Hi Raymond,

Here a couple of comments.

First, the gray coloring indicates that there is more than what is displayed. You can use InputForm to take a look like so

zA = 20 Quantity[1, "Feet"]
zA // InputForm
(* Should return: Quantity[20,"Feet"]*)

So, $zA$ is not equal to $20 ft$, but it is equal to Quantity[20,"Feet"] . So, dividing $zA$ by (20 ft) will not return 1, but the following:

zA/(20 ft) // InputForm
(* Quantity[1,"Feet"]/ft *)

Now, if you use the quantity framework, then you should not need to concern yourself with $g_c$. That was introduced before they had figured out that force could be expressed in terms of fundamental dimensions of Mass, Length, and Time. Mathematica knows that a pound force and a pound mass are different quantities. It looks like you dropped your gravity term in your energy expression. I reproduced Bernoulli's principle in pressure form so that it will be easier to convert to a standard unit like pressure.

$$\rho (\frac{{{v^2}}}{2} + gz) + p = {\text{Constant}}$$

Recall, that I suggested that we save the unit conversion for the end. Also, having incompatible units is Mathematica's way of telling you that your equation is dimensionally inconsistent. I rewrote your equations like so

zA = 20 Quantity[1, "Feet"]
prA = 10 Quantity[1, (("PoundsForce")/(("Inches")^2))]
velA = Quantity[7.761668264705552`, ("Meters")/("Seconds")]
\[Gamma]wtr = 62.4 Quantity[1, (("Pounds")/(("Feet")^3))]
g = Quantity["StandardAccelerationOfGravity"]
pressTA = \[Gamma]wtr ( velA^2/(2) + g zA) + prA
UnitConvert[pressTA, "PSI"]
UnitConvert[pressTA, "Pascals"]
(* 23.03350000868825`lbf/(in)^2*)
(*158810.39217209682`Pa*)

To summarize, gray text indicates the displayed is wrapped in Quantity, Mathematica eliminates the need for $g_c$, Incompatible units suggest that you check your works, and save UnitConvert until the end.

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.

In your notebook take a look at the syntax coloring in the gage = expression. Both lbf and in are blue, which means they are interpreted as undefined symbols. Try

atm = UnitConvert[Quantity[1, "Atmospheres"], "lbf/in^2"] // N
gage = Quantity[8.7, "lbf/in^2"]
atm + gage

You get the scalar by applying QuantityMagnitude at the result, e.g.:

ft = Quantity[1, "foot"]
(* output: 1 ft*)

QuantityMagnitude@ ft
(* output: 1*)

Look at QuantityUnit@atm it is ("PoundsForce")/("Inches")^2. So if you define

gage=Quantity[2.4, ("PoundsForce")/("Inches")^2]

everything is fine.