It’s true that Sutton uses v2 to represent the actual or average exhaust
velocity (Ve more commonly used) but the 2nd part of your interpretation is
inaccurate. Ve or v2 (in Sutton) is the average exhaust velocity *period* ie
irrespective of the balanced or imbalanced relationship of Pe with Pa. Ve (or
v2) is entirely dependent on the amount of expansion. The ambient pressure or
the pressure of the surroundings has no influence on the average exhaust
velocity in a theoretical sense.
Of course, in a practical sense, flow separation will provide a Pa
contribution to v2 for significant over expansion, but this discussion is based
wholly on the theoretical definitions.
The final sentence is true – that v2 = c only when p2=p3 because that’s the
point where your average exhaust velocity ‘v2’ is equal to your effective
exhaust velocity ‘c’ . Again, coz v2 is strictly your avg exhaust velocity and
‘c’ is v2 with effect of pressure thrust factored into it.
Troy
From: arocket-bounce@xxxxxxxxxxxxx [mailto:arocket-bounce@xxxxxxxxxxxxx] On ;
Behalf Of Ian Woollard
Sent: Wednesday, 16 November 2016 3:35 PM
To: arocket@xxxxxxxxxxxxx
Subject: {Spam?} [AR] Re: Pressure Thrust and Momentum Thrust
Sutton? Rocket Propulsion Elements uses 'V2' instead of Ve and specifically
defines it as the exhaust velocity when the exhaust pressure and ambient
pressure are the same:
https://www.ewp.rpi.edu/hartford/~ernesto/S2013/EP/MaterialsforStudents/Lee/Sutton-Biblarz-Rocket_Propulsion_Elements.pdf
See page 32.
The equation given is:
F = m_dot v2 + (p2 -- p3)*A2
where v2 is the effective exhaust velocity when p2=p3, and p2, p3 are the
exhaust plane pressures and the ambient pressures respectively.
On 5 November 2016 at 05:35, Norman Yarvin <yarvin@xxxxxxxxxxxx
<mailto:yarvin@xxxxxxxxxxxx> > wrote:
On Fri, Nov 04, 2016 at 05:37:49AM +0000, Ian Woollard wrote:
On 2 November 2016 at 19:52, Norman Yarvin <yarvin@xxxxxxxxxxxx
<mailto:yarvin@xxxxxxxxxxxx> > wrote:
if one
takes the textbook equation and compares it under vacuum conditions to
the conservation of momentum, there's an extra term ("pressure
thrust") on one side of the textbook equation,
Yes!
so the two equations disagree.
No!
Fans of Douglas Adams will appreciate the concept of having tea and no tea
simultaneously, but the resolution for this apparent paradox is far
simpler: the Ve/Isps are different! In one case it is the vacuum Isp, and
in the other, it is the effective Isp when Pe=Pa.
