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Question 1
In the picture you see a lock gate with at the left side freshwater and at the right side
saltwater. Suppose the gate is free to turn, in which direction will the gate turn, to the
left or to the right? Make your argument based on a calculation.
3,9
m
3,8 mfresh salt
Lock gate, width = 4 , Sluisdeur, breedte = 4m
Druk onderin door zoetwater = 1000 x 10 x 3,9 =39.000 N/m2
Gemiddelde druk = 19.500 N/m2
Kracht door zoet = 3,9 x4 x 19.500 = 304.200 N
Druk onderin door zoutwater = 1025 x 10 x 3,8 =38.950 N/m2
Gemiddelde druk = 19.475 N/m2
Kracht door zout = 3,8 x 4 x 19.475 = 296.020 N
De deur draait naar rechts
𝑦 =
𝑝
𝜌 ∙ 𝑔
p =
F
A
Question 2
An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Water
depth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is
0,023 s/m1/3
. At the end of the channel (downstream) a short crested weir is situated.
(Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2
/s). The water depth
downstream of the weir is 0,8 m The discharge is 5,1 m3
/s. The discharge does not
change over the length of the channel.
a. Calculate the bed slope of the channel. What is the difference in bottom height
between beginning and end of the channel.
b. Calculate froudes number. Indicate if the flow in the channel is sub-critical or
super-critical. As width you can use the average width of the channel. Give a
computation.
c. Calculate the level of the crest related to the bottom level of the channel. Assume
the weir has a free flow.
d. Check if the weir has a free flow. Give a motivation.
Gaan uit van een evenwichtssituatie
A = (3+2*1,3) x 1,3 =7,28 m2
O= 3 + 2 x (1,32
+ 2.62
)1/2
= 8,81 m
R= A/O= 7,28 / 8,81 = 0,83 m
u=Q/A = 5,1 / 7,28 = 0,70 m/s
𝑉 =
𝑅
2
3 ∙ 𝑆𝑓
1
2
𝑛
n=0,023 invullen geeft S=3,323 x 10-4
of 1:3009
Afschot over 4 km is 1.33 m
Question 2
An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Water
depth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is
0,023 s/m1/3
. At the end of the channel (downstream) a short crested weir is situated.
(Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2
/s). The water depth
downstream of the weir is 0,8 m The discharge is 5,1 m3
/s. The discharge does not
change over the length of the channel.
a. Calculate the bed slope of the channel. What is the difference in bottom height
between beginning and end of the channel.
b. Calculate froudes number. Indicate if the flow in the channel is sub-critical or
super-critical. As width you can use the average width of the channel. Give a
computation.
c. Calculate the level of the crest related to the bottom level of the channel. Assume
the weir has a free flow.
d. Check if the weir has a free flow. Give a motivation.
Onderdeel b
Gemiddelde breedte = 3 + 2*1,3 = 5.6 m
3
2
2
bg
q
y v
c


y kritisch = 0,44 m, Vc = 2,08 m/s
𝐹𝑟 =
𝑉
𝑔 ∙ 𝑦𝑐
2
=
𝑉
𝑉𝑐
=
0,7
2,08
= 0,34
Fr < 1, kortom we hebben te maken met stromend water
Question 2
An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Water
depth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is
0,023 s/m1/3
. At the end of the channel (downstream) a short crested weir is situated.
(Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2
/s). The water depth
downstream of the weir is 0,8 m The discharge is 5,1 m3
/s. The discharge does not
change over the length of the channel.
a. Calculate the bed slope of the channel. What is the difference in bottom height
between beginning and end of the channel.
b. Calculate froudes number. Indicate if the flow in the channel is sub-critical or
super-critical. As width you can use the average width of the channel. Give a
computation.
c. Calculate the level of the crest related to the bottom level of the channel. Assume
the weir has a free flow.
d. Check if the weir has a free flow. Give a motivation.
Onderdeel c
2
3
Hbmqv 
Overstortende straal = 0,74 m
Snelheidshoogte voor de overstort = v2/2g=0,72
/20 = 0,0245m
Hoogte stuw vanaf de bodem = 1,3 +0,0245 – 0,74 = 0,58 m
Onderdeel d
Hh 3
2
3
 H=0,74 2/3 H = 0, 49 h3 0,8 -0,58 = 0,22 m
Question 3
In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh water
a. Calculate the value of total
b. Calculate the water level upstream of the culvert.
c. Make a sketch of the energy and pressure line. Include the values
Channel u=0,3 m/s
Upstream
WL ??? Downstream
WL +2,0 m NAP
L= 45 m
3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s
Onderdeel a
R=A/O=(3 x 2)/(3+3+2+2)=0,60 m,
2
1
1








i =0,44
R
l
w


4
 =0,47
1u
ksi-totaal=1,91
Question 3
In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh water
a. Calculate the value of total
b. Calculate the water level upstream of the culvert.
c. Make a sketch of the energy and pressure line. Include the values
Channel u=0,3 m/s
Upstream
WL ??? Downstream
WL +2,0 m NAP
L= 45 m
3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s
2g
u
ξΔΗ
2
2
totaala  =0,187 m Q = u x A = 1,4 x 6 = 8,4 m3/s
Snelheidshoogte bovenstrooms = 0,32
/20 = 0,005 m
Snelheidshoogte benedenstrooms = 0,52
/20 = 0,013 m
Waterstand bovenstrooms = 2,0 + 0,013 + 0,187 – 0,005 = 2,195 m
𝑦1 + 𝑧1 +
𝑢1
2
2𝑔
= 𝑦2 + 𝑧2 +
𝑢2
2
2𝑔
+ ∆𝐻1−2
Question 3
In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh water
a. Calculate the value of total
b. Calculate the water level upstream of the culvert.
c. Make a sketch of the energy and pressure line. Include the values
Channel u=0,3 m/s
Upstream
WL ??? Downstream
WL +2,0 m NAP
L= 45 m
3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s
Onderdeel c
Snelheidshoogte duiker 1,42
/20= 0,098 m
Energieverlies instroom = 0,44 x 0,098 = 0,043 m
Energieverlies wrijving = 0,47 x 0,098 = 0,046 m
Energieverlies uitstroom = 0,098 m
Dip in druklijn tgv contractie.
Acontractie = A x 0,6 Q=v.a Ucontractie = 1,4/0,6 = 2,33 m/s
Snelheidshoogte 0,27 m
Question 4
In the picture above you see a sewer pipe designed for the discharge of rainfall directly
to surface water. The water level surface water is +2 m, the velocity is low. The water
level in the manhole is +2,95 m. The equivalent roughness k is 1 mm
a. Calculate the discharge in the sewer pipe. Assume the flow is turbulent.
b. Check if the flow is turbulent.
c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which is
connected to the sewer. As discharge use the result from question a.
d. Explain why the design of this sewer is actually bad.
Water level + 2 m
Surface level +3 m
Rainfall, discharge???
60 m
D=500 mm
Manhole
(inspectieput)
Onderdeel a
Waterstandsverschil is 2,95 – 2 = 0,95 m
I=dh/L=0,95/60=0,0158 1:63 R=D/4=0,500/4=0,125 m A=0,196 m2
k
R
C
12
log18 =57,17
𝑉 = 𝐶 ∙ 𝑅 ∙ 𝑆𝑓=2,55 m/s
Q=u x a =0,5 m3/s=500 l/s
Question 4
In the picture above you see a sewer pipe designed for the discharge of rainfall directly
to surface water. The water level surface water is +2 m, the velocity is low. The water
level in the manhole is +2,95 m. The equivalent roughness k is 1 mm
a. Calculate the discharge in the sewer pipe. Assume the flow is turbulent.
b. Check if the flow is turbulent.
c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which is
connected to the sewer. As discharge use the result from question a.
d. Explain why the design of this sewer is actually bad.
Water level + 2 m
Surface level +3 m
Rainfall, discharge???
60 m
D=500 mm
Manhole
(inspectieput)
b turbulent
]1[
4
Re

Rv
w

 =1275000 > 4000 ja
𝜐 = 1,00 ∙ 10−6
Question 4
In the picture above you see a sewer pipe designed for the discharge of rainfall directly
to surface water. The water level surface water is +2 m, the velocity is low. The water
level in the manhole is +2,95 m. The equivalent roughness k is 1 mm
a. Calculate the discharge in the sewer pipe. Assume the flow is turbulent.
b. Check if the flow is turbulent.
c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which is
connected to the sewer. As discharge use the result from question a.
d. Explain why the design of this sewer is actually bad.
Water level + 2 m
Surface level +3 m
Rainfall, discharge???
60 m
D=500 mm
Manhole
(inspectieput)
Onderdeel c
Q=500 l/s 500/60 = 8,33 ha
Onderdeel d
Het riool staat altijd vol water.

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Lock gate direction calculation

  • 1. Question 1 In the picture you see a lock gate with at the left side freshwater and at the right side saltwater. Suppose the gate is free to turn, in which direction will the gate turn, to the left or to the right? Make your argument based on a calculation. 3,9 m 3,8 mfresh salt Lock gate, width = 4 , Sluisdeur, breedte = 4m Druk onderin door zoetwater = 1000 x 10 x 3,9 =39.000 N/m2 Gemiddelde druk = 19.500 N/m2 Kracht door zoet = 3,9 x4 x 19.500 = 304.200 N Druk onderin door zoutwater = 1025 x 10 x 3,8 =38.950 N/m2 Gemiddelde druk = 19.475 N/m2 Kracht door zout = 3,8 x 4 x 19.475 = 296.020 N De deur draait naar rechts 𝑦 = 𝑝 𝜌 ∙ 𝑔 p = F A
  • 2. Question 2 An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Water depth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is 0,023 s/m1/3 . At the end of the channel (downstream) a short crested weir is situated. (Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2 /s). The water depth downstream of the weir is 0,8 m The discharge is 5,1 m3 /s. The discharge does not change over the length of the channel. a. Calculate the bed slope of the channel. What is the difference in bottom height between beginning and end of the channel. b. Calculate froudes number. Indicate if the flow in the channel is sub-critical or super-critical. As width you can use the average width of the channel. Give a computation. c. Calculate the level of the crest related to the bottom level of the channel. Assume the weir has a free flow. d. Check if the weir has a free flow. Give a motivation. Gaan uit van een evenwichtssituatie A = (3+2*1,3) x 1,3 =7,28 m2 O= 3 + 2 x (1,32 + 2.62 )1/2 = 8,81 m R= A/O= 7,28 / 8,81 = 0,83 m u=Q/A = 5,1 / 7,28 = 0,70 m/s 𝑉 = 𝑅 2 3 ∙ 𝑆𝑓 1 2 𝑛 n=0,023 invullen geeft S=3,323 x 10-4 of 1:3009 Afschot over 4 km is 1.33 m
  • 3. Question 2 An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Water depth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is 0,023 s/m1/3 . At the end of the channel (downstream) a short crested weir is situated. (Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2 /s). The water depth downstream of the weir is 0,8 m The discharge is 5,1 m3 /s. The discharge does not change over the length of the channel. a. Calculate the bed slope of the channel. What is the difference in bottom height between beginning and end of the channel. b. Calculate froudes number. Indicate if the flow in the channel is sub-critical or super-critical. As width you can use the average width of the channel. Give a computation. c. Calculate the level of the crest related to the bottom level of the channel. Assume the weir has a free flow. d. Check if the weir has a free flow. Give a motivation. Onderdeel b Gemiddelde breedte = 3 + 2*1,3 = 5.6 m 3 2 2 bg q y v c   y kritisch = 0,44 m, Vc = 2,08 m/s 𝐹𝑟 = 𝑉 𝑔 ∙ 𝑦𝑐 2 = 𝑉 𝑉𝑐 = 0,7 2,08 = 0,34 Fr < 1, kortom we hebben te maken met stromend water
  • 4. Question 2 An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Water depth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is 0,023 s/m1/3 . At the end of the channel (downstream) a short crested weir is situated. (Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2 /s). The water depth downstream of the weir is 0,8 m The discharge is 5,1 m3 /s. The discharge does not change over the length of the channel. a. Calculate the bed slope of the channel. What is the difference in bottom height between beginning and end of the channel. b. Calculate froudes number. Indicate if the flow in the channel is sub-critical or super-critical. As width you can use the average width of the channel. Give a computation. c. Calculate the level of the crest related to the bottom level of the channel. Assume the weir has a free flow. d. Check if the weir has a free flow. Give a motivation. Onderdeel c 2 3 Hbmqv  Overstortende straal = 0,74 m Snelheidshoogte voor de overstort = v2/2g=0,72 /20 = 0,0245m Hoogte stuw vanaf de bodem = 1,3 +0,0245 – 0,74 = 0,58 m Onderdeel d Hh 3 2 3  H=0,74 2/3 H = 0, 49 h3 0,8 -0,58 = 0,22 m
  • 5. Question 3 In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh water a. Calculate the value of total b. Calculate the water level upstream of the culvert. c. Make a sketch of the energy and pressure line. Include the values Channel u=0,3 m/s Upstream WL ??? Downstream WL +2,0 m NAP L= 45 m 3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s Onderdeel a R=A/O=(3 x 2)/(3+3+2+2)=0,60 m, 2 1 1         i =0,44 R l w   4  =0,47 1u ksi-totaal=1,91
  • 6. Question 3 In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh water a. Calculate the value of total b. Calculate the water level upstream of the culvert. c. Make a sketch of the energy and pressure line. Include the values Channel u=0,3 m/s Upstream WL ??? Downstream WL +2,0 m NAP L= 45 m 3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s 2g u ξΔΗ 2 2 totaala  =0,187 m Q = u x A = 1,4 x 6 = 8,4 m3/s Snelheidshoogte bovenstrooms = 0,32 /20 = 0,005 m Snelheidshoogte benedenstrooms = 0,52 /20 = 0,013 m Waterstand bovenstrooms = 2,0 + 0,013 + 0,187 – 0,005 = 2,195 m 𝑦1 + 𝑧1 + 𝑢1 2 2𝑔 = 𝑦2 + 𝑧2 + 𝑢2 2 2𝑔 + ∆𝐻1−2
  • 7. Question 3 In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh water a. Calculate the value of total b. Calculate the water level upstream of the culvert. c. Make a sketch of the energy and pressure line. Include the values Channel u=0,3 m/s Upstream WL ??? Downstream WL +2,0 m NAP L= 45 m 3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s Onderdeel c Snelheidshoogte duiker 1,42 /20= 0,098 m Energieverlies instroom = 0,44 x 0,098 = 0,043 m Energieverlies wrijving = 0,47 x 0,098 = 0,046 m Energieverlies uitstroom = 0,098 m Dip in druklijn tgv contractie. Acontractie = A x 0,6 Q=v.a Ucontractie = 1,4/0,6 = 2,33 m/s Snelheidshoogte 0,27 m
  • 8. Question 4 In the picture above you see a sewer pipe designed for the discharge of rainfall directly to surface water. The water level surface water is +2 m, the velocity is low. The water level in the manhole is +2,95 m. The equivalent roughness k is 1 mm a. Calculate the discharge in the sewer pipe. Assume the flow is turbulent. b. Check if the flow is turbulent. c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which is connected to the sewer. As discharge use the result from question a. d. Explain why the design of this sewer is actually bad. Water level + 2 m Surface level +3 m Rainfall, discharge??? 60 m D=500 mm Manhole (inspectieput) Onderdeel a Waterstandsverschil is 2,95 – 2 = 0,95 m I=dh/L=0,95/60=0,0158 1:63 R=D/4=0,500/4=0,125 m A=0,196 m2 k R C 12 log18 =57,17 𝑉 = 𝐶 ∙ 𝑅 ∙ 𝑆𝑓=2,55 m/s Q=u x a =0,5 m3/s=500 l/s
  • 9. Question 4 In the picture above you see a sewer pipe designed for the discharge of rainfall directly to surface water. The water level surface water is +2 m, the velocity is low. The water level in the manhole is +2,95 m. The equivalent roughness k is 1 mm a. Calculate the discharge in the sewer pipe. Assume the flow is turbulent. b. Check if the flow is turbulent. c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which is connected to the sewer. As discharge use the result from question a. d. Explain why the design of this sewer is actually bad. Water level + 2 m Surface level +3 m Rainfall, discharge??? 60 m D=500 mm Manhole (inspectieput) b turbulent ]1[ 4 Re  Rv w   =1275000 > 4000 ja 𝜐 = 1,00 ∙ 10−6
  • 10. Question 4 In the picture above you see a sewer pipe designed for the discharge of rainfall directly to surface water. The water level surface water is +2 m, the velocity is low. The water level in the manhole is +2,95 m. The equivalent roughness k is 1 mm a. Calculate the discharge in the sewer pipe. Assume the flow is turbulent. b. Check if the flow is turbulent. c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which is connected to the sewer. As discharge use the result from question a. d. Explain why the design of this sewer is actually bad. Water level + 2 m Surface level +3 m Rainfall, discharge??? 60 m D=500 mm Manhole (inspectieput) Onderdeel c Q=500 l/s 500/60 = 8,33 ha Onderdeel d Het riool staat altijd vol water.