# Rocket Propulsion

1. Mar 10, 2009

### rz_supra

1. The problem statement, all variables and given/known data

A rocket operating with combustion chamber pressure and temperatures of 14MPa and 2500K respectively, has a throat diameter of 0.3m, and a nozzle area ratio of 50:1.

Find the thrust and specific impulse developed by the motor with back pressures of 1 bar (10^5 Pa) and zero.

At what back pressure would the nozzle be correctly expanded?

Assume the combustion products behave as a perfect gas, with a constant specific heat ratio (y) of 1.4, and a constant specific gas constant (R) of 300 J/kg/K

2. Relevant equations

throat P/chamber P = ((y+1)/2)^(y/(1-y)) (for M=1) M = Mach number

throat T/chamber T = y/(y+1) (for M=1)

mdot = [(area of throat * throat P)/sqrt(throat t)]*[sqrt(y/R)]*((y + 1)/2)*((y + 1)/2(1 - y))

3. The attempt at a solution

chamber P = 14MPa
chamber T = 2,500K
throat diameter = 0.3m; throat diameter = 0.15m
y = 1.4 (constant specific heat ratio)
R = 300 J/kg/K

Calculate mdot --> through conditions @ throat M=M*=1

throat P/chamber P = ((y+1)/2)^(y/(1-y))

throat P = 7.4MPa

throat T/chamber T = y/(y+1)

throat T = 2,083.3K

mdot = [(area of throat * throat P)/sqrt(throat t)]*[sqrt(y/R)]*((y + 1)/2)*((y + 1)/2(1 - y))

mdot = 782.47

area of throat = 0.0707

This is where I get stuck, I need to incorporate these back pressures of 10^5 Pa and 0 Pa somehow. I know that for a correctly expanded nozzle, pressure of exit = ambient pressure

Any help would be greatly appreciated!

2. Mar 11, 2009

### D H

Staff Emeritus
You are missing some key relevant equations. One obvious one is the pressure at the end of the nozzle, which you need both for calculating the thrust and for determining the pressure at which the nozzle is correctly expanded. Another is an equation for the thrust, and to get that you might find you need to know the exhaust velocity.

3. Mar 11, 2009

### rz_supra

Thanks for the reply. Yes I thought I needed to calculate the velocity, but the only relevant equations for this that I could find were:

(velocity ^ 2)/2 = h chamber - h exit = Cp * (chamber T - exit T)?

4. Mar 11, 2009

### rz_supra

e = 50:1 = exit area/throat area = ambient P/exit P?
exit P = ambient P/50 = 280kPa?

EDIT:

Or are the exit pressures given as the back pressures?

ie. exit pressure 1 = 10^5 Pa & exit pressure 2 = 0 Pa.

AND...

The expansion ratio is used to calculate the exit pressure for correct expansion?

Last edited: Mar 11, 2009
5. Mar 14, 2009

### rz_supra

Bump. Can anyone help?

Is back pressure the ambient pressure?

I've now calculate mdot as 782.6kg/s & c* = 1264.76m/s & a = 3059.7m/s

Do these seem unreasonable?

6. Mar 14, 2009

### D H

Staff Emeritus
Question: Is this homework, or something you are doing on your own?

If this is homework your text almost assuredly has something on the velocity of the gas and the pressure at the nozzle.