Maximum flow rates through water turbines given power and head.

[PhysicsStudy]

Homework Statement

Calculate the maximum flow rates through the Francis and the Samson turbines under the conditions specified:

Table 1 Characteristics of some American water turbines, 1849–97, on the basis of 30-inch (760 mm) wheel and 12-inch (300 mm) head
Type | Maximum power output | Efficiency (%)
Francis | 0.15 kW / 0.20 horsepower| 79.7 at full power / 55.0 at half power
‘Samson’ | 1.38 kW / 1.85 horsepower | 82.0 at full power / 75.6 at half power

Homework Equations

Power = turbine efficiency * density of water * acceleration due to gravity * head * flow rate

so

Flow rate = Power / (turbine efficiency * density of water * acceleration due to gravity * head)

The Attempt at a Solution

Does anyone know if the efficiency in this equation should be given as a percentage or fraction? i.e. 79.7% or 0.797

Francis example...

Density of water = 1000kg/m^3
Acc due to gravity = 9.81 m/s^2

At full power:
Flow rate = 150/(0.797*1000*9.81*0.3)=0.064m^3/s

At half power:
Flow rate= (150/2)/(0.55*1000*9.81*0.3)=0.046m^3/s

However I have a feeling I'm missing a step. I'd be grateful for anyone who could take a look. :) Thanks.

 

[edgepflow]

I think your method and arithmetic is OK.

Check if you should add the radius of the wheel to the head.

 

[W R-P]

looks ok to me

 

[PhysicsStudy]

I think your method and arithmetic is OK.

Check if you should add the radius of the wheel to the head.

Ah that's a good suggestion. Looking at the context and previous questions of the textbook though I don't think I need to. Thanks. :)

 

[DeeNos]

Your approach to the problem is correct. However, you are correct in thinking that there may be a missing step.

In the given table, the maximum power output is given in kilowatts and horsepower. However, the equation you are using to calculate the flow rate requires the power to be in watts. So, you will need to convert the given power values to watts before plugging them into the equation.

1 kilowatt = 1000 watts
1 horsepower = 745.7 watts

So for the Francis turbine at full power:
Power = 0.15 kW = 0.15 * 1000 = 150 watts

Similarly, for the Francis turbine at half power:
Power = 0.20 horsepower = 0.20 * 745.7 = 149.14 watts

Once you have the power values in watts, you can use your equation to calculate the flow rates at full and half power for both the Francis and the Samson turbines. Make sure to use the correct efficiency values for each case (79.7% for full power and 55.0% for half power for the Francis turbine, and 82.0% for full power and 75.6% for half power for the Samson turbine).

I hope this helps!