Hydraulic gradient line:-

"The line which gives the sum of pressure head and datum head of fluid which is flowing in a pipe with respect to some reference line."

It can also be defined as the line joining all the top vertical coordinates, which shows the pressure head of fluid flowing in a pipe.

Total energy line:- (T.E.L)

The line which gives the sum of pressure head, datum head and kinetic head with respect to some reference line is known as Total Energy Line.

Numericals

Q1) The rate of flow of water pump into a pipe ABC, which is 200 m long and its diameter is 100 mm, while the length of the portion BC is also 100 m but its diameter is 200 mm. The change of diameter at B is sudden. The floor is taking place from A to C, where the pressure at A is $19.62N/cm^2$ and C is connected to a tank. Find the pressure at C and draw the hydraulic gradient and Total energy line. Take f= 0.008.

Solution: Given:-

Length of Pipe, ABC=200m

Discharge, $Q=20lit/s=0.02m^3/s$

Slope of pipe, i=1 in 40=$\dfrac 1{40}$

Length of Pipe, AB=100m

Diameter of pipe AB=100mm=0.1m

Length of pipe, BC=100m

Diameter of pipe BC=200mm=0.2m

Pressure of A, $P_A=19.62N/cm^2=19.62\times 10^4N/m^2$

f=0.008

Step 1: Velocity of water in pipe AB

$V_1=\dfrac Q{\text{Area of AB}}$

$V_1=\dfrac {0.02}{\dfrac \pi 4(0.1)^2}=2.54m/s$

Step 2: Velocity of water in pipe BC

$V_2=\dfrac Q{\text{Area of BC}}$

$V_2=\dfrac {0.02}{\dfrac \pi 4(0.2)^2}=0.63m/s$

Applying Bernoulli's equation to point A and C,

$\dfrac {P_A}{\rho g}+\dfrac {V_A^2}{2g}+z_A=\dfrac {P_C}{\rho g}+\dfrac {V_C^2}{2g}+z_C+\text{total losses from A to C}...............(1)$

Step 3: Total losses from A to C

• Loss of head due to friction in pipe AB,

$hf_1=\dfrac {4fLV^2}{d\times 2g}$

$hf_1=\dfrac {4\times 0.008\times 100\times (2.54)^2}{0.1\times 2\times 9.81}=10.52m$

• Loss of head due to friction in pipe BC,

$hf_2=\dfrac {4fLV^2}{d\times 2g}$

$hf_2=\dfrac {4\times 0.008\times 100\times (0.63)^2}{0.2\times 2\times 9.81}=0.323m$

• Loss of head due to enlargement at B

$h_e=\dfrac {(V_1-V_2)^2}{2g}=\dfrac {(2.54-0.63)^2}{2\times 9.81}=0.186m$

$\therefore$ Total losses from A to C $=hf_1+h_e+hf_2=10.52+0.186+0.323=11.03m$

Substituting the values in equation (1), we get

$\dfrac {P_A}{\rho g}+\dfrac {V_A^2}{2g}+z_A=\dfrac {P_C}{\rho g}+\dfrac {V_C^2}{2g}+z_C+11.03..................................(2)$

Taking datum line passing through 'A', we have

$z_A=0$

$z_C=\dfrac 1{40}\times \text{total length of pipe}$

$z_C=\dfrac 1{40}\times 200=5m$

Also, $P_A=19.62\times 10^4N/m^2$

$V_A=V_1=2.54m/s, V_C=V_2=0.63m/s$

Substituting the values in equation (2), we get

$\dfrac {19.62\times 10^4}{1000\times9.81}+\dfrac {2.54^2}{2\times 9.81}+0=\dfrac {P_C}{\rho g}+\dfrac {0.63^2}{2\times 9.81}+5+11.03$

$20.328=\dfrac {P_C}{\rho g}+16.05$

$\therefore\dfrac {P_C}{\rho g}=20.328-16.05=4.278m$

$P_C=4.278\times\rho\times g$

$P_C=4.278\times 1000\times 9.81N/m^2$

$P_C=\dfrac {4.278\times 1000\times 9.81}{10^4}N/cm^2$

$P_C=4.196N/cm^2$

Hydraulic Gradient and Total Energy Line

Q2) A pipeline, 300mm in diameter and 3200 M long is used to pump a 50 kg per second of oil whose density is $ 950kg/m^3$and whose kinematics viscosity is 2.1 Stokes. The centre of pipeline at the upper end is 40 m above than that at the lower end. The discharge at the upper and is atmospheric. find the pressure at the lower end and draw the hydraulic gradient line and Total energy line diagram.

Solution: Given:

Diameter of pipe, $d=300mm=0.3m$

Length of pipe, $L=3200m$

Mass, $m=50kg/s=\rho\times Q$

($\rho=950kg/m^3$)

$\therefore$ Discharge, $Q=\dfrac {50}\rho$

$Q=\dfrac {50}{950}=0.0526m^3/s$

• kinematic Viscosity, v=2.1 stokes

$v=2.1cm^2/s=2.1\times10^{-4}m^2/s$

• Height of upper end= 40m

• Pressure at the upper end = atmospheric = 0

-Reynolds number, $R_e=\dfrac {V\times d}{v}$

where $V=\dfrac QA=\dfrac {0.0526}{\dfrac \pi4\times(0.3)^2}=0.744m/s$

$\therefore, R_e=\dfrac {0.744\times 0.30}{2.1\times 10^{-4}}=1062.8$

As the flow of oil through the pipe is laminated,

$f=\dfrac {16}{R_e}$

$\therefore f=\dfrac {16}{1062.8}=0.015$

Step 1: Head lost due to friction,

$hf=\dfrac {4fLV^2}{d\times 2g}$

$hf=\dfrac {4\times 0.015\times 3200\times (0.744)^2}{0.3\times 2\times 9.81}$

hf=18.05m of oil

Step 2: Applying Bernoulli's equation at the lower end of pipe and taking datum line passing through the lower end

$\dfrac {P_1}{\rho g}+\dfrac {V_1^2}{2g}+z_1=\dfrac {P_2}{\rho g}+\dfrac {V_2^2}{2g}+z_2+hf$

But $z_1=0, z_2=40m, V_1=V_2$ (as diagram is same)

$P_2=0, hf=18.05m$

Therefore, substituting the values, we get

$\dfrac {P_1}{\rho g}=40+18.05=58.05m$ of oil

$\therefore P_1=58.05\times \rho g$

$P_1=58.05\times 950\times 9.81=540997N/m^2$

$\therefore P_1=54.099N/cm^2$

Step 3: Hydraulic gradient Line and Total energy line

$\dfrac {V^2}{2g}=\dfrac {(0.744)^2}{2\times 9.81}=0.0282m$

$\dfrac {P_1}{\rho g}=58.05m$ of oil

$\dfrac {P_2}{\rho g}=0$