Paddle Wheel Torque Calculations

[bertbcfm]

I'm designing a paddle wheel drive house boat and need to do some calculations to see if the parts I have on hand can do the job.

Paddle wheels will be 6' dia. with 8 paddles by 3' long . Paddles will be 3' long by 10" and will have the tops of the paddles submerged by 2" to 3" when the paddle is vertical.

I want the mid to high end RPM to be about 56 Rpm. I figure this will give me about 10 Mph with 20% loss from the hull and paddle efficiency loses. Hull is 3 - 25' aluminum pontoons no more than 50% submerged.

I'll be driving the paddle wheels with 2 Hydraulic motors that will Rpm at 556 with 458 lb-in torque. Probably chain drive reduced 10 to 1. Flow and pressure are not an issue to produce, but I'm wondering if the motors will be up to the task.

Any help or direction to resources would be appreciated.

Regards
Bert

 

[Baluncore]

Hi Bert.
First, a belated welcome to PF.

There is no mention of power or energy in your question.
Rule number one; follow the energy.

We will first need to know the power required to push the three displacement hulls through the water.
To estimate that power we need the hull profile, the displacement and the speed.
Convert 10 mph to v = 8.7 knots.
I will assume the 25 foot long hulls are hard chine, and the displacement is 3 tonne.

Based on my old diagram, that makes k = 2.4 where v = k * Sqrt( HP / d ); in units of knots, horsepower and tonnes displacement.

We now consider each hull independently, each displacing 1 tonne. HP = d * ( v / k )²;
HP = ( 8.7 / 3.4 )² = 6.55 HP per hull. Total HP = 19.64, call it 20 HP.
Each hydraulic motor will need to develop 10 HP, = 7.5 kW.

You specified motor torque = 458 lb·in and speed 556 RPM.
I convert that to 51.75 Newton and 58.25 rad/sec.
Power is the product = 51.75 * 58.25 = 3015 watt = 3 kW.

The power available from the motors is less than half that needed to push the boat through the water at 10 mph.

My displacement guess of 3 tonne is probably an underestimate, so the propulsion system you are planning is significantly under-powered.

If this question had been in an engineering forum it might have received a faster reply.

 

[sami173]

What is k in the HP formula and which diagram are you referring to

 

[berkeman]

What is k in the HP formula

k = 2.4 where v = k * Sqrt( HP / d ); in units of knots

 

[Baluncore]

Based on my old diagram, that makes k = 2.4

The 2.4 was an empirical coefficient for hard chine displacement hulls.

The diagram was in an old book on boat design, that gave a guide to hull speeds for powers, with different types of hull and mode of operation. I will see if I can find it again. There are probably better references now.

 

[Baluncore]

@sami173 Welcome to PF.

Fast Boats. A Guide to Speed Under Sail and Power.
By John Teale. 1961. Diagram is on page 27.
For a 25 foot water line length, the rounded hull had k = 2.4 while the hard chine had k = 2.9
I conservatively reduced the speed estimate for power by using the lower k value for a rounded hull.

 

[sami173]

Can i get value of k for a pontoon

 

[Baluncore]

Can i get value of k for a pontoon

It depends on the section, and how fast you push the pontoon.
At slow speeds, it will act like a rounded hull or a hard chine, depending on the section.
If it planes, with the nose out of the water, it becomes a sea sled.