Baixe Munson Fundamentals Of Fluid Mechanics 5th Chap12 e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
12
124 Water flows through a rotating sprinkler arm as shown
in Fig. P12.1 and Vídeo V12.2, Determine the flowrate if the
angular velocity. is 150 rpm. Friction is negligible. Is this a tur-
bine or a pump? What is the maximum angular velocity for this
flowrate?
T=mriVoa-r Va) =O since there
is no friction. Also, Va=0
fence, Voz =0
[sn
Ve wn
u 2 27rady mino
= 5a rev Lev =)
=916 É te
70 à Thus, H FIGURE P12.1
W smo = U,
Dr, Q16Ê act
b= ing “TSE
Hence, Q=2Wh Ay =2(Z(22 8) (9758)
= 0.00958 É = 4.3 gal/min
This às a turbne because the sprinider moves im response
to Fluiá Flow forces.
since trichon is negligróle the maximum angular velocity
Lo Pais Frowrate is the me corresprnding o 152 rpm
or
w (10 rev 6” 24) rad (eeiz 18.7
pin (> 5 ==
1200
12.2,
122
Water flows axially up the shaft and out through the
two sprinkler arms as sketched in Fig. P12.1 and as shown in
Video V 12.2, With the help of the moment-of-momentum equa-
tion explain why, at a threshold amount of water flow, the sprin-
kler arms begin to rotate. What happens when the flowrate in-
creases above this threshold amount?
a FIGURE Pi2.t 4
Tais sprinhler às similar fo the enc of Example 5:18.
7hus,
nata É “Yam
Prom the velvei ty Priangle showa tn the shetel, abwe,
we conclude Thar
VE (msm 70º - UV)
where
-r
UV = hã
Comb rning, we get
=-K (W, sin 70º. eo )
Tuatt 2 2 2
So when W is large enonçh with = 0 fo overiom
4
Z &º he sprinhler rotor begias £ rotate.
e
nhem Flowrate «ncreases further, w is no longey zero
: T negtrgi bh
And unth pato 9h grbly small,
Wo sm 7o rt
amd o vicreases with merease 0? W, anh) a? maximum,
vame of W, stwe alo have maximum value of (o.
12-14
(2.4 | (Cont)
Since the (ift force acting on each rotw blade sechon
e in the same direchon as the blade velocity we
conclude that his tfurtomachine is a FJurbine..
The energy hamkwred por unit mass is he shaft
work pOr unid mass, O alt » mhich we can determme
with Eg. HS. Thus
Vpals =- UM (1)
z &2
Fem the velocity triangles we otlam
TÁ = Ww smto" — 7
G2 a -
and
= = 2 z
W w, Y +U
Thus
Opa -U / V>+ ur snto - u)
“ -- (ob) HU foot sinto 20 tt) [LL
has ( +) (15%) + (zotf sinto" - 20 tt dE
-
» .— 330 ft
Shatd = slug
o”
Vai * -a30 Hb 1 + -poz ft
Sharf Slug (222 Ibm Hom
Slug
12.5
t2s
Sketch how you would arrange four 3-in.-wide by 12-
in.-long thin but rigid strips of sheet metal on a hub to create
a windmill fike the one shown in Videu V 124, Discuss, with
the help of velocity triangles, how you would arrange each blade
on the hub and how you would orient your windmill in the wind.
—————»
wind along votulion axis n
Va
jã-s
12, 6
12.6 — Sketched in Fig. P12.6 are the upstream (section (1))
and downstream [section (2)] velocity triangles at the arithmetic
mean radius for fiow through an axial-flow turbomachine rotor.
The axial component of velocity is 50 ft/s at sections (1) and
(2). (a) Label each velocity vector appropriately. Use V for ab-
solute velocity, W for relative velocity, and U for blade veloc-
ity. (b) Are you dealing with a turbine or a fan? (e) Calculate
the work per unit mass involved. (d) Sketch a reasonable blade
section. Do you think the actual blade exit angle will need to
be less or greater than 15º? Why?
FIGURE Piz.
(a) See figure above.
(b) 7= m(ra Vaz - ri Ver) = 37 Fen ( Vaz -Vo)
where Voz>0 and Vai <O (see figure above)
Thus, T>0. The machine is a fan,
3) ME subi = Uo Voz - U, Ver = V(Voa - Vos) where V-Uc=l
Since Va = Va = 508 » tt follows
from the figure lhat : Wi
Most 60 É Ls Ao
U
6
= St
aqu é NE W=U=U
Va coeso" = 50 É or Ve 577 É Y Vea
so tha!
Vo, =Y. sin/5º= 518 sin 5º = 134 É
and
Voze Vo sim30'= 28.9 Ê
Also, U=NVoiltVoa]= 42-2 É
Hence, pag = 402H( 20.08 (134 6) - 760 dh
- ” . a É
(d) From the figure tanB= da ares Jor 0=48 * , vm
Thus, the blade shape is as shown: é
(m Hs 8 48 Ve 84
Was SO ft É
va fm (cont)
4-6
12.8
A centrifugal water pump having an impeller diameter
of 0.5 m operates at 900 rpm. The water enters the pump parallel
to the pump shaft. If the exit blade angle, B,, (see Fig. 12.8) is
25º, determine the shaft power required to tum the impeller
when the flow through the pump is 0.16 m?/s. The uniform
blade height is 50 mm.
Te TN
Tite - p6 0 hi os ch ki) (£3. 10)
MM Ver Do Tohare = 20 ta Voz c1)
Frem Fig !2.8€ ess, ao:
so ak Vip = Doo Via CoÉ pf, (2)
Bor hs Zégm 2026 tia tw (roo Ba Je - 742 ad
trem 60 im
Dishão = (D28m (Pa 24)= 234 1
Since Mme flowrate 15 given, é follows Tuaé
P= 2mk bla
er Po (ont)
V. 2 Lies = Zon
tr” mb (2m/8.26m) (605) é
Thus, from fg 2)
hp: (236 2.64 Cotas) = 14,2 E
and from E. (13
Tomada T (19 EE o nt o.26m) 19.2 E )= TE Nem
t
o, o Cree Me) (om ted hoo Pr yo (o pe
Sao nt fo) (
Min Ss tew
12-49
12.9
12.9 Discuss the differences between a centrifugal pump
and a positive displacement pump (see Video V12.3 for an ex-
ample of a positive displacement pump impeller that loo%s like
a centrifugal pump impeiler).
A centrifiugal pump mmpeller pushes agamst The
fores developed by à Fluid Flowing over its wellel
surfaces and in So doing moves Fluid Forongh.
“
A positive dispincement urap “faps "o egohures*
a velume of Fluid and moves De entre volume
from one bcation o another.
f2-10
12.48
A centrifugal pump impeller is rotating at 1200 rpm
in the direction shown in Fig. P12.10. The flow enters parallel
to the axis of rotation and leaves at an angle of 30º to the radial
direction. The absolute exit velocity, Vo, is 90 ft/s. (a) Draw
the velocity triangle for the impeller exit flow. (b) Estimate the
torque necessary to tum the impeller if the fluid density is [
2.0 slugs/f?. What will the impeller rotation speed become if o
the shaft breaks?
+
pet
BM FIGURE P12.10
(a) The ext How velocity friangte can be Constructed
Graphicallo as mditates Lelo,
(1200 *% em Es &
. . = fo, Fa = 6.8 E
with Us 2% (0.542) (0 Ba ) 5
Bom tre veleerty Irangie = Vas
Eanho Y
Suco Vas Vestin ão! and Ny Th tossot di folluos
Tunt “UU - MU, simdo?
+
Cr tar E ws3be
u
HE FE) sm 30? o
tm” br.s GS (go SE) sm 30 = 12,9
(40 E) cos do?
( cen &)
ta-t
2.12
Q=0. Fr tt%ys
min. +
12.12. A centrifugal radial water pump has the dimensions
shown in Fig. Pi2./Z. The volume rate of flow is 0.25 fts,
and the absolute inlet velocity is directed radially outward. The
angular velocity of the impeller is 960 rpm. The exit velocity
as seen from a coordinate system attached to the impeller can
be assimed to be tangent to the vane at its trailing edge.
Calculate the power required to drive the pump.
0,75in.
E FIGURE P12.12.
From EG. tedt vm bzo, /
7
Wopare - QU Ver )
7% determne Vi we use
U= beto = a )(taotes Jar TE): He nm E
TT chten Vop we use e exct veloity Iriangle Shown below.
wu EA
wo 7
ss
Symee Voz= To — Vis fan 35º ; =
and GL e. (orst Now dei) Junte
fas A. kh amlssindlhssm) CS
st follows That ;
2 (mi Lar tn) = hold
Thus, from EM (1)
/
AE - class (o, 5 Nu 1 E) 45.188)
Wma /E
= toio fé
or Led
. tolo = o
Wsmgt * moteh TO tebthro
s-hp
12-14
(2.43 |
Water is pumped with a centrifugal pump, and meas-
urements made on the pump indicate that for a howrate of 240
gpm the required input power is 6 hp. For a pump efficiency of
62%, what is he actual head rise of the water being pumped?
From Ep. (2.23 The pump efficrenes 13 qrótu by Pa
efuator + ha 580
% o bhp
o 7n at
. CMC btp ese)
ha E 0
| = Coe thp (So Es
(624 fes tão Em) /Cr48 E oo bi)]
= bLId
(2-15
det
The performance characteristics of a certain centrif-
ugal pump arc determined from an experimental setup similar
to that shown in Fig. 12.10, When the flowrate of a liquid (SG
= 0.9) through the pump is 120 gpm, the pressure gage at (1)
indicates a vacuum of 95 mm of mercury and the pressure gage
at (2) indicates a pressure of 80 kPa. The diameter of the pipe
at the inlet is 110 mm and at the exit it is 55 mm. TÊ
zw — x = 0.5m, what is the actual head rise across the pump?
Explain how you would estimate the pump motor power re-
quirement.
Frem Es. 1217
- Wi" (1)
ha > BA + EE + 75 y
Since / p . lrrospm leon é am o) u
- do Eri dem o, LING
1 A = fo.Hoam)* s
and ph=hh
Vey om (e 797 )e)* -3 E
2 f Smam
. 3
Thus, frem Eq, vn p= “Cy On) =- (0.065 Msexid 8)
ana = 80x1D Man?
| sou La + (D095 Mia3x 10º = os (319 89". (018)
o 2 + OSmTt
< (0.9)/7.80x 10º), 2 (4.81)
ha = Sm
To estimate the pump motor power requirement use Gg. 12.23
-Y A ha
bhp (550)
to get
bhp -Y G ha
n6so)
Fw difhring values dm, a cVEsponding bhp can be caleulaho
12-46
(2.16
Né K is sometimes useful to have 4, — O pump per-
formance curves expressed in the form of an equation. Fit the
ha — OQ data given in Problem 12.4Sto an equation of the form
ho = h, — kQ? and compare the values of A, determined from
the equation with the experimentally determined values. (Hint:
Plot A, versus Q2 and use the method of least squares to fit the
data to the equation.)
Basta en fre data from Problem 1215, tre follewmg Enable
cam be ereated ano firm a Standard, linear regresssem
Curve Attriç Pregram he felewns Vesalfs are oblmuned.
Qlsem) | 20 | ho | bo | &o | too | io | 1%
(6 Gp] “exist teu1b?] aexID”| Gex ID | doox 10? fee gi] 196410”
Ace) | fes | fes | 874 | 485 | 173 [695 | SEE
Abe le) |- 1.00 | 0.81 | -027| é.26 | 6:39 | 2.33 [n087
A Sha - he (experimental ) - he Cpredictea )
The equation obtamed from re ata sms linear regressa às
bes PLG— 0.007 q” €1)
there he 5 sm PA torta Gm gem A plot showms
The Comparisom between The rxperimental dota ana tre
predicteo results (Frem Eg.!) +45 shown below.
fo
Flowrede, O, gpm
(2:14
“2,17
12.17 In Example 12.3, how will the maximum height, 2,
that the pump can be located above the water surface change if
(a) the water temperature is increased to 120 ºF, or (b) the fluid
is changed from water to gasoline at 60 “F?
(a) From Table Bl Te weter vapor pressure 5º [692 psiá
and E = br MEF Thus, wa Thes Change wm Eg(2
mm tExample (2.3
& nr
Co do (rm 84)
( Z So tl dell pat
ma 6LN És
ta p*
- [htg Eine 2) use
bu E,
so faut
(2) mau = Eus
Thus, Tere 13 à decrease ww herjné Som 74H to SISFE.
tb) bom Table Lt The gasolmne Vapor pressure af bo 4
So psie ana do 425 !effeS. Thus, és asove
o em Es
2 dm; th (se ds) - toa Lt
fes 3
té no
Vu Tfão Cm E
id a
Fes ds
so Tha
(E dm TALL
The negeéive sign tor E nes indecates nal under The
Conditens Specificd lhe pump Could not eperade tuiThod
cavitation unless if (5 located belour ne Surface level
of the Gasoline.
12- 20
(2.18
12.18 A centrifugal pump with a 7-in.-diameter impeller
has the performance characteristics shown in Fig. 12.12. The
pump is used to pump water at 100 ºF, and the pump inlet is
located 12 ft above the open water surface. When the flowrate
is 200 gpm the head loss between the water surface and the
pump inlet is 6 ft of water. Would you expect cavitation in the
pump to be a problem? Assume standard atmospheric pressure.
Explain how you arrived at your answer.
Frem Ef. /2.237
Pato
nesms Se -& - Ed - f “1
From Table BI Tre ader vapor pressure «É teo'F 13 OTEP3 psja
pr PS;
ana &= 62.00 E Thus, with Bim fe 7 ps, Zcizfe, gua
ZA, = 6 Ft, EG. qrelds
ra fe, Os tai)
Wes LTL -nte-ttt
62.00 le
FE é
4 -
(o vues reu têi)
b2.c0 És
= 139 fé
From Fig t2.12 at 200 pm
=—/2 Fé
VE,
Fhr Preper pump operation
MPsh, 2 NPSHr
Since fas ds Crue mm This se, toe expect Thaé cavitatios
m The permp would not be a problem. Ne.
f2-21
(2.20
12.20 | The centrifugal pump shown in Fig. P12.20 is not
selfrpriming. That is, if the water is drained from the pump and
pipe as shown in Fig. P12.20(9), the pump will not draw the
water into the pump and star! pumping when the pump is turned
on. However, if the pump is primed [i.e., filled with water as in
Fig. P12.20(h)), the pump does star pumping water when
tumed on. Explain this behavior.
ta) th)
W FIGURE P12.20
The heaa-tlowrate characterstiis for a Eypiial centntuga!
Pump are shown 1h Fig. de.tl. The maximum hega Juat The
Pump Cam add ocenrs iihen when Qxo (ee, «é start ap for example).
This heen vom terms el bre lui 1 lhe pump. Neglecêma losses
and She vetecrta head Cana cavrtation etfecds) the Pump cen
lift Me Fund à heght egua! to e head ander by ne pump.
However, 1f e Flu m e pump Às at (ce, not pejmed) e head
added 5 th terms É Ft or sm É ar, For example , ,£ hs do fd
de pump could raise tuater That Hugh vÊ 16 15 primed (filed
wita water) L£ fre Pump 5 not primed (filled vita ar) Then
de pump can enty raise taéer up to 4 distance
6o7es te
p= sofa Cb « sofa ed. 205 ft
Water es
Mente The water will mol qeé mt» e pump.
(A-24
(12.2)
Owing to fouling of the pipe wall, the friction factor
for the pipe of Example 12.4 increases from 0.02 to 0.03. De-
termine the new flowrate, assuming all other conditions remain
the same. What is the pump efficiency at this new flowrate?
Explain how a line valve could be used to vary the flowrate
through the pipe of Example 12.4, Would it be better to place
the valve upsircam or downstream of the pump? Why?
horta f=0.03, Eq) mm Example (2% becomes
rm (200 ft)
- , =D + (0,
to jo ft +| 0.05 CE TE) le 5415440)]
vê
2(322H) (3
si
V= LP = eg)
4 G)CÉr)*
Egtt! tam be wrben às
hp p + bon G*
or tutu Po sm el /min
hp= 10 + 3.00 XDÉ 7] (2)
The mbtrsection el Egl2) (the system equation) ba he
performance curve r Lhe pump, 45 shown below, !nilitates
Tuut ne new lowrate 15
- /
P= 400 de
ana e efficieney aÉ ris Elo ade ts approximately 79.0.
See,
100— = -
qt
0746 6601200] 16002605 2400
Cont) Flowrate, galmin
12-25
tez jo (Cont)
A me valve ack as a variable Írictonal resistance
to the ow, Closmg fhe valve is Gu ivalent % Adelino
Prietion and moving de system curve to the left
mnTersechng me read cume af an apembenal point
tavelvina less Plowrate than milh à more open valve sela,
This system curve is shetohtd tn the figure on fhe previous
Page and labeled “mare Fichon (Losi vale) * Openmp he
Vale às Similar Po removing frichem qnd moving fe
systêm curve fo fha vigh? 1 tersecfng he head curve
af an opera fina poiot rave bra more Favrate ham
with a fess ecpem valve sefing. is System carve is
stethed en lhe previous page. and tabela “ess herichon
Co Opersty va lve ). ”
zr wonid te generally beber do place the vatee
dowstre ara 2 the pero % avoid e low Suchon
pressure amd ecavitation possible wilh ups ihtam
placement of te vale.
12-26
(2.23 ( Cont)
(E) (E)! Ke $ do (8)
freh
do 0.089! garxm* 0.020 f7.o
po 4.178 LE5xn5] 00183 | 630
/20 0.267 2.78 x1| 00:87 | 437
dê o 0.357 371 408] o016H | Zipz
200 6d gersmnS| nois 373
ado | 0.535 | SsixW?] gor | 534
These data (Ro 15.0) are ploHed em Fig. 12.12 (reproduced
below), and Jhe lowrate ab e tabersectón of The system
Curve and The Pump Curve 15
- Al
£= (5 a
Smice at tuá Flowrate he pump eperates near peak efficrenes
ais fype of pump tould «ppear fo be & Good Chore sf
Me 168 9%/omo Fowrate 13 ak ov near The desired Pow rate,
500
400 [-7º
: 300
TF
5
= 200
100)-4
m- . nim -
: '
Li e JE8 DAR, =
o 40 Bo t20 160 200 240 280 320
Capacity, galímin
12-29
/2.24 ]
Determine the new fiowrate for the system described
in Problem 12.23 if the pipe diameter is increased from 3 in. to
4 in. Ts this pump still a good choice? Explain.
Peter to solution do Problem 12.22, lbitu D= Hz ft Eq.»
becomes (ooo ft) vz
4: 4 ER 1,
? (É te) coltsza É) (2)
ana vo £ q (E)
(E (Ea)
so ut hs = 2.45x10" £ Ig (Eh DN (3)
The Pe ynolds number becomes
pc $E = HO 73800 0º CL)
TDV O CH de) CEA É SE)
. . -+
br Commercial sjeel Sm dameber prpe (From Fis. 422), É = 43h.
Thus, fer a guie Q, E Con be obtaines fm Tre Moody charl, or the
Colebrook equation (Ep. 235), qnd hp determmed from Ep (3), Tabuladed
Values are quên mm me follows fable .
9 E) pt) Pe £ bo Cf)
| o nos | easun! | oo2H 41
| 80 eu78 | L39x/0 | cola 144
ro pu | Zogxp! | dD/83 32.6
| Ito 0.357 29240? | o.or7y 559
| 200 o su | B48NH85 | ooftb 55.3
| 2% 6.535 mz” | cola 22
| 280 ob2e | E8I4HÉ | avr (a
| 3260 0.7/3 EStunÊ Botto 2zj2
These data ( É, vs 0) are plofled on Fix 212 ( reproduced
on e fellowns page), and te Fewrate ué ne intersecton ef
fe system Curve and The pump Cuhve 13
P= 255 ae!
trun
Ceont)
12-30
(2, 2%
va
Lam E
o “o BO 120 160 200 240 280 320
Capacity, galímin
Smce at dus fluwrate The pump efficiency 45 dasrts
louvo (o BP) is pump
is mo longer a good chore.
(2-3!
(2.26 (tont)
Since Tre pump equation 4s
do* SEo-torxh 3 600)" (t)
EG (5) and Eq) can A autos bo de derme he Hwrate, Thus,
33 + 5 78X10 29? = S2.0-1.0/ xt “q?
mê
dna p= e0527 E
(b) IF fre Flowrte x to be é m half so rat
P= 00529/2 = 60265 mf, e head added 8y he
por 15 532
hp S2o- poxo (vozes )
= SE bm
Frem Eq.0t) torta k, un know
SdtmE Bm + (k, +60 326) (D.o25 2% 23 )*
So Tueé
dp 2%3
From Ea. /2.290b) fne valve ioonld be [8% epen £o
obtem Thiã k,
12-34
42.27
A centrifugal pump having an impeller diameter of
1 m is to be constructed so that it will supply a head rise of
200 m at a flowrate of 4.1 m?/s of water when operating at a
speed of 1200 rpm. To study the characteristics of this pump,
a 1/5 scale, geometrically similar model operated at the same
speed is to be tested in the laboratory. Determine the required
model discharge and head rise. Assume both model and pro-
totype operate with the same efficiency (and therefore the same
flow coefficient).
For simlarts The model Pump pmusÉ eperate aí ue same
Flow Wefficiemt, EF. 42.82, so Tua
2 ) (+
WD, wD3/p
where The subserpt Cm) reters to June model and €p3 db he
prototype. Thus
- Lim Da 3
Ga “p 5) %
ana desta dm, Dm /Dr= VE, and Go Flw Them
Q, = (EN ai?) = avsz8 E”
From EG. /2. 53
(ate) «($4s)
wo p*
has o £ (e) (22) +,
ana deita de= dm, lbm= py Dm Dp* Ve and ha* 200m
Then ? ? ?
so Pub
A OE lzoom) = 8.00 m
/2-35
[2.28
nos
Explain how Fig. 12.18 was constructed from test
data. Why is this use of specific speed important? Illustrate with
a specific example.
A variety of pump configurations like the ones shown in
Fig. 12.18 were tested over a range of flow rates.
Performance data like those shown in Fig. 12.17 were
acquired. For each pump configuration, the operation at
maximum efficiency was noted and the specific speed, Nç,
(Eq. 12. 43) was calculated for that condition of flow.
These specific speed values calculated at maximum
efficiency operation were then used to distribute the
different pump configurations as shown in Fig. 12.18.
Specific speed is important because from desired design
operational data (, Q, and h, ) a specific speed value can
be determined. With that value of specific speed and Fig.
12.18 the designer can decide what kind of pump
configuration to use for maximum efficiency operation.
For example, at lower values of specific speed, a
centrifugal pump is generally best. At higher values of
specific speed, an axial-flow pump may be best. In between
values of specific speed may suggest that a mixed-flow
pump would serve most efficiently.
12-36
/2.30
12.39 A centrifugal pump provides a flowrate of 500 gpm
when operating at 1750 rpm against a 200-ft head. Determine
the pump's flowrate and developed head if the pump speed is
increased to 3500 rpm.
br à Gwen Pump The etect of à change m speed on Q
ama he 5 Gên by Egs (2.36 ana 4257. Thus,
Po. 0%
A o (eg. s2.36)
and mia = Sbogem,t,= (7504pm, HH Ups Edo hm, Men
6,2 dee 6
“Lo,
1
= (800 q fm
Symilarto,
has o tor
E
so That tofu he, Zoo ft
tu
da
Pa
= 800 f
(3506 rpm)
Tizsarom) SUSPM
z
- 3500 pr 5 (200 ft)
(750 rpm
(Es. 12,37)
/2-34
12,81
12.37 A centrifugal pump with a 12-in.-diameter impelter
requires a power input of 60 hp when the flowrate is 3200 gpm
against a 60-ft head. The impeller is changed to one with a 10-
in. diameter. Determine the expected flowrate, head, and input
power if the pump speed remains the same.
for geometricalts similar pumps eperatim at Me same speed Tre
effect od a Change 10 mpeller diameter + Givên by Fps. 42.39
t2.40, J2pl. TAKS,
3
PA. (Eg, 12.39)
& 3
Ana durfa e, 3200 5pm , D
p= (2) PT (lee )f3200 3pm) = (850 (250 Sfim
= l2tmh., And D= fosh,
From Ego 12.40
2
fa, = É (Ep. l2to)
fas 2
So fnat witn ha,E to Et
(AY do o (Jintao tura
2) € - Md
4
da,
de ta
|
Smlarho From Es. (2.41
WO shege, = D” (
: Coe Es. srt
h Shutt z 2 í )
ana Wita Wma = £0 Ap
7
Vgafa,* (Bu Vspuga, É - (é ja. tp )=
12.40
/2.32
Do the head-flowrate data shown in Fig. 12.12 appear
12.32.
to follow the similarity laws as expressed by Egs. 12.39 and
12.402 Explain.
The data m Fig faiz show tre effect of champns impeller
diameter en heza-emrate charactenstas. fecordio do The
smilarita lows expressed by fg (2.37 Ino Eg. (2. +o
“em:
3
É & E (Eg. 12.39)
Z
dar = 2 (Ep. J2.t%o )
Az 2
Thus us me diameter E Imereastd from rh. do Tim to dim
e Flowrate Inlreeses ac cordins bo 7 1237 45
(Leom tin term) g(Bro = Jo = 1576,
and .
' = Bum. > - 2 37 q
Crom tun $yh.) Ps =) Ê) É !
Sum larly, rem Ég. (240
. 2
(Leom bin. do 74%.) fe,* (E) ha, 5 (Ze) he * 1 3b ha,
ana )
Cdram tm. £o $m.) am ) ha, E 1.78 ha,
Thus, fer Any Fivên ait, Such às (4) tuhere Qu zo gpm qnd
ha =2Zs0 fé ( see Ag. jr /jz em £o Hlowms pese) Ler Zne écin.
diameter Jmpellet, de Correspondis predicted Pont tvenlá be
at (RD) there
GQ = (s1)ltzogpm) (9! gro
Ap (136) (250 46) B%o Fé
Z
( Com £)
[2 44
12,3%
12.34 Ina certain application a pump is required to deliver
5000 gpm against a 300-ft head when operating at 1200 rpm.
What type of pump would you recommend?
Br = S000 Gp”, he = 800 Ft, and = (200hpm, the
specific speed 45
My = Co Cromo) VP Gem).
[Are]
(1200 nom) V voo gor
(300 fe)“
É“
!180
From Fig. (2.18, 44 Tys specific speed a radial
Low Pump Centritagal pump) would be recommended.
(2-4
(2.35
12.35 A certain axial-flow pump has a specific speeó of
Ng = 5.0. If the pump is expected to deliver 3000 gpm when
operating against a 15-ft head, at what speed (rpm) should the
pump be run?
da wo Cred) Vo CEL)
hs [g() Au l66)]
for MS So, / ta Ah) HP PE, ana uia
3000 fé!
o > EO tus té
? (us Fes) (bo tin) S
é olus ua
/
se Vo
otra) = (50) Lens Eca
| eta?
= 194 E
Hence do (rpm) = (144 mad ) (607)
= [900 rpm
JA s
(2.36
A certain pump is known to have a capacity of 3 m?/s
when operating at a speed of 60 rad/s against a head of 20 m.
Based on the information in Fig. 12.18, would you recommend
a radial-flow, mixed-flow, or axial-flow pump?
vs po ELraat) Veg
Co falho) de tm) ft
for lo=borad/s Qu Smmls, G= G8lmisi qua hs 2om
2 (toras ) S mts
Egor) om]
= 198
Ms
Bem Ely lulg corta Mp= 18 The pump 15
a imired-Alew Pump.
/2- &6
12:37
(cont)
Pe main Cauton m Placing te Pump vertreally m
the rituke pipe is % do so m a way de avoid
cavitatiom lin The pump. The collapse of cavitabom
bubbles o tne pump cam eredo pump blade and olher
metie! surtaces. Apply he energy equafim, Ba.5.8€,
between The Set suríce (1) ama the pump enhance (2) We
ge?
2 z
Biba Bem ye-h
+ 2 t+ 29
4h
29
and + Maximize f , we minimo 2,2, Po ochreve
fuis we Place pe puma high verticall, ma he miíxte
pipe. This will tend to lego Pá high enmagh % aviid
cavitatica which oceurs mhen p andfw velaleo pressares
dh tha pump become less than the vapor pressure of
the Fluid.
12-49
12:41
A Pelton wheel turbine is illustrated in Fig. P12.41.
The radius to the line of action of the tangential reaction force
on each vane is 1 ft. Each vane deflects fluid by an angle of
135º as indicated. Assume all of the flow occurs in a horizontal
plane. Each of the four jets shown strikes a vane with a velocity
of 100 ft/s and a stream diameter of [ in. The magnitude of
velocity of the jet remains constant along the vane surface.
(a) How much torque is required to hold the wheel stationary?
(b) How fast will the wheel] rotate if shaft torque is negligible
and what practical situation is simulated by this condition?
M FIGURE P+2.4)
T=nmra(U-VMlI-cosp) where n=%
ta) With the wheel stationary V=0 so that
T=-&m im Vi ll-cosf) where
m = AV= (ay sas) (HP (100 £) = sos ska
Thus, T=-4 (1057 8) (1 44) (100 HE) (I- cos 135º)
(9)
= - 722 f!lb
(b) From Eg.(1), when T=0, then Uh
Thvs
, vo?
Door oM or coro Mio 00088
=955 rem
The zero torque case represent a broken
shaft situation.
ta 50
12.42
12.42 Consider the Pelton wheel turbine illustrated in Figs.
12.24, 12.25, 12.26, and 12.27. This kind of turbine is used to
drive the oscillating sprinkler shown in Video V 12.4. Explain
how this kind of sprinkler is started, and subsequent]y operated
at constant oscillating speed. What is the physical significance
of the zero torque condition with the Pelton wheel rotatíing?
As shown on page 775 below Eg. 12.50
pato = mr. Cu-vuy- cos p )
So rw no rotafon ot the wheel or UzO, fre variafim
of Tuts tuto, changing m d near. When Laet ss
Just lavçer than Me resisting Yerque provieled by
the sprinkler, the Peltor wheel rotates and drives
He oserllatrom ot the sprinkler. Aflw wbgel
rotatim and sprinkier oscillafion begins, am
Comstam value of m and Toa a results o A
content valee of UT and thus rolnben speed
amd alto oscillabor period .
JF the shaft cmnechas e oscillabing sprinkdo
to Ffhe Rttm wiee) breaks during optratim ,
the sprinkler ml) cease oscilla bn; and the
Ritm wheel wilt rum af constan? rotatiom
speed comrespondras to UV=y.
t2-57
12.45
12.45 A sketch of the arithmetic mean radius blade sections
of an axial-flow water turbine stage is shown in Fig. PIZHS.
The rotor speed is 1500 rpm. (a) Sketch and label velocity tri-
angles for the fiow entering and leaving the rotor row. Use V for
absolute velocity, W for relative velocity, and U for blade ve-
locity. Assume flow enters and leaves each blade row at the
blade angles shown. (b) Calculate the work per unit mass
delivered at the shaft.
(a) U=ruw and
s that wah ra “ st,
D=tw where w=(1500 5a gos
Blade secticns
1 U | atthe arithmetic
mean radius
Pas
AV Nas
EB FIGURE P12.45
Lin ( 27 rad cad
(2 Emo) = 157 9º
- (0.5 H157 188 Lady. 78. st
e
a intet and exit velocity triangles 2º Va
are as shown. v/ [U=785 U, =78.5
Mote: U =U, (same radivs)
d .
PÊ W=Ula (trom continvily egr.) 7)
uso
nº
(ma HT U Voa -U Voy =U(Voz- Vos) tm)
From the figures? Vi cos 70º =W, cos 45º
and from law of Sines W 78.5 or W=635 E
smar S
(e3.8 8) costsº sin (zó=as")
so that ,= "5 — = (313 É
cos 70º
or g
Vo = cos20' = 123 E
Also,
Voozh-W sints" = 78.5 É - ess É sintrs"= 33.6 É
Hence, from Eq. (1)
Mepatt *
= 79.5 H[ 3248 12388] = 702084
/2- 5%
12.86
1246 An inward flow radial turbine (see Fig. P12.46) in-
volves a nozzle angle, &, of 60º and an inlet rotor tip speed,
U, of 9 m/s. The ratio of rotor inlet to outlet diameters is 2.0.
The radial component of velocity remains constant at 6 m/s
through the rotor and the flow leaving the rotor at section (2) is
without angular momentum. (a) If the flowing fluid is water
and the stagnation pressure drop across the rotor is 110 kPa,
determine the loss of available energy across the rotor and the
efficiency involved. (b) If the flowing fluid is air and the static
pressure drop across the rotor is 0.07 kPa, determine the loss
of available energy across the rotor and the rotor efficiency.
M FIGURE P12.46
(a) loss = Porcfm EM att o where Por foz = stagnation pressure
4 drop across reter =Ag,
an
Mshafi = UV, Voz Ui Vos =-U Voy since Voz =0
o m =
Thus, Mega * A92)(12Blc0s30) =-93,5 55 Vez Voe
so that 3
toxio
f mA m*
loss= É 4 (93.50) = /6.6ã
a,
Also, .
mM
= Meat Pas içã
Y ap: — QuoxZ) = 0.849
(999 3)
Cont)
12-65
12-46
(cont)
(b) loss = fede Mlpall “ where for” foz. = stagnalian pressure drop
across rotor = af
and
Mepaty = Voos -U Vos =- UV sinco Vea =0
Thos, Me pag = (EN 12 Ecos 30º) =- qu, LE V=12
Also, ON
28 = fra tipo) Ver = cosa
= 0.07 ka + (123 BO 02 2 (685) ( LÃ
=(0.07 +0.066%) kPa = 0.138%kP
Wn=6
Thus, 3y
loss = SittSO mê gas. 17.4 E
(123 da)
and
Meta past
0.843
ve (egg ES
123 +9
ms,
Ja- 56
12.50
12.59 A Peiton wheel has a diameter of 2 m and develops
500 kW when rotating 180 rpm. What is the average force of
the water against the blades? If the turbine is operating at max-
imum efficiency, determine the speed of the water jet from the
nozzle and the mass flowrate,
Wopatl E To = 2Fwor 5ooxid Mm =(Zm)F(igo Ler)( Amin) (20 rad)
cos
rev
Thus, F=26,600N
Also,
Wopaty = POUCU-VDlI-cosg) so thal a! maximum efliciency
with ê: =/80º and V=5 Ve Ihis gives
Wpaty = REC (2) = eU eu. ny ,
But
W=20=202 =wD= (1805 (giz) 2zraa oe) (am) = 37.6 &
Thos, from Eg tt):
o ZW - 2(500n0 12)
m tg
vê (37.6 E)
= 707 Ms =707 2
12-59
2.51 12.51 Water for a Pelton wheel turbine flows from the head-
water and through the penstock as shown im Fig. PI25/. The
effective friction factor for the penstock, control valves,and the
like is 0,032 and the diameter of the jet is 0.20 m. Determine
the maximum power.output.
Elevation = 975 m
o) VA
) “4 | Eleaion=250m
0.20m
E FIGURE Pt2.51
Wopaty = pRU(v-V)(I-cosg) or for maximum power B=/80º, V=
Thus, vê
What 2" e Z ti)
max a
But Beshesz = Golo en tr, here for po0, 2ee975m,
Z,=250m , and Vh=0
Zo =Z, +b bte where AV=AV mn
Pu = POW. Thatis V= (LV, «(EDU =0.0494V,
so thal £g. (2) becomes:
2
975 m = 250m + zaml! +o,032( Die) o. 0498)? | where Va 2
or V=i/43 E
Hence,
Q= AV =Hoazmtuas2)= as E
Therefore, from Eq:
Hence,
or
s
t mz
Wohart = os bes set ) (itta) =222x0º 2 «23,200 kW
max =
12-60
12.52
12.52 Waterto run a Pelton wheel is supplied by a penstock
of length € and diameter D with a friction factor f. If the only
losses associated with the flow in the penstock are due to pipe
friction, shown that the maximum power output of the tur-
bine occurs when the nozzle diameter, D,, is given by D, =
D/(re/D.
Wopati = egu( V-v) (- cosg) so the maximum power ovipul occurs
wilh p= 180º and vb - Thvs,
y to)
Wepatl = PO adere «1) ,
2 Db 4
t
But po=f1=0, o=0, and Po gch. Thvs,
z Z
hab Ê ubere since ALA or OM = Z0% ue have
n=(B)V
Vê emo
Therefore, h= all? tá 2 "25 3 —LrtdÃ
- LEOMh py qo —SBÊ
Wlopa = LRP - —Nagh
dO HED) O (ED 1d (140, D5)£ (a (3
For this problem £. 2, Dand h are constanis ; O, is variable.
Thus, from Egs.(2) and (3):
Winafy = Kd where K=const. and c<consl.= f Ê
(irc Diã
Note: Wopai —0 as D->0 and às D,-»00, To find the D, ihal gives
and Eq. €1) gives
maximum power over all, sel dWehaty =
P! 4 e 0
dVopatt 2KDy CEJKDZ 3
DO trens! Qrrenyrjra (040 =O
2kDr
“ trcoa [1-2 ]=0 =0,0r 14cbf=scD,*, er Dt=zE
D
Thus, Ds cine “qb
12-61
12. 55
12.55 A hydraulic turbine operating at 180 rpm with a head
of 170 feet develops 20,000 horsepower. Estimate the power
and speed if the turbine were to operate under a head of 300 ft.
4, =/80rpm , hy = 70H, Wpatt, * 22000hp
ma=P ma = 00H, Mpatg*?
Assume He efficiency remains constant 2h.) = 2h
so with D,=Da and q 92º “oa
170 300
(180% * Tua O ua = 239 rpm
W W
Assume Mhe same powercoetlicient: quédé. ao ué)
so with D =D. and p=fe:
2ovoo baga qr Vl" PÓU0O O
12.56
12.56 Draft tubes as shown in Fig. P1256 are often in-
stalled at the exit of Kaplan and Francis turbines. Explain why
such draft tubes are advantageous.
«)
Draft tube
ZA) GA
Le
ops
E FIGURE P12.55
Wilhovt the draft tube there would be a relatively high speed ext jet
(speed V,, pressure p,=0). With the drafl Jube (which act as a
ditfuser) the exit speed is much smaller (WxO , qo =0). From Gernoull"
equation if Tolhows Mal f,<O Cwilh he drati tube). Hence there is a
Jorger head available to lhe lurbine. More energy can be removed
from the Llid,
/2- 6+
12:57
12.57 Turbinesareto be designed to develop 30,000 horse-
power while operating under a head of 70 ft and an angular
velocity of 60 rpm. What type of turbines is best suited for this
purpose? Estimate the flowrate needed.
Wpati = 30000 hp 5 hy = 70; and w=60rpm so thai
Na = ria = dE =543 Fortis valve a Francis
, turbine wovld be appropiats.
Also, since Was = SQhr it follows that
; lb
q= “etutt - (32,000 tl sso 2/47 ”
E hr (62.4 Bs 70) SfEs
12-65
12.88
12.58 Show how you would estimate the relationship be-
tween feature size and power production for a wind turbine like
he one shown in Video VIZI.
%o estimate the vela ion shop belmeer feature size and pover
pro dnchom fr à wind turbine we use the dimensimaless
; £
pi terms 0É djs. 12.27 amd 12:30 which ave applicable fr
this jneompressible Flow. For similar durbines And epecát
” tendihons
Momutro o Womto a
pusD” 2 nº
fa d,
pmd
Ghas o 9 ha
Do CC wrDA
? t % >
Since PEA amd has = ha, we tombive and get
A 2
att ' - Db,
) =
Won > 2,
dr quer Varies with Lenture Size squaned,
12-66
72-87]
12.6 — The device shown in Fig. P12. 6 is used to investi-
gate the power produced by a Pelton wheel turbine. Water sup-
plied at a constant flowrate issues from a nozzle and strikes the
turbine buckets as indicated. The angular velocity, w, of the
turbine wheel is varied by adjusting the tension on the Prony
brake spring, thereby varying the torque, T.pans applied to the
output shaft. This torque can be determined from the measured
force, R, needed to keep the brake arm stationary as Tomar = Peltóh,
Ft, where € is the moment arm of the brake force. wheel
Experimentally determined values of « and R are shown
in the following table. Use these results to plot a graph of torque
as a function of the angular velocity. On another graph plot
the power output, Wan = Tonan, “2, AS à function of the angular
velocity. On each of these graphs plot the theoretical curves for
this turbine, assuming 100 percent efficiency.
Compare the experimental and theoretical results and dis-
cuss some possible reasons for any differences between them.
Brake shoe
H FIGURE P12.64
« (rpm) R (b)
0 247
360 1.91
450 1.84
600 1,69
700 1.55
940 1.17
1120 0.89
1480 0.16
(a) Experimental : T=Re=(o.5HR or T=0sR fl, where R=-h a)
and W, Mot = TU = Tu m 69) (eELad
” Wpaty = 01047 Tu Eb — uhere T=FtHb , core (2)
Vales of w,T, and Wjsy are given in the fable and graph below,
(b) Thewrefical: T=mr(U-V)I-cosg) where assume E =/80º,
Es
V = 2-ê = USE ss savÊ, and
7 (en
m =09=(19% do (assa) = 0,105 slas
Henco, with U=tw 2 = (Rtt)( PU Lad) = 9, ozs2w É w-rpm
T=(o.r105 sleep 2sDfo 0242) -53.7] É
7=/.4 [Éeexô'w -1] Ftlh , where to = rpm (3)
(con't)
or
/2-69
12,6% À (con't)
fls, Way = Tu = T($80) =0.to7Tw Eb where Te Bel, tu ep
Valves of Tand Wpatt from Egs.(3)and (4) are plobed in lhe gragh
below, ; H
experimen eory
o 1.235 0 14 Õ
360 0.955 36.0 1i6 43,8
450 0.920 43,3 1.100 51.8
600 0.845 53.1 2497 62.6
700 0.775 56.8 0,928 68.0
q40 0.585 57.6 0.763 75.1
t!z0 0.45 52.2 0.639 785.0
1480 0.080 12.& 0,392 60.7
Ló - “+80
La Wohaft
, te
Erlh ta E Je
Lo
0.8 Ju
0.6
o zo
0.2
0
/2- 70
APPENDIX A
Listing of Standard Programs
LINREG2.BAS
5 cls
10 print Stsoeaoeolotoiatetad ek sede dede de
20 print "xx This program determines the least squares fit **
30 print "** for a function of the form y=a+b*x ento
80 print MSM dd
50 dim x(101),y(101),ybar(101)
55 print
60 input “Number of points: “,n
70 print “Input X, Y!
80 for i=1 to n
90 input x(i),yti)
100 next i
101 sx=0
102 sy=0
110 sxy=0
120 sxsq=0
130 for 2=1 to n
131 sx=sxtx(i)
132 sy=syty(i)
140 sxy=sxytx(1)*y(i)
150 sxsq=sxsqtx(1)"2
160 next i
161 a=(sxsqtsy-sxy*sx)/(ntsxsq-sx"2)
170 b=(ntsxy-sx*sy)/(ntsxsq-sx"2)
180 print
190 print using "a
200 print using "b
210 print
220 print “ x Y Y(predicted)"
230 for i=1 to n
240 ybar(i)=a+btx(i)
250 print using "ABLAERDOSOO MRETOSTO AR RAC OO CGC D pi) ybarti)
260 next à
4-4
100
110
120
130
140
150
160
170
180
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
440
450
460
470
480
490
POLREG.BASº
cls
print 'Pedeeoeaaee aa ad esa ea aa a atoa dd aaa
print "** This program determines the least squares fit xx!
print "xx for any order polynomial of the form: a
print "xx y = dO + di*x + d2*x'2 + d3*X03 +... +
print esses ae ato aaa ta dO Rd df
print
dim b(21),d(21),5(21),x(101),y(101),f(101)
dim errf(101),p5j(101),pim1(101),ybar(101)
input "Enter number of terms in the polynomial: ",nterms
input “Enter number of data points: ",npoint
print:print "Enter data points (X , Y)"
for i=1 to npoint
input x(i),y(i)
d(i)=0
E(ti)=y(i)
next i
print
print "The coefficients of the polynomial are:"
for i=1 to npoint
f(i)=f(i)-dinterms+1)*x(i)"(nterms )
next i
for 5=1 to nterms
b(5
d(3
s(5)=0
next 3
c(1)=0
for 1 to npoint
d(1)=d(1)+f(i)
b(1 (1)+x(i)
s(1)=s(1)+1
next i
d(1)=d(1)/s(1)
for i=1 to npoint
errf(i)=f(i)-d(1)
next i
if nterms=i then goto 750
b(1)=b(1)/s(1)
(cont)
As
( cont )
POLREG.BASº
500 for i=1 to npoint
510 pômiti)=1
520 pjti)=x(i)-b(1)
530 next à
540 for j=2 to nterms
550 for i=1 to npoint
560 p=pi(i)
570 d(j)=d(j)+errf(i)*p
580 p=p*pi(i)
590 b(5j=btj)+x(i)*p
600 s(jj=s(j)+p
610 next i
620 dtj)=d(5)/st5)
630 for i=1 to npoint
640 errf(i)=errf(i)-d(j)*p(i) -
650 next i
660 if jsnterms then goto 750
670 b(5)=b(5)/st5)
680 cljl=s(5)/st3-1)
690 for i=1 to npoint
700 p=pjti)
710 piti)=(x(i)-b(5))*páli)-cls)*psmiti)
720 pimiti)=p
730 next à
740 next 5
750 print using "dk = +&.4HHH"""c";nterms-L, dinterms)
760 nterms=nterms-1
770 if nterms>0 then goto 300
780 print
790 print * x Y Y(predicted)"
800 for i=1 to npoint
810 print using "th. HWRRTCOS ap MPRRTTOO ABERRRO OO M xi) yli),yli)-errf(i)
820 next i
“This program is based on an algorithm described in Conte, S.D. and de Boor,
C., Elementary Numerical Analysis: An Algorithmic Approach, 3rd Ed., MeGraw-
Bill, New York, 1981, p. 259.
COLEBROO.BAS
100 cls
110 print 'Setseesso essas aeee se aaa aaa adota toe ta o dd tt
120 print '** This program determines the friction factor, £, for ***
130 print '*x pipe flow for the case of laminar or turbulent flow *X"
140 print "*x (solving iteratively Colebrook's equation), given om
150 print "xx the Reynolds number and the relative roughness of *xm
160 print "xx the pipe em
170 print 'Seseeesecese esse eea see a aa tata
180 print
190 input "Enter Reynolds number, Re = ",re
200 f=64/re
210 if re < 2100 then goto 260
220 input "Enter relative roughness, rr = ",rr
230 fp=f
240 £=1/(-2.0*log(rr/3.7+2.51/(re*fp”.5))/l0g(10))*2
250 if abs(1-f/fp)>0.001 then goto 230
260 print
270 print using "The friction factor is £ = +H.&HHEHTTTTULE
CUBIC.BAS
100 cls
110 Print ALAS ES
120 print "** This program determines the real roots of a aee
130 print "** cubic equation of the form x'3 + atx'2 + btx + c = 0 **"
140 print EA RR RR RR RR RR RR RR O e e e e a
150 print
160 input " a=",a
170 input " b=",b
180 input " c=",e
190 !
200 'Check if the equation has complex roots
210 p=(3*b-a”2)/3
220 q=(2%*a“3-9%aXb+27%0)/27
230 if q"2/4+p"3/27<=0 then goto 250
240 print:print "The equation has complex roots":stop
250 x0=-a/3+2/3*(a”2-3*b)"0.5
260 for i=1 to 20
270 x1=(2*x0"3+atx0"2-c)/(3*x072+2%a*x0+b)
280 if abs(x1/x0-1)<0.0001 then goto 310
290 x0=x1
300 next à
310 m=atx1
320 n=bratxl+x1"2
330 -m+(mº2-4*n)º.5)/2
340 x3=(-m-(mº2-4*n)7.5)/2
350 print
360 print "The roots of the cubic equation are:"
370 print using "xl=H.HftOTOO o x2=+BRRRROOOO x3=+H.RREROOO O" ;x1,x2,x3
4-7
FAN.RAY.BAS
1)
100 CLS
110 PRINT *Seteeee ease ses esa eee ae doa de ade
120 PRINT "** This program computes the one-dimensional Fanno or x!
130 PRINT "** Rayleigh flow functions for a gas with constant ee
140 PRINT "** specific heat and molecular weight. (NOTE: k > 1) e
150 PRINT '"eeeseaso ease eaeaa aos tea ea dd df
160 !
170 ' Fanno flow functions
180 DEF FNFTTSTAR (K, MA) = (K + 1%) / (28 + (K - 1%) * MA * 2)
190 DEF FNEVVSTAR (K, MA) = SQR(FNFTTISTAR(K, MA) * MA ” 2)
200 DEF FNFPPSTAR (K, MA) = SQR(FNFTISTAR(K, MA)) / MA
210 DEF FNFPOPOSTAR (K, MA) = (14 / FNFTISTAR(K, MA)) ” ((K + 14) / (24 * (K -
$))) 7 MA
220 DEF FNFLD (K, MA) = (18 - MA” 2) / (K X* MA * 2) + (K + 14) * LOG(FNFVVSTAR(
K, MA)” 2) / (26 * K)
230 !
240 ' Rayleigh flow functions
250 DEF FNRIMP (K, MA) = 1 + (K - 18) * MA * 2 / 2%
260 DEF FNRPPA (K, MA) = (1 + K) / (1 + K * MA * 2)
270 DEF FNRITA (K, MA) = (FNRPPA(K, MA) * MA) * 2
280 DEF FNRVVA (K, MA) = FNRPPA(K, MA) * MA ” 2
290 DEF FNRTOTOA (K, MA) = 24 * FNRPPA(K, MA) * 2 * MA” 2 * FNRIMP(K, MA) / (K
+ 18)
300 DEF FNRPOPOA (K, MA) = FNRPPA(K, MA) * (2% * FNRTMP(K, MA) / (K + 14)) * (K
7 4R — 18))
310 !
320 ! Get functions desired
330 LOCATE 8: PRINT "Program options"
340 LOCATE 9: PRINT " (1) Fanno flow calculations"
350 LOCATE 10: PRINT " (2) Rayleigh flow calculations"
360 LOCATE 11: INPUT "Enter the number of the option desired: ", OPT
370 IF (OPT <> 1) AND (OPT <> 2) THEN LOCATE 11: PRINT SPACE$(79): GOTO 360
380 !
390 '--- Display banner specifing which flow calculation is being performed
400 CLS
410 IF OPT = 2 GOTO 480
420 PRINT 'Seteeo eee aaa ee tao a et ooo tod dota ta dot foto dao oo tod e tdo tod dotada
430 PRINT "x* Computing the one-dimensional Fanno flow functions *x!
440 PRINT "*x for a gas with constant specific heat and molecular **!
450 PRINT "xx weight. (NOTE: k > 1) e
460 PRINT 'Sesesoesee desse ease tao a at ato e ota eta aaa toaa ar
470 GOTO 540
480 PRINT '“etetseeoese eee eae o ot aa ta tada ta dt e ted att at
490 PRINT "** Computing the one-dimensional Rayleigh flow functions **"
500 PRINT "xx for a gas with constant specific heat and molecular Mex
510 PRINT "** weight. (NOTE: k > 1) x
520 PRINT 'Setoeo ease dede aee tee o toa eae to od to od atado pol top eta ae
530 !
540 '--- Get the user specified specific heat ratio
SS0 LOCATE 7: INPUT "Enter the specific heat ratio, (k > 1): ", K
1
(cont )
Á-io
(cont )
FAN RAY.BAS
560 IF K <= 1 THEN GOTO 550
570 LOCATE 7: PRINT SPACE$(79): LOCATE 7
580 PRINT USING “The specific heat ratio is k=kf.H44; K
590 |!
600 * Get Mach number to solve for
610 FOR I = 8 TO 16: PRINT SPACÉ$(79): NEXT 1
620 LOCATE 16: PRINT SPACE$(79): LOCATE 16
630 INPUT "Enter a Mach number to solve for (999 to quit): “, MA
640 IF MA = 999 THEN END
650 IF MA > O AND OPT = 2 THEN GOTO 710
660 IF MA > O AND OPT = 1 THEN GOTO 830
670 LOCATE 9: FOR I = 1 TO 6: PRINT SPACES(79): NEXT I: LOCATE 14
680 PRINT “Valid Mach number range: Ma > O!
690 GOTO 620
700 !
710 '* Solve Rayleigh flow functions for specified k and Ma
720 LOCATE 9: FOR I = 1 TO 6: PRINT SPACE$(79): NEXT I: LOCATE 9
730 PRINT USING " Ma EMBROC OCL MA
740 PRINT USING " P/Pa «BERRO SOOU FNRPPA(K, MA)
750 PRINT USING " T/Ta BERACOCC; PNRTTA(K, MA)
760 PRINT USING " v/va
770 PRINT USING "To/Toa
780 PRINT USING "Po/Poa
790 !
800 ' Loop back for another Mach number
810 GOTO 620
820 !
830 ! Solve Fanno flow functions for specified k and Ma
840 LOCATE 9: FOR I = 1 TO 6: PRINT SPACE$(79): NEXT I: LOCATE 9
ARRROOSCU, FNRVVA(K, MA)
R$HECS CC, PNRIOTOA(K, MA)
«MERECCCCE, ENRPOPOA(K, MA)
MU!
HE dh ap NE st ae
850 PRINT USING “ Ma = &fETTOO; MA
860 PRINT VE(LA-1)/D = 4;
B65 PRINT USING "kh. &44Hº2"c"; ENFLD(K, MA)
870 PRINT " T/TX = tt;
875 PRINT USING FNFTITSTAR(K, MA)
880 PRINT
885 PRINT USING "p.&HHH"2"C"; PNEVUSTAR(K, MA)
890 PRINT " P/PX = H;
895 PRINT USING "k.gAHATTOCE; PNFPPSTAR(K, MA)
900 PRINT " Po/Po,* =;
905 PRINT USING "g.44H40"""!; PNFPOPOSTAR(K, MA)
910 PRINT
920 !
930 ! Loop back for another Mach number
940 GOTO 620
Ai