1. Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C
*All flows in [gpm] B
100
500
Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 E
400
QBE
3,600 C D
Find the flowrate through pipe BE, in gallons per minute. [pause] In this problem, ---
2,000
600
A) 2,340 C) 2,620
B) 2,480 D) 2,760

2. Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C
*All flows in [gpm] B
100
500
Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 E
400
QBE
3,600 C D
4,400 gallons per minute flow through a pipe network consisting of ---
2,000
600
A) 2,340 C) 2,620
B) 2,480 D) 2,760

3. Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C
*All flows in [gpm] B
100
500
Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 E
400
QBE
3,600 C D
Supply nodes A, B and C, and demand nodes, D, E and F, and the pipe network, which connect these nodes.
2,000
600
A) 2,340 C) 2,620
B) 2,480 D) 2,760

4. Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C
*All flows in [gpm] B
100
500
Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 E
400
QBE
3,600 C D
All flowrates listed in this problem are in gallons per minute.
2,000
600
A) 2,340 C) 2,620
B) 2,480 D) 2,760

5. Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C
*All flows in [gpm] B
100
500
Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 E
400
QBE
3,600 C D
The nodes are connected by pipes, whose roughness, lengths and diameters are provided. [pause] Network flows are governed by ----
2,000
600
A) 2,340 C) 2,620
B) 2,480 D) 2,760

6. Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 C D
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE
3,600 C D
the conservation of mass at each node,
2,000
600

7. Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
and the conservation of energy, around each loop.
2,000
600

8. Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
Our strategy will be to initially guess the flowrates through the network, such that the first equation is satisfied, ---
2,000
600

9. ? Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i ?
i=1 C D
and then solve for the headloss around each loop. Depending on the calculated headloss, the flowrates will be revised as necessary, and after enough iterations, ---
2,000
600

10. Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
of adjusting the flowrates, the sum of the headlosses around each individual loop will equal zero.
2,000
600

11. Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
Using the given data, we’ll want to develop an expression for the headloss through each pipe for a given flowrate.
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

12. Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
[ft]
In this equation, the headloss is in units of feet, ---
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

13. Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
[ft]
the length of the pipe is in units of feet, ---
[ft]
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

14. Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
[ft]
gal
the flowrate, is in gallons per minute, the friction coefficient ---
[ft]
2,000
min
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

15. Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600
*All flows in [gpm] B n
0 = Σ Q node i E
400
i=1
QBE n
3,600
0 = Σ h pipe i
i=1 C D
[ft]
gal
is unitless, and the diameter is in inches. [pause]
[ft]
2,000
min
600
10.44 * L * Q1.85 h=
C1.85 * d4.87
[in]

16. Find: QBE gal min h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 B E
400
3,600 C D
Returning to our given data, we notice we’ve been given ----
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

17. Find: QBE gal min h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE C
120
L[ft]
400
600 18
d [in] 12 8 B E
400
3,600 C D
the friction coefficient, the length and the diameter. For convenience, we’ll define the variable K, equal to ---
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

18. Find: QBE gal min K h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE
120
L[ft]
400
600 18
d [in] 12 8 B E
400
3,600 C D
10.44 times the length, divided by the friction coefficient to the 1.85 power divided by the diameter raised to the 4.87 power. K
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

19. Find: QBE gal min K h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE
120
L[ft]
400
600 18
d [in] 12 8 B E
400
3,600 C D
Plugging in the known values for L, C and d, the values of K --- K
2,000
600
10.44 * L * Q1.85 h=
C1.85 * d4.87

20. Find: QBE gal min K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400 BC
3.57*10-5
3,600 CD
4.17*10-5 DE
2.38*10-5 C D
are calculated. [pause] The pipe network has two loops, which we’ll define as a --- EF
3.57*10-5 K
2,000
600 FA
4.13*10-6
10.44 * L * Q1.85 BE
8.03*10-7 h=
C1.85 * d4.87

21. Find: QBE gal min 1 2 K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400BC
3.57*10-5
3,600 2 CD
4.17*10-5 DE
2.38*10-5 C D
Loop 1 and Loop 2. When calculating the headloss around a loop, --- EF
3.57*10-5 K
2,000
600 FA
4.13*10-6
10.44 * L * Q1.85 BE
8.03*10-7 h=
C1.85 * d4.87

22. Find: QBE gal min 1 2 K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400BC
3.57*10-5
3,600 2 CD
4.17*10-5 DE
2.38*10-5 C D
we’ll sum the headlosses in clockwise direction. [pause] Now it time to guess the initial flowrates --- EF
3.57*10-5 K
2,000
600 FA
4.13*10-6
10.44 * L * Q1.85 BE
8.03*10-7 h=
C1.85 * d4.87

23. Find: QBE gal min 1 2 K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400BC
3.57*10-5
3,600 2 CD
4.17*10-5 DE
2.38*10-5 C D
through each pipe in the network, using our best judgment, --- EF
3.57*10-5 K
2,000
600 FA
4.13*10-6
10.44 * L * Q1.85 BE
8.03*10-7 h=
C1.85 * d4.87

24. Find: QBE gal min 2,000 200 A F 8.03*10-7 Pipe AB BC CD DE EF FA BE K
3.30*10-6
3.57*10-5
4.17*10-5
2.38*10-5
4.13*10-6 B E
400
3,600 C D
and without violating the conservation of mass. [pause] To do so, it would be prudent to review the possible solutions, ---
2,000
600 n
0 = Σ Q node i
i=1

25. 2,000
200
gal
min
Find: QBE A F
2,340
2,480
2,620
2,760 B E
400
3,600 C D
for the correct flowrate through pipe BE. For the first iteration, we’ll guess ----
2,000
600 n
0 = Σ Q node i
i=1

26. Find: QBE gal min 2,000 200 A F 2,340 2,480 2,620 2,760 B E 400 guess
3,600 C D
2,620, gallons per minute, through pipe BE. [pause] After that, we’ll guess ---
2,000
600 n
0 = Σ Q node i
i=1

27. 2,000
200
gal
min
780
Find: QBE A F
1,220
580
2,340
2,480
2,620
2,760 B E
400
guess
2,620
3,600
1,000
400 C D
the flowrates for the remaining pipes. The headloss through each pipe, ---
1,000
2,000
600 n
0 = Σ Q node i
i=1

28. 2,000
200
gal
min
780
Find: QBE A F
1,220
580
2,340
2,480
2,620
2,760 B E
400
guess
2,620
3,600
1,000
400 C D
is equal to K times the flowrate, in gallons per minute, raised to the 1.85 power. Using our values of K ---
1,000
2,000
600
h = K * Q1.85

29. Find: QBE gal min 2,000 200 780 A F 1,220 580 Pipe K B E AB 3.30*10-6
400 BC
3.57*10-5
2,620
3,600
1,000 CD
4.17*10-5
400 DE
2.38*10-5 C D
we previously calculated, We’ll add a columns for flow and --- EF
3.57*10-5
1,000
2,000
600 FA
4.13*10-6 BE
8.03*10-7
h = K * Q1.85

30. Find: QBE gal min Pipe K Q h [ft] AB 3.30*10-6 1,220 1,220 BC
3.57*10-5
1,000
1,000 CD
4.17*10-5
1,000
1,000 DE
2.38*10-5
400
-400
headloss. Before calculating the headloss through each pipe, we’ll look at each loop individually. EF
3.57*10-5
580
580 FA
4.13*10-6
780
780 BE
8.03*10-7
2,620
2620

31. Find: QBE gal min Loop 1 2,000 200 780 A F 1,220 580 B E 400 2,620
3,600
Loop 1
Pipe K Q
h [ft] AB
For Loop 1, we notice pipes AF and FE flow in clockwise direction, and pipes ---
3.30*10-6
1,220
1,220 EF
3.57*10-5
580
1,000 FA
4.13*10-6
780
1,000 BE
8.03*10-7
2,620
-400

32. Find: QBE gal min CW flow CCW flow Loop 1 2,000 200 780 A F 1,220 580
400
2,620
3,600
Loop 1
Pipe AB K
3.30*10-6 Q
1,220
h [ft]
1,000
-400
8.03*10-7 EF FA BE
3.57*10-5
4.13*10-6
2,620
580
780
AB and BE flow in the counter-clockwise direction. Therefore, the headloss when traveling through these pipes is, ---

33. Find: QBE - + + - gal min CW flow CCW flow Loop 1 2,000 200 780 A F
1,220
CW flow
580 B
CCW flow E
400
2,620
3,600
Loop 1
Pipe K Q
h [ft] - AB
positive through the clockwise pipes and negative through the counterclockwise pipes.
3.30*10-6
1,220 + EF
3.57*10-5
580 + FA
4.13*10-6
780 - BE
8.03*10-7
2,620

34. Find: QBE gal min Loop 1 2,000 200 780 A F 1,220 580 B E 400 2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
The headlosses are calculated, and the sum of these headlosses equals ---
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

35. Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B
400
2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
2.16 feet. This means, that for the assumed flowrates, ---
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

36. Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B
400
2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
the flow through route AFE incurs 2.16 more feet of head, than the flow, through route ABE.
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

37. Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B
400
2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
In order to satisfy the conservation of energy equation, where ---
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

38. Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B
0 = Σ h link i E
400
i=1
2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
the sum of the headlosses around the loop equals zero, ---
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

39. Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B
0 = Σ h link i E
400
i=1
2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
we would have to increase the flow in the counter-clockwise direction ---
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

40. Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 +Q
0 = Σ h link i E
400
i=1
2,620
3,600
Loop 1
h = K * Q1.85
Pipe K Q
h [ft] AB
a certain amount, for all pipes in the loop. Before we do that, we’ll look at Loop 2.
3.30*10-6
1,220
-1.69 EF
3.57*10-5
580
4.62 FA
4.13*10-6
780
0.92 BE
8.03*10-7
2,620
-1.69

41. Find: QBE gal min Loop 2 B 2,620 E 400 3,600 1,000 400 C D 1,000 2,000
Pipe K Q
h [ft] BC
Loop 2 is calculated the same way as loop 1. The clockwise headlosses are positive, and ---
3.57*10-5
1,000 CD
4.17*10-5
1,000 DE
2.38*10-5
400 BE
8.03*10-7
2,620

42. Find: QBE + - - + gal min CW flow CCW flow Loop 2 B 2,620 E 400 3,600
1,000
400
CW flow C D
CCW flow
1,000
2,000
600
Loop 2
Pipe K Q
h [ft] + BC
and the counter-clockwise headlosses are negative. The headlosses are calculated ---
---
3.57*10-5
1,000 - CD
4.17*10-5
1,000 - DE
2.38*10-5
400 + BE
8.03*10-7
2,620

43. Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 400C D
1,000
2,000
600
Loop 2
h = K * Q1.85
Pipe K Q
h [ft] BC
and summed, and we learn that, for the given flowrates, ---
3.57*10-5
1,000
12.66 CD
4.17*10-5
1,000
-14.78 DE
2.38*10-5
400
-1.55 BE
8.03*10-7
2,620
1.69

44. Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 400C D
1,000
2,000
600
Loop 2
h = K * Q1.85
Pipe K Q
h [ft] BC
there is 1.98 feet more headloss incurred along route CDE, than along route ---
3.57*10-5
1,000
12.66 CD
4.17*10-5
1,000
-14.78 DE
2.38*10-5
400
-1.55 BE
8.03*10-7
2,620
1.69

45. Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 400C D
1,000
2,000
600
Loop 2
h = K * Q1.85
Pipe K Q
h [ft] BC
CBE. Therefore, to equalize the headlosses, ---
3.57*10-5
1,000
12.66 CD
4.17*10-5
1,000
-14.78 DE
2.38*10-5
400
-1.55 BE
8.03*10-7
2,620
1.69

46. Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 +QC D
1,000
2,000
600
Loop 2
h = K * Q1.85
Pipe K Q
h [ft] BC
the flowrates should be increased by a certain amount in the clockwise direction, for all pipes in the loop. [pause] If we look at our entire network again, ----
3.57*10-5
1,000
12.66 CD
4.17*10-5
1,000
-14.78 DE
2.38*10-5
400
-1.55 BE
8.03*10-7
2,620
1.69

47. 2,000
200
gal
min
780
Find: QBE A F
1,220
580 1
2,340
2,480
2,620
2,760 B E
400
guess
2,620
3,600
1,000 2
400 C D
we recall that Loop 1 should increase flows in the ---
1,000
2,000
600

48. 2,000
200
gal
min
780
Find: QBE A F
1,220
580 +Q
2,340
2,480
2,620
2,760 B E
400
guess
2,620
3,600
1,000 +Q
400 C D
counter-clockwise direction, and Loop 2 should increase flows in the clockwise direction.
1,000
2,000
600

49. 2,000
200
gal
min
780
Find: QBE A F
1,220
580 +Q
2,340
2,480
2,620
2,760 B E
400
guess
2,620
3,600
1,000 +Q
400 C D
Both of these changes would result in a higher flowrate in pipe BE.
1,000
2,000
600

50. 2,000
200
gal
min
780
Find: QBE A F
1,220
580 +Q
2,340
2,480
2,620
2,760 B E
400
guess
2,620
3,600
1,000 +Q
400 C D
Since our guess for the flowrate through pipe BE was 2,620 gallons per minute, and calculations from the first iteration indicate we increase the flowrates, through this pipe, ---
1,000
2,000
600

51. 2,000
200
gal
min
780
Find: QBE A F
1,220
580 +Q
2,340
2,480
2,620
2,760 B E
400
3,600
guess
1,000 +Q
400 C D
the flowrate through pipe BE must be 2,760, gallons per minute. The remaining pipe flows are guessed and adjusted until ---
1,000
2,000
600

52. 2,000
200
gal
min
670
Find: QBE A F
1,330
470
2,340
2,480
2,620
2,760 B E
400
2,760
3,600
1,030
370 C D n
the conservation of mass and the conservation energy equations are both satisfied.
0 = Σ Q node i
970
2,000
600
i=1 n
0 = Σ h pipe i
i=1

53. 2,000
200
gal
min
670
Find: QBE A F
1,330
470
2,340
2,480
2,620
2,760 B E
400
2,760
3,600
1,030
370 C D n
Since the flowrate through pipe BE equals 2,760 gallons per minute, ---
0 = Σ Q node i
970
2,000
600
i=1 n
0 = Σ h pipe i
i=1

54. Find: QBE gal min AnswerD 2,000 200 670 A F 1,330 470 2,340 2,480
2,620
2,760 B E
400
2,760
3,600
1,030
370 C D n
the answer is D.
0 = Σ Q node i
970
2,000
600
i=1 n
0 = Σ h pipe i
AnswerD
i=1

55. 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 A 50
200 B 20
400 C 40
400 D 50
500

56. ? 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