You cannot ask this question without defining the load.

If the load is such that the motor was sized to operate it when running in Delta, then running it in Star will result in the motor being severely overloaded, it will stall and hopefully a protective device will operate to disconnect it before it catches on fire. In either case it will cease to perform the work it was expected to perform and the application is a failure.

If the connected load was 1/3 (33%) of the designed power capacity of the motor, AND IT WILL NEVER INCREASE BEYOND THAT VALUE, then running it in Star will be sufficient to maintain the work being performed and the motor iron losses will be ever so slightly less than if it had been run in Delta, so you will consume a tiny bit less energy. The LOSSES in a motor can be divided into 4 main categories:

1. Iron losses, those that are associted with magnetizing the stator and rotor cores,
2. Resistance losses; those associted with resistance of the winding conductors;
3. Mechanical losses, those associted with issues like bearing frictuion and cooling fan windage; and finally
4. Stray losses, whatever remains, typically those associted with eddy currents in the steel, minor current imbalances etc.

Of those, the only one that is affected by the applied voltage are the iron losses. The iron losses comprise roughly 15% of the total losses, reducing the voltage can reduce those losses by maybe 5%. But keep it in perspective. That is 5% of the LOSSES, not 5% of the motor energy. If a motor has a rated efficiency of 90%, then 10% is losses. Of the 10% that are losses, 95% is not going to be affected by voltage reduction. So what you can have an affect on is 5% of 10% , which is only 0.05% of the power consumed.

So let's take an example. Say you have a 15kW motor that is 90% efficient and it is connected to load that only requires 5kW to operate with the desired effect (work performed). Connected in Delta, the 15kW motor uses the 5kW of the load, plus another 0.5kW in losses for a total absorbed power of 5.5kW from the power source. If you connect that in Star, the load still takes 5kW, but now the losses are lower by 0.05% or 0.4975kW so the absorbed power is 5.4975kW. You have saved 2.5W of energy. Hard to get excited by that and for that savings, you are risking that your application will stall and fail if the load ever changes.

But if the motor permanently only needs to deliver 33% of its rated power, then someone selected the wrong motor for the task in the first place.

If the motor runs OK for 24hours I assume the load is within its capability, of course.

If it runs OK in both Star and Delta configuration then the supply is matched to Delta, connected in Star the supply voltage is 1/√3 of nameplate e.g. 400V on 690V connection.

In this mode the rated power is this ratio squared so (1/√3)² = 1/3, so your load is less than 33% of nameplate power for the motor to be OK.

The slip curve flattens in Star connection so the motor runs slower which means the load is running slower which normally means power is lower too.

So, in Star, kW is similar (maybe slightly lower as explained), current is probably lower, RPM is slightly lower but similar, voltage is same(??), cos phi is improved, efficiency marginally improved.

The danger of this mode is that if there is a small increase in load, this can easily lead to an overload condition, the slip increases, pf and efficiency drop, current rises but maybe not so significantly and you can potentially and easily overheat the motor without noticing if not thermally protected by thermistors etc.

Hope this answers your query OK