MVAC Cooling to Shopping Mall Tenant

If the building is "not subject to outside thermal conditions" then all the heat dissipated within the space has to be rejected outside it to maintain a temperature, otherwise the temperature will rise without limit. This is not a good starting point, so reject the assumption.

Simple ventilation may be all that is required. Try opening a few windows while the space is in use, perhaps?

Space is in the internal of a shopping mall and tempeture of the adjoining common areas and rear corridors is maintained by the property (stated at 22 - 24 `C with 55% relative humidity). Cooling needs to be provided for any heat generated within the space so as not to effect the tempature of those common area.

I think once I can determine a baseline method of calculating the potential heat load I can then determine how much TR will be needed to cool the space in an unfitted state.

One might still duct it outside the building. One would need to do so were air conditioning plant to be installed, as well as a duct for the water that such a plant will generate.

What's the exterior temperature and how much heat is dissipated within the shop?

even if the shop is internal facing, it has some internal heat generating substances, like lights, equipment, and people. You need to calculate the number of these and how much heat is generated bu these. refer ASHRAE if it is in your reach.

Some basic load factors (adjusted for easy calculation by non-engineer)-

Bodies in space- use 125 W/occupant

If ceiling plenum is used for return air, figure total wattage of lamps installed

If ceiling plenum is NOT used for return air, use 75% of wattage of lamps installed

ADD total wattage of ALL equipment in space (registers, display illumination, display equipment, etc.)

If space HVAC system is connected to outside air with fresh air duct (likely) add 2 watts per SF of "occupied" space for Outside Air load.

If known, add total HP of supply fans times 750 watts. If unknown, add 5% to sub-total of loads.

Total "load" in watts from calculations divided by 200 is approximate Cubic Meters per minute required.

Total "load" in watts divided by 0.7 is Total HVAC cooling load in Watts

Load divided by 1,000 equals total cooling load in kW.

Cooling load in kW x 0.285 equals "tons of Refrigeration" required.