Calculating rocket nozzle throat area

Question

I'm trying to calculate a rocket nozzle throat area, $A_t$. My propellant is KNDX. I am trying to do this according to the following equations, taken from this reference:

where $P_t$ is

and $T_t$ is

So, according to Richard Nakka's web page the mass flow rate is:

where $A_b$ is the burning area, $P_p$ is the propellant mass density and $r$ is the propellant burn rate $r = a \times Pc^n$. According Richard Nakka's table:

I choose $a = 3.841, n = 0.688$. The heat ratio $k = 1.1308$ for KNDX and I choose chamber maximum pressure as $P_c = 850$ psi (or 5.86 MPa).

As you know $R$ constant is 8314 and $M$ is 42.39 (for KNDX). As Richard Nakka is saying chamber temperature for KNDX is 1700 i.e $T_c = 1700$ and the grain density $P_p = 1.785\ \frac{\text{g}}{\text{cm}^{3}}$. So my grain's total burning area $A_b = 55411 \textrm{ mm}^2$, because Grain's outer diameter is 76mm and core diameter 10mm.

Putting all this together I'm getting $A_t = 37209240 \textrm{ mm}^2$ which is wrong, because as Richard Nakka's calculations shows it's $1179 \textrm{ mm}^2$.

Here is Richard Nakka's official web page's file (you should download SRM_2014.1.ZIP), where he's calculating it.

What I'm doing wrong?

Answer 1

Probably units and the format of eq (7) are the problem. After looking at the referenced website by Nakka, I used the questions information given to get $\dot m= 1.187\ \frac{\text{kg}}{\text{s}}$ (using a rounded off $r=12\ \frac{\text{mm}}{\text{s}}= 0.012\ \frac{\text{m}}{\text{s}}$). Also, calculated throat pressure and temperature of 3421000 Pa and 1616 K. Equation (7) should have parentheses around the $ w_t/P_t$,because the pressure and temperature are divided. Eq (7) $w_t$ is $\dot m$. The $g_c = 1\ (no units)$ when using SI units. The $R = 8314/42.39 = 196.1 m^2/s^2 K$ (units simplified from $\frac{\text{J}}{\text{kg}\cdot\text{K}}$).

The $\dot m/P_t $ has units of $\frac{\text{kg/s}}{\text{Pa}}$, but $\text{Pa}$ (pascal) is $\frac{\text{N}}{\text{m}^2}$ and $\text{N}$ is $\frac{\text{kg}\cdot\text{m}}{\text{s}^2}$ so $\frac{\text{m}^2\cdot \text{s}}{\text{m}}$. The square root of $RT$ has units of $\frac{\text{m}}{\text{s}}$, so the result is $m^2$.

Using values above I got $1.837\cdot 10^{-4}\ \text{m}^2$ or $183.7\ \text{mm}^2$

Added information:

The mass burning rate = $\rho r A$ where $\rho$ is the solid propellant density, $r$ is the burning rate (how fast the solid is consumed) and $A$ is the burning surface area.

As given above, $r=a P_c^n$, for design chamber pressure of 5.86 MPa, a and n are 3.84 and 0.688 so $r=3.84\cdot(5.68^{0.688})=12.69\ mm/s$. A trick here is to realize the pressure is in MPa. Using the English units, the rate is near the same (850 psi): $0.005\cdot(850^{0.688})=0.5181\ \frac{\text{in}}{\text{s}}=13.16\ \frac{\text{mm}}{\text{s}}$.

The mass burning rate goes out the nozzle, so calculate $\rho r A$ in SI units: $\rho = 1.785\ \frac{\text{g}}{\text{cm}^3} = 1785\ \frac{\text{kg}}{\text{m}^3}$, $r= 0.012\ \frac{\text{m}}{\text{s}}$, $A=55411\ \text{mm}^2=0.055411\ \text{m}^2$, mass flow = $1.187\ \frac{\text{kg}}{\text{s}}$. This mass flow goes to the nozzle so it is used in eq (7).

The units on $\sqrt{RT}$ are $\frac{\text{m}}{\text{s}}$. If we multiply eq (7) by $P_t$ the left side is a force (pressure times area), the right side has mass flow rate times velocity and that is a force so the units are correct. Remember, the basic rocket thrust force equation is mass flow rate time exhaust velocity.

Just for completeness, eqs (7), (8) and (11) can be combined to show flow rate, chamber pressure and temperature together. From Rocket Propulsion Elements by Sutton (7th edition, but 1st edition has the same with same eq number):

Sutton uses k for $\gamma$, other symbols you should be able to figure out.