FLUID MECHANICS MODULE B
Sample Problem #1
A new cast iron pipe must carry 1.2cu.m/s at a headloss
of 5m per km length of pipe. Compute the diameter of the
pipe using:
a. Hazen Williams formula with c = 120
b. Manning’s formula with n = 0.012
c. Darcy Weishback formula with f = 0.021
Solution:
𝑉 = 0.8492 𝐶 𝑅0.63 𝑆 0.54
5
𝑆=
= 0.005
1000
𝐷
𝑅=
4
𝑄
𝑉=
𝐴
1.2
1.53
𝑉= 𝜋 = 2
𝐷
4(𝐷2 )
1.53
𝐷
= 0.8492 (120) ( )0.63 (0.005)0.54
2
𝐷
4
D = 0.838m
1 2 1
𝑉 = 𝑅3𝑆 2
𝑛
1
1.53
1
𝐷 2
=
( )3 (0.005)2
2
𝐷
0.012 4
D = 0.853m
0.0826𝑓𝐿𝑄 2
𝐻𝐿 =
𝐷5
0.0826(0.021)(1000)(1.2)2
5=
𝐷5
D = 0.870m
Sample Problem #2
Three concrete pipes are connected in series as shown. If
the rate of flow in the pipe is 0.10cu.m/s. What is the total
head loss of the given pipe?
Solution:
0.0826𝑓𝐿𝑄 2
𝐻𝐿 =
𝐷5
0.0826(0.0248)(360)(0.10)2
𝐻𝐿1 =
= 23.05𝑚
0.205
2
0.0826(0.0242)(300)(0.10)
𝐻𝐿2 =
= 78.97𝑚
0.155
2
0.0826(0.0255)(600)(0.10)
𝐻𝐿3 =
= 12.94𝑚
0.255
𝐻𝐿 = 𝐻𝐿1 + 𝐻𝐿2 + 𝐻𝐿3 = 114.96𝑚
Sample Problem #3
Three pipes A, B and C are connected in parallel. If the
combined discharged of the 3 pipes is equal to
0.61cu.m/s and assuming they have equal values of
friction factor “f”, compute the following using the
tabulated data.
PIPELINE
LENGTH
DIAM
A
600m
150m
B
480m
200m
C
750m
100m
a. Compute the rate of flow of pipeline A in liters/sec
b. Compute the rate of flow of pipeline B in liters/sec
c. Compute the rate of flow of pipeline C in liters/sec
Solution:
𝐻𝐿𝐴 = 𝐻𝐿𝐵 = 𝐻𝐿𝐶 = 𝐻𝐿
0.0826𝑓𝐴 𝐿𝐴 𝑄𝐴2 0.0826𝑓𝐵 𝐿𝐵 𝑄𝐵2 0.0826𝑓𝐵 𝐿𝐵 𝑄𝐵2
=
=
𝐷𝐴5
𝐷𝐵5
𝐷𝐵5
𝑄𝐴 = 0.436𝑄𝐵
𝑄𝐵 = 2.295𝑄𝐴
𝑄𝐴 = 3.08𝑄𝐶
𝑄𝐶 = 0.325𝑄𝐴
𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 = 0.61
𝑄𝐴 + 2.295𝑄𝐴 + 0.325𝑄𝐴 = 0.61
𝑄𝐴 = 0.17 𝑚3 /𝑠 = 170𝑙𝑖/𝑠
𝑄𝐵 = 2.295𝑄𝐴
𝑄𝑏 = 0.39 𝑚3 /𝑠 = 390𝑙𝑖/𝑠
𝑄𝐶 = 0.325𝑄𝐴
𝑄𝑐 = 0.05 𝑚3 /𝑠 = 50𝑙𝑖/𝑠
Sample Problem #4
A pump draws water from reservoir A and lifts it to
reservoir B as shown. The loss of head from A to1 is
5times the velocity head in the 250mm dia. pipe and the
loss of head from 2 to B is 25times the velocity head in
the 150mm dia. pipe. When the discharge is 0.056cu.m/s,
a. Compute the pressure at point 2 in KPa. (5pts)
b. Compute the power being given up by the water to the
pump in horsepower (hp). (5pts)
c. Compute the total head loss from reservoir A to
reservoir B (5pts)
B
ELEV. 200
A
ELEV. 10
1
250mm∅
P
2
150mm∅
ELEV. -30
Conventional Solution:
𝑣 = 𝑄/𝐴
𝑄1
0.056
𝑣1 =
=
= 1.141𝑚/𝑠
𝐴1 𝜋 (0.252 )
4
𝑄2
0.056
𝑣2 =
=𝜋
= 3.169𝑚/𝑠
𝐴2
(0.152 )
4
𝐻𝐿𝐴→𝐵 = 𝐻𝐿𝐴→1 + 𝐻𝐿2→𝐵
5(𝑣1 )2 25(𝑣2 )2
𝐻𝐿𝐴→𝐵 =
+
2𝑔
2𝑔
5(1.141)2 25(3.169)2
𝐻𝐿𝐴→𝐵 =
+
2(9.81)
2(9.81)
𝐻𝐿𝐴→𝐵 = 0.33177 + 12.79633 = 𝟏𝟑. 𝟏𝟐𝟖𝟏𝒎
General Bernoulli’s Equation
(𝑉𝐴 )2 𝑃𝐴
(𝑉𝐵 )2 𝑃𝐵
+ + 𝑍𝐴 + 𝐻𝐴 =
+
+ 𝑍𝐵 + 𝐻𝐸 + 𝐻𝐿
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
𝐵𝐸𝐸 𝐴 → 𝐵 : (Assume Datum to be @ el.-30)
(𝑉𝐴 )2 𝑃𝐴
(𝑉𝐵 )2 𝑃𝐵
+ + 𝑍𝐴 + 𝐻𝐴 =
+
+ 𝑍𝐵 + 𝐻𝐿𝐴→𝐵
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
0 + 0 + 40 + 𝐻𝐴 = 0 + 0 + 230 + 13.1281
HA = 203.1281m
𝐵𝐸𝐸 𝐴 → 𝐵 : (Assume Datum to be @ el.0)
(𝑉𝐴 )2 𝑃𝐴
(𝑉𝐵 )2 𝑃𝐵
+ + 𝑍𝐴 + 𝐻𝐴 =
+
+ 𝑍𝐵 + 𝐻𝐿𝐴→𝐵
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
0 + 0 + 10 + 𝐻𝐴 = 0 + 0 + 230 + 13.1281
HA = 203.1281m
𝐵𝐸𝐸 𝐴 → 𝐵 : (Assume Datum to be @ el.10)
(𝑉𝐴 )2 𝑃𝐴
(𝑉𝐵 )2 𝑃𝐵
+ + 𝑍𝐴 + 𝐻𝐴 =
+
+ 𝑍𝐵 + 𝐻𝐿𝐴→𝐵
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
0 + 0 + 0 + 𝐻𝐴 = 0 + 0 + 190 + 13.1281
HA = 203.1281m
𝑃=
𝑄𝛾𝐸 0.056(9810)(203.1281)
=
= 𝟏𝟒𝟗. 𝟓𝟖𝟓𝒉𝒑
746
746
BEE 2 →B: (Assume Datum to be @ el. -30)
(𝑉2 )2 𝑃2
(𝑉𝐵 )2 𝑃𝐵
+ + 𝑍2 =
+
+ 𝑍𝐵 + 𝐻𝐿2→𝐵
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
2
(3.169)
𝑃2
+
+ 0 = 0 + 0 + 230 + 12.79633
2(9.81) 9.81
𝑷𝟐 = 𝟐𝟑𝟕𝟔. 𝟖𝟏𝟎𝑲𝑷𝒂
Another Solution:
BEE A → 2: (Assume Datum to be @ el.-30)
(𝑉𝐴 )2 𝑃𝐴
(𝑉2 )2 𝑃2
+ + 𝑍𝐴 + 𝐻𝐴 =
+ + 𝑍2 + 𝐻𝐿𝐴→1
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
0 + 0 + 40 + 203.1281
(3.169)2
𝑃2
=
+
+ 0 + 0.33177
2(9.81) 9.81
𝑷𝟐 = 𝟐𝟑𝟕𝟔. 𝟖𝟏𝟎𝑲𝑷𝒂
Another Solution: (Assume Datum to be @ el.-30)
BEE A →1 then BEE 1 → 2
(𝑉𝐴 )2 𝑃𝐴
(𝑉1 )2 𝑃1
+ + 𝑍𝐴 + 𝐻𝐴 =
+ + 𝑍1 + 𝐻𝐿𝐴→1
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
(1.141)2 𝑃1
0 + 0 + 40 + 203.1281 =
+ + 0 + 0.33177
2(9.81) 𝛾𝐿
𝑃1
= 39.601875
𝛾𝐿
(𝑉1 )2 𝑃1
(𝑉2 )2 𝑃2
+ + 𝑍1 + 𝐻𝐴 =
+ + 𝑍2
2𝑔
𝛾𝐿
2𝑔
𝛾𝐿
(1.141)2
(3.169)2 𝑃2
+ 39.601875 + 𝑍1 + 𝐻𝐴 =
+ + 𝑍2
2𝑔
2𝑔
𝛾𝐿
𝑷𝟐 = 𝟐𝟑𝟕𝟔. 𝟖𝟏𝟎𝟕𝑲𝑷𝒂