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section 3.4 Resistance & Propulsion source: WoudChapter 3

3.4.1 Hull Resistance

R:= c1c1vs2

physics(3.1)

P

E

:= R vRs

P

E

= effective_power defined(3.2)

y c 0 vs physics

PEc1vs3

substitution

(3.3)

c1:= y ( ) (3.4)

y= f fouling displacement_variations( , ,sea_state,water_depth) essentially time and operations (3.5)

speed dependency of c1.... nondimensional resistance coefficient

RRCT:=

1 2CT= non_dimensional_total_resistance

defined(3.6)

As

vs

2

As(ship surface area) not readily available, so use volume proportionality ... As~ Vol^2/3

PEPECE:= CE= specific_resistance defined (3.7)2

3 3 Vol vs since = Vol Vol:=

PE PE PEPECE:= CE CE:= (3.8)

2 2 1 2

VolVol3

vs3

3

vs3

3

3vs

3

CE= f Re Fr( , ,Ro,Hull_form ,external_factors) dimensional analysis, physics (3.9)

vsLenRe:= Re= reynolds_number (3.10)

vvssFr:= Fr= froude_number (3.11)

gLe

kkRo:= Ro= non_dimensional_roughness defined (3.12)

Len

CE= f v( s, ,fouling,Hull_form ,sea_state ,water_depth)

3PEPE

PE R vs PEc1vs c1:= R :=3

vs

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1 2 1 2

and from (3.8)PE:=

3

3vs

3CECE c1:=

PEPEc1

3

3CE (3.13)

3vs

1 2

PE:= 3

3

vs3CECE

shows dependency of PEon speed and displacement

e.g. if CEand vsare assumed constant ... a change in from nominal changes effective power

2

3

PE:= PE_nom (3.14)nom

3.4.2 Propulsion need to deliver thrust T to overcomeR vs

N.B. I am assuming one

resistance R at speed vsPE:= R (3.2) propeller. Woud uses kp=

number of propellers.

power delivered by propeller in water moving at vA

PT:= T vAA PT= thrust_power defined (3.15)v

Thrust deduction factor

required thrust T normally exceeds resistance R for two main reasons:

propulsor draws water along the hull and creates added resistance

conversely, the advance velocity is generally lower that the ship's speed, due to operating in the wake

t= thrust_reduction_factor= difference_between_thrust_and_resistance_relative_to_thrust defined

T RR RR=>

T 1 t (3.16)t:= R:= (1 tt) T T:=

"The term thrust deductionwas chosen because only part of the thrust produced by the propellers is

used to overcome the pure towing resistance of the ship, the remaining part has to overcome the addedresistance: so going from thrust T to resistance R there is a deduction. The term is somewhat

misleading since starting from restance R the actual thrust T is increased." page 55

Wake fraction

propeller generally in boundary layer of ship where velocity is reduced; vAis then < vs

vsvs vAw:= w= wake_fraction defined

(3.17)vs

w= difference_between_ship_speed_and_advance_velocity_in_front_of_propeller_relative_to_vs

"(Note that as a result of the suction of the propeller, the actual water velocity at the propeller entrance is muchhigher than the ship's speed: the advance velocity, however is equal to the water velocity at the propeller disc

area if the propeller would not be present In other words it is the far field velocity that is felt by the propeller

located in the boundary layer of the hull.)" page 56

thus ...vA:= (1 ww) vs

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Hull efficiencywith these two factors the thrust power does not equal

the effective power. The ratio of effective power to thrust PE

PE

power is defined as the hull efficiency.H:=

PT

(3.18)

RR R vs 1 tredefine R

T:= := (1 ww := H (3.19)vA ) vs H 1 t 1 wT vA

Propeller efficiency

to deliver the required thrust at a certain ship's speed, power must be delivered to

the propeller as torque Q and rotational speed p.

Po:= QQp defined Po= open_water_power (3.20)

since ... p:= 2 npnp Po:= Qpp Po 2 Q np

o:=PP

PTdefined o= open_water_efficiency o

1T

vA (3.21)

oo 2 Q np

"In reality, i.e. behind the ship, the torque Mpand thus the power delivered Pp actually delivered to the propeller

are slightly different as a result of the non-uniform velocity field in front of the propeller." page 58

PNA vol II page 135 says: " Behind the hull, at the same effective speed of advance V A, the thrust T and revolutionsT VAT

n will be associated with some different torque Q, and the efficiency behind the hull will be B:= (34)2 nQ

The ratio of behind to open efficiencies under these conditions is called the relative rotative efficiency, being given by

T VT A T VT A B 1B:= o:= R:= R Qo (35)

2 nQ 2 nQo oo Q

Thus we define Ppas power delivered. (per propeller)

2 Mp(3.22)Pp:= Mppp Pp np

and ... the ratio between open water power and actually delivered power is

PoPo QR:= R (3.23)

P Mp p

Propulsive efficiency combining all these effects .. looking forward to design/evaluation at model (open water) scale

PEPE effective_power PE for kp= 1D:=PD

defined D=power_delivered

=Pp

(3.24)

rewriting ... PE PT Po PE PT Po

D

= = Pp PT Po

PT Po Pp

using definitions of efficiency from above ...

PE 1 t PT PoPo 1 ttH= = o:= R:= D:= HH o R D:= o R

PT 1 w PoPo Pp 1 w

(3.25) (3.26)

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Actuator Disk

u

in

s p

r

r

e

e

ss

ub

t

t

re

e

am

,D0 A0

(10.2)

(10.3)

(10.3a)

(force = mass flow * delta velocity)

VA, p0, D0

VA+v, D1p0

V

stream tube

actuator disk A

assume: propeller is a disk with

diameter D and area A

frictionless

no rotation - upstream or downstream

model propeller as thin "actuator disk"

causing instantaneous increase in

pressure

A1,D1,VA+ v ,D A,V VA,

Thrust= T= Ap (10.1)

continuity ... VA= constant

m_dot= VAA0= (VA )V A + v

A1 VAD02

2 ( v)

2

= V D +=

VA D1=

2 V 2 2 V 2D DD

0

D

1

= = VA

VA

+ v

VVD0:= D

VAVA D1:= D+ vvVA

_in_momentum= thrust_on_disk= T= m_dotout(VA+ v) m_dotinVA

T= A1(VA+ v)2

A0VA2

2 2 (10.4)

T:= D1D1

(VA+ v)2

D0

VA2

4 4

1 2Tsimplify V D using (10.3a) above (10.5) v

4

now using Bernoulli equation p+1

v2

= constant2

on both sides of the disk (a force is applied at the disk)

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ahead ... p+1

V2

= p0+1

VA2

aft ... p+ p+1

V2

= p0+1

(VA+ v)2

2 2 2 2

subtract ahead from aft ... p=1

2 (VA+ v)

2 VA

2

=1

2 v(2 VA+ v) (10.6)

(VA )2 VA22 VA

1 v 2 VA vv 2

+ v result ... simplify + v p:= ( + )

2

now using (10.1) and equating to (10.5)A:=

4DD

TT:= Ap1

D2 ( + v) v 2 VA

8v

1 2

from which ... V:= VAVA+(10.5) T:= V D 2 v

4

so .... T 1 D2 VA+1v v T:= 4

DD2

VA

+ 2v

v (10.9)

4

2

define a thrust loading coefficient ...

TTCT:= 1 vsubstitute (10.9) 2 VA+ v a quadratic in v (10.10)1 2 2

CT D VA 2 VA

2

2 4

Given

1

v 1 1

CT

= 2 VA

+

v

Find

v

2 2

( ) 2 VA

2 1 +(1+ CT) ( ) (1+ CT) ( ) 1 VA

1

taking only positive rootv

1 + 1+ )2

= ( ) ( CTVA

useful_work_from_disk PT T VAI= ideal_efficiency=

work_done_on_fluid_by_thrust_per_unit_time=

Padded

=T V

T VA

1II:= VA uses relationship for V above (10.9) (10.11)T V 1

VA

+ v2

1

with ... v:= VA( ) +(1+ CTCT)2

:=1

1

v

2

1

(10.12) 1 II simplify

1+ 2 2VA 1+(1+ CT)

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create plot with loading

0 1

CT:=

1

2

3

4

+ CTi

I=

0.828

0.732

0.667

0.618

as shown in PNA

i:= 0..4 Ii:=

1+

2

1

0

1

2

3

4

0

0.5

1

I

CT

Observations: 1). Propeller at high load coefficient CT less efficient

2). I:=1

=> efficiency maximum when v small1 vv

1 +2 VA

3) for given thrust T, T 1

4

2 D

VA+1

2v

v v small => D large => propellerdiameter large

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Propeller Testing

Screw propeller replaced paddle wheel ~1845 in Great Britain (vessel) - Brunel

In test;

independent variables are

velocity of advance VA

shaft rotation speed n (rev/sec), N (rev/min)

dependent variables are:

torque Q

thrust T

i.e. we build a propeller, rotate it a a given speed in a given flow and measure thrust and torque

(at this point - conceptually - not practical at full scale)

are considering propeller in general, no ship present, => open water

velocities relative to blade:

VA

VR VA

2**n*r=*n*d*n*d

test at given n, vary VA, measure thrust (T), torque (Q) and calculate efficiency ( )

Q

T

VA

Q

To

typical performance curve at

given rotaion speed, note zero

efficiency at VA= 0 and T = 0

o

Obviously, testing at full scale impractical, hence use model scale and apply to geopmetrically similar propeller.

Expect performance to depend on:

VA velocity of advance

D diameter of propeller

n rotational speed

fluid densitydynamic viscosity (=/= kinematic viscosityp - pvpressure of fluid (upstream static pressure) compared to vapor pressure

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First non-dimensionalize: using n and D

TThrust KT :=

2 4n D

Torque KQ

:=QQ

2 5n D

VAVAJ :=

advance_velocity n D

DVAReynold's number based on diameter: ReD :=

pppvnominal cavitation index (presure) N :=

1 2

VA2

dimensional analysis would show:f J ReD, = f J ReD,= , ,KT ( N) KQ ( N)

Typical propeller: fully turbulent, hence only weakly dependent on Re D

deeply submerged, N not influential, hence:

KT f J( )= KQ f J( )=

substituting the above coefficients ...

recall open water efficiency efficiency o :=T VA

2 n Qo

To

1

2KT

J

KQ o :=

1

2

KKTT

KQ

J

so now we test a model scale propeller ~ 12 inches diameter measuring thrust and torque and plotting

non-dimensionally: (10 * KQis used for similar scales, KQhas extra D when non-dimensionalized)

10*KQ

KT

o

o

10*KQ

KT

J=VA/(n*D)

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- unyielding fluid - chord defined as line between nose and tip

Propeller Series Testingref: PNA pg 186 ff

Early series done by Taylor, Gawn, Schaff, NSMB

For design purposes NSMB became standard

NSMB = Netherlands Ship Model Basin; now MARIN Maritime Research Institute Netherlands

first series designated A were airfoil shapes had some cavitationrevised shapes to avoid cavitation:

widened blade tips

circular section near tip

airfoil near hub, etc.

designated B series see figure 48 in PNA for geometry

Propeller pitch

Pitch = distance moved along axis of propeller by an imaginary line parallel to the blade chord line for

one rotation of the blade

P/(2)

r

typically use at r =0.7*R if variable

( )usually non-dimensionalized by D tan =

P=

P

r 2 D

D D r( ) = (= D radius)

B series is family of curves of open water performance at model scale for numbers of blades and area ratio

Blade area ratio AE/A0

Number of

blades Z

2 0.30 . . . . . . . . . . . . . .

3 . 0.35 . . 0.5 . . 0.65 . . 0.80 . . . .

4 . . 0.40 . . 0.55 . . 0.70 . . 0.85 . 1.0 .

5 . . . 0.45 . . 0.6 . . 0.75 . . . . 1.05

6 . . . . 0.5 . . 0.65 . . 0.80 . . . .

7 . . . . . 0.55 . . 0.7 . . 0.85 . . .

above performance curve (KT, KQ, vs. J shown for particular number of blades, P/D A E/A0member designated as: B.5.50 =>

B series

5 blades

0.50 area ratio

This introduced Expanded area ratio =

consider section along cylindrical surface at radius r using helix of pitch P

flatten helix

rotate to show cross section at radius r

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sum expanded section over radius = expanded area of blade * number of blades Z =

expanded area

EAR (Expanded area ratio) = Expanded area / disk area

Expanded_area AEEAR = =

disk_area D2

4

can also express developed area and projected area see hydrocomp report

Troost published a set of these curves in "notebook"

later Oosterveld and Van Oossanen published a set of curves based on an empirical curve fit

ref: "Further Compiuter - Analyzed Data of the Wageningen B-Screw Series", International Shipbuilding

Progress, Volume 22

P AE t P AE t

KT = f1

J,D

,A0

,Z,Rn ,c

and .... KQ = f2

J,D

,A0

,Z ,Rn ,c

the coefficients for Re = 2*10^6 without t/c in the fit are listed in Table 17 page 191 of PNA

corrections for t/c and Re can be added later

this provides a set of curves as indicated. e.g.

regressioncoefficients Re=2*10^6

plot for B.5.75 for single value of P/D P_over_D := 0.6 EAR := 0.75 z := 5

38

n

sKtn tKtn uKtn vKtn

Kt J P_over_D( , ) := a J P_over_D EAR z

n =046

Kq J P_over_D( , ) := b

nJ

sKqnP_over_DtKqnEAR

uKqnzvKqn

n =0

( J(J P_over_D) :=

Kt J P_over_D, ),

2 Kq J P_over_D( , )

trust_power T VA revolutions

= = n =propeller_power Q 2 second n

1 2 4 T VA T 2

n D D VA Kt J

=

= Q 2 n 1

n2D

4D Q 2 n 2 Kq

2

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http://www.hydrocompinc.com/knowledge/whitepapers/HC135-BladeAreaRatio.pdfhttp://www.hydrocompinc.com/knowledge/whitepapers/HC135-BladeAreaRatio.pdfhttp://www.hydrocompinc.com/knowledge/whitepapers/HC135-BladeAreaRatio.pdf

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we have some data problem with polynomials as they calculate some values beyond real data (K T0 , ( , is made positive definite, ( J P_over_D) ,0)

eliminate negative segments - make

Kt(J P_over_D) := if Kt J P_over_D( , ) >0 , ( , ) ,0) positive definite, ( Kt J P_over_D

Kq(J P_over_D) := if Kq J P_over_D( , ) >0 , ( , ) ,0), ( Kq J P_over_D

plottingconstructs

EAR := 0.75 z := 3 P_over_D := 1.2

0 0.13 0.25 0.38 0.5 0.63 0.75 0.88 1 1.13 1.25 1.38 1.5 1.63 1.75 1.88 2

Advance ratio J=VA/(n*D)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

Kt,10*Kq,e

fficiency(eta)

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Kt,Kq*10,efficiency

Plot for P/D = 1.4, 1.2, 1.0, 0.8, 0.6 calculated using regression relationships

B_series z := 3 EAR := 0.75

1.5

1.4

1.3

1.2

1.1

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6

Advance Ratio J=VA/nD

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Using KTand KQfor design

we have seen in general the development of the Wageningen B series. The performance curves are available either in c

polynomials:

regressioncoeff. Re=2*10^6

polynomialrepresentation

use in design

A typical design problem calls for designing a propeller that will provide the required thrust at a given speed of advanc

from applying thrust deduction and wake fraction to resistance and ship velocity respectively. Design will imply selectin

plot that will maximize open water efficiency.

For now we will arbitrarily pick a number of blades and expanded area ratio. Later we will address the criteria in their s

non-dimensional forms of the parameters associated with thrust and speed:

T VAK

T= J = we have independent variables n and D. Normally one of these is determined by

n2D4 n D by hull form, or n by the propulsion train design, so we will look at two cases, one

and the other where n is fixed determine D

case 1 given: VA,T ,D find n and P/D for maximum efficiency

only thing unknown is n, eliminate ... from ratio of K Tand J

Kt T n2D

2T

= = this says that propeller (full scale and model) must match this ratio w

J2

n2D

4VA

2 D

2VA

2 VA, D and

Kt_over_J_sq :=T

we can plot a curve of KTvs J2and determine the points (values of

2 2D VA match.

the design point for a particular propeller (B.n.nn) i.e. n is determined from the value of J that satisfies: Kt J( )

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for example, let Kt_over_J_sq := 0.544 what n i.e. J will satisfy the relationship for a B 5.75 propelle

Kt_design J( ) := Kt_over_J_sq J2

select using B_series := 0.75z 5:= EAR

determineintersection

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Kt,Kq*10,efficiency

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

Advance Ratio J=VA/nD

VAintersection occurs at JJ =0.64 so ... n = where VAand D are known as described abo

JJ D

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selection of the optimum n for this B z.EAR propeller is a matter of comparing similar curves for a range of

P/D and choosing the maximum open water efficiency o

B series z= 5 EAR= 0.75

Kt,Kq*10,e

fficiency

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

Advance Ratio J=VA/nD

busy plot of Kt, Kq, oand Kt = constant * J^2. see breakdown below. P/D not labeled but ~ J at Kt = 0

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1

intersectionsolution

plot with only Kt but vertical lines at J for Kt/J^2 = Kt to show points which satisfy the design requirements

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

Advance Ratio J=VA/nD

P/D = 0.6 P/D = 0.8 P/D = 1.0 P/D = 1.2

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Kt,efficiency

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Kt,efficiency

note the oat each J intersection and select the maximum (P/D curves not well labeled, P/D ~ = J at K T=0. lef

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

Advance Ratio J=VA/nD

Plot for P/D = P_over_DT = ( 1.4 1.2 1 0.8 0.6 ) calculated using regression re

this case appears to have maximum at J_ans =0.64 P_over_D_ans =1 ( , ,z ,P_ J_ans EAR

VAso ... n = where VAand D are known as described above

J_ans D

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case 2 given: VA,T ,n find P/D and D for maximum efficiency

only thing unknown is D, eliminate ... from ratio of K Tand J

Kt T n4D

4T n

2

= = this says that propeller (full scale and model) must match this ratio w

J4 n2D4 VA4

VA4 VA, n and

Kt_over_J_4 :=T

we can plot a curve of KTvs J4and determine the points (values of

2 2D VA match.

for example, letKt_over_J_4 := 0.544

Kt_design( )J := Kt_over_J_4 J4

select using B_series z := 5 EAR := 0.75

the design point for a particular propeller (B.n.nn) i.e. n is determined from the value of J that satisfies: Kt (

since the process is identical to case 1, only the final result is shown

intersectionsolution

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Kt,efficiency

note the oat each J intersection and select the maximum (P/D curves not well labeled, P/D ~ = J at K T=0. lef

J_ans n

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

Advance Ratio J=VA/nD

Plot for P/D = P_over_DT = ( 1.4 1.2 1 0.8 0.6 ) calculated using regression re

this case appears to have maximum at J_ans =

0.74 P_over_D_ans =

1 (J_ans EAR,

,

z

and ... D =VA

where VAand n are known as described above

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Propeller Design (Detail Stage)

To this point we have developed the KTvs J2(J4) design approach. Most references present the series data in

alternative format. One version is curves of constant efficiency and 1/J on a scale of P/D vs B P1or P/D vs

Kq1/4*J-5/4. These are due to D. W. Taylor plotting the series data using B Pwhich we will define later. The best

description of use of the forms of the charts I found was in PNA, page 159 Propulsion and Propellers Section 10:

Propeller Design, by Karl Schoenherr. The following is excerpted from the text:

The procedure in using these charts depends on the nature of the problem to be solved; that is, on which

data are known and which are unknown. In general, propeller design problems belong to one of the

following categories:

1. Preliminary Design.

a. Given: The designed speed of the ship, the corresponding ehp and the propeller diameter .Required:

The propeller pitch and the rpm for best efficiency.

b. Given: The designed speed of the ship the corresponding ehp and the engine rpm. Required: The

propeller pitch and the propeller diameter for best efficiency.

2. Final Design.

Given: The ehp curve as a function of the ship speed, the propeller diameter, and the power output of

the engine at the designed rpm.

Required: The propeller pitch, the efficiency and the ship speed obtainable under the given conditions.

3. Analysis.

Given: The propeller dimensions, the ship speed, power, thrust and rpm.

Required: The true slip, wake fraction and thrust deduction.

The KTvs J2(J4) design approach we have done to date is directed at 1. Preliminary Design. At this stage,

the power required is determined based on a reasonable first estimate of propeller efficiency determined with

this approach. The propulsion plant is then sized accordingly. The propulsion plant may have discrete

incremental sizes and thus may not exactly match the first estimate exactly. The ship design proceeds,

perhaps a new resistance (close to preliminary design) etc, is obtained and then Final Design takes place. At

this point, PD(power delivered) to the propeller is known. It may not match (exactly), the preliminary estimate,

hence the V may be different.s

Taylor selected two parameters for plotting information for design work:

BPn= N * P1/2/VA

5/2 where N = rpm, P = power delivered (hp) ( = Q*2* *N) and VA= speed of

advance (kts), n = number of blades

and ..

BUn= N * P1/2/VA

5/2 where N = rpm, U = useful power (hp) ( = T*VA) and VA = speed of advance

(kts), n = number of blades

These are not non-dimensional but Taylor thought that was ok "since propellers work in water of practically

constant density, which will be taken care of by the constants used". S&P page 100

This motivated NSMB to present the data on plots of P/D vs K Q^1/4*J^(-5/4) = BP^1/2* constant which can be

shown to be equivalent as follows:

see B_series_units_US.mcd

1 5 1

4 4 2KQ J = 0.17279 BP

1

P in hp

n in RPMsimilarly, not

1

3 1

4 4 2KQ J = 1.75 BP

2

P in hpUK

VAin kts developed here...D in ft

VAin kts

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12

1 51

4

4

5

PDhp550lbf ft

2

4 in lbf sec

4

4 4 Q VA hp s 2 min ftKQ J =

= rpm n Dn

2D

5

sec2

VA kt

ft 5

sec2

n PD0.5

2 lbf 1.688 60

ft4 sec kt min BP 2.5

=

PD VAPD = Q 2 n Q =

2 n

1

1 21 1 1

1 5 4 1 1 1 1

KQ4

J4

=

PDhp

5rpm

2

4

5502

4

=

PDn

5

2

4

PD2

5

n

4= BP

2

4=

VAkt 2 lbf sec

41.688 60

2( ) 5 2 VA

ft VA

removing units 1another approach which accommodates other

550 4

units for PD, VAand n is shown in := 1.99 = 0.1728 5 2 B_series_units_conversion.xmcd2 1.688 60

regressioncoeff. Re=2*10^6

details

form of plot shown in PNA: P/D vs Kq1/4*J-5/4. Curves are constant and1/J. These are derived from the same data as our previous K TKQcurves.

1.4

1.2

1

these are fixed;

EAR 0.40 z 4

P/D

0.8

0.6

0.40 0.2 0.4 0.6 0.8 1 1.2 1.4

Kq^1/4*J^-5/4

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bp1i j,

i j,

nn

min knot2.5

0.5 0.6,

another form of the same information constant

efficiencies

158.871 0.5conversion of Kq1/4*J-2.5to BP. note 2abs1 139.697 0.55abscissa is log scale := 601.6889=

=

124.652

112.533

102.562

=

0.6

0.65

0.7

0.17279 J

EAR 0.4 z= 4=higher to right

= 0.7

= 0.65

= 0.6

= 103

= 0.55

= 112

= 0.5

= 125

= 140

= 159

P/D

1.2

1

0.8

0.6

0.4

1 10 100

BP1

an example of use of these curves ...

100say we have a design such that: P

D:= 16000hp n := V

A:= 16knot

min

n PD0.5

VA2.5

BP1 :=BP1 12.353

hp0.5 =

BP_ans := BP1 v_line := ..1.5

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we will plot that vertical line on the curves and determine the maximum efficiency, P/D and

P/D

= 0.7

1.2

= 0.65

1

= 103

0.8

= 112

= 125 = 0.6 = 0.550.6

= 140 = 0.5

= 1590.4

1 10 100

BP1

it appears that for the max is 0 := 140 0 := 0.67 P_over_D0 := 0.98 approximately

n D

VA=

ft D :=0VA

ft

min knot n min knotD =22.4 ft

let's say the diameter is limited to 20 ft by another constraint D1 := 20ftn D1

VA

1 :=ft 1 =

125then the best situation is

min knot1 := 0.65 P_over_D1 := 1.25 approximately

N.B. The shape of the developed curves is generally OK. I'm not completely confident in the exact values. The

validation is not as close as I would like.

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P R O P E L L E R N O T E

B O O K

B

S E R I E S PROPELLERS

D e f i n i t i o n :

S e r i e s No. o f B l a d e s

N o t a t i o n :

E x p a n d e d Area : R a t i o ,

P e r c e n t

P D

= P i c h / D i a m e t e r

T = T h r u s t , I b f

= T o r q u e , I b f f t

=

M a s s d e n s i t y , 1 . 9 9 0 5 l b f

s / f t

n = R o t a t i o n a l s pe e d , r e v s / s

D = M ax im um d i a m e t e r , f t

V A =

V e l o c i t y o f a d v a n c e , f t / s

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Cavitation Notes ref: PNA pages 181-183handout

p0= uniform_stream_total_pressure p1= pressure_at_arbitrary_point

V0= uniform_stream_velocity V1= velocity_at_arbitrary_point

1 2q= V

0

= dynamic_or_stagnation_or_ram_pressure

2

1 2p0+ V0 = constant Bernoulli

2

1 2 1 2p0:= constantconstant V0 p1:= constantconstant V1

2 2

for propeller immersion, measured at radius r, minimum p0is obtained from ...

p0= pa+ gh grpa

= atmosphere

h= shaft_centerline_immersion

gr

accounts for minimum when r vertical up

V0estimated as (VA^2+(*r)^2)0.5

if p1=> pv= vapor pressure, cavitation occurs

pa+ gh grpvdefine: L

= local_cavitation_number= and if pressure REDUCTION / q

2 2 2

>= Lcavitation occurs2

VA + r

early criteria (Barnaby) suggested limiting average thrust per unit area to certain values (76.7

kN/m2 = 10.8 psi) for tip immersion of 11 in increasing by 0.35 psi (unit conversions don't

match up)

kN76.7

=

11.124 psi

earlier PNA (1967) stated Barnaby suggested 11.25 psi

2m

can calculate pressure distributions around blade so can calculate local cavitation situation

early in propeller design, want blade area to avoid cavitation (more blade area, less pressure

per unit area for given thrust)

Burrill ((1943) "Developments in Propeller Design and Manufacture for Merchant Ships",

Trans. Institute of Marine Engineers, London, Vol. 55) proposed guidance as follows:

limit thrust (coefficient) to a certain value depending on cavitation number at the 0.7 radius

T

c

= coefficient_expressing_mean_loading_on_bladesAP

=

T= thrust = water_densityc

1 VR

2

2

AP= projected_area VR= relative_velocity_of_water_at_0_7_radius

can estimate projected area from AP Pfrom Taylor S & P page 91 P/D

= 1.067 0.229 from 0.6 to 2.0 elliptical bladedAD D prop, hub = 0.2 D

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and as usual ... T=

PE=

PDD PD= delivered_power RT= T 1( t)

1 t) V (1 t ( ) V

PE= effective_power PE= RTV

PE 1 tD= quasi_propulsive_coefficient= = H R o H=

PD

1 w

this parameter is plotted versus 0.7 cavitation number at 0.7*r using relative velocity at0.7*r and pressure at CENTERLINE

:= 1.0259103 kg

3 = 1.99057slug

m 3ft

mVA= h= m

p0+ gh 188.2 sec

units in PNA (61)pv + 19.62 h= =0.7

1VA

2+( n )

2

2+ 4.836 (

2 = n= sec

1 approximation SI 0.7 D

VA n D) D m

pv apparently ~2

0.69 psi (90 degF)

2026+ 64.40.7 = in US units

2 2VA + 4.836( nD)

Carmichael correlationcc+ 0.3064 0.523

0.2C= cavitation % c as above

C:=0.2 at 0.7 radius as

0.0305 0.0174 above (centerlineimmersion)

example numbers for C

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1

c(30 , )

c(20 , )

c(10 , )

c(5 , )

c(2 , )

0.10.1

Carmichael correlation valid only for C

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( ) :=T

APAP C1 2 0.2 0.2 = 1.067 0.229 assume AD ~ AE

VR C(0.0305 0.0174)+ 0.523 0.3064 D

P

2AD

AP C( ) ( ) :=AE C 1.067 0.229 P_over_D

cavitation % estimated minimum EAR to avoid

( ) C = ( ) = ( ) = =

AE CAP C AE C

5

10

15

20

25

20.367

16.333

13.632

11.698

10.245

m2 23.045

18.48

15.425

13.236

11.592

m2

D

2

4

1.404

1.126

0.94

0.806

0.706

supercavitating c to the left. very low

cavitation % is 100

h= 3.048 m D= 4.572 m m 1VA= 7.203

sn= 218

min 0.7 nD= 36.531m

s

mAV := 10 n:= 1000rpm 0.7 nD= 167.573

ms

s

p0+ ghpv0.7r =

1

p0

+ ghpv

3

2 VA

2+(0.7 nD)2

:=

=

8.8

101

VA2

+(0.7 nD)2

2

to avoid 25% cavitation C:= 25

0.2 0.2 off the scale hencec (C, ):= C(0.0305 0.0174)+ 0.523 0.3064 c(C, )= 0.243 supercavitating propellers

correlation not valid but

trend is ok

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Waterjetfirst draft 9/23/04 from Prof. Carmichael notes.

9/17/06: modified to reflect w (V=>VA) and separate inlet

and outlet pressure loss (in addition to drag) to reflect paper

VA velocity inlet

w wake fraction

Vs ship velocity

Vj nozzle (outlet) velocity

VA:= VsVs(1 w)

T= m_dot(VJ VA)

m_dot= mass_flow_rate

d

hVj

VA

g dat inlet centerline ... plocal= patmos +

at this point ... total pressure (pitot tube) poin= plocal+1

VA2

= patmos + gd+1

VA2

2 2

pressure at inlet to pump ... 1 2

total pressure (pitot tube) ...pop= poin g(d+ h)= patmos gh+

2 VA

pressure at pump exit... total poj= patmos +1

2 Vj

2

pressure (pitot tube) ..

total pressure increase

across pump ... pojpop= patmos +2

1 Vj

2

patmos + gh2

1 VA

2

=2

1

Vj2

VA2

+ gh

energy rise across the pumpp

oj

pop

1

2 2

per unit mass flow is ... =

2

Vj VA + g h

power absorbed by ideal pump is pojpop 1

2 2 therefore ...

Ppi= m_dot = m_dot

2

Vj VA + g h

ideal efficiency is then ... and quasi propulsive coefficient is ...

) T PTi effective_power PE R Vs (1 t Vs 1 t T VA 1 t PTi 1 t i= D= = = = = = = i

Ppi power_delivered Ppi PPi PPi 1 w PPi 1 w Ppi 1 w

VJ

PTi T VA m_dotVA VJ ) 2 VA VJ ) 2 VA VJ VA)2

VA

1

( VA ( VA ( i= = = = = =

Ppi Ppi m_dot

1 Vj

2

VA2

+ g h

Vj2

VA2

+ 2g h Vj2

VA2

+ 2gh

Vj2

g h2 1+ 2

VA V2

http://stellar.mit.edu/S/course/2/fa06/2.611/courseMaterial/topics/topic3/readings/hybrid_propulsion_asne_2004/hybrid_propulsion_asne_2004.pdfhttp://stellar.mit.edu/S/course/2/fa06/2.611/courseMaterial/topics/topic3/readings/hybrid_propulsion_asne_2004/hybrid_propulsion_asne_2004.pdf

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if h = 0 VJ 2

VA

1

2same as propeller (we

i=

= developed following in

Vj2

Vj+ 1

actuator disk

VA

1 VA

from actuator_disk.mcd using new variables to avoid duplication

v:= VVVVjj VVA

vvVV:= VVA+ T VVA VVA

what are implications of2 II:=

simplify 2 I=

2VVj= VVA?

T V VVA+ VVj VVj1+

VVAh cannot be negative (would be ducted prop, h limits efficiency

Real waterjet with losses

= T T= m_dot net thrust of waterjet

Tnet Draginlet (Vj VA)

conventional drag coefficient Cd=Drag

1 2

v A2

= = =drag coefficient of inlet

CD1

Drag

2 1

Drag

1

Drag

v A VAAVA m_dotVA2 2 2

net thrustTnet T Draginlet= m_dot(Vj VA) CD

2

1m_dotVA=

V

V

A

j 1

CD2

1= m_dotVA

net thrust power PT_net= TnetVA= 2

V

V

A

j 1

CD

1

2

m_dotVA

delta p across pump must be increased to account for losses ...

we'll assume separate inlet and outlet losses

assume internal losses are ... ~ 1/2* *v^2 ploss= pin_loss+ pout_lossand the pump pressure rise is ...

non-dimensional pressure loss coefficient is .... =pin_loss pout_loss

=and the real pump pressure rise is ...

Kin1

VA

2

Kout1

Vj2

2 2

pojpop=2

1 Vj

2

VA2

+ gh+ ploss=2

1 Vj

2

VA2

+ gh+ Kin2

1 VA

2

+ Kout2

1 Vj

2

g h Vj1 2

Vj2

2

pojpop= VA 1+ 2 + Kin+ Kout 2

VA VA

2 VA

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ideal pump power is ... Ppi=

=

1

2

2

Vj

VA

2

1+ 2g h

2V

Kin Kout

2

Vj

VA

++

2

2

pop

m_dotVA

m_dotVA

poj

1

2

m_dot

pop=

m_dot VA

Pp= actual_pump_power

Ppi

poj

define psuch that =pPp

2

2PPi Vj

CD

+ Kout

Vj

VA

m_dot+ 2

g h

2V

Pp

=

1 Kin+=

PT_net

= VAp p

=

p

Vj

1

2

2

1 VA

realPp 2

2

Kin

for a different form ... multiply numerator and

denominator by (VA/Vj)^2 ...

2m_dotVA

p

VjVj

VA

g h

2VA

1 1 2 Kout+ + +

VA2

Vj VA VA

VA

2 Vj

CD

22p

=

Kout+ Vj VA

2

1

Kin

1 12p

1

1 1 CD

VA

VA Vj Vj

=real2

2

2Vj

VA

VA

g h

2VA

g h

2

2 Kin 2 Kout+ + + + + Vj VjVj

1

2

1

2

(

g h2

(

g h2

)2

2

1 CD

Kout

2

Kout

1

)

CD

Kin

g h2

p

pand substitute for VA/Vj... =real =2

( ) ( Kin 1 + 1 1

)

1

CD

Kin

+ + + + 2

Vj Vj

and the quasi propulsive coefficient

is then ..

Vj

1 1(

)

2p

1

1 CD

Kout

2 1 VA 21 t

1 t

CD

2=D = p p

1 w 1 w2 2 2(

Vj

VA

Vj

VA

1 Kout 1+g h

2

VA

2 Kin+ + + Vj

as from above ...

Vj

1

2

2

m_dotVAnet thrust power PT_net TnetV 1 = =

VA

)

+

2

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first some comments to relate to previous lecture/notes version and Wrtsil paper

with Kout= 0 (N.B. this just means lumping all the pressure losses into a factor of 1/2* *VA^2 and

accounting for a drag increase due to the inlet ...

Vj 1

D=1 t

p

2p

VA

1

CD2

this is the form previously1 w 2

Vj

1

+

2

g h +

K

VA VA2

and ... with CD= 0 and assuming h = 0 (i.e. head loss is small compared to other terms ...

this is the form in the paper with

1 t 2 (1 ) pout_lossD= p Kout= = nozzle_loss_coefficient Kout=

1 w 2

1 21+ Kout (1 Kin)

2 Vj

Kin= = inlet_loss_coefficient Kin

1

pin_loss

2

=

VA2

at this point, assuming K, CD, and pare constant, could differentiate wrt Vj/VA(or

) and determine Vj/VAfor max propulsive coefficient, but minimum weight usually

determines parameters.

pumpbackground(Wislicenus)

example

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First Law Sept 2005: changes reflect text: WoudSept 2006: added examples

first law: during any cycle a system undergoes, the cyclic integral of the heat is proportional to the cyclic

integral of the work

pg 83 van Wylen & Sonntag Fundamentals of Classical Thermodynamics 3rd Edition SI Version

first law for cycle

1 dQ =

1 dW(5.2)

The net energy interaction between a system and its environment is zero for a cycle executed by the system.

pg 2 Cravalho and Smith

1 dQ

1 dW = 0 where integral are cyclic and dQ = Q dW = W

plot data

Pressure Volume plot for Processes

0.5

1

1.5

2

2.5

p

1 2 1

0.5 1 1.5 2 2.5

v

A process

B process

1

2 2 2 2

C process 2 1 2 1

2

2

1 dQ =

1 dW apply first law to cycle A B 1 dQA 1 dQB 1 dWA 1 dWB+ +=

2 1 2 1

1

1

2

1

apply first law to cycle A C 1 dQA 1 dQC 1 dWA 1 dWC+ +=

1 1subtract A C from A B

1 dQB 1 dQC 1 dWB 1 dWC=

rearrange ...

1

i.e. Q -W is a point function ... only dependent

2

9/21/2006 19/21/2006

dE Q

1 dQ_WB 1 dQ_WC=upon the end points => define as ...

energy, point function W=

2

1

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first law for system (Woud: Closed system) - change of stateN.B. Woud starts with rate

equation and obtains this

assuming steady staterearrange and integrate ...

Q1_2 is the heat transferred TO system

= +Q dE + W Q1_2 = E2 E1 W1_2 E1 E2 are intial and final values ofenergy of system and ...

(5.5)

W1_2 is work done BY the system

energy E consists of internal energy + kinetic energy + potential energy

E = U +KE +PE dE = dU +dKE +dPE

and first law can be restated ... Q dE + dU +dPE +(5.4)

= W = +dKE W

Closed System

dU = Q_dot W_dot dU = Q W m_dote = m_doti = 0 (W 2.3)

dt

d and : VW&S: page 62 Woud page 11d = differential of point functions state variables

difference between d and = differential of path functions - amount depends onpath/process: diminutivesee discussion of cyclic process below

cycle may be considered a closed system; initial state and final state are identical, For example (detailed

discussion later)

set up limits and calculations

p-v plot of Brayton cycle

0 1 2 3 4 50

2

4

6

pressure

volume

adiabatic compression

heat addition

adiabatic expansion in turbineheat rejection

closed (cycle)t

dU

d= 0 Q_dot W_dot= Q_dot = W_dot Qcycle Wcycle= (W 2.6)

this is where we started above using

different approach to first law

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example 5.3

Sonntag example 5.3: vessel with volume 5 m3contains 0.05 m3of saturated liquid water and 4.95 m3of

saturated water vapor at 0.01 MPa. Heat is added until the vessel is filled with saturated vapor. Determine Q.

State 1: V := 5m3

V := 4.95m3

MPa := 106Pa 3vap kJ := 10 J

3Vliq := 0.05m p := 0.1MPa

3m kJ

constant volume and := 0.001043 := 417.36 page 616 1Q2

mass => constant vvf

kguf

kg Sonntag

steam tables at p = m3

kJ kJ

0.1 MPavg := 1.694 ug := 2506.1 ufg := 2088.7

kg kg kg

State 2: V2 = V v = vg u = ug

Liq H2O

Vap H2O

first law: Q1_2 = E2 E1 +W1_2 W1_2 = 0 Q1_2 = U2 U1 V 0 z = 0 E= = U

Vfhave volume, determine mass of each V = m vn massf =n n vf

Vliq Vvapmass1_liq := mass1_liq =47.9386 kg mass1_vap := mass1_vap =2.9221 kg

vf vg

mass1_vapx = quality of_steam) x1 := x1 =0.0575 x%1 := x1100 x%1 =( 5.7453

mass1_vap +mass1_liqintensive property = not dependent on

4 mass (x, v, u, )U1 := mass1_liquf +mass1_vapug U1 =2.7331 10 kJ extensive does (U, V, mass)

or ... using average specific properties

kJ 4u1 := uf +x1ufg u1 =537.3611

kgU11 := u1(mass1_vap +mass1_liq) U11 =2.7331 10 kJ

which can be shown by ...

3 kJ 3 kJU1 = mass1_liquf +mass1_vapug ug = uf +ufg ufg =2.0887 10

kgug uf =2.0887 10

kg

u

1a

=

mass1_liquf +mass1_vapugsubstitute for ug

mass1_liq +

mass1_vap

+ + + + mass1_liq uf mass1_vap (uf ufg) (mass1_liq mass1_vap)uf mass1_vap ufg= = uf +x1ufg

mass1_liq +mass1_vap mass1_liq +mass1_vap

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state 2: need 2 properties, e.g. quality = 100%, can

calculate v (specific volume). T and p will be rising as heat

is added.

mass_total := mass1_liq +mass1_vap

3V m

v2 := v2 =0.0983 and ... is saturated vapor so we need to look up mass_total kg (interpolate) steam tables for v

g= v

2

vx x1my_interp x2 x1 ,y2,y1,vx) := y1 + (y2 y1) an interpolation statement( ,

x2 x1 v is value between x1and x2

result is y at v

could use mcd function linterp(vx,vy,x) where:

vx is a vector of real data values in ascending order.vy is a vector of real data values having the same number of elements as vx .

x is the value of the independent variable at which to interpolate a result. For best results, this

should be in the range encompassed by the values of vx.

using Table A.1.1 T vg ug pg

3 kJ

values at 1 T1 := 210 v1g := 0.10441m

kg

u1g := 2599.5kg

p1g := 1.9062MPa

3 kJvalues at 2 m := 2601.1 := 2.104MPa

T2 := 215 v2g := 0.09479kg

u2gkg

p2g

interpolated values at vx := v2

T2 ( T2 T1 ) T2:= my_interp v2g,v1g, , ,vx =213.1717

:=

my_interp v2g,

v1g , , ,

vx =

2.0317 MPap2 (

p2g p1g )

p2

3 kJ:= my_interp v2g,v1g , , ,vx =2.6005 10u2 ( u2g u1g ) u2

kg

total internal energy at state 2: U2 := mass_total u2 U2 105

kJ =1.3226

heat added ... Q1_2 := U2 U1 Q1_2 =104933 kJ

same result can be obtained using Table A.1.2 Pressure Tables

vx x1my_interp x2 x1 ,y2,y1,vx) := y1 + (y2( , y1)

x2

x1

p vg ug T

3m kJ

values at 1p1g 1g:= 2.0MPa v := 0.09963

kg1gu := 2600.3

kg1T := 212.42

3m kJ

values at 2 p2g 2g:= 2.25MPa v := 0.08875kg

2gu := 2602.0kg

2T := 218.45

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interpolated values at vx := v2

T2 ( T2 T1 ) T2:= my_interp v2g,v1g, , ,vx =213.1529

p2 ( p2g p1g ) p2

3 kJ

:= my_interp v2g,v1g , , ,vx =2.0304 MPa

u2 ( u2g u1g ) u2kg

total internal energy at state 2:2U := mass_total u2 U2 =1.3226 10

5

:= my_interp v2g,v1g , , ,vx =2.6005 10

kJ

heat added ... Q1_2 := U2 U1 Q1_2 =104933 kJ

example 5.3

first law as a rate equation

from above ... Q1_2 = E2 E1 +W1_2 = U2 U1 +KE2 KE1 +PE2 PE1 +W1_2

Q = U + KE + PE + W in small time interval t

Q U KE PE W divide by t= + + +

t t t t t

first law as a rate equationd

dtQ =

d

dtU +d

dtKE +d

dtPE +d

dtW =

d

dtE +d

dtW (5.31 and 5.32)

first law as a rate equation - for a control volume (Woud: system boundary)

Q1_2 = E2 E1 +W1_2 QE2 E1

W (5.38)=> = +

t t

Et= energy in control volume at time t

Et_t= energy in control volume at time t +dt

system consists of control

volume and differential entitiesE1 := EtEt+eimi = the energy of the system at time t mieach with ei, vi, Ti, pi

where i = input

and differential entities meE2 := Et_tEt_t+eeme = the energy of the system at time t +dt each with ee, ve, Te, pewhere

e = output

E2 E1 Et_t +eeme Et eimi E2 E1 = Et_t Et +eeme eimi (5.39)

eeme eimi represents flow of energy across boundary during t as a result ofmiand mecrossing the control surface

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now consider work associated with masses miand me

work done ON mass miis ... pivimi ON as work must be done to make it entersystem

work done BY mass m is ... peveme BY as leaving represents work donee

work done BY system in t is then ... W +peveme pivimi(5.41)

Q E2 E1divide by t and substitute into first law ... (5.38) and combining and rearranging = + W (5.38)

t t

Q+

mi(ei +pivi)=

Et_t Et+

me(ee +peve)+

Wc_v(5.42)

t t t t t

= V

2V

2 H U +p V

enthalpy defined - is (5.43)e +p v = u +pv + +g z = h + +g z

a property (5.12)2 2 h = u + p v

therefore ...

Q mi Vi

2

Et_t Et me Ve

2

Wc_v (5.44)+ hi + +g zi = + he + +g ze +t t

2 t t

2 t

mi and me are summed over all inputs and outputs ...

example 5.4

Sonntag example 5.4 a cylinder fitted with a piston has volume 0.1m3 and contains 0.5 kg steam at 0.4 MPa.

Heat is transferred until the temperature is 300 deg_C while pressure is constant

What are the heat and work for this process?

V1 := 0.1m3

mtot := 0.5kg I think by definition, steam = water + vapor at quality = x

2T := 300 deg_C p := 0.4MPa => v1 :=V1

v1 =0.2m

3

mtot

kg

quality =mass_of_vapor

= x an intensive property 1 x =mass_of_liquid

total_mass total_mass

3 3

3

m m

mfv := 0.001084 gv := 0.4625 vfg := vg vf vfg =0.4614

kg kg kg

x vfgv1 = vf + x :=v1 vf

x =0.4311vfg

constant pressure

p V2 = W1_2 = ( V1) p mtot (v2 v1)Q1_2 = E2 E1 +W1_2

E2 E1 = U2 U1 as V^2 and z = 0

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= = E2 E1 U2 U1 mtot (u2 u1)

= E2 E1 + = u2 u1 + v2 v1 = +p v 2 u1 +p v1Q1_2 W1_2 mtot ( ) p mtot ( ) mtot u2 ( )

kJ kJ:= 604.74 := 2133.8 := + 3hf

kghfg

kgh1 hf x hfg h1 =1.5246 10

kJh2 := 3066.8

kJ

kg kg

Table A.1.2 Saturated:=

Q1_2 mtot (h2 h1) Q1_2 =771.0904 kJ Steam Pressure Table

3m 3

2v := 0.6548kg

Table A.1.2 Saturated Steam Pressure Tablev1 =0.2

m

kg

:= W1_2 p mtot (v2 v1) W1_2 =90.96 kJ

Q1_2 = E2 E1 +W1_2 = U2 U1 +W1_2

= = U2 U1 Q1_2 W1_2 Q1_2 p mtot (v2 v1)

U := Q1_2 p mtot (v2 v1) U =680.1304 kJ

example 5.4

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first law as a rate equation - for a control volume

2 2Vi Ve (5.45)

Qc_v = E h + W++ ++m_doti g zid

n n

this is where Woud starts

d

g z c_v

= heat_flow

elevationz =Q_dot

velocity mass_systemv = m =

h specific_enthalpy=U internal_energy_system=

i = inletW_dot = work_flow

e = exitm_dot = mass_flow

d

d+ + hi m_dotec_v e e2 2d dt t t

m_dote, ve, he

Q_dot

ze

m_doti, vi, him, U

zi

W_dotsystem boundary

First law: change of energy within the system equals the heat flow into the system, minus the work flow delivered

by the system, plus the difference in the enthalpy, H, kinetic energy E kinand potential energy Epot of the entering

and exiting mass flows.

and ...assuming energy E U Ekin+ Epot Ekin = Epot = 0 E = U+=

2

+U p v

2

N.B. dot => rate not d( )/dtVi VedU Q_dot W_dot

U p V+

m_doti hi m_dote h+ ++ ++= g zi

enthalpy=

h

g z

a defined property

e e2 2dt W (2.1)

enthalpy=

H= =

steady state, steady flow process ... Woud: open systems steady state (stationary)

assumptions ...

1. control volume does not move relative to the coordinate frame

2. the mass in the control volume does not vary with time

3. the mass flux and the state of mass at each discrete area of flow on the control surface do not vary with time

and .. the rates at which heat and work cross the control surface remain constant.

tmc_v

d

d= 0

tEc_v

d

d= 0

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steady state, steady flow process ...m_dot = flow_rate

m_doti = m_dote (5.46)n nn n

m_doti g zi

n n

m_dote

he g ze

+ d

dtWc_v

(5.47)

2 2Vi Ved

d+

Qc_v hi ++ ++2

= 2t n n

m_dot hi g zi

d d

dt dtq = w = are heat transfer and work per unit mass

m_dot m_dot flow (5.51)

steady state steady flow ... - single flow stream

(

Q Wc_v c_v

2 2Vi Ve

g zi g ze

)

2 2Vi Ve (W 2.8) see text for examples of

W_dot Q_dot he+ ++= ze application to steam turbine,2boiler or heat exchanger, nozzle

and throttle

hi he+ ++ ++2

+wq =(5.50)2

uniform state, uniform flow process USUF

1. control volume remains constant relative to the coordinate frame2. state of mass within the control volume may change with time, but at any instant of time is uniform

throughout the entire control volume - I define this as f(t) but not of space

3. the state of mass crossing each of the areas of flow on the control surface is constant with time although the

mass flow rates may be changing

at time t, continuity equation ... tmc_v

d

d+m_doten m_dotin = 0

n n

t

0

d

dtintegrating over time gives change in mass of the control volume mc_v dt = m2_c_v m1_c_v

mass entering and leaving

t

0

t

0

m_doti dn t = t = m_dote dmi men n nn n n n

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+ continuity for USUF process ... m2_c_v m1_c_v = 0me mi (5.53)n nn n

apply first law at time t (5.45)

m_doti g zi

d

2

m_dote

Vid

d

+

+

Qc_v hi Ec_v+ + =

2 dt t

he+

2

g ze

Ve d

d

+2

+

n n

since at time t c.v. is uniform ...

2

(5.45)Wc_v

t

m_doti g zid

m uc_v

m_dote

now integrate over time t n

2

...

Vi Vc_vd

dQc_v + hi ++ + + = g zc_v

2Ve

2 2dt t

n

g ze

d

dt+ he Wc_v++ 2 +

ddt

Qc_v dt = Qc_v

t

0

t

t

0

m_doti g zin n

2

mi

u1

2

Vi Vi

= hi d hi++ t ++

g zi

g z1

2 2

dm uc_v g zc_v

m_dote g ze

n n

2

u2

2

m1

2Vc_v V2 V1

dt+ + ++2

++2

= m2 g z2

2

2dt

0

t

me

uniform state, uniform flow process USUF

0

2

g ze

t

0

Ve Ve d

dt

=

he d he

Wc_v d Wc_v

g z1

++2

t ++ t =2

mi g zi

n n

2

me

m2

2

g ze

g z2

Vi VeQc_v + hi he++ ++ = ...2 2n n (5.54)

2

2

V2 V1 Wc_v+ ++

2++ + u2 m1 u1

2

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Summary of Thermo from: first_law_rev_2005.mcd,second_law_rev_2005.mcd,

First Law availability.mcd

first law for cycle

1dQ=

1dW (5.2)

ref: van Wylen & Sonntag (eqn

#s) Woud (W nn.nn)

first law for system -change of stateQ1_2 is the heat transferred TO system

Q1_2

E2 E1+ W1_2

E1

E2 are intial and final values of

energy of system and ...(5.5)

=

W1_2 is work done BY the system

(5.4)Q dE = + W dU+ dKE+ dPE+ W=

Closed System d U (W 2.3)Q_dot W_dot dU Q W m_dote= m_doti= 0= =

dt

dQ

dU

d+

d

dKE

t+ d PE+ d W d E+ d Wfirst law as a rate equation (5.31 and 5.32)= =

d

d dt dt d

p V

t t t t

H U= +enthalpy defined - is

first law as a rate equation - for a control volumea property (5.12)

h= u+ p v

g zm_doti g zid

n n

2

m_dote

h

2Vi Ved

d

d

d+ + (5.45)Qc_v hi E Wc_v++ ++ 2 + = c_v e e2 dt t t

Woud assuming energy and ...E U

Ekin

Epot

Ekin

= Epot

= 0 E

= U+ +=

2

2 N.B. dot => rate not d( )/dtVi VedU Q_dot W_dot m_doti hi m_dote h+ ++ ++g zi

steady state, steady flow process ...

=

g z

m_doti = m_dot m_dot= flow_rate (5.46)e

dg z c_v

e e 2 2dt W (2.1)

open stationary n nn n

m_doti g zi m_dotc_v

n n

steady state steady flow ... - single flow stream

2 2Vi Ve

g zi g z

2

2

Vi Ved

d+ (5.47), (W2.8)Q hi h W++ ++ + = e e e2 2 dt tn n

this on per unit massq+ hi ++

2= he ++

2e+ w basis q = Q/m_dot (5.50)

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uniform state, uniform flow process

mi g zi

n n

2

n

he

2

g ze

g z2

Vi VeQc_v+ hi ++ ++ = ...2 2n (5.54)me

m2

2

2 V2 V1

Wc_v+ ++ ++ +g z1

Carnot cycle most efficient, and only function of temperature thermal= 1

u2 m1 u1

2

2

Second Law

TL

TH

1Entropy integrals are cyclicinequality of Clausius ... dQ0

T

1dQ= 0

1 1=> for all reversible

heat engines ...

=> all irreversiblesdQ= 0 1dQ0 dQ is a porperty of2 the substance. entropy is an extensive property and entropy

per unit mass is = s

1

1dQrev. S2 S1= (7.3)T

two relationships for simple compressible

2 reversible and when appicable to rev & irrev processes1

equality holds when substance - Gibbs equations

dQ

irreversible, the change ofT dS pV T ds

T dS V dp T ds

1

S2 S1 T

(7.5)entrpy will be greater than pvthe reversible (7.7)

vdp

dU du+ += =

(7.6)dH dh= =second law for a control volume

Q_dotd

dt

c_v( ) ( )+ (7.49) = when reversiblem_dotese m_dotisi

n n c_v

steady state, steady flow process dSc_v = 0

(7.50)dt

Sc_vT

Q_dotc_v

T (7.51) = when reversible( ) ( )

m_dotese m_dotisin n c_v

uniform state, uniform flow process

tQ_dotc_v (7.56) = when reversible

m2s2 m1 s1 mese misi

0

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+ ( ) ( ) dtTn n

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Availability

reversible work (maximum) of a control volume that exchanges heat with thesurroundings at To

g zi m g z

+ g z2 g z1

2

2V

i

V

e

(8.7)...Wrev

hi T h T++mi

m2

si s= o

e e

o

e

e2 2n n

n nlatter [..] is total for

2

2V2 V1 c.v.

T T++ u2 s2 m1 u1 s1o o2 2

system (fixed mass

2 2 Wrev_1_2 V1 V2(8.8) T T++ g z1

steady-state, steady flow process - rate form

g zi

= =wrev_1_2 u1 s1 u2 s2

g z2

g z

Ve

o o

2

2m

2

me

h

2

Vi Ve

(8.9)W_dot hi T T++ 2mi

single flow of fluid

= si srev o e o e e2n n

n n

2

2 W_dot Virev

hi T h T g zi

g z

availablity

steady state, steady flow process...(e.g. single flow ...availability (per unit mass flow)

= w =rev

si

s (8.10)o

e

o

e

e

m_dot

2

2

2

2 Vi Vo(8.16) h T h T++ g z g z

reversible work between any two states = decrease in availablity between them

= s so o o o o2 2

s1 h2+ Tos2=(h1 h2) To s1 s2) (8.17) extendedw =rev i e= h1 Tos1 h2+ Tos2= h1 To

can be written for more than one flow ...

++

++

++

++

++++

++

(

en n

n n

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W dot i

= mi m rev en n (8.18)

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availability w/o KE and PE per unit mass of system

= (u+p v T s)(u +p v T s )= u u +p (v v ) T (s s ) (8.21)o o o o o o o o o o o o

and reversible work maximum between states 1 and 2 is ...

2 2

wrev_1_2 1 2po(v1 v2)+V1

2

V2+ ( z2) (8.22)= g z1

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for the reversible heat engine ... Wc Qo Qc_v_rev=

from second law S for Qoand Qc_v_revare the same, Qoat constant temperature

t

d and Qo To

0

Q_dotc_v_revd that is express as integral of rateS= Sc_v_rev Qo ToS Sc_v_rev t= =

T uniform state => T constant in c.v.

t

0

t

t

Q_dotc_v_rev

0

Qo Q_dotc_v_rev

T

Q_dotc_v_revdt t= =

T To

0

substituting into ... (8.4)Wc Qo Qc_v_rev To dt Qc_v_rev= = T

t

Q_dotc_v

0

+

from second law

m2s2 m1s1 mese misi

n

n

substituting (8.5) into (8.4) ... Wc Qo Qc_v_rev= m2s2To=

( ) ( )=+

(7.56) = whenalso for USUF dtT

reversible (8.5)

( ) ( )

m1s1 mese

so ... the bottom line, substitute (8.3) rearranged and (8.6) into (8.1) Wrev Wc_v_rev

me g ze

Qc_v_revmisi

Wc

n n

+=

mi

2

2

2

Vi VeWrev Qc_v_rev+ hi he++ ++= g zi

u1

...2 2n

n

n

n

2

} (8.3)

+ m2

To m2s2

V2

V1 m1++

2

m1s1

++ u2 g z2

+ g z1

Qc_v_rev

...

( 2

misi( )

) + mese

n n

} (8.6)

Qc_v_revcancels and rearranging (moving Toand s terms into mass flow terms) ...

reversible work (maximum) of a control volume that exchanges heat with thesurroundings at To

Tosi g zi me Tose g ze

n

2

2

V

i

V

e

(8.7)Wrev hi he++ ++=

mi

m2

...

2 2n

n

n

latter [..] is total for

u2

2

m1

u1

2

V2 V1 c.v.++

2++

2

+ Tos2 g z2 Tos1 g z1

two special cases: a system (fixed mass) and steady-state, steady flow process for a control volume

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system (fixed mass)

2

= 0 me

n he Tose

2Vi Ve

hi Tosi e = 0

g z2

m1= m2= m++2 2n

++mi g zin n

system (fixed mass

g z

2

Wrev_1_2

m= =wrev_1_2

u1 Tos1

2

V1 V2(8.8) Tg z1

s1

u2 s2o2 2

steady-state, steady flow process

2

2

V2 V1 T T = 0++ g z2

steady-state, steady flow process - rate form

g zi m

n

m2 u2 s2 m1 u1 g z1

T

o o2 2

2

2

Vi

Ve

(8.9)W_dot hi T h ++

g z

g zi g z

above represents maximum work for given change of state of a system

what is maximum work that can be done by system in a given state???

answer: when system is in equilibrium with the environment, no spontaneous change of state can occur, and isincapable of doing work. therefore if system in a given state undergoes a completely reversible process until it

is in equilibrium with the environment, the maximum reversible work will have been done by the system

mi

single flow of fluid

= si srev o e e o e e2 2n n

n

2

2

W_dotrev Vi Ve(8.10)hi T h T++

2= w =rev si so e o e e

2m_dot

2

2

Vi Vesteady state, steady flow (8.10)hi T h T g zi e g z

maximum when mass leaving c.v. is in equilibrium with environment. define = availability (per unit mass flow)

steady state, steady flow process ...(e.g. single flow ...availability (per unit mass f low)

w =rev si so e o e2 2process ...(e.g. single flow)

2

2V Vo

(8.16) h T h T++ g z g z

reversib le work between any two states = decrease in availablity between them

= s so

o

o

o

o2 2

++

++

++

++

++

++++

++

)(8.17) extended

extension

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( ) ( i h1 T h2 T h1 T h2 T h1 h2 T + +w =rev = s1 s2= s1 s2= s1 s2e o o o o o

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can be written for more than one W dotrev=

min

in

men

en

(8.18)flow ... n n

for a system (no flow across the contro l surface)

. need to account for work done by system against the surroundings ...

... assume kinetic and potential energy changes negligible ...

Wrev_1_2 V12 V22 = Tos1 g z1 g z2

(8.8)wrev_1_2= u1 + + u2 Tos2+ + m

2

2

becomes ... =wrev_1_2 (u1 Tos1)(u2 Tos2)

wrev_max= (u Tos)(uo Toso) (8.19)

availability per unit mass is then ... this maximum work - that done against the surroundings

Wsurr po(Vo V)= m po(vo v) (8.20)=

= availability_w_o_KE_PE = wrev_max wsurr= (u Tos)(uo Toso)+po(v vo)

availability w/o KE and PE per unit mass of system

= (u+pov Tos)(uo+povo Toso)= u uo+po(v vo) To(s so) (8.21)

and reversible work maximum between states 1 and 2 is ...

2 2

g z1 (8.22)= + + wrev_1_2 1 2 po(v1 v2)V1

2

V2( z2)

check ...

1:= u1u1 uo+po(v1 vo) To(s1 so) 2:= u2u2 uo+po(v2 vo) To(s2 so)

2 2

wrev_1_2:= 11 2po(v1 v2)+V1

2

V2+ g z( 1 z2)

1 2 1 2

wrev_1_2simplify u1 + Tos2+ V1 + g z1 Tos1 u2 V2 g z22 2

matches ...

Wrev_1_2

V12

V22

g z1 g z2(8.8) from above

m

2

2

= wrev_1_2= u1 Tos1+ + u2 Tos2+ +

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geothermalwellexample

define some units ... kPa:= 103

Pa kJ:= 103

J

example ... geothermal well

water as saturated liquid issues from a process at 200 deg C.

What is maximum power if the environment is at 10^5 N/m^2 at 30 deg CT1_C:= 200

V2

Vo

2

Tos (8.16) = h + + g z ho Toso+ + g zo 2 2

kJ kJ1 saturated T1:= (273+ T1_C)K T1= 473 K p1:= 1.5538MPa h1:= 852.45

kgs1:= 2.3309

kgK

environment p0:= 105

NT0_C:= 30 T0:= (273.16+ T0_C)K T0= 303.16K p0= 100kPa

(dead state) m2

water at this state is "compressed liquid" as pressure exceeds saturation pressure at 30 deg C

ref: water saturated liquid at 30 deg C p

sat_30

:= 4.246kPa

33 m kJ kJ

vf_30:= 1.004 10 hf_30:= 125.79 sf_30:= 0.437kg kg kgK

limited values for compressed liquid are in Table A.1.4 well beyond this pressure

water is ~ incompessible

values for u, v and s can be estimated to be the saturation values at the T so... (see example validation below)

v0:= vf_30 s0:= sf_30

but work must be done to compress to higher pressure than saturated

estimate from definition of enthalpy:2

2 2 2

h

= u

+

dh

= du

+

+

v dp 1

dh

= h2 = 1du+ pdvp v p dv h1 + vdp 1

1 1 1

we'll see later 2

1du= cv(T2 T1) dv ~ 0

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to show example of estimates u, v, and s of compressed liquid = saturation, but not h consider value in Table vs

stauration at T

i:= 1..4

Table A.1.1.4

p=10MPaT=40 deg C

Table A.1.1.1

saturationt=40 deg C difference

(col 2 - col 3)

--------------------

data:=

6 37.384 10 col2

datai 1

datai 2,,

1010

0.0010034 0.001008

p

v 100 =u 166.35 167.56

h 176.38 167.57

s 0.5686 0.5725

datai 1,

-0.5

-0.7

5

-0.7

differences all < 1% for 106pressure

difference

h differs by 5%

,

using estimate from definition of enthalpy

,

3

data3 2

data1 2

data0 1,,

data3 1

kJh= 177.643

kJ

kg

m( )h ,

,

data0 2

kg100

data3 1

Pa Pa:= + kgkg

kJ h

and difference is ... difference now < 1%0.716=

kJ

kg

geothermalwellexample

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ref: Keenan, Joseph H., Thermodynamics, The MITKeenan AvailabiltyPress, paperbook 1970, original J. Wiley & Sons,

1941, QC311K26 1970 Science library

Keenan's definition:

the maximum work which can result from interaction of system and medium when only cyclic changes occur inexternal things except for the rise of a weight. page 290

medium: environment, atmosphere of infinite extent in which system operates;- in most stable state- all parts at rest relative to each other- homogeneous in temperature and composition- uniform in pressure at any height in gravitational field

to be shown: Availability= (E+p V T S) (E +p V T S )o o o o o o o

system at energy E, volume V and entropy S

po, Topressure and temperature of medium (and of system at dead state - no more possibility of obtaining work)

Eo, Vo, So; energy, volume and entropy of system in dead state

first show when exchange of heat can occur between system and medium only, the maximum amount of work

which can be delivered beyond the boundaries of the medium when the system changes from one state to

another is the work which is delivered when the change is in every respect reversible

assume such process exists: consider complementary reversed process with both rev and irrev - it (irrev) will

violate second law

in same way can show same work occurs between any reversible processes

next determine work delivered when system goes from state 1 to 2 ininfinitesimal step t

Q= heat_flow_TO_system + if into, - if out of

To TWstate_change= Q

from Carnot type cycleT

as ... Qmed:= ToSS Q:= TSS S:=

T

QQ

work:= (To T)SS work (To T)

T

Q

W 0 as both Q and (To - T) must be same sign>

not only work done by system ... volume can expand, etc

any change in volume dV is resisted by the medium with pressure p otherefore work done by

system - the amount of which can be received by things other then the medium is then ...

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W= all_work_done_by_system

WpodVand net work delivered is then ... substitute W:= Q dEdE

To TWnet

:= W+T

Qpo

dVdVWnetcollect T dEpodV+ To

Q,

T

Q cancels Wnet= dE+ To

T

QpodV

and since process is reversibleQ

= dST

Wnet= dEpodV+ TodS

it follows that the maximum amount of work that can be

delivered by each step is then the decrease in ...E+poV ToS

maximum decrease is to the dead state Availability_at_state= E+poV ToS(Eo+poVo ToSo) [145]

if system state changes from 1 to 2

increase in availability is ... E2+poV2 ToS2(E1+poV1 ToS1) [146]

which is negative unless work or heat is supplied TO the system from a source other than the medium

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Second Lawfirst draft 9/23/04, second Sept Oct 2005

minor changes 2006, used spell check,

expanded example

Kelvin-Planck: It is impossible to construct a device that will operate in a cycle and produce no effect other than the

raising of a weight and the exchange of heat with a single reservoir.

Clausius: It is impossible to construct a device that operates in a cycle and produces no other effect than the

transfer of heat from a cooler body to a hotter body.

Woud: used to: 1) predict the direction of processes

2) establish the conditions of final equilibrium

3) determine best possible theoretical performance of a process

if it is impossible to have a heat engine with 100% efficiency, how high can it go??

define ideal process, termed reversible process: a process that, once having taken place, can be reversed

without changing either the system or surroundings

examples irreversible; piston expanding against stop

reversible; piston expanding by removing and replacing weights; excerpt from VW&S page 166 good

description of reversible and irreversible processes

Let us illustrate the significance of this definition for a gas

contained in a cylinder that is fitted with a piston. Consider first

Fig. 6.8, in which a gas (which we define as the system) at high

pressure is restrained by a piston that is secured by a pin. When

the pin is removed, the piston is raised and forced abruptly

against the stops. Some work is done by the system, since the

piston has been raised a certain amount. Suppose we wish to

restore the system to its initial state. One way of doing this

would be to exert a force on the piston, thus compressing the gas

until the pin could again be inserted in the piston. Since the

pressure on the face of the piston is greater on the return stroke

than on the initial stroke, the work done on the gas in this reverseprocess is greater than the work done by the gas in the initial

process. An amount of heat must be transferred from the gas

during the reverse stroke in order that the system have the same

internal energy it had originally. Thus the system is restored to

its initial state, but the surroundings have changed by virtue of

the fact that work was required to force the piston down and heat

was transferred to the surroundings. Thus the initial process is an

irreversible one because it could not be reversed without leaving a

change in the surroundings.

In Fig. 6.9 let the gas in the cylinder comprise the system and let

the piston be loaded with a number of weights. Let the weights

be slid off horizontally one at a time, allowing the gas to expand

and do work in raising the weights that remain on the piston. As.

the size of the weights is made smaller and their number is

increased, we approach a process that can be reversed, for at each

level of the piston during the reverse process there will be a small

weight that is exactly at the level of the platform and thus can be

placed on the platform without requiring work. In the limit,

therefore, as the weights become very small, the reverse process

can be accomplished in such a manner that both the system and

surroundings are in exactly the same state they were initially.

Such a process is a reversible process.

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Carnot cycle

example steam power plant - working substance steam

boiler - heat transferred from high T (constant) reservoir to

steam - steam T only infinitesimally lower than reservoir

=> reversible isothermal heat transfer process. (phase

change fluid - vapor is such a process

turbine - reversible adiabatic (no heat transfer) T

decreases from THto TL

condenser - heat rejected from working fluid to T Lreservoir

(infinitesimal T) some steam condensed

pump - temperature raised to THadiabaticly

can reverse and act as refrigerator

Carnot cycle four basic processes:

1. reversible isothermal process in which heat is transferred to or from the THreservoir

2. reversible adiabatic process in which the temperature of the working fluid decreases from THto TL

3. reversible isothermal process in which heat is transferred to or from the TLreservoir

4. reversible adiabatic process in which the temperature of the working fluid increases from TLto TH

Carnot cycle most efficient, and only function of temperature

efficiency (in heat engine)

W = energy_sought QH QL QLthermal =

QH = energy_that_costs=

QH

= 1 QH

temperature scale (arbitrary but defined in terms of Carnot efficiency)

f TH thermal = 1

QL= (TL ,TH)

QH=

( )=

THproposed by TL

QH QL ( TL Lord Kelvin thermal = 1 THmost efficientf TL)

at this point have ratio of absolute temperatures

derive scale from non-Carnot heat engine operating at steam T Hand ice temperature TL

if we could measure it would find THto be 26.80% th = 0.2680 = 1

TLif want difference to be 100 as on the Celsius scale T := 100

TH := 100 TL := 200Given

TLinitial values 0.2680 = 1 TH = TL + T

TH

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:= )

TH

TL

373.134

273.134

=TH

Find TH TLTL

T_deg_C +273.134 = T_deg_K VW&S has 273.15 changed to 273.16 tocorrespond to triple point of water 0.01 deg_C

,(

Entropyinequality of Clausius ...

1dQ 0

T

for fig 7.1

1 dQ QH QL >0=

from definition of absolute temperature scale and T Hand TLconstant

QH

QL1Q = 0d =

T TH TL

1dQ = 0

T

if .. 1 dQ approaches 0, THapproaches TL, while reversible

1 dQ 0 and ...1

dQ = 0T

=> for all reversible heat engines ...

if irreversible, with TH, TL, and QHsame ... Wirrev Wrev QL_irrev QL_rev< >

and ...1 dQ QH QL_irrev >0=

QH

QL_irrev1dQ 0..

as ...

1 dQ = 0

1

TdQ all irreversible engines

11 dQ 0 dQ

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examplefigure7.3pg188VW&S

example fig 7.3 - simple steam power plant cycle - not

typical - pump handles mixture of liquid and vapor in such Saturated vapor, 0.7 MPa

proportions that saturated liquid leaves the pump and enters

the boiler. The pressures and quality at various points are

given in the figure. ? Does this data satisfy the inequality of

Clausius?

inequality of Clausius ...

T

1dQ 0

heat is transferred in boiler and condenser, both at constant T

1

1

1 1 1 dQ = dQboiler +

dQcondenser = 1 dQboiler + 1 dQcondenser T T T Tboiler

Tcondenser

on a per unit mass basis mass := 1kg := 106Pa kJ := 10

3J kPa := 10

3Pa

kJboiler ... p1 := 0.7MPa hfg := 2066.3

kgT1 := 164.97 deg_C steam tables Table A.1.2

kJq1_2 := hfg q1_2 =2066.3

kgq = h from first law

kJ kJpcondenser = p3 = p4 = 15kPa hf := 225.94

kgfgh := 2373.1

kgT3 := 53.97 steam tables Table A.1.2

x3 := 0.9 h3 := hf +x3hfg

kJx4 := 0.1 h4 := hf +x4hfg q3_4 := h4 h3 q3_4 = 1898.48

kg

T_deg_C +273.15 = T_deg_K

q1_2 q3_4 kJint_dQ_over_T := + int_dQ_over_T = 1.087 deg_K is < 0

T1 +273.15 T3 +273.15 kg

examplefigure7.3pg188VW&S

Boiler

Turbine

1 - saturated liquid, 0.7 MPa

2

W

390% quality, 15 kPa

Condenser

4

10% quality, 15kPaPump

MPa

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entropy

plotdata

2.5

two reversiblecycles

from 1 to 2 (not labeled)

A - B2and ... A - C

1.5

1

0.50.5 1 1.5 2 2.5

1

A processA process

B processB processT

dQ = 0 reversible ...C processC process

A -B 2 1

2

2

+

1

2

A - C

1

subtract second from first =>

1

1 1 1 1 1 1dQ = 0 dQA+ dQB dQ = 0 = dQA dQC=

T TA TB T TA TC

1 1

2

Qrev

2

reversible ...

1 11

T

dQB

dQC

=

so as we did for energy E (e) in first law dQ is independent ofTB TC

path in reversible process => is a property of the substance

dS =

(7.2), W (2.13)T entropy is an extensive property and

entropy per unit mass is = S

2

1

1dQrev. S2 S1= (7.3)T

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entropy change in a reversible processexample Carnot

Carnot cycle four basic processes:

1. reversible isothermal process in which heat is transferred to or from the THreservoir

2. reversible adiabatic process in which the temperature of the working fluid decreases from THto TL

3. reversible isothermal process in which heat is transferred to or from the TLreservoir

4. reversible adiabatic process in which the temperature of the working fluid increases from TLto TH

plotdata

1 to 2i := 1 ..2

21 Q1_2

dQrev. = S2 S1 =T TH

2 to 3 - adiabatic Si := 2 ..3

1

T

T

T

T

3

3 to 4S

2

1 dQrev. = 0 S3 =S2 S3 S2=T

4 i := 3 ..41 Q3_4

dQrev. = S2 S1 =T TL

4 to 1 - adiabatic

3

1 S

4

total cycle ...i := 1 ..5

1dQrev. = 0 = S1 S4 S1 = S4 i := 4 ..5

T

T

S

S

in general .. for reversible process, area under T S curve represents Q

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from first law ..two relationships for simple compressible

.substance: Gibbs equations in Woud

p d

(5.4)Q dE= W+

+without KE or PE Q dU= W

reversible ... W W = = p AdsQ T d= = =

dU+

p dV

QED (7.5)

as .. e.g. a piston ... F d

du

p dV

(7.7)

substitute ... T dS

since ... H U=

=

T ds p v

T dS V dp T ds v dp

applicable to BOTH reversible and irreversible processes as they are relationships between state variables

entropy change for irreversible process

p

dH

dH dU p dV V dp

QED (7.6)mass ...

+ + + += =on a per unit

substitute ... dh= =

0.5 1 1.5 2 2.5

plotdata

reversible cycle from 1 to 22.5

to 1 (not labeled) A - B

and ...

irreversible cycle from 1 to 2 2to 1 (not labeled) A - C

1.5

1

0.5

A process reversibleA process reversible

A - B reversible ...B process reversibleB process reversible

C process irreversibleC process irreversible

1

2

1 1Qd = 0 QAd +=

T TA

2

11

dTB

QB

A - C irreversible2 1

1

subtract second from first and rearrange ...

1 1 1d 0+ +TA TB TA TC

21

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11

1 1 11 1 1

dQB dQC >0 dQB> dQCTB TC TB TC

2 2 2 2

is a property and although calculated for reversible1

2 2

2

1 1process, is identical between states for irreversible.1

dQB_rev 1 dSB_rev 1 dSC= = substitute into inequality above ...TB_rev

11

2

2 equality holds when2 1

principle of increase in entropy

consider system at T and surroundings at To, Q transferred from surroundings to system

than the reversible

due to above

1reversible and whenQ 11 dSC dQC or in general ...> (W 2.14)dS S2 S1 dQTC irreversible, the change ofT Tentropy will be greater

Qfor the surroundings, Q is negative thereforedSsystem

T

dSsurr

T

0

Q total net change in entropy is ...=

Q Q 1 1

=dSnet dSsystem+ dSsurr Q

if T > To reverse signs and result

=T T0 T T0

since heat is transferred FROM surroundings, To > T therefore ...

Q1 1

dSnet 0 holdsT T0

thus ...

princ iple of increase in entropy

for all processes that a system and its surroundings can

undergo=dSnet dSsystem+ dSsurr 0

dSisolated_system 0

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second law for a control volume

not developed but second law stated in terms of lost work LW

Q LWdS = +

T TS2 S1 1 Q 1 LW

during t change in entropy is ... t=

t

T

+

t

T

(7.43)

St entropy_in_c_v_at_time_t=

St_t entropy_in_c_v_at_time_t_plus_t=

S1 St simi+= entropy_of_system_at_time_t=

S2 St_t se me+= entropy_of_system_at_time_t_plus_t=

S2 S1 St_t St se me+ si mi=(7.44)

etc .....

second law for a control volume

dSc_v + (m_dotese) (m_dotisi)

Q_dotc_v(7.49) = when reversible

dt Tn n c_v

steady state, steady flow process

dSc_v = 0

(7.50)dt

Q_dotc_v

(m_dotese) (m_dotisi) T (7.51) = when reversiblen n c_v

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uniform state, uniform flow process

Q_dotc_v

T(7.54) = when reversiblerewrite 7.49 as ... d

dt(m s)

c_v ( ) ( )+ m_dote se m_doti sin n c_v

t

=

0

d

dt

in control volume

mi si

and integrate ... (m s) dtc_v

= m2 s2

t

m1 s1

m_doti si

( )t

0

( ) = dt( ) ( )m_dote sen n

dt me se

n n

0

)

tQ_dotc_v

( ) (+ therefore for time t dt (7.55) = when reversible m2 s2 m1 s1 me se mi si0

since the temperature over the control volume is uniform at any instant of time

T

n n c_v

t t

0

and second law for a uniform state, uniform process is ...

uniform state, uniform flow process

0 0

t in first integral T can be a function ofQ_dotc_v Q_dotc_v1

space (location in c. v.) U(niform) S(tate)d Q_dotc_v d dt t t= =T T T => T only dependent on timec_v c_v

tQ_dotc_v (7.56) = when reversible

m2 s2 m1 s1 me se mi si

0

steady state, steady f low processassumptions ...

1. control volume does not move relative to the coordinate frame

2. the mass in the control volume does not vary with time

3. the mass flux and the state of mass at each discrete area of flow on the control surface do not vary with

time and .. the rates at which heat and work cross the control surface remain constant.

example: centrifugal air compressor, operating at constant mass rate of flow, constant rate of heat transfer to

the surroundings, and constant input power.

uniform state, uniform flow process USUFassumptions:

1. control volume remains constant relative to the coordinate frame

2. state of mass within the control volume may change with time, but at any instant of time is uniform

throughout the entire control volume - I define this as f(t) but not of space

3. the state of mass crossing each of the areas of flow on the control surface is constant with time

although the mass flow rates may be changing

example: filling a closed tank with a gas or liquid, discharge from a closed vessel.

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(

)

(

)

+

dtT

n n

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B series z =5 EAR =0.75 P_over_DT=( 1.4 1.2 1 0.8 0.6

Kt_over_J_sq =1.111

Kt,Kq*10,efficiency

1.5

1.4

1.3

1.2

1.1

1

0.9

0.80.7

0.6

0.5

0.4

0.3

0.2

0.1

0

P/D = 1.4

P/D = 1.2P/D = 1.2

P/D = 1.0P/D = 0.8

P/D = 1.0

P/D = 0.8

P/D = 0.6

P/D = 0.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

Advance Ratio J=VA/nD

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UNITS (Propulsors)

Quantity SI U.S.

Mass, m kg slug

Mass Flow Rate, m kg/s slug/s

Thrust, T N ( or kN) lbf

Torque, Q Nm (or kNm) lbf ft

Density , kg/m3 slugs/ft3

lb s2/ft4

Velocity , V m/s ft/s

Rotational speed,n rps rps

Useful Values: 1 knot = 1.688 ft/s = 0.5144 m/s

1 HP = 550 ft lbf/s = 0.7456 kW

density (p) for sea water