Suppose my room is x degrees C. The air in my AC ductwork is a constant y degrees C and is blowing at z m/s. We assume there is no heat transfer from the outside. How can I figure out how long it will take for my room to reach the desired temperature?
You are missing some key params, like volume of the room, humidity or energy consumption.
Simplifying the assumptions:
Steady state conditions 1D heat transfer from AC to room Fluid properties are uniform Adiabatic system 0 thermal radiation (Shades drawn) Cooling happens everywhere simultaneously No new energy enters or leaves system
The last one is the least likely, since the AC is bringing in energy to power it, and heat will be lost to surroundings heating the room.
You need Newton's law of cooling, since you say temps are constant and adiabatic system.
Since you give z in m/s, I would propose assuming an area, A, that the air flows through. You then need to assume that the system expels mass at an equal rate.
Then you can say:
q/A = h * deltaT
Where q is the convective heat flux and h is convection heat transfer coefficient (assuming air). I don't know how you would get q without statements about the energy the AC needs, but this is solvable if you determine a relationship between h and z, through either AC manufacturer or estimate (text gives
h = 10.9 W * s^0.8/m^2.8 * K * z^0.8 for a similar problem…determining that can be involved). Then you can determine q, the energy needed.
Using q would be more accurate, since you're looking for change in internal energy over time, dU/dt. Since U is only dependent on the volume, density (rho), specific heat (c) and temperature of the solution (assuming air), you get…
dT/dt = q/(rho * V * c)
Then you use newton's law of cooling,
T(t) = Ti + (T_AC - Ti)exp(-kt)
Solve for t.
This again is my assuming a number of key params: Volume of room, quality of air/humidity, properties of fluid (assuming air), and a relationship with velocity from your manufacturer as well as the AC's opening area.
There are complicated ways to go about driving at this accurately, but even those require simulations.