Two reservoirs are connected by a pipeline. the difference in the water levels is 6 m. the pipe is 1,2 m diameter and 720 m long. the pipe rises for the first 240 m to a height 3 m above the level of the first reservoir. take f =0.004 and ignore shock losses. determine the flow rate and gauge pressure at the highest point in the line.

the flow rate is 3.958 m³/s and gauge pressure at the highest point in the line is - 55.1814 kPa

Explanation:

Given data

pipe diameter (d) = 1.2 m

pipe length (Pl) = 720 m

f = 0.004

pipe rise (Pr) = 240 m

difference in the water levels =  6 m

height (Z(c)) =  3 m

to find out

the flow rate and gauge pressure

solution

we know flow rate i.e.

flow rate = volume × area     .................1

here area = × d² / 4 = × 1.2² / 4 = 1.130

and volume will be find out by bernoulli equation i.e

P(a)/ρg + v(a)²/2g + Z(a)  =   P(b)/ρg + v(b)²/2g + Z(b) + head loss (a to b )

we know V(a) = V(b) = 0 and head loss = 4f(Pl)v² / 2gd

and P(a) = P(a) = atmospheric pressure so equation will be

P(a)/ρg + v(a)²/2g + Z(a)  =   P(b)/ρg + v(b)²/2g + Z(b) + head loss

0 + 0 + Z(a) - Z(b) -0 -0 = 4f(Pl)v² / 2gd

we know difference in the water levels =  6 m so  Z(a) - Z(b) = 6m

4f(Pl)v²/ 2gd = 6

4×0.004 (720) v² / 2×9.81×1.2 = 6

v = 3.50 m/s

put volume and area in equation  1 we get

flow rate = volume × area

flow rate = 3.50 × 1.130

flow rate = 3.958

so flow rate is 3.958 m³/s

now for pressure at high point (c)  we apply bernoulli equation ( a to c) i.e

P(a)/ρg + v(a)²/2g + Z(a)  =   P(c)/ρg + v(c)²/2g + Z(c) + head loss (a to c )

we know P(a) and volume will be zero and pressure height is 3 m and volume at b is 3.50 m/s

so equation will be

0 + 0 + 0 -P(c)/ρg - v(c)²/2g - 3 =  4f(Pr)v² / 2gd

-P(c)/1000×9.81 - 3.50²/2×9.81  - 3 =  4× 0.004 (240) 3.50² / 2×9.81×1.2

P(c) = - 5.6250 × 1000 × 9.81 Pa

P(c) = - 55.1814 kPa

so pressure at the highest point is - 55.1814 kPa

1

explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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