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.