing (s)
-(ft/veh)
density-concentration(k)
S2-3
k=5280/(s)
- S(veh/mi)
1-2
(2)
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Key equations
1
1
N
N

1
1
vi
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Key equations
q=u.k
(4)
flow=u(SMS) * density
q=ku
(veh/hr) = (veh/mi)  (mi/hr)
h=1/q
(sec/veh) = 1 / (veh/hr)  (3600)
s=1/k
(ft/veh) = 1 / (veh/mi)  (5280)
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Example 1
Data obtained from aerial photography showed six
vehicles on a 600 ft-long section of road. Traffic data
collected at the same time indicated an average time
headway of 4 sec.
Determine
(a) the density on the highway,
(b) the flow on the road,
(c) the space mean speed.
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Solution
• Given:
• h=4 sec, l=600 ft, n=6
• Part (a):
– Density (k):
K= (n)/(l)= 6/600= 0.01 veh/ft
k=0.01*5280= 52.8 veh/mile
• Part (b):
– flow (q):
q= 1/h= ¼= 0.25 veh/sec
q = 0.25*3600= 900 veh/hour
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Solution
• Part (c):
– Space Mean Speed (U(sms)):
U(sms)= q/k
=900/52.8
U(sms)= 17 miles/hour
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Traffic Flow Curves
Maximum Flow, Jam Concentration, Freeflow Speed
u
u
uf
q
kj k
qmax
k
qmax
q
qmax - maximum flow
kj - jam concentration
u = 0, k = kj
uf - free flow speed
k = 0, u = uf
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Example 2
Assume that :
u=57.5*(1-0.008 k)
Find:
a) uf free flow speed
b) kj jam concentration
c)
relationships q-u,
d) relationships q-k,
e) qmax capacity
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Solution
• Part a):free flow speed?
i)
when k=0
uf
ii) u=57.5*(1-0.008 k)
i+ii)
uf
kj
u=57.5*(1-0.008 k)= 57.5*(1-0.008*0)
uf =57.5 miles/hour
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Solution
• Part b): kj jam concentration?
i)
when u=0
kj
ii) u=57.5*(1-0.008 k)
i+ii)
uf
kj
0=57.5*(1-0.008 k)---0.008k=1
kj = 125 veh/miles
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Solution
• Part c): relationships q-u?
i)
q=u.k
ii) u=57.5*(1-0.008 k)---u/57.5=1-0.008k
1-u/57.5=0.008K-----K=125-2.17u) (iii)
i+iii)
q= u.k= u.(125-2.17u)
q= 125u-2.17u^2
q
u
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Solution
• Part d): relationships q-k?
i)
q=u.k
u=q/k
ii) u=57.5*(1-0.008 k)
i+ii)
q/k=57.5*(1-0.008k)
q=57.5k -0.46k^2
q
k
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Solution
• Part e): qmax capacity ?
q
qmax
dq
i) To find qmax , set
 0:
dk
ii)
q=57.5k -0.46k^2
k
d(q)/d(k)=0
57.5-0.92k=0
km=62.5
qmax =1796 veh/hour
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Example 3
• The data shown below were obtained on a highway.
Use linear regression analysis to fit these data and
determine
Speed (mi/h) Density (veh/mile)
– (a) the free speed,
– (b) the jam density,
– (c) the capacity,
14.2
24.1
30.3
36.8
40.1
50.6
55.0
85
70
55
47
41
20
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– d) the speed at maximum flow.
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Plot data
60
y = -0.57x + 62.92
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Speed
40
30
20
10
0
0
20
40
Density
60
80
100
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Solution
• from plot: u = -0.57k + 62.92
• Part a)
uf=62.92 miles/hour
• Part b) kj = 110.8 veh/mi
• Part c) qmax = 1736 veh/hr
• @ qmax, u = 31.5 mph
and k = 55.2 veh/mi
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Shockwave Analysis
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Example 4Length of Queue Due to a Speed Reduction
• The volume at a section of a two-lane highway is 1500
veh/h in each direction and the density is about 25
veh/mi.
• A large dump truck from an adjacent construction site
joins the traffic stream and travels at a speed of 10
mi/h for a length of 2.5 mi.
• Vehicles just behind the truck have to travel at the
speed of the truck which results in the formation of a
platoon having a density of 100 veh/mi and a flow of
1000 veh/h.
• Determine how many vehicles will be in the platoon by
the time the truck leaves the highway.
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Solution
Approach Conditions(Case1)
Platoon Conditions (Case2)
q1=1500 veh/hr
K1=25 veh/mi
q2=1000 veh/hr
K2=100 veh/mi
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Solution
• 𝑆𝑡𝑒𝑝 1: 𝑢𝑤 =
1000−1500
=
100−25
-6.7 mile/hour
• Step 2: Growth rate of platoon:
𝑢𝑤𝑝 = 10+6.7=16.7 mile/hour
• Step 3: Time(truck)= distance/u= 2.5/10= 0.25 hour
• Step 4: Length of platoon= time*𝑢𝑤𝑝
= 0.25*16.7= 4.2 mile
• Step 5: Queue length= density*distance
=100*4.2= 420 vehicle
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Distance
Time
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Example 5Length of Queue Due to Stop
• A vehicle stream is interrupted and stopped by policeman.
• The traffic volume for the vehicle stream before the
interruption is 1500 veh/hr and the density is 50 veh/mi.
• Assume that the jam density is 250 veh/mi. After four
minutes the policeman releases the traffic.
• The flow condition for the release is a traffic volume of
1800 veh/hr and a speed of 18mph.
 Determine :
• the length of the queue
• and the number of vehicles in the queue after five minutes.
• how long it will take for the queue to dissipate after the
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policeman releases the traffic.
Solution
Approach conditions
Shockwave 1
State 1
q = 1500 veh/hr
k = 50 veh/mi
Platoon Conditions Release Conditions
Shockwave 2
State 2
q = 0 veh/hr
k = 250 veh/mi
State 3
q = 1800 veh/hr
u = 18 mi/hr
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Solution
• 𝑢𝑤12 =
0−1500
=
250−50
-7.5 mi/hr
• Shockwave 1 is moving upstream at -7.5 mph
• Length of the queue after 4 minutes
Length = u*t = 7.5 mph * 4/60 hr = 0.5 mile
• Vehicles are in the queue after 4 minutes
# of vehicles = k * L = 250 *0.5 = 125 vehicles
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Solution
𝑢𝑤23 =
1800−0
=
250−100
-12 mi/hr
• Shockwave 2 is moving upstream at 12 mph
• usw1 = - 7.5 mph
usw2 = - 12 mph
• The queue will dissipate at rate of 4.5 mph
Time to dissipate a 0.5 mile queue is L/speed
=0.5 mile / 4.5 mph = 0.012 hr = 6.6 minutes
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