# Calculating rocket nozzle throat area

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 Tc = 1700 and the grain density Pp = 1.785 g/cm3. 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. Nozzle calculation file

What I'm doing wrong?

• What is the source of the equations? Sep 4, 2021 at 1:25
• Sep 4, 2021 at 1:36
• Consider editing your question to include the sources of the information you quoted. Sep 4, 2021 at 2:05
• @OrganicMarble everything is in that SRM_2014.1.ZIP file. You can download it from official Richard Nakka's web page. Sep 4, 2021 at 2:06
• What I mean is, you should put the source infomation in the question not in comments. Sep 4, 2021 at 3:16

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 $$\dot m= 1.187 kg/s$$ (using a rounded off r=12mm/s= .012m/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 J/kg K).

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

Using values above I got 1.837 10^-4 $$m^2$$ or 183.7 $$mm^2$$

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 .688 so $$r=3.84 (5.68^.688)=12.69 mm/s$$. A trick here is to realize the pressure is in MPa. Using the english units, the rate is near the same (850 psi): $$.005(850^.688)=.5181 in/s=13.16 mm/s$$/

The mass burning rate goes out the nozzle, so calculate $$\rho r A$$ in SI units: $$\rho = 1.785 g/cm^3 = 1785 kg/m^3, r= .012 m/s,A=55411mm^2=0.055411 m^2$$, mass flow = 1.187 kg/s. This mass flow goes to the nozzle so it is used in eq (7).

The units on $$\sqrt( RT)$$ are m/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 ed, but 1st edition has the same with same eq number):

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

• how did you get mass flow m=1.187 kg/s? or r=12mm/s? Sep 5, 2021 at 22:52
• @W H G can you explain your answer more detailed please? Sep 5, 2021 at 23:01
• @W H G Yes I converted solid propellant density into kg/m3, burn rate into the m/s and burning area into the m2, and now is q = 1.25 kg/s. then I calculated √RT, then divided it by k(heat ratio 1.1308,) g is 1 as you said and finally multiplied it by (q/Pt). throat area At is 202 mm2, why did you mention Pc ? Pc is important only when we are calcaulating Pt. Sep 6, 2021 at 10:42
• @Alatriste I was thinking of chamber pressure rather than throat pressure. I have changed to P_t. Your eq (7), (8) and (11) can be combined. I'll put that in the answer. Sep 6, 2021 at 14:04
• One more question about pressure unit , you said that "Pa (pascal) is N/m^2 and N is kg m/s^2 so m^2 s/ m" but this is my calculation ibb.co/rv7JZPs Sep 6, 2021 at 14:39