Page: Minor and Major energy losses
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The loss of energy due to friction in a pipe is known as a major loss.

"The loss of energy due to the changes of velocity of the fluid in the magnitude is called a minor loss of energy."

Loss of energy include the following cases:-

1) due to sudden enlargement:-

$h_e=\dfrac{(V_1-V_2)^2}{2g}$

$h_e$ = loss ofhead due to sudden enlargement

2) loss due to sudden contraction:-

$h_c=0.5.\dfrac{V_2^2}{2g}$

### Numericals:

Q1) Find the loss of head when a pipe of diameter 200 mm is suddenly enlarged to a diameter of 400 mm. The rate of flow of water through the pipe is 200 lit/s.

Solution: Given:

Diameter of smaller pipe

$D_1=200mm=\dfrac {200}{1000}=0.20m$

Area, $A_1=\dfrac \pi 4\times D_1^2$

$A_1=\dfrac \pi4\times (0.20)^2=0.03141m^2$

Diameter of larger pipe

$D_2=400mm=\dfrac {400}{1000}=0.40m$

Area, $A_2=\dfrac \pi 4\times D_2^2$

$A_2=\dfrac \pi4\times (0.40)^2=0.12564m^2$

Discharge, $Q=250 lit/sec=\dfrac {250}{1000}=0.25m^3/s$

Velocity, $V_1=\dfrac Q{A_1}=\dfrac {0.25}{0.03141}=7.96m/s$

Velocity, $V_2=\dfrac Q{A_2}=\dfrac {0.25}{0.12564}=1.996m/s$

Now to find the loss due to enlargement

$h_e=\dfrac{(V_1-V_2)^2}{2\times g}$

$h_e=\dfrac{(7.96-1.99)^2}{2\times 9.81}$

$h_e=1.816m$ of water

Q2) The rate of flow of water through a horizontal pipe is $0.25m^3/s$. The diameter of the pipe which is 200 mm is suddenly enlarged to 400 mm. The pressure intensity in the smaller pipe is $11.772N/cm^2$.

Find:-

1) loss of head due to sudden enlargement

2) pressure intensity in large pipe

3) power lost due to enlargement

Solution: Given:

Discharge, $Q=0.25m^3/s$

Diameter of smaller pipe

$D_1=200mm=\dfrac {200}{1000}=0.20m$

Area, $A_1=\dfrac \pi 4\times D_1^2$

$A_1=\dfrac \pi4\times (0.20)^2=0.03141m^2$

Diameter of larger pipe

$D_2=400mm=\dfrac {400}{1000}=0.40m$

Area, $A_2=\dfrac \pi 4\times D_2^2$

$A_2=\dfrac \pi4\times (0.40)^2=0.12566m^2$

Pressure in smaller pipe,

$P_1=11.772N/cm^2=11.772\times 10^4N/m^2$

Velocity, $V_1=\dfrac Q{A_1}=\dfrac {0.25}{0.03141}=7.96m/s$

Velocity, $V_2=\dfrac Q{A_2}=\dfrac {0.25}{0.12566}=1.996m/s$

(i) Loss of head due to sudden enlargement

$h_e=\dfrac{(V_1-V_2)^2}{2\times g}$

$h_e=\dfrac{(7.96-1.99)^2}{2\times 9.81}$

$h_e=1.816m$ of water

(ii) Let pressure intensity in large pipe = $P_2$

Apply Bernoulli's equation,

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

Since $z_1=z_2$

$\dfrac {P_1}{\rho g}+\dfrac {V_1^2}{2\times g}=\dfrac {P_2}{\rho\times g}+\dfrac {V_2^2}{2\times g}+h_e....................(1)$

Rewrite as

$\dfrac {P_2}{\rho g}=\dfrac {P_1}{\rho g}+\dfrac {V_1^2}{2g}-\dfrac {V_2^2}{2g}-h_e........................(2)$

Putting values in equation (2), we get

$\dfrac {P_2}{\rho g}=\dfrac {11.772\times 10^4}{100\times 9.81}+\dfrac {7.96^2}{2\times 9.81}-\dfrac {1.99^2}{2\times 9.81}-1.816$

$\dfrac {P_2}{\rho g}=12.0+3.229-0.2018-1.8160$

$\dfrac {P_2}{\rho g}=15.229-2.0178=13.21m$ of water

$\therefore P_2=13.21\times \rho g$

$P_2=13.21\times 1000\times 9.81$

$P_2=12.96N/cm^2$

(iii) Power lost due to sudden enlargement

$P=\dfrac {\rho.g.Q.h_e}{1000}$

$P=\dfrac {1000\times 9.81\times 0.25\times 1.816}{1000}$

$P=4.453kW$

Q3) A horizontal pipe of diameter 500 mm is suddenly contracted to a diameter of 250 mm. The pressure intensities in large and smaller pipe are given as $13.734N/cm^2$ and $11.772N/cm^2$. Find loss of head due to contraction if $C_c=0.62$. Also determine the rate of flow of water.

Solution: Given:-

Diameter of large pipe

$D_1=500mm=\dfrac {500}{1000}=0.50m$

Area, $A_1=\dfrac \pi 4\times D_1^2$

$A_1=\dfrac \pi4\times (0.50)^2=0.1963m^2$

Diameter of small pipe

$D_2=250mm=\dfrac {250}{1000}=0.25m$

Area, $A_2=\dfrac \pi 4\times D_2^2$

$A_2=\dfrac \pi4\times (0.25)^2=0.04908m^2$

Pressure in large pipe,

$P_1=13.734N/cm^2=13.734\times 10^4N/m^2$

Pressure in smaller pipe,

$P_2=11.772N/cm^2=11.772\times 10^4N/m^2$

To find the head loss due to contraction,

$=\dfrac {V_2^2}{2g}\left[\dfrac 1{C_c}-1.0\right]^2$

$=\dfrac {V_2^2}{2g}\left[\dfrac 1{0.62}-1.0\right]^2$

$=0.375\dfrac {V_2^2}{2g}$

From continuity equation,

$A_1V_1=A_2V_2$

$\therefore, V_1=\dfrac {A_2V_2}{A_1}=\dfrac {\dfrac \pi 4\times (D_2)^2\times V_2}{\dfrac \pi 4(D_1)^2}$

$V_1= \left[\dfrac {D_2}{D_1}\right]^2\times V_2$

$V_1=\left(\dfrac {0.25}{0.50}\right)^2\times V_2$

$V_1=\dfrac {V_2}4$

Apply Bernoulli's equation, ($Z_1=z_2$)

$\dfrac {P_2}{\rho g}=\dfrac {P_1}{\rho g}+\dfrac {V_1^2}{2g}-\dfrac {V_2^2}{2g}-h_c$

But $h_c=0.375\dfrac {V_2^2}{2g}$ and $V_1=\dfrac {V_2}4$

Putting values in equation, we get

$\dfrac {13.374\times10^4}{9.81\times 1000}+\dfrac {\left(\dfrac{V_2} 4\right)^2}{2\times 9.81}=\dfrac {11.772\times 10^4}{1000\times 9.81}+\dfrac {V_2^2}{2g}+0.375\dfrac {V_2^2}{2g}$

$14.0+\dfrac {V_2^2}{16\times 2\times 9.81}=12.0+1.375\dfrac {V_2^2}{2\times 9.8}$

$\therefore, 14-12=1.375\dfrac {V_2^2}{2\times 9.81}-\dfrac 1{16}\dfrac {V_2^2}{2\times 9.81}$

$2=1.3125\dfrac {V_2^2}{2\times 9.81}$

$\therefore V_2=\sqrt{\dfrac{2\times2\times9.81}{1.3125}}=5.467m/s$

$\therefore h_c=0.375\times\dfrac {V_2^2}{2g}$

$h_=\dfrac {0.375\times (5.467)^2}{2\times 9.81}=0.571$

Loss of head at the entrance of pipe:-

This loss is similar to loss due to sudden contraction. This loss occurs when a liquid enters A pipe which is connected to a large tank.

This loss is also dependent on the type of entrance.

$h_i=0.5\dfrac {V^2}{2g}$

Loss of head at the exit of pipe:-

This loss is mainly due to the velocity of liquid at the outlet of the pipe.

$h_o=\dfrac {V^2}{2g}$

Loss of head due to an obstruction in a pipe:-

This loss occurs when there is a reduction in the cross-section area of the pipe due to some obstruction. The area suddenly increases or enlarges after the obstruction.

$=\dfrac {(V_c-V)^2}{2g}=\dfrac {\dfrac {A\times V}{C_c(A-a)}-V}{2g}$

$\therefore$ Loss of head due to obstruction

$=\dfrac {V^2}{2g}\left(\dfrac A{C_c(A-a)}-1\right)^2$

Loss of head due to bend in a pipe:-

This loss occurs when there is a bend in a pipe which causes a change in velocity which leads to separation of the flow from the boundaries and it also forms eddies.

Thus the energy is lost.

$h_b=\dfrac {KV^2}{2g}$

Loss of head in various pipe fittings:-

$=\dfrac {KV^2}{2g}$

### Numericals

Q4) A horizontal pipeline 40m long is connected to a water tank at one end and discharges freely into the atmosphere at the other end. For the first 25 m of its length from the tank, the pipe is 150 mm diameter and its diameter is suddenly enlarged to 300 mm. The height of the water level in the tank is 8 m above the centre of the pipe. Considering all losses of head which occurs, determine the rate of flow. Take f=0.01 for both sections of the pipes.

Solution:

Given:

Total length of pipe, L=40m

Length of first pipe, $L_1=25m$

Diameter of first pipe, $d_1=150mm=0.15m$

Length of second pipe, $L_2=40-25=15m$

Diameter of second pipe, $d_2=300mm=0.30m$

Height of water, $H=8m$

Coefficient of friction, f=0.01

Applying Bernoulli's Theorem,

$0+0+8=\dfrac {P_2}{\rho g}+\dfrac {V_2^2}{2g}+0+\text{all losses}$

$8=0+\dfrac {V_2^2}{2g}+h_i+hf_1+h_e+hf_2................(A)$

where,

loss at entrance, $h_i=0.5\dfrac {V_1^2}{2g}$

Head lost due to friction in pipe 1, $hf_1=\dfrac {4\times f\times L_1\times V_1^2}{d_1\times 2g}$

Loss due to sudden enlargement, $h_e=\dfrac {(V_1-V_2)^2}{2g}$

Head lost due to friction in pipe 2, $hf_2=\dfrac {4\times f\times L_2\times V_2^2}{d_2\times 2g}$

But continuity equation,

$A_1V_1=A_2V_2$

$\therefore, V_1=\dfrac {A_2V_2}{A_1}$

$=\dfrac {\dfrac \pi 4 (d_2)^2\times V_2}{\dfrac \pi 4(d_1)^2}$

$=\left(\dfrac {d_2}{d_1}\right)^2\times V_2$

$=\left(\dfrac {0.3}{0.15}\right)^2\times V_2$

$V_1=4V_2......................(1)$

Substituting the values of $V_1$ in different losses, we get

$h_i=0.5\dfrac {V_1^2}{2g}=0.5\dfrac {(4V_2)^2}{2g}=\dfrac {8V_2^2}{2\times9.81}$

$hf_1=\dfrac {4\times f\times L_1\times V_1^2}{d_1\times 2g}=\dfrac {4\times 0.01\times 25\times (4V_2)^2}{0.15\times 2\times 9.81}=106.67\dfrac {V^2_2}{2\times 9.81}$

$h_e=\dfrac {(V_1-V_2)^2}{2g}=\dfrac {(4V_2-V_2)^2}{2g}=\dfrac {9V_2^2}{2\times 9.81}$

$hf_2=\dfrac {4\times f\times L_2\times V_2^2}{d_2\times 2g}=\dfrac {4\times 0.01\times15\times V_2^2}{0.3\times 2\times 9.81}=2.0\dfrac {V^2_2}{2\times 9.81}$

Put the values in equation (A), we get

$8=0+\dfrac {V_2^2}{2g}+\dfrac {8V_2^2}{2\times9.81}+106.67\dfrac {V^2_2}{2\times 9.81}+\dfrac {9V_2^2}{2\times 9.81}+2.0\dfrac {V^2_2}{2\times 9.81}$

$8=\dfrac {V_2^2}{2\times9.81}[1+8+106.67+9+2]$

$\therefore 8.0=126.67\dfrac {V_2^2}{2\times9.81}$

$\therefore V_2=\sqrt{\dfrac {8.0\times 2\times 2.91}{126.67}}$

$\therefore V_2=\sqrt{1.2391}$

$V_2=1.113m/s$

Therefore the rate of flow,

$Q=A_2\times V_2$

$Q=\dfrac \pi4\times (0.3)^2\times 1.113$

$Q=0.07867m^3/s$

or, $Q=78.67$ litres/sec

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