I’m using the ideal rocket equations and isentropic flow but cannot determine what is wrong in my understanding about total pressure. Can anyone explain where I am wrong using the following example?

**Example:**
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$\gamma = 1.4 $

Converging-diverging nozzle Area ratio $A/A^\star = 12.0$

Inside the rocket chamber $P\_0 = P\_t = 12 MPa$

Exit area $A\_e = 0.002 m^2$

Mass flow rate $\dot m = 2.5kg/s$

$Thrust=5000N$

Supersonic flow

Atmospheric pressure $P\_{atm} = 101.3 kPa$

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**Solution:**
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Based on isentropic equations - (e.g. <https://www.grc.nasa.gov/www/k-12/airplane/isentrop.html>)

Mach No. $M = 4.127$

$P/P\_t = 0.0056$

$T/T\_t = 0.227$

Substituting in $P\_t$ gives:
$P = P/P\_t \* P\_t = 0.0056\*12 MPa = 67200 Pa $

Then based on the ideal rocket equation (i.e. <https://www.grc.nasa.gov/WWW/K-12/rocket/thrsteq.html>)

$Thrust = v\_e \* \dot m + A\_e (P-P\_{atm}) $

Rearranging for exit velocity:

$v\_e = \frac {Thrust - A\_e (P-P\_{atm})}{\dot m} $

$v\_e = \frac {5000 - 0.002 (67200 - 101300)}{2.5} = 2027.28 m/s$

Then the exit density can be found by rearranging this (also from the isentropic equations website):

$a = \sqrt{\gamma \frac{P}{\rho}}$

Rearranges into:

$\rho = \frac{\gamma P}{a^2}$

Speed of sound
$a = v\_e / M = 2027.28 / 4.127 = 491.22 m /s $

Substituting into the density equation:

$\rho = \frac{1.4 \* 67200}{491.22 ^ 2} = 0.390 kg/m^3$

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**And here is the problem/confusion:**
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From the question the total pressure is $P\_0 = P\_t = 12 MPa$
This is also supposed to be the total pressure at the nozzle exit because it is reversible/isentropic flow.

But $Total Pressure = Static Pressure + Dynamic Pressure$
$Total Pressure = P + 0.5 \rho v\_e^2 $
$Total Pressure = 67200 + 0.5 \* 0.390 \* 2027.28^2 = 868624 Pa$

And this is nowhere near 12 MPa... so what part am I not understanding correctly?