1. Find: QC [L/s] 3.4 34 340 3,400 Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
3.4 34
340
3,400
Find the flowrate through pipe C, in liters per second. [pause] In this problem, --- A 50
200 B 20
400 C 40
400 D 50
500

2. Find: QC [L/s] 3.4 34 340 3,400 Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
3.4 34
340
3,400
water flows from Tank 1 to Tank 2 through a pipe network, consisting to pipes A, B, C and D. The diameters and lengths --- A 50
200 B 20
400 C 40
400 D 50
500

3. Find: QC [L/s] 3.4 34 340 3,400 Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
3.4 34
340
3,400
of all 4 pipes are provided in the data table. The total headloss --- A 50
200 B 20
400 C 40
400 D 50
500

4. Find: QC [L/s] 3.4 34 340 3,400 Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
3.4 34
340
3,400
between the tanks, and the friction coefficient for all pipe, is provided. To begin, --- A 50
200 B 20
400 C 40
400 D 50
500

5. Find: QC [L/s] 3.4 34 340 3,400 Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
3.4 34
340
3,400
we’ll make some general observations about this particular pipe network, --- A 50
200 B 20
400 C 40
400 D 50
500

6. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
which will guide us, as we solve the problem. [pause] From the conservation of mass, we note that --- A 50
200 B 20
400 C 40
400 D 50
500

7. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
the flowrate through pipe A, must equal the flowrates through through pipes B and C, added together, which must also equal the flowrate though pipe D. A 50
200
QA = QB+QC = QD B 20
400 C 40
400 D 50
500

8. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
We’ll call this our first guiding equation. [pause] Since pipes B and C are parallel, --- A 50
200
QA = QB+QC = QD
(1) B 20
400 C 40
400 D 50
500

9. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
the headlosses through these two pipes are equal, and this will be our second --- A 50
200
QA = QB+QC = QD
(1) B 20
400
hL,B = hL,C C 40
400 D 50
500

10. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
guiding equation. [pause] Lastly, we know the total head loss between the two tanks equals 20 meters, --- A 50
200
QA = QB+QC = QD
(1) B 20
400
hL,B = hL,C
(2) C 40
400 D 50
500

11. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
So 20 meters equals the headloss through pipes A, C and D, will be our third guiding equation. [pause] Since the problem provides the --- A 50
200
QA = QB+QC = QD
(1) B 20
400
hL,B = hL,C
(2) C 40
400
20 [m] = hL,A + hL,C + hL,D D 50
500
(3)

12. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
friction factor, f, we’ll first solve for the --- A 50
200
QA = QB+QC = QD
(1) B 20
400
hL,B = hL,C
(2) C 40
400
20 [m] = hL,A + hL,C + hL,D D 50
500
(3)

13. Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
flowrates through each pipe. A 50
200
QA = QB+QC = QD
(1) B 20
400
hL,B = hL,C
(2) C 40
400
20 [m] = hL,A + hL,C + hL,D D 50
500
(3)

14. Find: QC [L/s] hL=f * Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
The flowrate divided by the area is substituted --- A 50
200
QA = QB+QC = QD
(1) B 20
400
L * V2 C 40
400
hL=f *
d * 2 * g D 50
500

15. Find: QC [L/s] hL=f * Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
Guiding Equations
into the headloss equation, for the velocity. [pause] A 50
200
QA = QB+QC = QD
(1) B 20
400
L * V2 Q C 40
400
hL=f *
d * 2 * g A D 50
500

16. Find: QC [L/s] hL=f hL=f * * Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
L * Q2
pipe
d [cm]
L [m]
hL=f
If we revise this equation, --- *
d * 2 * g * A2 A 50
200 B 20
400
L * V2 Q C 40
400
hL=f *
d * 2 * g A D 50
500

17. Find: QC [L/s] hL=f hL=f * * Δh=20 [m] Tank pipe A 1 pipe B Tank 2
pipe C
(all pipes)
pipe D
L * Q2
pipe
d [cm]
L [m]
hL=f
and then solve for the flowrate, Q, we can calcualte --- *
d * 2 * g * A2 A 50
200 B 20
400
L * V2 Q C 40
400
hL=f *
d * 2 * g A D 50
500

18. Find: QC [L/s] hL=f hL * d * 2 * g * A2 * Q= Δh=20 [m] Tank pipe A 1
pipe B
Tank 2
f=0.03
pipe C
(all pipes)
pipe D
L * Q2
pipe
d [cm]
L [m]
hL=f
the flowrate, for any pipe. For example, if we want the flowrate pipe A, --- *
d * 2 * g * A2 A 50
200
hL * d * 2 * g * A2 B 20
400 Q=
f * L C 40
400 D 50
500

19. Find: QC [L/s] hL=f hL,A * dA * 2 * g * A2 * Q= Δh=20 [m] Tank pipe A1
pipe B
Tank 2
f=0.03
pipe C
(all pipes)
pipe D
L * Q2
pipe
d [cm]
L [m]
hL=f
we’ll substituted in the headloss, diameter, area, and length, unique to pipe A. [pause] This brings up the issue of units. *
d * 2 * g * A2 A 50
200
hL,A * dA * 2 * g * A2 B 20
400 A Q=
f * LA C 40
400 D 50
500

20. Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
For this problem, we’ll use the following units, --- A 50
200
hL * d * 2 * g * A2 B 20
400 Q=
f * L C 40
400 D 50
500

21. Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B
pipe C
(all pipes)
pipe D
pipe
d [cm]
L [m]
[m]
head loss in meters, diamter in --- A 50
200
hL * d * 2 * g * A2 B 20
400 Q=
f * L C 40
400 D 50
500

22. Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B
pipe C
(all pipes)
pipe D m
pipe
d [cm]
L [m]
[m]
[m] s2
meters, acceleration in meters per second squared, area in -- A 50
200
hL * d * 2 * g * A2 B 20
400 Q=
f * L C 40
400 D 50
500

23. Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B
pipe C
(all pipes)
pipe D m
pipe
d [cm]
L [m]
[m]
[m] s2
[m2]
meters squared, length in meters. And the flow rate in --- A 50
200
hL * d * 2 * g * A2 B 20
400 Q=
f * L C 40
400
[m] D 50
500

24. Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B
pipe C
(all pipes)
pipe D m
pipe
d [cm]
L [m]
[m]
[m] s2
[m2]
meters cubed per second. For ease of calculations, let’s expand the data table to include --- A 50
200
hL * d * 2 * g * A2 B 20
400 Q=
f * L C 40
400 m3
[m] D 50
500 s

25. Find: QC [L/s] hL * d * 2 * g * A2 Q= [m] [m] [m4] f * L [m] f=0.03
pipe
d [cm]
d [m]
A2 [m4]
L [m]
a column for diameter, in meters, and a column for area squared, in meters to the 4th power. A 50
200 B 20
400 C 40
400 D 50
500

26. Find: QC [L/s] hL * d * 2 * g * A2 Q= [m] [m] [m4] f * L [m] f=0.03
pipe
d [cm]
d [m]
A2 [m4]
L [m]
These values are calculated based on the given diameters, for each pipe. We’ll remove the first --- A 50
0.50
200 B 20
0.20
9.872*10-4
400 C 40
0.40
400 D 50
0.50
500

27. Find: QC [L/s] hL * d * 2 * g * A2 Q= [m] [m] [m4] f * L [m] f=0.03
pipe
d [m]
A2 [m4]
L [m]
diameter column, and solve for the flowrate, --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

28. Find: QC [L/s] hL,A * dA * 2 * g * A 2 QA= f * LA f=0.03 pipe d [m]
A2 [m4]
L [m]
Q, for pipe A. [pause] Substituting in the known values for --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

29. Find: QC [L/s] hL,A * dA * 2 * g * A 2 QA= 9.81 f * LA f=0.03 pipem
9.81 s2
hL,A * dA * 2 * g * A 2 A
QA=
f * LA
f=0.03
pipe
d [m]
A2 [m4]
L [m]
diameter, gravitational acceleration, area squared, friction coefficient, and length. The flowrate through pipe A equals, --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

30. Find: QC [L/s] hL,A * dA * 2 * g * A 2 QA= =0.2510 * hL,A 9.81 f * LAm
9.81 s2
hL,A * dA * 2 * g * A 2 A
QA=
= * hL,A
f * LA
f=0.03
pipe
d [m]
A2 [m4]
L [m]
times the square root of the headloss through pipe A. [pause] If we calculate --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

31. Find: QC [L/s] hL,B * dB * 2 * g * A 2 hL,C * dB * 2 * g * A 2
QA= * hL,A
QB=
f * LB
hL,C * dB * 2 * g * A 2
hL,D * dB * 2 * g * A 2 C D
QC=
QD=
f * LC
f * LD
pipe
d [m]
A2 [m4]
L [m]
the flowrate through the other three pipes, in the same manner, --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

32. Find: QC [L/s] QA=0.2510 * hL,A QB=0.0180 * hL,B QC=0.1017 * hL,C
QD= * hL,D
pipe
d [m]
A2 [m4]
L [m]
we’ll find they are all a function of the square root of the headloss, through that particular pipe. [pause] To find all flowrates and all headlosses, --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

33. Find: QC [L/s] QA=0.2510 * hL,A QB=0.0180 * hL,B QC=0.1017 * hL,C
20 [m] = hL,A + hL,C + hL,D
(3)
QB= * hL,B
QC= * hL,C
QD= * hL,D
pipe
d [m]
A2 [m4]
L [m]
we’ll turn to our third guiding equation. [pause] If we represent the headloss --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

34. Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QB= * hL,B
QC= * hL,C
QD= * hL,D
pipe
d [m]
A2 [m4]
L [m]
across pipes C and D, as a function of the headloss across pipe A, --- A
0.50
200 B
0.20
9.872*10-4
400 C
0.40
400 D
0.50
500

35. Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QB= * hL,B
QC= * hL,C
QD= * hL,D
we could solve for the headloss across pipe A, explicitly. The headloss across pipe D, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

36. Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QB= * hL,B
QC= * hL,C
QD= * hL,D
as a function of the headloss across pipe A, can be computed since ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

37. Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,B
QC= * hL,C
QD= * hL,D
the flowrate through pipe A equals the flowrate through pipe D. [pause] Substituting in the flowrates, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

38. Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,B
QC= * hL,C
0.2510* hL,A=0.1587* hL,D
QD= * hL,D
for pipes A and D, we find that ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

39. Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) hL,D=2.5 * hL,A QA=0.2510 * hL,A
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,B
QC= * hL,C
0.2510* hL,A=0.1587* hL,D
QD= * hL,D
hL,D=2.5 * hL,A
the headloss across pipe D equals 2.5 times the headloss across pipe pipe A.
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

40. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QB= * hL,B
QC= * hL,C
QD= * hL,D
Next we’ll solve for the headloss across ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

41. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QB= * hL,B
QC= * hL,C
QD= * hL,D
pipe C in terms of the headloss across pipe A. To do so, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

42. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,B
QC= * hL,C
QD= * hL,D
we note the flowrate through pipe A equals the flowrate across pipes B and C, combined. And we also know ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

43. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,B
hL,B = hL,C
(2)
QC= * hL,C
QD= * hL,D
the headloss across pipe B equals the headloss across pipe C. This means, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

44. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
hL,B = hL,C
(2)
QC= * hL,C
QD= * hL,D
we can substitute headloss C, in for headloss B.
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

45. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
hL,B = hL,C
(2)
QC= * hL,C
QB + QC = * hL,C * hL,C
and the flowrate, through pipes B plus C, equals, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

46. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
hL,B = hL,C
(2)
QC= * hL,C
QB + QC = * hL,C * hL,C
QB + QC = * hL,C
times the square root of the headloss across pipe C.
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

47. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
QC= * hL,C
QB + QC = * hL,C
Plugging in the flowrates to equation 1, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

48. Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
* hL,A
=0.1197* hL,C
QC= * hL,C
QB + QC = * hL,C
the headloss across pipe C is equal to ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

49. Find: QC [L/s] ƒ(hL,A ) hL,C=4.397 * hL,A 2.5 * hL,A QA=0.2510 * hL,A
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
* hL,A
=0.1197* hL,C
QC= * hL,C
hL,C=4.397 * hL,A
QB + QC = * hL,C
4.397 times the headloss across pipe A. [pause] Now, equation 3 ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

50. Find: QC [L/s] hL,C=4.397 * hL,A 2.5 * hL,A 4.397 * hL,A
QA= * hL,A
20 [m] = hL,A + hL,C + hL,D
(3)
QA = QB+QC = QD
(1)
QB= * hL,C
* hL,A
=0.1197* hL,C
QC= * hL,C
hL,C=4.397 * hL,A
QB + QC = * hL,C
reduces to 1 unknown variable, the headloss across pipe A.
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

51. Find: QC [L/s] 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A 2.5 * hL,A
20 [m] = hL,A + hL,C + hL,D
(3)
20 [m] = hL,A * hL,A+ 2.5 * hL,A
[pause] The headloss across pipe A equals ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

52. Find: QC [L/s] 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A
20 [m] = hL,A + hL,C + hL,D
(3)
20 [m] = hL,A * hL,A+ 2.5 * hL,A
hL,A = 2.53 [m]
2.53 meters. [pause] This makes ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

53. Find: QC [L/s] hL,B,hL,C =4.397 * hL,A hL,D=2.5 * hL,A
20 [m] = hL,A + hL,C + hL,D
(3)
20 [m] = hL,A * hL,A+ 2.5 * hL,A
hL,A = 2.53 [m]
hL,B,hL,C =4.397 * hL,A
hL,D=2.5 * hL,A
the headloss through pipes B, C and D equal to ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

54. Find: QC [L/s] hL,B,hL,C =4.397 * hL,A hL,B,hL,C =11.14 [m]
20 [m] = hL,A + hL,C + hL,D
(3)
20 [m] = hL,A * hL,A+ 2.5 * hL,A
hL,A = 2.53 [m]
hL,B,hL,C =4.397 * hL,A
hL,B,hL,C =11.14 [m]
hL,D=2.5 * hL,A
hL,D=6.33 [m]
11.14 meters, meters, and 6.33 meters, respectively. Substituting these headloss ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

55. Find: QC [L/s] hL,B,hL,C =11.14 [m] hL,D=6.33 [m] hL,A = 2.53 [m]
QA= * hL,A
QB= * hL,B
QC= * hL,C
hL,A = 2.53 [m]
hL,B,hL,C =11.14 [m]
QD= * hL,D
hL,D=6.33 [m]
values into our flowrate equations, we learn, the flowrates through pipes A, B, C and D are ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

56. Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14m3 s
QA,QD=0.399
QB= * m3 s
QB=0.060
QC= * 11.14 m3 s
QC=0.339
QD= * 6.33
0.399 meters cubed per second, meters cubed per second, meter cubed per second, and meters cubed per second, respectively.
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

57. Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14m3
QA,QD=0.399 s
QB= * m3
QB=0.060 s
QC= * 11.14 m3 s
QC=0.339
QD= * 6.33
Since the question asks to find the flowrate through pipe C,
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

58. Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14m3
QA,QD=0.399 s
QB= * m3
QB=0.060 s
QC= * 11.14 m3 s
QC=0.339
QD= * 6.33 L
1,000 m3
in liters per second, we multiply by 1,000, and the flowrate through pipe C equals, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

59. Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14m3
QA,QD=0.399 s
QB= * m3
QB=0.060 s
QC= * 11.14 m3
QC=0.339 s
QD= * 6.33 L
QC=339 [L/s]
1,000 m3
339 liters per second. [pause]
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

60. Find: QC [L/s] 3.4 34 340 3,400 QA,QD=0.399 QB=0.060 QC=0.339m3
QA,QD=0.399 s m3
QB=0.060 s m3
QC=0.339 s L
QC=339 [L/s]
1,000 m3
When reviewing the possible solutions, ---
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

61. Find: QC [L/s] AnswerC 3.4 34 340 3,400 QA,QD=0.399 QB=0.060 QC=0.339m3
QA,QD=0.399 s m3
QB=0.060 s m3
QC=0.339 s
AnswerC L
QC=339 [L/s]
1,000 m3
The answer is C.
Tank
Δh=20 [m]
pipe A 1
pipe B
Tank 2
pipe C
pipe D

62. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4