The decision between HV and LV distribution is usually driven by capital cost and safety considerations, with energy efficiency being a lesser objective.
The choice depends on:
- the size of the building
- the size of the loads
- the expense of electrical equipment with appropriate continuous current rating and fault current rating, including:
- transformers
- cables
- copper busbars
- circuit breakers.
If your "medium sized industrial building" contains a number of arc furnaces, each of which requires thousands of amps, you are likely better off having high voltage distribution.
The high fault currents involved in a low-voltage solution are difficult to deal with - while it is possible to build one big transformer feeding an 8,000 amp switchboard, it is expensive to build an 8,000 amp switchboard to safely withstand a 100 kA fault level.
It is cheaper and safer to have two transformers feeding 2 × 4,000 amp switchboards rated for 50 kA fault level, or four transformers feeding 4 × 2,000 amp switchboards rated for 25 kA fault level.
If your "medium sized industrial building" contains things like lathes, welding sets, mills, and so on, which are comparatively light loads, you are likely to be better off with a low-voltage distribution system.
It is an electrical engineer's job to make this judgement on a case-by-case basis, to deliver a safe, lowest-cost, fit for purpose design.
This might include doing enough engineering design on both options to evaluate their relative safety and cost.
I always considered that the currents are in phase but looking at your schematic I can see that on the primary current is entering at the dot end while on the secondary it is leaving at the dot end and feeding the load.
I suspect that this is what causes them to be out-of-phase in the analysis.
I think that you have correctly explained the other aspects in your question.
Best Answer
Look at the transformer model you have in your textbook. Or also here (taken from wikipedia).
In an ideal transformer, the magnetizing reactance Xm is infinite, the core losses are zero, i.e. Rc = infinite (by the way, Rp and Xp, X's and R's are 0 in an ideal transformer, but this information is not useful here). Therefore, since no current flows on the secondary (open circuit), from the transformer equations you get also no current flowing on the primary (Is = 0). Since Rc and Xc = infinite, Io = 0. Hence Ip = 0 too.
In practice, Xc and Rc are not infinite, (in particular at least Xc, since the number of turns is finite, as well as the magnetic permettitivity of the core) therefore you'll have a small current under even no load.