CH.12: COMPRESSIBLE FLOW It required an unhesitating boldness to undertake such a venture …. an almost exuberant enthusiasm…but most of all a completely.

CH.12: COMPRESSIBLE FLOW It required an unhesitating boldness to undertake such a venture …. an almost exuberant enthusiasm…but most of all a completely unprejudiced imagination in departing so drastically from the known way. J. Van Lonkhuyzen, 1951, discussing designing Bell XS-1 1947 – Chuck Yeager breaks sound barrier in Bell XS-1 Mach number = 1.06 = 700 mph at 43,000ft At that time no wind tunnel data on transonic flight!

Compressible flow is a fun subject.
John Anderson

(note if isentropic stagnation properties do not change)
Ch.12 - WHAT CAUSES FLUID PROPERTIES TO CHANGE IN A 1-D COMPRESSIBLE FLOW FLOW? (note if isentropic stagnation properties do not change)

Flow can be affected by: area change, shock, friction, heat transfer
Ch.12 - COMPRESSIBLE FLOW Flow can be affected by: area change, shock, friction, heat transfer shock Q friction Q Area2 shock Area1

One-Dimensional Compressible Flow
Rx P1 P2 dQ/dt Surface force from friction and pressure heat/cool (+ s1, h1,V1,…) (+ s2, h2, V2, …) Shock, heating, cooling, area change and friction affect 1-D flow. Could be T(s), p(s), .. Look at each one separately though in real world more than one may occur simultaneously. What can affect fluid properties? Changing area, normal shock, heating, cooling, friction. - need 7 equations -

Affect of Area Change Assumptions = ?

ASSUMPTIONS ALWAYS ~ MOSTLY ~ ISENTROPIC Steady Flow Ideal Gas
Ignore Body Forces only pressure work (no shaft, shear or other work) Constant specific heats “One – Dimensional” MOSTLY ~ ISENTROPIC

“quasi-one-dimensional”
V(s) V(x,y)

“quasi-one-dimensional”
Flow properties are uniform across any given cross section of area A(x), and that they represent values that are some kind of mean of the actual flow properties distributed over the cross section. NOTE – equations that we start with are exact representation of conservation laws that are applied to an approximate physical model

7 Variables = T(x), p(x), (x), A(x), v(x), s(x), h(x)
One-Dimensional Compressible Flow f(x1) f(x2) + s1, h1 +s, h2 Shock, heating, cooling, area change and friction affect 1-D flow. Could be T(s), p(s), .. 7 Variables = T(x), p(x), (x), A(x), v(x), s(x), h(x) 7 Equations = mass, x-momentum, 1st and 2nd Laws of Thermodynamics, Equations of State (3 relationships)

Variables = T(x), p(x), (x), A(x), v(x), s(x), h(x) Equations = mass, momentum, 1st and 2nd Laws of Thermodynamics, Equation of State (3 relationships) Cons. of mass (steady / 1-D) Cons. of momentum (& no FB) Cons. of energy (& only pressure work) 2nd Law of Thermodynamics . Ideal gas Ideal gas & constant cv, cp Shock, heating, cooling, area change and friction affect 1-D flow. Could be T(s), p(s), .. (steady) Eqs. of State

Want to find qualitative relationships for:
dT, dV, dA, d ISENTROPIC

Affect of Change in Velocity on Temperature
Isentropic, steady, cp&cv constant, no body forces, quasi-one-dimensional, ideal gas, only pressure work

For isentropic flow if the fluid accelerates
what happens to the temperature?

what happens to the temperature? if V2>V1 then h2<h1

what happens to the temperature? if V2>V1 then h2<h1 if h2<h1 then T2<T1 Temperature Decreases !!!

Affect of Change in Velocity on Temperature
Velocity Increases, then Temperature Decreases Velocity Decreases, then Temperature Increases NOT DEPENDENT ON MACH NUMBER Isentropic, steady, cp&cv constant, no body forces, quasi-one-dimensional, ideal gas, only pressure work

Isentropic, steady, no body forces, quasi-one-dimensional
Affect of Change in Velocity on Pressure Isentropic, steady, no body forces, quasi-one-dimensional

Affect of Change in Velocity on Pressure

If density is a constant get Bernoulli’s Eq
If density is a constant get Bernoulli’s Eq. Which assumes isentropic flow Euler’s equation was derived from differential form of the general momentum equation in three dimensions, holds along a streamline.

= pdA + dAdp/2 + pA – pA – pdA– dpA - dpdA
FSx = (p + dp/2)dA + pA – (p+dp)(A + dA) = pdA + dAdp/2 + pA – pA – pdA– dpA - dpdA (dm/dt)(Vx+dVx) - (dm/dt)Vx = = (Vx + d Vx) {VxA}- Vx{VxA} = Vx VxA + dVx VxA – VxVxA -dpA = VxdVxA or dp/+d{Vx2/2}=0 If density is a constant get Bernoulli’s Eq. Which assumes isentropic flow Euler’s equation was derived from differential form of the general momentum equation in three dimensions, holds along a streamline.

dp/+d{Vx2/2}=0 Velocity Increases, then Pressure Decreases Velocity Decreases, then Pressure Increases NOT DEPENDENT ON MACH NUMBER Isentropic, steady, no body forces, quasi-one-dimensional

Affect of Change in Area on Pressure and Velocity
Isentropic, steady, no body forces, quasi-one-dimensional

EQ b EQ.12.1a {d(AV) + dA(V) +dV(A)}/{AV} = 0 steady isentropic, steady

isentropic, steady

EQ. 12.5 EQ.12.6 isentropic, steady

Affect of Change in Area on Velocity
dV/V = - (dA/A)/[1-M2] M <1 Velocity and Area change oppositely M > 1 Velocity and Area change the same Isentropic, steady, no body forces, quasi-one-dimensional

Affect of Change in Area on Pressure
dp/(V2) = (dA/A)/[1-M2] M <1 Pressure and Area change the same M > 1 Pressure and Area change oppositely Isentropic, steady, no body forces, quasi-one-dimensional

M<1 isentropic, steady, ~1-D dVx > 0 or dVx < 0 dp > 0 or
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1. If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A

M<1 isentropic, steady, ~1-D dVx > 0 dp > 0 or or dVx < 0

M >1 isentropic, steady, ~1-D dVx > 0 dp > 0 or or dVx < 0

M >1 isentropic, steady, ~1-D dp > 0 dVx > 0 or or dp < 0

M<1 isentropic, steady,~1-D dp > 0 dVx > 0 or or dp < 0

M>1 isentropic, steady,~1-D dp > 0 dVx > 0 or or dp < 0

isentropic, steady, ~1-D M<1 If M < 1 then [ 1 – M2] is +,
Subsonic Nozzle Subsonic Diffuser (dp and dV are opposite sign) M<1 If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1. If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A If M < 1 then [ 1 – M2] is +, then dA and dP are same sign; and dA and dV are opposite sign qualitatively like incompressible flow

isentropic, steady, ~1-D M>1 If M > 1 then [ 1 – M2] is -,
Supersonic Nozzle Supersonic Diffuser (dp and dV are opposite sign) M>1 If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1. If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A If M > 1 then [ 1 – M2] is -, then dA and dP are opposite sign; and dA and dV are the same sign qualitatively not like incompressible flow

And this is the reason!!!!!!! If  is constant then dA and dV
must be opposite signs, but for compressible flows all bets are off, e.g. for M>1 both dV and dA can have the same sign

Affect of Change in Area on Density
Isentropic, steady, no body forces, quasi-one-dimensional

-(dA/A)/(1-M2) = -dA/A - d/
Affect of Change in Velocity on Pressure -(dA/A)/(1-M2) = -dA/A - d/ d/ = (dA/A)[1/(1-M2) - 1] d/ = (dA/A)[M2/(1-M2)]

d =? d/ = (dA/A)[M2/(1-M2)]
I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa. d/ = (dA/A)[M2/(1-M2)]

d/ = (dA/A)[M2/(1-M2)] d/ > 0 d/ < 0 d/ > 0
I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa.

dp/(V2) = (dA/A)/(1-M2); dV/V = -(dA/A)/(1-M2);
WHAT HAPPENS AT M = 1 ? ? dp/(V2) = (dA/A)/(1-M2); dV/V = -(dA/A)/(1-M2); d/ = (dA/A)[M2/(1-M2)]

dp/(V2) = (dA/A)/(1-M2); dV/V = -(dA/A)/(1-M2);
If M = 1 have a problem, Eqs. blow up! Only if dA  0 as M  1 can avoid singularity. Hence for isentropic flows sonic conditions can only occur where the area is constant. I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa. dp/(V2) = (dA/A)/(1-M2); dV/V = -(dA/A)/(1-M2); d/ = (dA/A)[M2/(1-M2)]

What happens to dp and d across the throat in a supersonic
nozzle with steady, isentropic flow? I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa.

dp/ = - d{Vx2/2} d/ = (dA/A)[M2/(1-M2)]
What happens to dp and d across the throat in a supersonic nozzle with steady, isentropic flow? dp/ = - d{Vx2/2} Vx continues to increase so p continues to decrease I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa. d/ = (dA/A)[M2/(1-M2)] (dA/A)[M2/(1-M2)] is always negative so  continues to decrease

What happens to dp and d across the throat in a supersonic
diffuser steady, isentropic flow? I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa.

dp/ = - d{Vx2/2} d/ = (dA/A)[M2/(1-M2)]
What happens to dp and d across the throat in a supersonic diffuser with steady, isentropic flow? dp/ = - d{Vx2/2} Vx continues to decrease so p continues to increase I think from 12.2c can infer that as velocity goes down, then temperature goes up and vice versa. d/ = (dA/A)[M2/(1-M2)] (dA/A)[M2/(1-M2)] is always positive so  continues to increase

Same shape, but in one case
accelerating flow, and in the other decelerating flow

Goddard realizes (1917) that De Laval nozzle could be used for rocket
De Laval designed a turbine (1888) whose wheel was turned by jets of steam flowing through a contraction - expansion section.

Sketch a Supersonic Wind Tunnel

SUPERSONIC SONIC

Sneeze Breath (180 mph)

Affect of Area Change - Qualitative
7 Variables = T(x), p(x), (x), A(x), v(x), s(x), h(x) 7 Equations = mass, x-momentum, 1st and 2nd Laws of Thermodynamics, Equations of State (3 relationships)

Variables = T(x), p(x), (x), A(x), v(x), s(x), h(x) Equations = mass, momentum, 1st and 2nd Laws of Thermodynamics, Equation of State (3 relationships) Cons. of mass Cons. of momentum Cons. of energy 2nd Law of Thermodynamics . Ideal gas Ideal gas & constant cv, cp Shock, heating, cooling, area change and friction affect 1-D flow. Could be T(s), p(s), ..

Have found qualitative relationships for dT, dV, dA, d
Now want to find quantitative relationships. IDEAL GAS, ISENTROPIC, QUASI-1-D, FBX=0, STEADY, only PRESSURE WORK

Coupled, nonlinear equations hard to solve.
Much easier to express flow variables in terms of stagnation quantities and local Mach number.

stagnation properties)
Eqs a,b,c = Eqs. 12.7a,b,c EQ b + EQ c P / k = constant Knowing M1 and P1 and M2 and want to know P2, can find P0 from P1 and M1 and P2 from P0 and M2! (po, To refer to stagnation properties) isentropic, steady, ideal gas, constant cp, cv

using stagnation properties
Constant Area - Isentropic Example ~ using stagnation properties

Example ~ Air flows isentropically in a duct. At section 1: Ma1 = 0.5, p1 = 250kPa, T1 = 300oC At section2: Ma2= 2.6 Then find: T2, p2 and po2

p2 = p02/(1 + {[k-1]/2}M22)k/(k-1)
Example ~ Air flows isentropically in a duct. At section 1: Ma1 = 0.5, p1 = 250kPa, T1 = 300oC At section2: Ma2= 2.6 Then find: T2, p2 and p02 Equations: T2 = T02/(1 + {[k-1]/2}M22) T01 = T02; T1 = T01/(1 + {[k-1]/2}M12) p2 = p02/(1 + {[k-1]/2}M22)k/(k-1) p01 = p02; p1 = p01/(1 + {[k-1]/2}M22)k/(k-1)

T2 = T02/(1 + {[k-1]/2}M22) T01 = T02 p2 = p02/(1 + {[k-1]/2}M22)
Know: Ma1=0.5, p1=250 kPa, T1=300oC, Ma2=2.6 T2 = T02/(1 + {[k-1]/2}M22) T01 = T02 p2 = p02/(1 + {[k-1]/2}M22) p01 = p02 Then find: p02, p2, T2 p01 = p02 = p1(1 + ((k-1)/2)Ma12)k/(k-1) = 250[ (0.5)2]3.5 ~ 297 kPa p2 = p02/(1 + ((k-1)/2)Ma21)k/(k-1) = 297/[ (2.6)2]3.5 ~14.9 kPa T01 = T02 = T1(1 + {[k-1]/2}M12) = 573[1 +0.2((0.5)2] ~ 602oK T2 = T02/(1 + {[k-1]/2}M22) = 602/[1 +0.2((2.6)2] ~ 256oK

FROM STAGNATION REFERENCE
Knowing M1 and P1 and M2 and want to know P2, can find P0 from P1 and M1 and P2 from P0 and M2! CAN NOT GET AREA INFO FROM STAGNATION REFERENCE

Calculating Area

Missing relation for Area since stagnation state
does not provide area information*. So to get Area information use critical conditions as reference. *mathematically the stagnation area is infinity

If M = 1 the critical state; p*, T*, *….
Critical conditions related to stagnation Local conditions related to stagnation EQ c = [kRT]1/2 = [kRT*]1/2 isentropic, steady, ideal gas

Want to Relate Area to Mach Number
and Critical Area

EQ c = [kRT]1/2

EQ. 12.7c EQ b EQ. 12.7b EQ. 12.7b

1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1)
AxAy = Ax+y A = [1 + (k-1)M/2]1/(k-1) 1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1) = (k+1)/(2(k-1))

Provide property relations in terms of local Mach
EQs. 12.7a,b,c,d Provide property relations in terms of local Mach numbers, critical conditions, and stagnation conditions. NOT COUPLED LIKE Eqs. 12.2, so easier to use. isentropic, ideal gas, constant specific heats

Not built this way because of separation
* * A* Looks like converging-diverging section for accelerating subsonic flow to supersonic – but not. For example diverging section too large an angle, may result in separation. Then no longer isentropic. Note two different M #’s for same A/A* SUPERSONIC TUNNELS Not built this way because of separation

For accelerating flows, favorable pressure gradient,
the idealization of isentropic flow is generally a realistic model of the actual flow behavior. For decelerating flows (unfavorable pressure gradient) real fluid tend to exhibit nonisentropic behavior such as boundary layer separation, and formation of shock waves. accelerating

Sneeze Breath (180 mph)

What happens to p(x) as lower pb?
Isentropic Flow In A Converging Nozzle What happens to p(x) as lower pb?

As lower pb by vacuum pump, how does p(x)/po change?
IDEAL GAS, ISENTROPIC, QUASI-1-D, FBX=0, STEADY, only PRESSURE WORK, Ignore gravity effects, cp & cv are constant

M < 1 throat dp/ = - d{Vx2/2}

i: valve closed, no flow. Stagnation conditions everywhere e. g
i: valve closed, no flow. Stagnation conditions everywhere e.g. p = p0 everywhere. ii: = pe= pb < p0 iii: = pe= pb << p0 iv: = pe = p* = pb <<< p0 choked flow v: = pe = p* = pb <<< p0 Shock After Throat Not Isentropic

Given stagnation conditions and throat area
Isentropic Flow In A Converging Nozzle Given stagnation conditions and throat area What is mass flow when choked?

o/* = [1 + (k-1)/2] 1/(k-1) = [(k+1)/2]1/(k-1)
dm/dtchoked = ? = eVe Ae = e* Ve* Ae o/ = [1 + (k-1)M2/2] 1/(k-1) o/* = [1 + (k-1)/2] 1/(k-1) = [(k+1)/2]1/(k-1) * = e* = o [2/(k+1)]1/(k-1)

Ve* = c* = {kRT*}1/2 To/T = 1 + (k-1)M2/2 To/T* = (k+1)/2
dm/dtchoked = ? = eVe Ae = e* Ve* Ae Ve* = c* = {kRT*}1/2 To/T = 1 + (k-1)M2/2 To/T* = (k+1)/2 T* = To2/(k+1) Ve* = {kR2To/(k+1)}1/2

Ve* = c* = {2kRTo/(k+1)}1/2 * = e* = o [2/(k+1)]1/(k-1)
dm/dt = ? = eVeAe Ve* = c* = {2kRTo/(k+1)}1/2 * = e* = o [2/(k+1)]1/(k-1) [2/(k+1)]1/2 [2/(k+1)]1/(k-1) = [2/(k + 1)][k-1+2]/(2[k-1)] = [2/(k+1)][k+1]/(2[k-1])

* = e* = o [2/(k+1)]1/(k-1)
dm/dt = ? = eVeAe Ve* = c* = {2kRTo/(k+1)}1/2 * = e* = o [2/(k+1)]1/(k-1) o = po/RTo

dm/dtchoked = 0.04 Aepo/To1/2 dm/dtchoked = 76.6 Aepo/To1/2
For air in kg/s: dm/dtchoked = 0.04 Aepo/To1/2 For air in lbs/s: dm/dtchoked = 76.6 Aepo/To1/2

GIVE ME TWO REASONS WHY WE
CAN NOT INCREASE THE MASS FLOW RATE ABOVE:

GIVE ME TWO REASONS WHY WE CAN NOT INCREASE THE MASS FLOW RATE ABOVE:
Downstream conditions-influences propagate at the speed of sound so can’t move upstream past throat (2) Can’t exceed sonic conditions at throat cause that would require the flow to become sonic upstream in the converging section

Schematic Ts diagram for
choked flow through a converging nozzle.

Affect of Area Change in a Converging-Diverging Nozzle

CONVERGING-DIVERGING
S E N T R O P C NONISENTROPIC

Isentropic supersonic nozzle flow Isentropic subsonic nozzle flow
Single solutions Infinite number of solutions

??? – IF FRICTIONLESS WOULD WE FIND
COMPLETE PRESSURE RECOVERY - ???

!!! – NO – ASSUMING FRICTIONLESS COMPRESSIBILITY AFFECT - !!!

i – if flow is slow enough, V<0. 3M, then  incompressible so B. Eq
i – if flow is slow enough, V<0.3M, then  incompressible so B. Eq.  holds. ii – still subsonic but compressibility effects more apparent, B.Eq. Not Good. iii – highest pb where flow is choked; Mt=1 i, ii, iii (M<1) and iv (M>1) are all isentropic flows 0.3 M ~ 100 m/s But replace Ae by At.

* * 0.3 M ~ 100 m/s Note: diverging section decelerates subsonic flow, but accelerates supersonic flow. What is does for sonic flow depends on downstream pressure, pb. There are two Mach numbers, one < 1, one >1 for a given C-D nozzle which still supports isentropic flow when M=1 at throat.

Flow can not contract isentropically to pb so contracts through a shock. Flows are referred to as being overexpanded because pressure pe < pb. Lowering the back pressure, pb, below condition iv has no effect on the flow in the nozzle. The flow is isentropic from the plenum chamber to the nozzle exit and then undergoes a 3-D irreversible expansion (shock?) to the lower back pressure. A nozzle operating under these conditions is said to be underexpanded because additional expansion takes place outside the nozzle. If supersonic and isentropic and pe=pb then referred to as design conditions. C-D nozzles designed to achieve supersonic flow I.e. case iv. Flow leaving a C-D nozzle is supersonc if back pressure, pb, is at or below design pressure. The exit M# is fixed once the area ratio Ae/A* is specified. All other exit plane properties for isentropic flow are uniquely related to stagnation properties by Me. The assumption of isentopic flow for C-D nozzles is good for pb at or lower than design conditions. Because back pressure will change for an ascending rocket, impossible to be at design conditions everywhere.

When M >1 and isentropic then flow is said to be at:
design conditions. Lowering the back pressure, pb, below condition iv has no effect on the flow in the nozzle. The flow is isentropic from the plenum chamber to the nozzle exit and then undergoes a 3-D irreversible expansion (shock?) to the lower back pressure. A nozzle operating under these conditions is said to be underexpanded because additional expansion takes place outside the nozzle. If supersonic and isentropic and pe=pb then referred to as design conditions. C-D nozzles designed to achieve supersonic flow I.e. case iv. Flow leaving a C-D nozzle is supersonc if back pressure, pb, is at or below design pressure. The exit M# is fixed once the area ratio Ae/A* is specified. All other exit plane properties for isentropic flow are uniquely related to stagnation properties by Me. The assumption of isentopic flow for C-D nozzles is good for pb at or lower than design conditions. Because back pressure will change for an ascending rocket, impossible to be at design conditions everywhere.

Lowering pb further will have no effect upstream, where flow remains
isentropic. Flow will go through 3-D irreversible expansion. Flow is called underexpanded, since additional expansion takes place outside the nozzle. Lowering the back pressure, pb, below condition iv has no effect on the flow in the nozzle. The flow is isentropic from the plenum chamber to the nozzle exit and then undergoes a 3-D irreversible expansion (shock?) to the lower back pressure. A nozzle operating under these conditions is said to be underexpanded because additional expansion takes place outside the nozzle. If supersonic and isentropic and pe=pb then referred to as design conditions. C-D nozzles designed to achieve supersonic flow I.e. case iv. Flow leaving a C-D nozzle is supersonc if back pressure, pb, is at or below design pressure. The exit M# is fixed once the area ratio Ae/A* is specified. All other exit plane properties for isentropic flow are uniquely related to stagnation properties by Me. The assumption of isentopic flow for C-D nozzles is good for pb at or lower than design conditions. Because back pressure will change for an ascending rocket, impossible to be at design conditions everywhere.

Affect of Area Change Example

Consider a rocket engine:
hydrogen Me = ? Ve = ? Ae = ? 3517 K 0.4 m 25 atm 1.174 x 10-2 atm oxygen dm/dt = ? = 1.22; Molecular Weight = is 16, R = 8314/16 = J/kg Calorically perfect gas, isentropic flow

Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? Me = ?

po/pe = {1 + [(k-1)/2]Me2}k/(k-1)
Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? po/pe = {1 + [(k-1)/2]Me2}k/(k-1)

Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? Ve = ?

Ve = Mece = Me(RTe)1/2 To/Te = 1 + [(k-1)/2]Me2 Ve = Me(RTe)1/2
Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? Ve = Mece = Me(RTe)1/2 To/Te = 1 + [(k-1)/2]Me2 Ve = Me(RTe)1/2

Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? Ae = ?

Ae/A* = [1/Me][(1 + {(k–1)/2}Me2/{(k+1)/2}](k+1)/[2(k-1)]
Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? Ae/A* = [1/Me][(1 + {(k–1)/2}Me2/{(k+1)/2}](k+1)/[2(k-1)]

? Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? dm/dt = ?

o/e = [1 + (k-1)/(2Me2)]1/(k-1)
eAeVe = constant Given: To, p0, A*, pe Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ? dm/dt = eAeVe po = oRTo o/e = [1 + (k-1)/(2Me2)]1/(k-1)

With Numbers At= 0.4 m2 hydrogen Me=? Ve =? Ae=? dm/dt=? oxygen
pe = atm To =3517K po= 25 atm MW = 16  = 1.22 Me=? Ve =? Ae=? dm/dt=? At= 0.4 m2 oxygen

po/pe = (1 + ½ [k-1]Me2) k/(k-1)
Know: To; po; At; pe Me=? Ve =? Ae=? dm/dt=? po/pe = (1 + ½ [k-1]Me2) k/(k-1) (Eq a) Me = 5.21

To/Te = 1 + ½ (k-1)Me2 = [po/pe](k-1)/k
Know: To; po; At; pe Me=? Ve =? Ae=? dm/dt=? Ve = Mec = Me(kRTe)1/2 R = 8314/16 = J/(kg-K) To/Te = 1 + ½ (k-1)Me2 = [po/pe](k-1)/k (Eq b) Te = K Ve = Me(kRTe)1/2 = 1417 m/s

dm/dt = * A* V* dm/dt = * A* V* V* = c = (kRT*)
Know: To; po; At; pe Me=? Ve =? Ae=? dm/dt=? dm/dt = * A* V* */ o = (2/[k+1])1/(k-1) (Eq c) = 0.622 o = po/(RTo) = kg/m3 * = kg/m3 dm/dt = * A* V* V* = c = (kRT*) T*/To = 2/(k+1) (Eq b) T* = 3168K V* = (kRT*)1/2 = 1417 m/s dm/dt = * A* V* = kg/s

[1 + Me2(k–1)/2)](k+1)/[2(k-1)]
Know: To; po; At; pe Me=? Ve =? Ae=? dm/dt=? dm/dt = e Ae Ve e = pe/(RTe) = kg/m3 Ae = (dm/dt)/(eVe) = 487.4/( )(3909) = 48.5 m2 Ae/A* = [1/Me] x [1 + Me2(k–1)/2)](k+1)/[2(k-1)] [(k+1)/2](k+1)/[2(k-1)] Eq. 12.7d Ae = 48.5 m2

the end – class 12 1947 – Chuck Yeager breaks sound barrier in Bell XS-1 Mach number = 1.06 = 700 mph at 43,000ft At that time no wind tunnel data on transonic flight! It required an unhesitating boldness to undertake such a venture …. an almost exuberant enthusiasm…but most of all a completely unprejudiced imagination in departing so drastically from the known way.

CH.12: COMPRESSIBLE FLOW It required an unhesitating boldness to undertake such a venture …. an almost exuberant enthusiasm…but most of all a completely.

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CH.12: COMPRESSIBLE FLOW It required an unhesitating boldness to undertake such a venture …. an almost exuberant enthusiasm…but most of all a completely.

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Presentation on theme: "CH.12: COMPRESSIBLE FLOW It required an unhesitating boldness to undertake such a venture …. an almost exuberant enthusiasm…but most of all a completely."— Presentation transcript:

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