1 turbomachinery lecture 3 - compressibility - isentropic - area, mass flow functions - c-d nozzle
TRANSCRIPT
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Turbomachinery Lecture 3
- Compressibility- Isentropic- Area, Mass Flow Functions- C-D Nozzle
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Turbomachinery• Definition:
– A turbomachine transfers energy to or from a fluid flowing continuously through a casing by the dynamic action of a rotor and by the flow conditioning of a stator.
• Works on a fluid to produce power or flow (and pressure rise)
• Adds energy to fluid................Pump or Compressor– Fan: pressure rise up to 1 lbf/in2
– Blower: pressure between 1 - 40 lbf/in2
– Compressor: pressure rise > 40 lbf/in2
• Extracts energy from fluid............Turbine
– Pressure changes due to motion of parts or displacement of boundaries
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Compressible Flow• Density varies making continuity & momentum more
difficult to solve.
because varies with velocity.
• Also, can't integrate Bernoulli directly
• Compressible flow problems can be solved iteratively using continuity, state et. al.
cosm AC
.2
2
constVdP
4
• Example:m = 50 lb/sec A = 200 sq.in.
P0 = 14.7 psia = 30
T0 = 519 R
• GuessC = 646.8 ft/sec
2
0
22
2
2 2
2
646.8 / sec519
. .2 32.174 778.16 .24
.sec
6008.8 / sec
484.19deg
p
CT T
C
ftT
ft lbm ft lbf BTUlbf BTU lbm R
ft R
T R
Compressible Flow
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Compressible Flow
• Pressure
/ 1
00
TP P
T
3.5484.19
14.7 11.529519
P psi
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Compressible Flow• Density can now be found from state:
11.529 144
53.349 484.19
0.06427 / .lbm cu ft
RT
P
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Compressible Flow• Mass Flow:
• note:
• 19%>
cos
0.06427 646.8 cos30 200 /144
50.00 / sec
m AC
m
m lb
0 0.0765 / .lbm cu ft
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Compressible Flow
• Mach Number Functions:
– Easily calculated & clarify physics
• Mach number & acoustic speed are critical concepts!
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Compressible Flow
0
1
p
dPIsentropically TdS dh
dPc dT RT
PdT dP
T P
State P RT
dP RdT RTd
T
dTd
P
dP
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Compressible Flow• Using isentropic relation between pressure &
temperature derivatives:– Use adiabatic state law
P
dPd
P
dP
1
d
P
dP
1
2dP Pa RT
d
1P CT
11
Compressible Flow
• Using equation of state, acoustic speed in an ideal gas is [from kinetic theory]:
• By definition Mach Number is:2
2
2
V dynamic pressure
p static pressureV VM
a a V kinetic energy
RT thermal energy
1716
287
Ta gRT
T
12
Compressible Flow
• Static & Total properties as functions of Mach number: 2
0 2
Vh h
g
20
20
12
12
p
p
T V
T gc T
T R V
T c gRT
0 211
2
TM
T
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Compressible Flow – Critical Velocity
• What does subscript * mean? It means value of variable when M=1 [sonic]
• Vcr is only function of gas [] and stagnation props.
2
0
2 2 22 220 0
0
2
1
1 1 2 1 2 2 1 1
2
1
cr crcr
cr
Vh h
a V V RTa VV
V RT
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Compressible Flow
• The relation between static & stagnation properties is isentropic. Then:
/ 120 1
12
PM
P
1/ 10 21
12
M
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Compressible Flow• The relation between compressible and Bernoulli [B-p.55]
12
0
2 2
2 40
2 2 22
2
1/ 1
2
1(1 ) 1 ( 1) / 2 ... &
1 2
2/ 1 ...
2 2
2 2 2 / 2
n
p p M
Binomial expansion for small x is x nx n n x n x M
For small M one gets p p M M
V V VBut since pM p p
a p
The
2 24
0
2
0
21 ...
2 4 2
0.3, 2.3% ( ).2
V Mexpanded isentropic equationbecomes p p M
Vfor M p is in error from Bernoulli p
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Compressible Flow Relationships
• Mass Flow parameter [=0]
0
0
m VA
dm VdA AdV VAd
dm dA dV d
m A V
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Compressible Flow Relationships
• Area-Mach number differential relation
• Area-Mach number integral relation
2
22
11
MdA dV dpM
A V M p
1
2 121 2 1
11 2
AM
A M
More on next chart
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Compressible Flow Relationships
• What does subscript * mean?– For all flow variables it means value of
variable when M=1 [sonic]
– For area A* this is reference area for choking flow [M=1]
• Note this area is a minimum or throat
1
2 121 2 1
11 2
AM
A M
More on next chart
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Compressible Flow Relationships
Flow textbooks
-www.engr.uconn.edu/barbertj- Compressible
- Aero Calculator- calcbody2
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Compressible Flow Relations
21
2
22
11
MdA dV dpM
A V M p
Of interesthere
Of interesthere
22
2
22
11
MdA dV dpM
A V M p
Over-expanded
23
24
25
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Compressible Flow Examples
0 00 0
: 450 1890 1.5
3.671 6938 1.45 652.5
1.4 1716 450 1040
1.5 1040 1560
s s
s s
s s
Given T R p psf M
p Tp psf T R
p T
a a RT fps
V fps
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Compressible Flow Examples01 01 *
*
0 0
*
0
10 300 / 6
/ 6 0.097
1.006 9.94 1.002 299.4
1.4 287 299.4 346.8
33.6
/ 6 3.368
63.13 0.
s ss s
s s
ss
Consider isentropic flow in C D nozzle
p atm T K A A
Subsonic A A M
p Tp atm T K
p T
a a RT mps
V mps
Supersonic A A M
pp
p
0
1 /2 1
0 *
0
1584 3.269 91.77
192 646.7
2
1
ss
s
Tatm T K
T
a a mps V mps
p Am VA if choked
RT
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Compressible Flow
• Mass Flow Parameters:
VRT
P
A
m
AVm
cos
cos
1/ 2
0
0cos
Tm V g
PA RT TgRT
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Compressible Flow Relationships
• Mass flow parameters
0 0
0 0
00 1
0 2 12
1/ 2
0 2
( , )
11
2
11
2ss
m VA
m p V pV RT M
A RT a RT
m T TpM f M
p A p T R
m T R MFP
p AM
m T RFP M M
p A
Note: FPo, FPs are similar, but different f[M] powers
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Compressible Flow Relationships• Mass flow parameters
00 1
0 2 12
01
0 2 12
11
2
11
2
m T R MFP
p AM
p A Mm
RTM
How to get more mass flow, i.e. greater thrust, more power?
1
2 10
0
, 1
2
1
if choked at throat M
p Am
RT
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Compressible Flow Relationships
• Mass flow parameters units– m in lbm/sec
– p0A in lbf [spatial dimensions cancel]
– T0 in degs. Rankine
– A is sometimes frontal area Acos
00
0
0
0
0
0
0
0
1
1716 /1.4
32.2
1.0888
RTmFP
p A g
m T gR
p A g
m T
p A
m T
p A
For air
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Compressible Flow Examples
01 01 *
*
0 0
1 /2 1
0 *
0
10 300 / 6
/ 6 3.368
63.13 0.1584 3.269 91.77
192 646.7
2
1
s ss s
s
Consider isentropic flow in C D nozzle
p atm T K A A
Supersonic A A M
p Tp atm T K
p T
a a mps V mps
p Am VA if choked
RT
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Example2
0 0
2*
*
00
0
2 ,
0.5 1 300
1.4, 0.5 1.340 1.49
( , ) 353.6 / sec
Air in duct of A m has flow such that
M p atm T K
AFor M A m
A
area to choke
p Am FP M kg
T R
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Static Pressure Mass Flow Parameter
• Defining: FP = Flow parameter=f(M)
• For Air
• Can be inverted
1/ 220 1
1cos 2s
RTmFP M M
PA g
01.0883coss
m TFP
PA
2/1
2
1
1211
sFPM
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Total Pressure Mass Flow Parameter• Introducing P0:
• No explicit solution for M • FPs is single valued, FPo is not• FPo max = 0.5787 for =1.4• FPo max always at M=1
1/ 2
00
0 0cos
Tm P gP M
A P RT T
1/ 2
0 00
0 0cos
RT Tm PFP M
P A g P T
1
2 12
0
11
2FP M M
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Calculate FPo• From Previous Example:
m = 50 lb/sec A = 200 sq.in.
P0 = 14.7 psia = 30
T0 = 519 R
• Rearrange FPo
00
0
0.4869cos
RTmFP
P A g
1 /2/ 12
0
11 0.5997
2calc guessM FP M M
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Mass Flow Parameters
Be careful: FPs single valued, FPo double values
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Total Pressure Mass Flow Parameter
• Consider FPt:
• For fixed , a fixed value of
produces the same Mach number - regardless of the level of pressure, temperature or molecular weight (R).
1
2 120
00
11
cos 2
RTmFP M M
P A g
0
0 cos
m RT
P A
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Total Pressure Mass Flow Parameter
• Defines common flow parameters.
• Valid for flow with one gas.
• Corrected flow.
0
0 cos
m RT
P A
0
0
m T
P
0
0
/ 519
/14.7
m T m
P
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Other Parameters[Covered in Lecture 4]
• Ideal gas equation for Mach number leads to speed parameters, also for a single gas.
• Speed parameter
• Corrected speed
0
N
T
N
0 0 14.696 518.7
in inP T
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Significance of Flow & Speed Parameters
• A device operating at the same speed parameter and flow parameter has the same Mach numbers, velocity diagrams, flow angles etc, regardless of the level of physical speed, pressure & temperature.
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Flow and Speed Parameters• Conditions: same gas, high Reynolds number, same
clearances, and same • Speed and Flow parameters are used for turbine
maps
0
5
10
15
20
25
30
35
1.0 1.5 2.0 2.5 3.0
Exp Ratio
Mrt
T/P
N/rtT
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Corrected Flow & Speed Parameters
• Corrected Flow and Corrected speed used for compressor maps
3
4
5
6
7
8
9
50 60 70 80 90 100 110
Corrected Flow lb/sec
Pre
ssu
re R
atio
Corrected Speed
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Flow Parameter
• Again, Consider FP0:
• Unlike P, T & R; cannot be "corrected".
• Changing , changes relation between FP0 and Mach number!
1
2 120
00
11
cos 2
RTmFP M M
P A g
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Flow ParameterGamma Effect On Continuity
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
1.0 1.2 1.4 1.6 1.8
Gamma
Ma
ch N
um
eb
er
Air
Helium
Butane
FPT = .560
Message: More complex gasses choke at a lower Mach number
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Example Solution to Mass Flow Parameter
0 1.0719 484.18degT
T RT
646.87 / secV M gRT ft
/ 1
0 0 11.53P T
P psiP T
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Area Ratio
• A/A* is flow area / flow area at M = 1.0
0 1
0
1 / 2 / 12
1 / 2 / 1
*
11
2* 1
12
MFPA
A FP
MA
AM