Shocks

Question 1

Mass flow rate

Answer 1

-Presence of a shock implied supersonic flow in the region, therefore Ma = 1
Rearrange Fliengner’s number for mass flow rate

Question 2

Mass flow rate equation

Answer 2

ṁ= Atps*√(γ/(RTs) * ((γ-1)/2)^((-γ+1)/(2(γ-1))

Question 3

Fliengner’s number rearranged

Answer 3

Fn= γ((γ+1/2)^-(γ+1)/(2(γ-1)) (At/Ashock)

Question 4

Finding mach number from table

Answer 4

If downstream from shock, it must be Subsonic. M<1

If upstream from shock it must be Supersonic. M>1

Question 5

Finding mach number upstream/downstream of shock

Answer 5

Ma1 or Ma2 in equation

Question 6

velocity downstream

Answer 6

Ma2√ (γRTs2(p/ps)^(( γ-1)/ γ))

Question 7

Boyle’s law

Answer 7

-states that pressure of a given mass of an ideal gas is inversely proportional to its volume at constant temperature

Question 8

Why, within the thickness of the shock, the poisson adiabat does not apply even though the process is “adiabatic” in that there is no net heat transfer into or out of that parcel of fluid.

Answer 8

-Temperature is rapidly varying, shock region is so thin, that the Pedet number for shock is comparable with 1.
-Everywhere else the pedet number is enormous.
-Within the shock, significant conduction of heat through the fluid can take place
-Everywhere else, conduction of heat cannot happen as fast as the fluid passes through the system.
-Heat transfers mean that change in internal temperatures not determined by just the work done, but but also by conductive heat transfer.
-pVT relationship cant be determined by poisson adiabat, as it doesn’t factor in heat transfer just work.
All conductive heat transfers add up to 0 as a parcel passed through the shock,
-B is at same temperature as A, but C is hotter, so there is heat transfer from C to B
-Althrough there is a series of instantaneous heat transfers that disturb PVT relationship away from Poisson adiabat, they all add up to 0 net heat transfer

Question 9

pedet number

Answer 9

Pe= u*t/K
where
t= length of scale which T changes significantly)
K= thermal diffusivity

Question 10

Explain why , at mass flow rate choked conditions, further reductions in the pressure at a location downstream of the throat makes no difference to mass flow rate

Answer 10

-Once choked conditions are reached, any further increase in velocity, in the region near the downstream control associated with a reduction in pressure there, is cancelled out by the decrease in density.

Question 11

As=At(Fnt/Fns)

Answer 11

=0.83(1.4+(Ms^2/2))^3

Question 12

Boyle

Answer 12

= ps/Rts = kg/s

Question 13

velocity

Answer 13

u=ṁ/(ρsAs)

= ps*Fns/√(γRTs) ρs