I have a 24v square wave (0 to 24v ) feeding a solenoid. The frequency is 10hz with a 4% duty cycle, so that's 95ms off 5ms on.
Now I believe that Vrms is Vp* square root of on-time/ period 24* square root(4/95)= 4.9v
If the solenoid says it has a power rating of 2.5W and 10 ohms Do I say that Ip is 24/10 2.4A so peak power is 24* 2.4 5.76W
OR can i say that I rms = IP* squareroot of 4/95 =0.550A So Prms= 0.49* 4.9= 2.4Wrms?
So the 2.4Wrms is within the 2.5W spec of the solenoid

 

Welcome to AAC!
Your maths don't take into account the inductance of the solenoid. The inductance prevents the current reaching a maximum value instantaneously. It is also dependent on whether the solenoid armature is extended or retracted. Are you aware that suddenly switching off the solenoid will result in a large back-emf voltage spike, which will fry your square-wave generator unless you suppress the spike?

 

so peak power is 24* 2.4 5.76W

Recheck the maths here and following.

 

Welcome to AAC!
Your maths don't take into account the inductance of the solenoid. The inductance prevents the current reaching a maximum value instantaneously. It is also dependent on whether the solenoid armature is extended or retracted. Are you aware that suddenly switching off the solenoid will result in a large back-emf voltage spike, which will fry your square-wave generator unless you suppress the spike?

yes I have a diode and a transorb

 

Recheck the maths here and following.

Ok you got me there, but you haven't really answered my question

 

Welcome to AAC!
Your maths don't take into account the inductance of the solenoid. The inductance prevents the current reaching a maximum value instantaneously. It is also dependent on whether the solenoid armature is extended or retracted. Are you aware that suddenly switching off the solenoid will result in a large back-emf voltage spike, which will fry your square-wave generator unless you suppress the spike?

Also because the frequency is so low 10hz ,the inductive part is insignificant only the resistive element of 10R

 

Ok you got me there, but you haven't really answered my question

Check the rest of the maths.
if the signal is 5ms on and 95ms off, that is 5% duty cycle and that is 2.88W RMS which will be more than pushing your luck with a component rated at 2.5W.

 

Also because the frequency is so low 10hz ,the inductive part is insignificant

My LTspice simulation says otherwise, for a duty cycle of only 5%, unless the inductance is also insignificant. An inductance as little as 15mH has a noticeable effect, for example. Can you provide a link to the solenoid spec?

 

My LTspice simulation says otherwise, for a duty cycle of only 5%, unless the inductance is also insignificant. An inductance as little as 15mH has a noticeable effect, for example. Can you provide a link to the solenoid spec?

 

https://www.geeplus.com/wp-content/uploads/2019/08/VIBRO1.pdf

 

Recheck the maths here and following.

Usually the power rating for a solenoid is for constant duty, 100% on time. So even if the current is not constant, and your 4% on time peak power is 5.76 watts, the AVERAGE power over the whole time will be far less, and so heating will not be a problem at all.

 

My LTspice simulation says otherwise, for a duty cycle of only 5%, unless the inductance is also insignificant. An inductance as little as 15mH has a noticeable effect, for example. Can you provide a link to the solenoid spec?

the inductance is also small at 0.6mH

 

the inductance is also small at 0.6mH

Then at 5% duty cycle the solenoid will dissipate ~2.84W. The rated 2.5W assumes a massive heatsink at 20C. Do you have that?
As Albert says, you're pushing your luck.