## Converting Pressure to Power

Discussions on classical and modern physics, quantum mechanics, particle physics, thermodynamics, general and special relativity, etc.

### Converting Pressure to Power

I am curious how I can calculate the amount of Power - maybe in Watts, that it takes to propel a model boat.

My first thoughts are that if I can measure the pressure applied to the hull by the motor, and say the Propeller is say, a square centimeter in area, if the pressure produced was .1 kilograms,

does that provide measure off .1 Kilograms per square centimeter, that converts to a Force measurement of 0.980665 newtons per square centimeter (http://www.allconversions.com/)

If I can apply that Force Unit x the mass of the boat , does that give me the Power required measurement ?
Dnomyar

### Re: Converting Pressure to Power

Dnomyar » Thu Sep 11, 2014 4:25 pm wrote:...
If I can apply that Force Unit x the mass of the boat , does that give me the Power required measurement ?

Force times speed = power

Say the force is 1 Newton, and the speed attained with that force is 1 meter/second.

Then the power is what you get by multiplying the two: 1 Newton meter per second = 1 Joule per second
= 1 Watt.

However this does not take account of INEFFICIENCIES. For example inefficiency of the electric motor rotating the shaft. Inefficiency of the propeller converting rotation into forward thrust.

Energy may be lost thru friction and thru TURBULENCE caused by the propeller, so some power may end up as turbulence that just heats the water.

But ignoring losses, the power (work per unit time, joules per second, newton meter per second) is going to be equal to the force pushing the boat multiplied by the speed the boat is going.
Marshall

### Re: Converting Pressure to Power

That's great to confirm that

Force times speed = power

Say the force is 1 Newton, and the speed attained with that force is 1 meter/second."

Then the power is what you get by multiplying the two: 1 Newton meter per second = 1 Joule per second
= 1 Watt."

The only confirmation that remains is :

the part that calculates the Force component, would you say that can be calculated as pressure per area ?

To my mind, the 'gotchas' of prop inefficiency etc should be able to be ignored if I take the actual Pressure being supplied by the motor, while the boat is travelling at the actual speed.

This is the reason I didn't opt for just measuring the Wattage from the electric motor, as if I was using 1 Watt of Power on the Motor,I didn't know what proportion of the that power was being used by controller inefficiencies, prop performance and motor friction etc.
Attachments
Calculating pressure from electric outboard
Dnomyar

### Re: Converting Pressure to Power

That strikes me as a neat idea. But you should get Canady (full name CanadysPeak) to discuss it with you. He's more the mechanical engineer type.

You are trying to measure how much mechanical power the electric outboard is actually delivering.

You can measure the speed, so you need to measure the force that the outboard is applying to the boat.

You have a kind of see-saw set up on the rear of the boat.

You need to accurately estimate the LEVER ARMS between the gauge connection (up) and the propeller (down).

if the two distances were equal then the force measured by the gauge would be the same as the forward force of the propeller, they would balance.

But if the up lever arm to the gauge is TWICE as long as the down lever arm to the propeller than it will take only HALF as much gauge force resistance to balance the forward thrust force of the propeller.

so the relation between the forces, the ratio, depends on the ratio of the effective lever arms.

That makes it more complicated. In your picture the lever arms are not equal, the little blue "fulcrum" is not in the middle of the see-saw.

Let me know if the words I'm using are unfamiliar (I'll try to say it a different way.) Some of our members speak American English and some British. I'm in the US.
Marshall

### Re: Converting Pressure to Power

Well observed Marshall.

That diagram is not fully explanatory, (better one attached) but the concept is that the outboard motor has to have a 'tilt arm' to control the Trim of the outboard anyway, eg [youtube]qsePLSOpZNQ[/youtube]

and this may be a convenient way to incorporate a pressure sensor.

If I build a pressure sensor into this arm, it will be directly linked to the amount of thrust produced by the propeller.

Yes, the leverage effect will have a direct affect on the amount of pressure, but I am going to have to manually calibrate the indicator, so if that is done with the controls hooked up, that should be taken into account.

Attachments
Improved diagram.
Dnomyar

### Re: Converting Pressure to Power

Dnomyar » Thu Sep 11, 2014 7:25 pm wrote:I am curious how I can calculate the amount of Power - maybe in Watts, that it takes to propel a model boat.

My first thoughts are that if I can measure the pressure applied to the hull by the motor, and say the Propeller is say, a square centimeter in area, if the pressure produced was .1 kilograms,

does that provide measure off .1 Kilograms per square centimeter, that converts to a Force measurement of 0.980665 newtons per square centimeter (http://www.allconversions.com/)

If I can apply that Force Unit x the mass of the boat , does that give me the Power required measurement ?

Neither pressure nor power will answer your question. The quantity you seek is thrust, measured in either pounds or Newtons. Since you are looking at a model boat, simply set a tank on a table, put the boat in the tank, tie a string to the bow, run the string level, use a pulley to turn the string and add weights till you get the performance you want. That's the thrust you need.

The amount of thrust, in general, depends on the size of the boat, the speed, and the drag. The weight of the boat will show up through draft. All this is rather difficult to calculate, easy to measure by trial and error. There are all sorts of trade offs in propeller size, pitch, and speed, but you probably will end up choosing a motor based on torque; when you know the torque and the prop specs, you can figure the thrust.

Once you have the desired torque, and the speed, you'll be able to derive the power (more or less). If you need a starting place, see if you can find a copy of Model Boats from around 2011 that covered a lot of this.

http://www.modelboats.co.uk/

### Re: Converting Pressure to Power

I'm sorry - I am more confused by your post. THRUST = The force acting perpendicular to a surface is called thrust.

THUST - the propulsive force of a jet or rocket engine.
"the engine was a Russian-built Nene of higher thrust than the original models"
synonyms: force, motive force, propulsive force, propulsion, drive, driving force, actuation, impetus, impulse, impulsion, momentum, push, pressure, power

This is how I understand the process at the moment
Pressure x Area = Force (to be measured by this experiment )
Force x Velocity = Power. ( How much the motor had to produce )

I need to know how much definable Power is required to achieve each level of Velocity.

The limitations on the project are
1) It is an independently operating model - table top experiments will not provide the info, as hull performance ( planing, non planing, wave action ) are all part of the results.
2) I cant simple measure the Electrical Energy required, as losses through the electric motor, speed controller, shaft drive will lead to errors
Dnomyar

### Re: Converting Pressure to Power

Dnomyar » Fri Sep 12, 2014 11:11 pm wrote:I'm sorry - I am more confused by your post. THRUST = The force acting perpendicular to a surface is called thrust.

THUST - the propulsive force of a jet or rocket engine.
"the engine was a Russian-built Nene of higher thrust than the original models"
synonyms: force, motive force, propulsive force, propulsion, drive, driving force, actuation, impetus, impulse, impulsion, momentum, push, pressure, power

This is how I understand the process at the moment
Pressure x Area = Force (to be measured by this experiment )
Force x Velocity = Power. ( How much the motor had to produce )

I need to know how much definable Power is required to achieve each level of Velocity.

The limitations on the project are
1) It is an independently operating model - table top experiments will not provide the info, as hull performance ( planing, non planing, wave action ) are all part of the results.
2) I cant simple measure the Electrical Energy required, as losses through the electric motor, speed controller, shaft drive will lead to errors

I gave you the answer. You want some other answer. Good luck to you.

### Re: Converting Pressure to Power

No, you didnt give me any answer - I need the Math, not a concept, and within the limits of the experiment which cant be done on a table top.

Force x Speed = Power
I have "The power is what you get by multiplying the two: 1 Newton meter per second = 1 Joule per second = 1 Watt.""

Now I would like to confirm I can arrive at the Force component by measuring Pressure, or any other practical way.

Maybe someone else has a better handle on the problem.
Dnomyar

### Re: Converting Pressure to Power

Oh - I found the formula for Thrust ( or Force as it is more commonly known )
thrust F must overcome both the weight (m*g) of the rocket and add to that the acceleration force (m*a):

F = m*g + m*a
= m*(g+a)
= 6(9.8+3.6)
= 80.4 N

If we can get the Power needed by the Engine out of that, that would be great, but I cant see how.
Dnomyar

### Re: Converting Pressure to Power

Dnomyar » Sat Sep 13, 2014 7:18 am wrote:No, you didnt give me any answer - I need the Math, not a concept, and within the limits of the experiment which cant be done on a table top.

Force x Speed = Power
I have "The power is what you get by multiplying the two: 1 Newton meter per second = 1 Joule per second = 1 Watt.""

Now I would like to confirm I can arrive at the Force component by measuring Pressure, or any other practical way.

Maybe someone else has a better handle on the problem.

I'm quite sure that many have a better handle on the problem. Good luck to you.

### Re: Converting Pressure to Power

So that you can solve this, here is a white paper on the problem,

http://www.fao.org/docrep/x0487e/x0487e04.htm

### Re: Converting Pressure to Power

Its a very interesting paper, thank you but it doesn't provide me with a solution.

Irrespective of whether my propeller is efficient or not, my hull shape is efficient or not, whether my engine size is optimal or not , I need to know how much Power is required of the existing engine to propel the boat at whatever speed it reaches.

I am pretty sure I can measure how much 'push' the engine is providing at a given speed.

I am pretty sure that this 'pressure' can be converted to an equivalent Power measurement,(be it Watts or HP )if I know the speed of the craft at the time I know the Pressure being exerted.

If I change the hull shape, Prop size, boat weight I should be able to graph the effect on the amount of Power the engine has to use in different configurations.
Dnomyar

### Re: Converting Pressure to Power

I may be wrong, but I have the impression you can solve your problem by monitoring the velocity and knowing the mass of the boat.

Starting with the boat still, give a fixed amount of gas (set the power regimen of the motor to a certain value).
Record such velocity as a function of time.
Square it, multply by the mass of the boat (m = weight/9.8) and divide by two.
Plot the resulting function of time = Ek(t): the kinetic energy of the boat.
Now consider the time derivative of Ek: dE/dt = W (power). This is the power delivered by the motor and translated into an increase in kinetic energy.

If we assume that the motor is continually delivering the same power to the boat, then such power initially fully goes into kinetic energy (call it "kinetically" active power, Wk), then gradually shifts to contrasting the forces that oppose to the advancing boat ("resistively dispersed" power, Wr), until, at steady state, it fully goes in contrasting Wr.

Thus, the total power delivered by the motor to the boat (at this particular regime) is Wt = Wk(0), and
Wr(t) = Wk(o)-Wk(t).

Notice that such value of Wr(t) is the power needed to maintain the velocity of the boat (against shear, frictions and all that) at the value v(t), with no change in kinetic energy.

Thus, you can plot Wr vs. v (see the next figure), and this should give you the wanted answer: power needed by the boat to mantain velocity v, or velocity that can be maintained by delivering to the boat a power Wr.

If this theoretical model is right, you should be able to repeat the experiment and measurements at several values of gas (several power regimens of the motor), and the Wr vs. v plots should superimpose.

(note that the shapes of the curves plotted here are totally arbitrary)

neuro
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### Re: Converting Pressure to Power

Hi Neuro

Many thanks for that analysis. It is definitely food for thought. I may need to investigate further if my current ideas dont work out. I need to say that I have an electric powered boat, so the fuel component may need to be thought about a bit.

In a couple of other discussions, I have been discussing some suggestions provided earlier in this thread.

I have come up with a simplified illustration (attached) of what I think will work, and a formula which seems to fit.

The idea is to measure the Force provided by the outboard motor, as Weight.
Lets say, the Force ( Thrust, Push etc ) can be measured as say, 1 Kilogram.

The 'earth' value of a Newton is 9.81.

So the Force equivalent to a Kilo is the force of 9.81 Newton,
say the speed attained with that force is .5 meter/second. ( 2 klm/hr)

Then the power is what you get by multiplying the two:
4.9 Newton meter per second
= 4.9 Joule per second
= 4.9 Watt.

That gives me the Power needed to drive the boat at 2 Kilometers per hour as 4.9 Watts in theory.

Does this seem a a sound method ?
Attachments
Dnomyar

### Re: Converting Pressure to Power

Well, if you actually have a measure of the force applied to the boat, this is obviously a sound method.
But it is - in a sense - the starting point of the thread: obviously, power = force x velocity, but it seemed to me that your problem was that it is not easy to measure the force applied to the boat (because not all the power developed by the motor is conveyed into the thrust applied to the boat...).

neuro
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Posts: 2618
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Location: italy

### Re: Converting Pressure to Power

Thanks for the confirmation Neuro, it is helpful. I hope you thought caefully about the principle of substituting Newtons in place of Weight, and not just assumed that was correct.

Just thinking the process through from your comment.

"(because not all the power developed by the motor is conveyed into the thrust applied to the boat...)."

I understand that if the motor will use X Watts, but only X * .7 say will be applied to boat propulsion due to transmission losses, generation inefficiencies etc.

I have no trouble finding the Total Power sent to the Motor, as I can put a Meter on the current from the Battery.

What I need is the amount of Power that the boat requires to achieve certain speeds. This will tell me how big a motor I need on a full size boat. It will also allow me to make changes to the scale boat hull and observe improvements or reductions in performance under power.

The big trick will be to accurately measure the Force on the boat exerted by the motor. In an ideal world I could put a digital pressure sensor between the boat and the hull, but at small scale its not possible to get accurate enough readings.

So, I think I can arrange an Analogue setup - and get a mechanical indicator to display the Force results. This mechanical meter will be calibrated using set weights, hence the need to convert Weight to Newtons.

It seems that i have a workable formula now, so that allows me to go ahead and design the mechanism.

Thanks again
Dnomyar

### Re: Converting Pressure to Power

If you are proposing to extrapolate a small model to a big boat, I'd not be so sure that you can do it easily.

If the problem were to slide something on the floor, you'd simply compute the friction force as the weight of the object times the coefficient of friction. This tells you that such coefficient of friction is easily computed, and then everything is simple and linear.

(to clearly understand this, if you push something to speed 10 km/h and then stop pushing, it will continuously and constantly decelerate, the speed decreasing linearly to zero)

On the other hand, you know well that if you stop the motor of your boat it wll abruptly decelerate, change its position with respect to the water, and then gradually continue to decelerate until it finally - very slowly - it stops.
This tells you that the forces that oppose to the boat are quite different depending on the speed, the position of the boat (how much it is immersed, whether the front s elevated or not, and considering the possibility that at high speed, if the bottom is flat, the boat can swhitch to a condition of "sliding" on water surface as if on ice), and the amout of vorticosity it creates in water.

I'm quite sure that all this cannot be exactly computed, even by hydrodynamic engineers, due to the problem that laminar fluxes are easily modeled but vortices are not.

This, however, also suggest that if 2W are needed to drive a 2Kg boat to 10 km/h, there is no equation that lets you extrapolate from this the power needed to drive a 200 kg boat to the same speed, because the position of the big boat will not be the same as that of the small one (at the same speed) and the resistance opposed by water will not be the same.

neuro
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Posts: 2618
Joined: 25 Jun 2010
Location: italy

### Re: Converting Pressure to Power

I understand what you are saying Neuro, but luckily I dont think i will have to compute something so difficult.

All I intend to do is measure the force between the propulsive unit, and the boat.

In the previous diagram, I show a green arrow. Imagine that is a piston. The pressure of the motor propelling the boat can be measured, and the resultant force used in the computation.
Dnomyar