Video

3:35 The 'c_p' value is given for air, not fuel. We therefore need to find 'c_p' of fuel (butane) separately and then, add the fuel and air contributions together; as the heat released will increase the temperature of both air and fuel (butane). What you did here is assume that c_p of air is the same as that of butane, which is clearly wrong... Am I right?

As we saw in the graph, there are two theoretical Shock Wave Angles possible for one Flow Deflection Angle, then why is the higher shock wave angle not visible in the Schileren images? Can someone please explain this?

4:00 Shuttle was a fake rocket 🚀

Yes thank you.

Need to add in mg and drag to model a rocket launch

at 7:36 the temperature ratio is incorrect. It should be 2/(y+1)

Your great video helped me a lot. Thank you so much

Excellent 👍

Could you elaborate more on heat transfer when it comes to re-entry? -Nice table at the end.

I believe you kind of answered it, yet my question is regarding pressure increase?

Say I have an engine with 60 seconds of specific impulse, does that basically mean the same as "If I expend all of the energy that this rocket has, I will get 60 seconds of positive boost in a given direction?"

Literally answered my question thank you!!

Amazingly intuituve video! Deepest gratitude 🙏🙏

Well the guy running might be an idiot but he's capable of running 5 km/s.. that's pretty impressive

Wonderful explanation Sir. I want to join in your group Sir. Could you please suggest me?

5:45

This is sooooo cool!

I couldn't understand well, why is he using Helium... It's not clear from the recording.

Amazing explanation and the Aussie accents makes it even better

A little hilarious to see your calm and gentle look after the last section ^^

0:45 Burrrrrn

How your operating at different mach numbers

for what does the "d" in "pmp" stand for?

Prof needs to drink more water

Hypersonic🚀

Please link the previous video youre talking about jet engines.

La claridad del curso es fantastica, muchas gracias por compartirlo

Choked flow always talks about reducing the back pressure won’t increase m dot once you get sonic flow in the throat. What about increasing the inlet pressure? Surely that will, if you are ramming more in.

Navy

This is actually fun.

need to do it again as iam back of vital information needed to solve here

Does the Shockwave reduce the airflow friction on the surface? (and so less heat is generated?)

Which shape is the ramjet? There's upper lower and ?

But in a pipe (as one example) velocity flow eventually should go into a parabola shape.

Why does the oldest video start with, "Let's continue ..." Where is the starting video?

Thank you so much for this well put together video!

Is there any weird heat conductance at hypersonic speeds? Since it seems like the craft is performing an exothermic reaction (splitting air molecules) which is then heating up the aircraft, is there much loss of heat or is the craft moving so fast that there is little chance for the expulsion of excess heat into the atmosphere?

Amazing video, i Have a questione. If I want to reach mach 12 on an experimental plane I have to start with normal jet engines, then switch to ram jets and then finally turn on the scramjets?

The 40 degree wedge example with the curved shock due to expansion fan interaction is wonderful. It finally helped me understand how a strong oblique shock can transition back to the Mach angle in the far field away from the body.

very nice i like thank you everyone. hellovryone

are u indian ? xx

Well explained. Thank you.

I don't agree with your assessment that the space shuttle's solid rocket boosters were reliable. Was this filmed before Jan. 1986?

But how do we know the max. exhaust velocity and max specific impulse?

Why do you plug in the value for L2* in the calculator? If we know that L2* is the point at which the flow becomes choked, why isn't M=1 when you plug in that value?

Av

Claim insurance.

Simple explain of deep concept fabulous

I can't make the algebra work for the equation you have for m-dot, and I can't reproduce the plot for m-dot with either your equation or the one I derived. I get m-dot/p_0 a_0 A_e = P_rat^[g+1/2g] * {2/(g-1) * [P_rat^(1-g/g) - 1]}^1/2 where gamma = g, P_rat is the pressure ratio. Plotting in Desmos, this is only defined for [-1,1] with what I derived. Plotting your equation, it never reaches a maximum. I'm typing up my notes and wanted to include the plots; that's the only reason I noticed.

😷