Bow thruster motor rewinding

[Ian MacAulay]

Afternoon.

I have a underpowered bow thruster that is not up to the job but biggest available for my tunnel size.

Has anyone tried having the motor rewound in order to gain more power from the motor to give a bit more pushing power ?


Regards

Ian

 

[vas]

how big is the tunnel and how big is the motor?
for 43 ft 185mm dia tunnel is I think the typical size.
5hp, 7hp motor?
How many V, what's the voltage drop ?

To answer the question, doubt you're going to get much out of a rewind (assuming it's successful and the motor works as well as with the original winding).

cheers

V.

 

[Ian MacAulay]

It's a 50kgf vetus thruster that I believe uses a 150mm tunnel on the targa 43. I think the thruster is spec is 4hp

One load I see 12v at the motor

I've spoke to a rewind company who maybe able to recon it and wind it to 9v to give around 25% increase in power output


Regards

Ian

 

[vas]

hm,

that's outside my expertise, sorry cannot help.

Do all targa 43s have 150dia tunnel?
Will you get anything back out of a more powerful motor on a small tunnel?
Is the battery(ies) at the bow by the thruster?
I'd try to make sure I get all the power that the existing motor can give before embarking in such a job.

see what others's say.

cheers

V.

 

[petem]

Afternoon.

I have a underpowered bow thruster that is not up to the job but biggest available for my tunnel size.

Has anyone tried having the motor rewound in order to gain more power from the motor to give a bit more pushing power ?


Regards

Ian

Where's the battery Ian?

 

[superheat6k]

Will they guarantee it will work rewound for 9v. Applying 12v to a 9v motor will considerably increase the amps, and yes with it the power, but the heat losses are a factor of the square of the current, so there is a severe risk of burning the motor out in short order. Bear in mind also that you are already likely running the motor closer to 14v when the engine is running.

Assuming the original is say 750 w that is 62.5amps at 12volts. Reducing the voltage to 9v means reducing the winding resistance so the current flow will increase at the lower voltage to produce the same original power. So rewound and to produce the same power, the motor will now draw 83 amps (at 9 volts). So now you apply 12volts to a reduced resistance motor.

The motor resistance was originally 0.19 ohms. Now it has reduced to 0.11 ohms. At 12v the current will now be 109 amps. The power will have increased to 1308 watts. The overload being applied will be 109/83 or 31%.

Possibly OK in very short bursts. However, the cable rating will also need to be considered to handle the considerable increase in operating current, including volts drop due to the length of the overall cable run to and from the battery, and the capability of the battery will also need to be checked.

Might work, but perhaps not for very long.

 

[Ian MacAulay]

Where's the battery Ian?

They are under the garage mate and 95mm2 cable runs up to the front. I did use a battery at the front with no effect.

I have 3 270ah mastervolt batterys at the back

Ian

 

[Ian MacAulay]

Will they guarantee it will work rewound for 9v. Applying 12v to a 9v motor will considerably increase the amps, and yes with it the power, but the heat losses are a factor of the square of the current, so there is a severe risk of burning the motor out in short order. Bear in mind also that you are already likely running the motor closer to 14v when the engine is running.

Assuming the original is say 750 w that is 62.5amps at 12volts. Reducing the voltage to 9v means reducing the winding resistance so the current flow will increase at the lower voltage to produce the same original power. So rewound and to produce the same power, the motor will now draw 83 amps (at 9 volts). So now you apply 12volts to a reduced resistance motor.

The motor resistance was originally 0.19 ohms. Now it has reduced to 0.11 ohms. At 12v the current will now be 109 amps. The power will have increased to 1308 watts. The overload being applied will be 109/83 or 31%.

Possibly OK in very short bursts. However, the cable rating will also need to be considered to handle the considerable increase in operating current, including volts drop due to the length of the overall cable run to and from the battery, and the capability of the battery will also need to be checked.

Might work, but perhaps not for very long.

Wow a top answer there ��. I think the motor is on the way out so may be worth a gamble

Having said that they are used in short bursts but currently with its lack of power I have to hold my hand on it constantly to try getting it to do something.

Thanks

Ian

 

[Ian MacAulay]

hm,

that's outside my expertise, sorry cannot help.

Do all targa 43s have 150dia tunnel?
Will you get anything back out of a more powerful motor on a small tunnel?
Is the battery(ies) at the bow by the thruster?
I'd try to make sure I get all the power that the existing motor can give before embarking in such a job.

see what others's say.

cheers

V.

Not 100% on tunnel size as different places give different dimensions for the vetus motor. I will measure when the boats lifted next Tuesday

Thanks

Ian

 

[Ian MacAulay]

Will they guarantee it will work rewound for 9v. Applying 12v to a 9v motor will considerably increase the amps, and yes with it the power, but the heat losses are a factor of the square of the current, so there is a severe risk of burning the motor out in short order. Bear in mind also that you are already likely running the motor closer to 14v when the engine is running.

Assuming the original is say 750 w that is 62.5amps at 12volts. Reducing the voltage to 9v means reducing the winding resistance so the current flow will increase at the lower voltage to produce the same original power. So rewound and to produce the same power, the motor will now draw 83 amps (at 9 volts). So now you apply 12volts to a reduced resistance motor.

The motor resistance was originally 0.19 ohms. Now it has reduced to 0.11 ohms. At 12v the current will now be 109 amps. The power will have increased to 1308 watts. The overload being applied will be 109/83 or 31%.

Possibly OK in very short bursts. However, the cable rating will also need to be considered to handle the considerable increase in operating current, including volts drop due to the length of the overall cable run to and from the battery, and the capability of the battery will also need to be checked.

Might work, but perhaps not for very long.

Can I ask a favour and to apply the maths to my 3kw 4hp motor at 12v?

Regards

Ian

 

[zt260]

Can you get an upgraded propeller for it,a mate of mine changed the prop on his bowthruster on a targa 40 an said it seemed to make a differance


Ash

 

[vas]

Can I ask a favour and to apply the maths to my 3kw 4hp motor at 12v?

Regards

Ian

are you sure you want that???
Trevor, you have to tell us what size the cable will have to be for a 2X10m run (my estimate, but cannot be much shorter!)
I recon it's going to be scary big! and difficult to route.

How about re winding for 22V and fitting 2X12V 200Ah batteries at the bow with a thin wire to the back in order to charge them through an inverter or something? A kludge but may get more power out of such a conversion.

For the record on an 11ton, 43ft f/b I have 185mm tunnel and a 7HP or 5kW 24V thruster with two 3blade props. Works fine

V.

 

[Andy Bav]

Maybe a basic question, but has it been like it ever since you have had her?

On ours, the bowthruster is very efficient at the beginning of season, when you can feel a real jolt when it starts, mid season and with a lot of growth, it was more of a gentle nudge.

Last year we burnt out a 300 amp fuse in mid summer with about two 5 second bursts. Rather than a crane out, a go pro on a selfie stick revealed a very badly fouled tunnel, which was sorted with a Heath Robinson arrangement of a jet washer, lashed to a broom handle triggered by the the painter from the tender... worked a treat after that... and with a new fuse of course !

Definitely not as scientific an answer as above but it solved a similar problem for us

 

[penberth3]

Can you get an upgraded propeller for it,a mate of mine changed the prop on his bowthruster on a targa 40 an said it seemed to make a differance


Ash

A neat bit of lateral thinking there. The motor spins the prop, shouldn't be any problem with that, then the prop moves the water.

 

[exec2k]

You are right in saying the 5 is not man enough to move the bow of a T43 especially with the slightest bit of cross wind . I did try the upgraded propeller but to be honest it made no difference so last year I upgraded my Targa 43 to a 7.5 with slight alteration it took me one days work and the difference is incredible. Your tunnel will be 185 so this upgrade will be straight forward. I will see if I have pictures of the fitted motor to show layout.

 

[superheat6k]

Can I ask a favour and to apply the maths to my 3kw 4hp motor at 12v?

Regards

Ian

It is simply Ohms Law: V=IR and Power formula: P=IV where V is volts; I is amps or current; R is ohms; P is watts.

Also 1 HP = ~ 750 watts hence 3 kW = ~ 4HP (HP is the non metric method of measuring power)

So 3kW is 3,000 watts P = IV V=IR so P also equals I I R or IsqR (sorry can't do superscript on here).

P = 3000 V = 12 so I = P/V or 3000/12 = 250 a

R = P/Isq so 3000/(250x250) = 0.048 Ohms

You can see for any decent size motor on a low voltage that I (amps) becomes very high. Applying such amps to even fairly large conductors will result in massive heat generation. Even a properly designed motor used for high power low voltage applications will get quite hot quite quickly. The thing is that the heat must not build up a significant temperature within the windings exceeding the melt / burn temperature of the insulations, bearing in mind for most winding wires this insulation is just a few micron thick. The safe temperature limit for most motors (and indeed bearings within) is ~ 120oC. Things become fatal for the motors ~ 150oC. Once the insulations disintegrate things start going crazy as local short circuits develop as the insulations disintegrate.

A phenomenon of any induction based electric motor (most of them) producing significant power (except those on soft starters or inverters) is that they all experience overload conditions when they first start up and draw what is called Locked Rotor Current. This initial in rush causes a brief heating as the motor gets going, and then cools to normal temperatures. But if the motor is repeatedly started, each start introduces a fresh burst of heat into the windings, and after a few repetitive starts the motor internal temperatures can start to reach dangerous levels as far as the internal insulations are concerned, and bow thruster motors are particularly susceptible to this by the way they are used in bursts (repetitive starts), especially those rewound to a lower voltage.

Please note I have avoided here entirely the subject of inductive reactance and impedance, because the subject of induced emf and reactance within induction motors becomes far to complex for a thread like this. At standstill and as the motor first starts, however, it is the pure resistance only at play and hence the locked rotor current effect.

Well you did ask !

 

[Ian MacAulay]

It is simply Ohms Law: V=IR and Power formula: P=IV where V is volts; I is amps or current; R is ohms; P is watts.

Also 1 HP = ~ 750 watts hence 3 kW = ~ 4HP (HP is the non metric method of measuring power)

So 3kW is 3,000 watts P = IV V=IR so P also equals I I R or IsqR (sorry can't do superscript on here).

P = 3000 V = 12 so I = P/V or 3000/12 = 250 a

R = P/Isq so 3000/(250x250) = 0.048 Ohms

You can see for any decent size motor on a low voltage that I (amps) becomes very high. Applying such amps to even fairly large conductors will result in massive heat generation. Even a properly designed motor used for high power low voltage applications will get quite hot quite quickly. The thing is that the heat must not build up a significant temperature within the windings exceeding the melt / burn temperature of the insulations, bearing in mind for most winding wires this insulation is just a few micron thick. The safe temperature limit for most motors (and indeed bearings within) is ~ 120oC. Things become fatal for the motors ~ 150oC. Once the insulations disintegrate things start going crazy as local short circuits develop as the insulations disintegrate.

A phenomenon of any induction based electric motor (most of them) producing significant power (except those on soft starters or inverters) is that they all experience overload conditions when they first start up and draw what is called Locked Rotor Current. This initial in rush causes a brief heating as the motor gets going, and then cools to normal temperatures. But if the motor is repeatedly started, each start introduces a fresh burst of heat into the windings, and after a few repetitive starts the motor internal temperatures can start to reach dangerous levels as far as the internal insulations are concerned, and bow thruster motors are particularly susceptible to this by the way they are used in bursts (repetitive starts), especially those rewound to a lower voltage.

Please note I have avoided here entirely the subject of inductive reactance and impedance, because the subject of induced emf and reactance within induction motors becomes far to complex for a thread like this. At standstill and as the motor first starts, however, it is the pure resistance only at play and hence the locked rotor current effect.

Well you did ask !

I love that answer. ��

Ashamed to admit I should remember all these formulas. I'm a Cable Jointer by trade and joint every thing up to 33kv but not using formulas daily it begins to slip the mind but my understanding is still great on the subject.

Being so involved and interested in electrical engineering and so on brings me to why I've not also just gone and got a bigger thruster also. The thought of solving a problem and making use of what's already there had given me the drive to give it a go but looks like I may have to settle on trying to get a bigger thruster in there rather than throwing money at something that may just not last five minutes.

Thanks again


Ian

 

[superheat6k]

are you sure you want that???
Trevor, you have to tell us what size the cable will have to be for a 2X10m run (my estimate, but cannot be much shorter!)
I recon it's going to be scary big! and difficult to route.

How about re winding for 22V and fitting 2X12V 200Ah batteries at the bow with a thin wire to the back in order to charge them through an inverter or something? A kludge but may get more power out of such a conversion.

For the record on an 11ton, 43ft f/b I have 185mm tunnel and a 7HP or 5kW 24V thruster with two 3blade props. Works fine

V.

Now look I had a couple of ciders after my visit to our Air Con chiller at site at the National Maritime Museum Greenwich this afternoon, however, I will try my best to include not only the operating current but fair compensation for the volts drop over an extended run.

So 5 kW at 24volts (thats assuming you are not suffering too much in the way of cable losses).

5000 / 24 = 208 volts

Cable rating for 208 volts continuous is

50mm2 Max 204 a with a volts drop of 0.45 mV / a / m 0.45 x 208 x 20 = 1.87 v reducing the effective voltage to 22 v and with it the power to 22 x 208 = 4.5 kW

This would be fine for a short run, but not 20m IMHO.

70mm2 Max 259 a with VD 0.32 mV / a / m 0.32 x 208 x 20 = 1.3 v

95mm2 Max 321a PD 0.24 0.24 x 208 x 20 = 1v

The cable rated amps here are less significant, because these will take the cables to 80oC continuous, but for a bow thruster even a difficult berthing manoeuvre will only see the thruster working for a few minutes intermittently.

Also bear in mind that any cable joints, lugs, bolted connections, etc may not be up to scratch and are subject to corrosion and less than perfect transfer of the cable CSA across the joint. Most cable failure occur at joints and connections. A couple of bad joints and cable on its limit could see the overall circuit resistance badly affecting the actual voltage at the motor.

My personal view of a 20m run as described would be 70mm2, with minimal joints and connections. 50mm2 is under doing it and 95mm2 if it is a Superyacht.

 

[superheat6k]

I love that answer. ?

Ashamed to admit I should remember all these formulas. I'm a Cable Jointer by trade and joint every thing up to 33kv but not using formulas daily it begins to slip the mind but my understanding is still great on the subject.

Being so involved and interested in electrical engineering and so on brings me to why I've not also just gone and got a bigger thruster also. The thought of solving a problem and making use of what's already there had given me the drive to give it a go but looks like I may have to settle on trying to get a bigger thruster in there rather than throwing money at something that may just not last five minutes.

Thanks again


Ian

I still have to use these simple formula for the day job, selecting and sizing Inverters for large 3 phase compressor motors and then having to size the cabling and connectors correctly. Too big costs money, too small causes electrical overheating with consequent risk of electrical fire, so need to get this fairly right ! Conveniently power and ohms law calculations don't vary too much from 12vdc to 415vac 3ph.

I have noted that some 12 vdc equipment suppliers refer to lower rating charts for heavy cabling, but I think they work this on the basis that normally these are for starter motors rarely operating more than a few seconds. These may not be so suitable for more continuous loads, and I prefer to use my trusty industrial tables for calculations, then if I am challenged I can show my selections are valid, mind you not yet had any issues of under selection / cable overheating yet !

 

[vas]

Now look I had a couple of ciders after my visit to our Air Con chiller at site at the National Maritime Museum Greenwich this afternoon, however, I will try my best to include not only the operating current but fair compensation for the volts drop over an extended run.

So 5 kW at 24volts (thats assuming you are not suffering too much in the way of cable losses).

5000 / 24 = 208 Amps

Cable rating for 208 Amps continuous is

50mm2 Max 204 a with a volts drop of 0.45 mV / a / m 0.45 x 208 x 20 = 1.87 v reducing the effective voltage to 22 v and with it the power to 22 x 208 = 4.5 kW

This would be fine for a short run, but not 20m IMHO.

70mm2 Max 259 a with VD 0.32 mV / a / m 0.32 x 208 x 20 = 1.3 v

95mm2 Max 321a PD 0.24 0.24 x 208 x 20 = 1v

The cable rated amps here are less significant, because these will take the cables to 80oC continuous, but for a bow thruster even a difficult berthing manoeuvre will only see the thruster working for a few minutes intermittently.

Also bear in mind that any cable joints, lugs, bolted connections, etc may not be up to scratch and are subject to corrosion and less than perfect transfer of the cable CSA across the joint. Most cable failure occur at joints and connections. A couple of bad joints and cable on its limit could see the overall circuit resistance badly affecting the actual voltage at the motor.

My personal view of a 20m run as described would be 70mm2, with minimal joints and connections. 50mm2 is under doing it and 95mm2 if it is a Superyacht.

Trev,

you may want to replace volts with Amps on a couple of spots when the cider effect is over
didn't realise I have a superyacht, remember to buy you a couple more ciders if we ever meet!
I do have 95mm2 cables for the 10m run to my thruster, wasn't easy to route them around though.

cheers