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.

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

Hi Raymond,

Have you tried the built-in unit discovery using ctrl+=? Mathematica understands units so you can focus on the relations such as force = pressure x area. At the end, you can use UnitConvert to convert to whatever is required. A screen capture is shown below where you could use almost any oddball English unit and easily convert to SI at the end.

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.