### CE 2710 Tutorial Traffic Flow Shock Wave March 27

```Introduction to
Transportation Engineering
Instructor Dr. Norman Garrick
Hamed Ahangari
27th March 2014
1
Traffic Stream Analysis
2
Key equations
Flow
- (s/veh)
Flow rate (q)
- (veh/h)
q=3600/(h) (1)
T=3sec
T=0
sec
h1-2=3sec
3
Key equations
Density
Spacing (s)
-(ft/veh)
density-concentration(k)
S2-3
k=5280/(s)
- S(veh/mi)
1-2
(2)
4
Key equations
1
1
N
N

1
1
vi
5
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)
6
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
Determine
(a) the density on the highway,
(b) the flow on the road,
(c) the space mean speed.
7
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
8
Solution
• Part (c):
– Space Mean Speed (U(sms)):
U(sms)= q/k
=900/52.8
U(sms)= 17 miles/hour
9
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
10
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
11
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
12
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
13
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
14
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
15
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
16
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
15
– d) the speed at maximum flow.
17
Plot data
60
y = -0.57x + 62.92
50
Speed
40
30
20
10
0
0
20
40
Density
60
80
100
18
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
19
Shockwave Analysis
20
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.
21
Solution
Approach Conditions(Case1)
Platoon Conditions (Case2)
q1=1500 veh/hr
K1=25 veh/mi
q2=1000 veh/hr
K2=100 veh/mi
22
Solution
•  1:  =
1000−1500
=
100−25
-6.7 mile/hour
• Step 2: Growth rate of platoon:
= 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*
= 0.25*16.7= 4.2 mile
• Step 5: Queue length= density*distance
=100*4.2= 420 vehicle
23
Distance
Time
24
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
25
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
26
Solution
• 12 =
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
27
Solution
23 =
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
28
```