Examples

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Examples

Energy Head

Problem. Water flows through the turbine in the figure at the rate of 0.214m³/s and the pressures at A and B, respectively, are 147 KPa and -34.5 KPa. Determine the power delivered to the turbine by the water.

Solution:
We need to determine first the Total Head and next calculate the Power.

1. In the main form tab the Pipes button and next the Energy Head button. You will be prompt to the Energy Head form, which contain the variables in the Bernoulli equation.

2. After finding the velocities in A and B using V=Q/A where Q is the rate of flow and A is the area. We obtain VA=3.03 m/s and VB=0.758 m/s. Choose the variable correspondent to the Turbine Head in this case E. extract (Energy Extracted by the turbine). Enter the variables known.

Fill all the fields, the unknown fields enter 0 (zero). If you have the Pressure Head in m or ft, µ=1 µ is the specific weight of the fluid. So P/µ is the pressure head. It shows that µ must not be zero 0

3. Finally tab Calc to get the result.

4. Know replace the Turbine Head HT in the equation of Power

Power = µ Q HT

Power = (9.79)(0.214)(20) = 41.9 kW

Remember ! Water Flow is a helpful application. But, just your analysis combined with this application will bring the right results.

Hydraulic Jump

Problem. A rectangular channel 16 ft wide carries a flow of 192 cfs. The depth of water on the downstream side of the hydraulic jump is 4.2 ft. What is the depth upstream?

Solution:

1. In the main menu tab the Hydraulic Jump button. You will be prompted to a form with different variables in a hydraulic jump.

2. Before of finding the depth after the jump. We should find the critic depth. So, tab the Y critic button and next the button of the variable known, in this case tab Flow Known. Fill out all the fields and press Calc to get the result of Y critic.

3. Finally we can find the depth after the jump. Tab the back icon on the top right of the form and tab again. Next tab the Go button next to Y2. Enter the values and the Y critic of 1.64754

4. To find Y1 start iterating to find Y1 and compare with Y2= 4.2 ft choose Y1=1 ft. replace it in the form. Yc = 1.6475

Y1 = 1 then Y2 = 2.7511 ft

Y1 = 0.4 then Y2 = 4.63 ft
Y1 = 0.479 then Y2 = 4.2 ft Okay!