is immersed in water. If the atmospheric pressure head is 10.3 m of water and loss of head due to friction in the draft tube is equal to $0.2\times $ velocity head at outlet of the tube find pressure head at inlet and efficiency of the draft tube.

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 08

Years: MAY 2015

Given :

Diameter at inlet $D_1 = 1.0 m$

Diameter at outlet $ D_2=1.5m$

Velocity at outlet $V_2 =2.5m/s$

Total length of tube $H_s+y =6.0m $

Length of tube in water $y=1.3 m \\ H_s = 6.0-1.3 \\ =4.7m $

Atmospheric pressure head $\dfrac {P_a}{pg}=10.3 m $

Loss of head due to friction $hf = 0.2\times $ velocity head at outlet

$=0.2\times \dfrac {V_2^2}{2g}$

Discharge through tube $Q=A_2V_2\\ = \dfrac \pi4D_2^2\times 2.5\\ = \dfrac \pi4 \times (1.5)^2 \times 2.5 \\ =4.4178 m^3/s$

Velocity at inlet $V_1 = \dfrac Q{A_1} = \dfrac {4.4178}{\dfrac \pi4\times 1^2}= 5.625m/s $

i) Pressure head at inlet $(\dfrac {P_1}{pg})$

$\dfrac {P_1}{Pg}=\dfrac {Pa}{pg}- H_s - (\dfrac {V_1^2}{2g}-\dfrac {V_2^2}{2g}-hf)\\ =10.3-4.7-(\dfrac {5.625^2}{2\times 9.81}-\dfrac {2.5^2}{2\times 9.81}-0.2\times \dfrac {V_2^2}{2g})\\ =10.3-4.7-(1.6126-0.3185-0.0637)\\ = 4.369=4.37m$

ii) Efficiency of Draft tube $(N_d)$

$N_d = \dfrac {(\dfrac {V_1^2}{2g}-\dfrac {V_2^2}{2g})-hf}{\dfrac {V_1^2}{2g}} = \dfrac {\dfrac {V_1^2}{2g}-\dfrac{V_2^2}{2g}- \dfrac {0.2V_2^2}{2g}}{\dfrac {V_1^2}{2g}} \\ = \dfrac {V_1^2-1.2V_2^2}{V_1^2}= 1-1.2 (\dfrac {V_2}{V_1})^2 \\ =1-1.2(\dfrac {2.5}{5.625})^2 \\ = 1-0.237 \\ =0.763 \space or \space 76.3\%$