That is the output watts I think. But I want the input watts consume so that I can measure the time the transformer will run on a 1200VA UPS.
Transformers are pretty efficient, perhaps 98% unloaded dropping to maybe 80% to 90% fully loaded. It depends on the resistive losses in the windings and core losses (from magnetic hysteresis, mainly). Your best bet would be to temporarily load up the secondary to get 50 A at 30 V using a temporary nichrome heating wire load and measure the primary current and line voltage and secondary current and line voltage. It helps to have two meters: one to measure RMS voltage, a second clamp-on type to measure RMS current. Since the secondary has a purely resistive load, you don't have to worry about a power factor correction.
Maybe you can salvage the length of nichrome wire you need for this test from a discarded space heater. You will need a length that measures one ohm or less to provide a 1500 watt load at 50 A with 30 V on the secondary. The nichrome increases in resistance as it gets hot. When it is dissipating 1500 watts (the rating of your transformer) it should have a resistance of 30/50 or 0.6 ohms. You may have to parallel several strands of nichrome to get a resistance this small when the wire is hot. I would try to start out with one ohm or more and then add parallel strands until you reach 1500 watts dissipation. You will also need to figure out a way to support the nichrome wire well away from combustible objects. This will not be a permanent load, so neatness doesn't matter. You can attach the ends of the nichrome wire to a barrier strip, preferably ceramic but phenolic willl do for short-term (a few seconds) operation. Allow a few minutes for everything to cool down between voltage and current measurements. You need to energize the transformer only long enough to measure the current and voltage on the primary side and the current and voltage on the secondary side. The ratio of these two power products will approximate the efficiency of the transformer, which you can then use in your calculations to estimate the primary current as a function of the secondary current. Or, do what I would do: guess and leave a margin for error in the guess. So, if the maximum transformer load is 1500 watts, I am going to guess the worst-case primary power dissipation is 2000 watts or 75% efficiency. Then, if I have a 1200 watt UPS, I am going to have to lower the load on the secondary from 1500 watts to 900 watts to keep the primary power dissipation under 1200 watts. Or get a larger UPS.
Most UPS devices have a very limited hold-up time under their full rated load, typically no more than ten or fifteen minutes. Some are a lot less, maybe five minutes. This gives you time for an orderly shut-down of computer disk drives and not much else. Some UPS have provision for user to attach an external battery, but I wouldn't go there because it gets expensive very quickly. If you want to continue to record data from sensors during a power outage, you might want to separate the relays for loads that you will drop from the relays for the sensors that must continue to record. Power the latter from a separate power supply and connect that power supply to another UPS. And try to minimize the load the UPS must drive when the power goes out.
I don't understand why you want to use such a large transformer to drive eighty relays. That would provide more than eighteen watts for each relay (1500/80 = 18.75)! What's going one here?
It sounds to me like you plan to buffer each Arduino output with a transistor. This is a good plan. You will need only place the relay coil in series with the collector of a small NPN transistor, ground the emitter, and insert about a 1 kΩ resistor between the base and the Arduino output. A diode will have to be mounted in parallel with the relay coil, and this should be done at the relay, not on the board with the driving transistor. It will help with troubleshooting if you also include, on the circuit board, a 10 kΩ resistor from each collector to the positive relay power rail. That will allow you probe the collectors to see which ones are turned on. If you want to get really fancy, use an LED with a suitable current-limiting resistor in place of the 10 kΩ resistor. For this project you need to find a source of really inexpensive relays with a low-current requirement for the coil. To some extent, your choice will be determined by the loads you want to switch. Some high-current loads (an air conditioner for instance) will require a relay with a high current rating which in turn may mean the relay must be bigger to handle the increased weight of the contacts... all the way up until the relay is no longer called just a relay, but a relay contactor! So nail that down first. Determine how much current each of the relays must switch and then try to find a single relay that covers the entire range of current. It would be better to select relays with 12 V DC coils because these are the most common and easily available.
Now, where is that block diagram of what you propose to build?