Tim, I tried using "Cntrl+=" with Bernoulli Equation using the English system,but after two hours I still can't get it to work. Mathematica will not put the gravitational constant "32.2 lbm ft / (lbf s^2)" into units and any units described by "Cntrl+=" will not compute with 'variables' of the same names

zA = 0 Quantity[1, "Feet"]
prA = 10 Quantity[1, (("PoundsForce")/(("Inches")^2))]
velA = UnitConvert[
  Quantity[7.761668264705552`, ("Meters")/("Seconds")], "ft/s"]
\[Gamma]wtr = 62.4 Quantity[1, (("Pounds")/(("Feet")^3))]
g = UnitConvert[Quantity["StandardAccelerationOfGravity"], 
   "Ft/s^2"] // N
gc = 32.2  *(lb ft)/(lbf s^2)
energTA = prA/\[Gamma]wtr + velA^2/(2 gc) + zA/gc

zA = 0 Quantity[1, "Feet"]
prA = 10 Quantity[1, (("PoundsForce")/(("Inches")^2))]
velA = UnitConvert[
  Quantity[7.761668264705552`, ("Meters")/("Seconds")], "ft/s"]
\[Gamma]wtr = 62.4 Quantity[1, (("Pounds")/(("Feet")^3))]
g = UnitConvert[Quantity["StandardAccelerationOfGravity"], 
   "Ft/s^2"] // N
gc = 32.174 Quantity[1, "Pounds"]*
  Quantity[1, (("Feet")/("PoundsForce" ("Seconds")^2))]
energTA = prA/\[Gamma]wtr + velA^2/(2 gc) + zA/gc

My result

the result should be something close to 32.7 ft lbf / lbm

Tim, thank you kindly for all your work, I am slowly learning, didn't know that you could ignore the gravitational constant when using Quantity, does make life a little easier.

Thanks again, I tried..

atm = UnitConvert[Quantity[1, "Atmospheres"], "PSI"] // N
atmN = QuantityMagnitude@atm * QuantityUnit@atm
gage = 8.7 lbf/in^2
atmN + gage

I got

units in Quantity are different than the units in QuantityUnits and I still can not add them, Is here a possible work around????

Beautiful Tim -- didn't know about cntrl+= or understand it. thank for the demonstration, I think that is about as simple as I can get it, instead of using units like variables and confusing Mathematica in the process. I will work with what you suggested and see how well that works out.