1. PROBLEM
The discharge pipeline of water system consists of a 5500 m of 300 mm pipe joined by 3660
m of 450 mm pipe that connects to the base of a huge water reservoir. The difference in
elevation between the water surface in the reservoir and the center of 300 mm pipe at the
end of the line is 45.7 m. Neglecting velocity head and other losses in head due to
entrances and using coefficient of friction of 0.02 for both pipes.
➀ Compute the velocity of the water in the bigger pipeline.
➁ Compute the velocity of the water in the smaller pipe.
➂ Evaluate the rate of flow in the pipe.
Solution: w.s.
➀ Velocity of the bigger pipeline hf 1
hf1 + hf2 = 45.7 1
366
0 2 hf 2
0.0826 f L Q 2 450 m
hf1 = mm
550
D5 0
300 m
0.0826(0.02)(3660) Q12 mm
hf1 =
(0.45)5
hf1 = 327.66 Q12
0.0826(0.02)(5500) Q 22
hf2 =
(0.3)5
hf2 = 3739.09 Q 22 ➁ Velocity of smaller pipe
Q1 = Q2 Q 2 = A 2 V2
327.66 Q12 + 3739.09 Q22 = 45.7 Q 2 = Q1
4066.75 Q12 = 45.7
π
Q1 = 0.106 m3/sec. 0.106 = (0.3)2 V2
Velocity of bigger pipeline 4
V2 = 1.5 m / s
Q1 = A1 V1
π ➂ Rate of flow
0.106 = (0.45)2 V1
4 Q1 = Q2 = 0.106 m3/sec.
V1 = 0.666 m / s
, 2. PROBLEM
For his construction project in Southern Luzon, Engr. Santos commissioned a barge to
transport a 20 mm diameter by 6 m long steel reinforcing bars he purchased in Metro
Manila. When unloaded at the Pasig River (where the water is observed to have a sp.gr. of
1.01) the barge has a draft of 2.41 m and the portion of the barge at the water line measures
300 mm x 60 m. Loaded with the reinforcing bars and sailing Metro Manila Bay (where
seawater is observed to have a specific gravity of 1.03), the vessel chiefmate recorded a
draft of 4.97 m. If it costs five pesos to transfer a bar of the given dimension, determine how
much would it cost Engr. Santos for the transfer in thousands of pesos. Assume the bar to
have a constant horizontal section.
Solution:
Wt. of barge: W
W = BF1
W = 2.41(300)(60)(1.01)(1000)
W = 43,813,800 kg
Empty
2.41 m
W + Wb = BF2
W + Wb = (4.97)(300)(60)(1.03)(1000)
W + Wb = 92,143,800 kg sp.gr. = 1.01
BF1
Wb = 92,143,800 - 43,813,800
Wb = 48,330,000 kg (wt. of bars)
Unit weight of steel bar = 7850 kg/m3 W + Wb
48,330,000
No. of bars =
⎛ π⎞
7850 ⎜ ⎟ (0.02)2 (6)
⎝ 4⎠
4.97 m
No. of bars = 3,266,225 bars
steel bars
Total cost in thousands of pesos
sp.gr. = 1.03
5(3266225)
= BF2
1000
= 16331 (thousands of pesos)
The discharge pipeline of water system consists of a 5500 m of 300 mm pipe joined by 3660
m of 450 mm pipe that connects to the base of a huge water reservoir. The difference in
elevation between the water surface in the reservoir and the center of 300 mm pipe at the
end of the line is 45.7 m. Neglecting velocity head and other losses in head due to
entrances and using coefficient of friction of 0.02 for both pipes.
➀ Compute the velocity of the water in the bigger pipeline.
➁ Compute the velocity of the water in the smaller pipe.
➂ Evaluate the rate of flow in the pipe.
Solution: w.s.
➀ Velocity of the bigger pipeline hf 1
hf1 + hf2 = 45.7 1
366
0 2 hf 2
0.0826 f L Q 2 450 m
hf1 = mm
550
D5 0
300 m
0.0826(0.02)(3660) Q12 mm
hf1 =
(0.45)5
hf1 = 327.66 Q12
0.0826(0.02)(5500) Q 22
hf2 =
(0.3)5
hf2 = 3739.09 Q 22 ➁ Velocity of smaller pipe
Q1 = Q2 Q 2 = A 2 V2
327.66 Q12 + 3739.09 Q22 = 45.7 Q 2 = Q1
4066.75 Q12 = 45.7
π
Q1 = 0.106 m3/sec. 0.106 = (0.3)2 V2
Velocity of bigger pipeline 4
V2 = 1.5 m / s
Q1 = A1 V1
π ➂ Rate of flow
0.106 = (0.45)2 V1
4 Q1 = Q2 = 0.106 m3/sec.
V1 = 0.666 m / s
, 2. PROBLEM
For his construction project in Southern Luzon, Engr. Santos commissioned a barge to
transport a 20 mm diameter by 6 m long steel reinforcing bars he purchased in Metro
Manila. When unloaded at the Pasig River (where the water is observed to have a sp.gr. of
1.01) the barge has a draft of 2.41 m and the portion of the barge at the water line measures
300 mm x 60 m. Loaded with the reinforcing bars and sailing Metro Manila Bay (where
seawater is observed to have a specific gravity of 1.03), the vessel chiefmate recorded a
draft of 4.97 m. If it costs five pesos to transfer a bar of the given dimension, determine how
much would it cost Engr. Santos for the transfer in thousands of pesos. Assume the bar to
have a constant horizontal section.
Solution:
Wt. of barge: W
W = BF1
W = 2.41(300)(60)(1.01)(1000)
W = 43,813,800 kg
Empty
2.41 m
W + Wb = BF2
W + Wb = (4.97)(300)(60)(1.03)(1000)
W + Wb = 92,143,800 kg sp.gr. = 1.01
BF1
Wb = 92,143,800 - 43,813,800
Wb = 48,330,000 kg (wt. of bars)
Unit weight of steel bar = 7850 kg/m3 W + Wb
48,330,000
No. of bars =
⎛ π⎞
7850 ⎜ ⎟ (0.02)2 (6)
⎝ 4⎠
4.97 m
No. of bars = 3,266,225 bars
steel bars
Total cost in thousands of pesos
sp.gr. = 1.03
5(3266225)
= BF2
1000
= 16331 (thousands of pesos)
