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?
Or, how about this, exhaust exit pressure is determined by the expansion ratio given by:
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...
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)?
Homework Statement
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...