HP-19C & 29C Solutions Mechanical Engineering 1977 B&W

7/29/2019 HP-19C & 29C Solutions Mechanical Engineering 1977 B&W

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INTRODUCTION

This HP-19C/HP-29C Solutions book was written to help you get the most from your calculator.The programs were chosen to provide useful calculations for many of the common problemsencountered.They will provide you with immediate capabilities in your everyday calculations and you willfind them useful as guides to programming techniques for writing your own customized software.The comments on each program listing describe the approach used to reach the solution andhelp you follow the programmer's logic as you become an expert on your HP calculator.You will find general information on howto key in and run programs under "A Word about ProgramUsage" in the Applications book you received with your calculator.We hope that this Solutions book will be a valuable tool in your work and would appreciate yourcomments about it.

The program material contained herein is supplied without representation or warranty of any kind.Hewlett-Packard Company therefore assumes no responsibility and shall have no liability,consequential or otherwise, of any kind arising from the use of this program material or any partthereof.

HEWLETT-PACKARD COMPANY 1977


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1.

2.

3.

4.

5.

TABLE OF CONTENTS

RPM/TORQUE POWER This program provides interchangeable solutions for RPM/Torque,and power.

CRITICAL SHAFT SPEED This program finds the fundamental critical speed of a rotatingshaft.LINEAR PROGRESSION OF A SLIDER CRANK I This program calculates the displacement, velocity, and accelerationof the slide in a slider crank mechanism.

14

8

SPUR GEAR REDUCTION DRIVE I I . 12This program provides interchangeable solutions for reduction, dis-tance between centers, diametral pitch, and number of pinion teeth.The program also outputs values for the pitch diameters of thepinion and the gear, and the number of gear teeth.BELT LENGTH I This program computes the belt length around an arbitrary se t ofpulleys.

. . 16

6. REVERSIBLE POLYTROPIC PROCESS FOR AN IDEAL GAS 20This program provides interchangeable solutions between pressureratio, volume ratio, temperature ratio, and density ratio.7. ISENTROPIC FLOW FOR AN IDEAL GAS I 23This program replaces isentropic flow tables for a specified8.

specific heat ratio.HEAT TRANSFER THROUGH COMPOSITE CYLINDERS AND WALLS This program calculates the overall heat transfer coefficient fromindividual section conductances and surface coefficients.

I 27

9. BLACKBODY THERMAL RADIATION. I I I I I 31This program calculates the wavelength of maximum emissive power fora given temperature (or vice versa), and the emissive power (total,from A, to A2, or at AI).

10. CONSERVATION OF ENERGY I I I 36This program converts kinetic energy, potential energy, and pressure-volume work to energy, sums all the energy contributions, and convertsthe total to an equivalent velocity, height, pressure, or energy perunit mass.


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RPM/TORQUE/POWER

This program provides an interchangeablesolution for RPM, torque, and power inboth Systeme International (metric) andEnglish units.RPM

SIRPM

English'RPM

Torque nt-m ft-lbPower watts hp

EQUATIONS:RPM x Torque = Power1 hp = 745.7 watts1 ft-lb = 1.356 joules1 RPM = TI/30 radians/sec1 hp = 550 ft-lbsec

EXAMPLE 1:Calculate the torque from an enginedeveloping 11 hp at 6500 RPM. Findthe SI equivalent.EXAMPLE 2:A generator is turning at 1600 RPM witha torque of 20 nt-me If i t is 90%efficient, what is the power inputin both systems?

SOLUTIONS:(1)

(2)

1

GSB46588.88 EHIt

8.88 EHH11.88 GSB58.89 --- Torque, ft-l bR/S12.85 u* Torque, nt-mGSB3

1688. 88 ENH28.88 EHTt8.988.88 GSBS3723.37 *uR/S Power, watts4.99***

Power, hp


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2 User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS1. Key in the program 0 02. Choose sytem of units: 0 0Metric [Jo=J 1.36

or 0 1 IEnglish I GSB II 4 I 5251.41

3. Enter variables (the unknown quantity, x, 0 0must be input as zero) and compute x. I II IRPM RPM I ENTtII ITorque Torque I ENH IOPower Power I GSB II 5 I x

0 04. (Optional) to convert torque or power to 0 1 I

other system ~ O x convertedI II I0 00 00 00 1 I0 0I II I0 0I II I0 00 00 0

L ~ I II J I


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ProgralD Listings 348 RCL381 tL8L3 4982 3 Set up for metric 58 RCL?83 e units 51 x84 Pi 52 R.lS ** RPM8S 53 tL8L186 STO!54 RCL487 ., RCL25588 4 5689 S S7 RCL!18 , 58 x11 7 59 R.,S12 ST05 68 RCL6 *** Torque13 1 6114 , 62 R.lS ** Torque convertedS oj 63 tL8L816 5 64 RCL217 6 65 RCL318 STOG 66 x

19 RTH 67 Ret?28 tLBL4 Set up for English 68 *** Power1 I;S83 units 69 R..'S22 I/X 78 RCL523 STOG 7124 X:Y .,., R/S ** Power converted'c- I/X26 STO'52? -28 .1


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4

CRITICAL SHAFT SPEED

Suppose a rotating shaft is simplysupported at both ends and has a seriesof n weights, WI, . . . , W, attached.Then there are critical speeds at whichthe shaft will become dynamically unstable. This program finds the fundamental critical speed from the formula

f =_127T

where

n9LWy. 1 1 11= cycles/sec

g = Acceleration due to gravityYi = Static deflection of weight Wi

The program is set up to accept thestatic deflections Yi as inputs. Ifthe static deflections are not known,i t calculates Yij' the static deflection of weight 1 due to Wj. Then thetotal deflection of weight i is thesum of the deflections from al l theWj's. That is,

nYi= L Yij.j=lThe individual Yi 's are added to provide the Yi's w h 1 ~ h the program acceptsas inputs. The Yij'S are calculatedas follows:If xi


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EXAMPLE: A 2 inch diameter steel shaftof total length 40 inches has a fly-wheeland a gear located respectively 15 and25 inches from the end. The flywheelweights 60 pounds and the gear 45 pounds.Assume the modulus of elasticity of thesteel is 30 x 10 6 psi. Find the fundamental critical speed of the shaft.

eo lb.- 45 lb.I -IIIII '-I L....-I

in.Ir -25 n.------+I

1f in.

I 40 In . , - - - - -

SOLUTION:5

2.88 EHTt38.+86 EHTt2.88 GS86 d48.88 (;S8168.88 EHTt15.88 (;S8245.88 EHTt25.88 (;S8345.88 EHTt25.88 (;S8268.88 EHTt15.88 GSB3386.888 GS8S44.15

*** f,cyc1es/sec68.88 \;"2648.85*** f, RPM


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6 User InstruetionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITSl. Key in the program DD2. If Yi are known, go to step 10 DD3. If the Yi are not known, input n G JD

Modulus of elasticity E CLJD EMoment of inertia** I I t II I ILength of shaft JI, ~ [ L J n

4. Repeat steps 5-7 for i = l , .. ,n I II I5. Input Wi Wi I t II I Wi

Xi Xi [iJ[LJ Index6. Repeat step 7 for all j 'f i I II I7. Input W CLJD w.OJ '" '"[iJ[LJ; where j 'f i Xj Index or WHOJ DD. Input acceleration of gravity and calculate

critical sneed g I SB II 5 I f{cycles/sec9. For a new case, go to step 2 DD10. If the Yi are known, input length of shaft JI, ~ I 1 Il l. Repeat step 12 for i = 1, . . . , n DD12. Input Wi W I NTt II I W'1 1

Yi Yi ] 'CO WiYi 213. Input acceleration of gravi ty and compute I II I

critical speed g ][IJ f{cycles/se14. For a new case go to step 2 I II I** If I is not known, i t may be calculated d ]ITJ I

from the diameter (solid cylinderical DDshaft only). DD

L ~ I II II I


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ProgralD Listings 781 ,LBU 48 RCL882 CLR(; 49 X1Y? Index 'f n?83 ST08 5/, SB R/S84 x SI 88S x S2 ST0886 6 S3 RCL4 Wi87 x S4 RCL3 Yi88 ST09 GEl5/, SS fLBL489 RJ 56 x y . ~ J .1 118 51.8 n 57 ST+l11 R ~ S 58 LSTX12 ,LBL2 59 x y.2W13 ST05 x 68 ST+2 1 114 xty 61 R/S15 ST04 Wi 62 ,LBL516 8 63 RCL117 5T03 64 x18 RJ. 65 RCL219 xty x \.J.1 1 6628 ,LBL3 xJ J 67 IX21 rsz 68 Pi22 RCL5 6923 X)Y? x'1 78 224 xty 7125 ST06 72 R/S **f26 '1.2 73 'LBLt d27 Xty 74 428 ST07 75 vxi29 '1.2 76 Pi38 + .,...I ( x31 RCL8 78 632 RCL? 79 433 x 8834 " 81 R..5 I.35 x36 -37 x38 RCL639 x **"Printx" may be inserted before "RIS I48 RCL741 RCL842 -43 x44 RCL94S Yij46 ST+347 RC.8

REGISTERS0 Index 1 2 V 3 LV.' , - : v ~ 4 56 Min(x; .x;) 7 Max (x . x; ) 8 5/, 9 GEIR, .0 .1n.2 v .3 " .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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8

LINEAR PROGRESSION OF SLIDER CRANK

This program calculates the displacement, velocity, and acceleration of theslider in a slider crank mechanism,(e.g. the piston wrist-pin in an internal combustion engine) given crank radius, connecting rod length, slider offset,crankshaft speed, and crank position.The maximum and minimum displacementsand the stroke are also calculated.N= Crankshaft speed, RPME =Slider offsetL = Connecting rod lengthR =Crank radius

EQUATIONS:_ TINw - 30

x = R cose + L cos$xmax = (R + L) cos [ s i n ~ l ( R L)]'m1n = (L - R) cos [s;n- 1(L R)J

$ = s;n-1 (E + s;na)_ dx _ R (-sin (e+


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EXAMPLE:Find the displacement, velocity andacceleration of the wrist-pin in theslider of a slider crank mechanismhaving a crank radius of 2.0 inchesand connecting rod length of 7.0 inches,turning at 4800 RPM. Calculate valuesfor

Assume the slider crank mechanism isin-line (E=O). Also find the maximumand minimum displacements and the stroke.

SOLUTION:

4888.88 EHH8.88 EHTt7.88 EHTt2.88 GSB1582.65 ftt8.88 GSB29.88 fttJ?/S9.88 tU

rus5.88 Ut4.88 ftt

GSBJ8.88 tUt;SB4-649781.96 Ut15.88 GSB28.91 U tGSBJ-3J2.28 U tGSB4-614226.44 U*

225.88 t;SB2~ . 4 4 tU

GSB:]~ 6 4 . 2 2 U t

t;SB4J54181.29 U t

9

w rad./secX in.xmax ' in.xmin ' in.6.x, in.v, in./sec.a, / 21n sec.x, in.v, in. /sec.a, i n . / s e c ~

x, in.v, in./sec.a, in./sec. 2


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10 User InstructionsSTEP INSTRUCTIONS INPUT KEYSDATA/UNITS

l . Key in the program DI I2. Input the slider crank parameters N IENTtlDE IENTtID

L IENH II Iand display the crank angular velocity R I GSB II 1 I

3. Input the crank angle and compute the DDslider displacement e I GSB II 2 IOptional: calculate the maximum displace- I II I

ment [MJDOptional: calculate the minimum I II Idisplacement ~ D

Optional: calculate the stroke DD4. Calculate the slider velocity ~ c = J5. Calculate the slider acceleration IGSB II 4 I6. Repeat step 3 for all desired values of e. DD

Steps 4 and 5 may be executed (in order) DDas required. DD

7. For a new case, go to step 2. DDDDI II IDDI II IDDDDDD

L ~ I II II I

OUTPUTDATA/UNITS

NELw

x

xmaxxmint.xva


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ProgralD Listings 1148 PCU81 tLBL1 49 PCL282 ST01 R 58 PCL183 Rl 51 CHS84 ST02 L 52 GT08 Calculate xmin85 P-l- 53 tLBU86 STC3 E 54 PCL687 R-l- 55 PCL588 Pi 56 +89 x 57 PCL118 3 58 CHS -R 8+


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12

SPUR GEAR REDUCTION DRIVE

+~ P i n i O n

For a spur gear meshing with a plnl0n,this program performs an interchangeablesolution among the variables reduction(f), distance between the centers (C.D.),diametral pitch (P), and number of pinion teeth (Ne. Once these four basicvariables ha e been determined, theprogram will also output values for thepitch diameters of the pinion and thegear (Dp and Dg) and the number of gearteeth (Ng).The basic formula used in all solutionsis :

f + 1 = 2P x C.D.Np (1 )The calculations for f,P, and C.D. arestraightforward. The solution for Npis more complicated since i t must be aninteger. Because of this constraint,there may not be a gear-pinion combina-tion that will give exactly the desiredreduction. In this case, the programfinds the closest integer value for Npby the formula

N = INT (2P x C.D. + 0 5)P f + 1

where INT (x) = the integer portion of x.Then a new value for the reduction, f ' ,is found by substituting this N intoequation (1) above. The next s ~ e p is tocompute the number of gear teeth (alsoan integer) byNg = INT (f'Np + 0.5).

Finally the true value of the reductionis found byf = ~Np

This modified value for f is stored inRl and may be recalled by the user ifdesired.REMARKS:The program assumes that the reductionwi 11 be expressed as a decima1 numbergreater than 1. For instance, a reduction of 9:2 should be input as or 4.5.If f


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SOLUTION:1888. 88 ENTt6S8.882.77 t** f design9.88 ENTt8.88 ENTt8.88 (;58138.88 tU : Np(;S824.75 *U:R/S186.88

***IUSB.2S *** OgRCL12.79 u* f

13


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14 User InstructionsINPUT KEYSTEP INSTRUCTIONS DATA/UNITS

l . Key in the program DD2. Enter variables (the unknown quantity, x, DDmust be input as zero) and compute x. DD

Reduction f I ENT t lDCenter distance C.D. I ENTt II IDiametral pitch P I ENT t lDNumber of pinion teeth No I GSB II 1 I

3. Display the following variables: I II IPitch diameter of pinion ~ [ i ]Number of gear teeth I R/S II IPitch diameter of gear C B L U D

4. To display any of the basic variables: DDReduction ~ [ L JCenter distance I RCL II 2 I

CLJDDiametral Ditch GCLJwNumber of pinion teeth ~ ~

5. To change any inputs, go to step 2 DDDDI II IDDI II IDDDDDDI II II II I

OUTPUTDATA/UNITS

x

OnNnOn"

f

C.D.p

No


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ProgralD Listings 1548 581 ,LBL182 ST04 49 +83 IU- 58 IHT Ng84 ST03 51 RCL485 N 52 86 ST02 53 STD1 f

87 ST+2 54 RCL488 N 55 R/S **NP89 STDI 56 ,LBL818 X=8? f=O? 57 RCL411 bTD8 58 RCL112 N 59 113 X=8? N =O? 68 +14 bTD9 P . 61 x15 N 62 RCL216 X=81) OJ17 br08 P=O? 64 STD318 bTD6 65 R/S **p19 ,LBL8

CD=O 66 ,LBL628 RCL2 67 RCL121 RCB 68 122 x 69 +23 RCL4 78 RCL424 71 x25 1 72 RCB26 - 7J 27 STa1 74 ST0228 R,?S ***f 75 229 ,LBL9 76 38 RCL2 77 R/S **C.O.31 RCL3 78 ,LBL232 x 79 RCL433 RCL1 88 RCL334 1 81 35 + 82 ST0536 83 R/S ***0P37 . 84 RCL438 5 85 RCL139 f 86 x48 !NT 87 STD641 STD4 Np 88 R.I'S ***N42 CHS 89 RCL5 943 RCL2 98 RCL144 RCB 91 x45 x 92 STD746 + fiN 93 R/S **0947 . P ** II Pri ntxll may bE inserted before IIRI SII***IIPrintxll may lace IIRISII.

REGISTERS0 1 f 2 C.O. 3 P 4 Nil 5 On6 Ng 7 Og 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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16

BELT LENGTH

This program calculates the belt lengtharound an arbitrary set of pulleys. Itmay also be used to calculate the totallength between any connected se t ofcoordinates. The program assumes thecoordinates of the first pulley to be(0,0).(Xi'Yi,Ri) = x,y coordinates andradius of pully i

Ro = Radius of first pulleyC.D. = Center to center distanceof consecutive pulleys

L = Total length of beltEQUATIONS:

L12 = It.D. 12 2 - (R 2-R 1 )2Arc

y

This program generates accurate resultsfor any convex polygon, i .e. , a line between any two points within the regionbounded by the center-to-center linesegments is entirely contained withinthe region.

Concave

In some cases, there are two physicallypossible directions for the belt to take:

-The program chooses the upper side if themiddle pulley center lies above the lineconnecting the previous and followingpull eys: Case 1-


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The program chooses the lower side if themiddle pulley center lies below the lineconnecting the previous and followingpulleys: Case 2

The program generates inaccurate answersin the second case. Note the figurebounded by the center-to-center linesegments for the second case is notconvex.REMARKS:The calculator is set and left in radiansmode.EXAMPLE 1:Assume three pulleys are positioned asshown below with the following coordinates and radii:

Pulley 1 (0,0,4 inches)Pulley 2 (-8,15,1.5 inches)Pulley 3 (9,16,1 inches)

Find the belt length around the threepulleys. (66.53 inches)

17EXAMPLE 2:Find the length of line connecting thepoints (0,0), (1.5,7), (3.2,-6), (0.0.5),(0,0). (L = 28.01). Let the radius ofeach "pulley" be 0.SOLUTION:1.

2.

4.88 I;S81-8.8e ENTtIS.88 EHTt!.58 (;S829.88 ENIt

16.88 ENT1'1.88 fS828.88 ENTte.88 ENP'4.88 1;.:'B2

GS8366. S3 H.. L

8.88 I;SS!1.58 EHTt7.88 EHTt8.88 GS823.28 PiTt

-6.88 ENIt8.88 GS828.88 PITt8.58 E!


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18 User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITSl . Key in the program 0 02. Enter the initial pulley radius Ro ~ [ I J 1.003. Enter the next pulley coordinates and xi I ENTt IO

radius Yl I ENTt IORi I GSB II 2 I R;

4. Repeat step 3 for al l pulleys. 0 0(the last pulley to be entered is the I II Iorigin pulley). I II I

5. Compute belt length [iJLLJ LI II I0 00 000I II I0 00 1 I000 1 I000 1 I0 0I II I0 00 00 0

L ~ I II II I


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PrograDl Listings 19tLBll 4e PCL4

82 PAD 49 -83 CLPG 58 xt'r'84 STOt Ro 5185 f 52 TP.N-I -13.86 STC8 Set flag 53 ST07 ex87 R. 'S 54 +88 .t.lBL2 5S PCL!89 fT04 x. /) Arc length18 "f v 57 ST+8 ii . . ; i11 RCD S8 PCL412 xtV 59 STOl17 STIJ3 68 R'", .'14 - 61 tLBL815 xty -16 PCl2 1 Y0.'17 xt'r' 64 Resolves to less18 ST02 6'5 than 2 radians19 - 66 x28 ABSI.21 gz C.D.e 68 pel? ex22 xty 69 -23 >O8? RTN24 'bTIJO 71 tLBL325 2 72 pel626 Pi 7J RelS27 x 74 GSB828 + 75 ReL!29 tlBL8 76 .x.'38 DSZ Test flag PCl8I31 GT09 78 +

ST05 7C1 DE;... I" Restore II nonna1 I mod33 STaG 88 p... :, ** L34 tLBl93S peL636 xtV3' ST0638 GSB839 xtr C.D. 248 peL1 ** IIPrintx ll may inserted before IIR SII.41 RCL442 -43 g244 -45 rL'... ,\46 ST+8 Li_l'i47 RCL1

REGISTERS0 Flag 1 Ri-l 2 Xi 3 Yi 4 Ri 5 e"6 ei 7 8 length 9 .0 .1ex E.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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20

REVERSIBLE POLYTROPICPROCESS FOR AN IDEAL GAS

This program may be used to solve inter-changeably between pressure ratio, volume ratio, temperature ratio, and densityratio for polytropic processes involvingideal gases. Polytropic processes aredefined by the relationPVn = C

which is shown graphically in Figure 1.

IogP ~ - . - "' - - - - - - - - - - - - - -109 V

Figure 1.

Isentropic processes are special casesof polytropic processes. For isentropicprocesses, k, the specific heat ratio,is equal to n.EQUATIONS:

~ = (V2)-n = (T2) 1 =(P2) nPI VI TI PIwhere

P2/P I is the final pressure dividedby the inital pressure;V2/V I is the final volume divided

by the inital volume;T2/TI is the final temperaturedivided by the initialtemperature;P2/PI is the final density dividedby the initial density.

EXAMPLE: A compressor has a compressionratio of 8.5 (V I/V2). The polytropicconstant is 1.43. If inlet air is at300K, what is outlet temperature? Whatis the pressure in atmospheres if theinlet pressure is one atmosphere?SOLUTION:

1.4J &SBl8.50 1/,1(

&SB3PPST (for 29C manuallyreview the stack)8.58 T pdpI8.12 Z V2/V I2.51 r T2/TI21.33 ,I( P2/P IpeL8388.88 )(

752.96 Ut Outlet temp. (K)peLS1. Be .x:

21.33 u* Pressure (atm)


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21User Instructions

STEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program DD2. Enter polytropic constant n ~ [ I J n3. Enter one of the following: DD

Pressure ratio PdP l IGs!J[IJ PdP lVolume ratio VdV l [GSB II 3 I PdP1Temperature ratio TdT1 I GSB ICU PdP1Density rati 0 pdp1 I GSB Ie P2/P1

4. The four parameters are now in the stack I II Ior may be recalled from storage: DDPressure ratio (x-register) I RCL Il s i PdP1Temperature ratio (y-register) ~ [ i ] T2/T1Volume ratio (z-register) ~ [ L J VdV1Denisty ratio (t-register) ~ W pdp1

I II IDDD [ ~DDD C ]DD --C J [ ~ ---[ = ~ [ ~ - ~ - - - - - - -

L ~ [ ~ = - ~ J ---- ---- -- - ---

[--] [----]-- --- - - - - - - - - - - - -[ - ~ [---]-- ----- ---- - -- - -f - - - - ~ ~ - -

[ - ~ [---l----- - - - --- -- - -------

-- - - - - [ ~ ~ = = J [ ~ ~ ~ ] -- - - --r----l 1----1


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22 PrograDl Listings61 tLBLl n62 STD283 184 -8S STD386 RCL2e7 5H388 RIS89 fLBL2 P2/ Pl18 111 bTDB12 fLBL:! V2/V l13 RCL214 CHS1S bToe T2/Tl16 tLBL417 RCL318 11K19 bToe28 tLBLS P2/P l21 RCL222 .LBLe23 rx24 ST05 P2/Pl25 RCL226 II)!;27 '(I( Pz/Pl28 STOG29 EHTt V2/V l38 11>:31 STD732 RCLS'J3 RCD34 y'X T IT 1JS STa836 RCL5'37 RIS ***P2 T2 V2 P2

P 7 ; T ~ - ' V1 , ~

*** IIPrint Stack ll may be insertedbefore IIR/S IIREGISTERS

0 1 2 n 3 (n-l)/n 4 5 P?/P,6 Pz/Pl 7 Vz/Vl 8 T IT 1 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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23

ISENTROPIC FLOW FOR IDEAL GASES

This program can be used to replaceflow tables for a specified specificheat ratio, k.IQYATIONS:

2T/To= 2 + (k-l) M2P/P o= (T/To)k/(k-l)

whereM the mach number;T/To the ratio of flow temperature

T to static or zero velocitytemperature To;P/P o the ratio of flow pressureP to static pressure Po;p/po the ratio of flow density p

to static density po;A/A* b' and A/A* are thesu supratios of flow area A to the

throat area A* in convergingdiverging passages. A/A*subrefers to subsonic flow whileA/A*sup refers to supersonicflow.

M2 is determined using- Newton's method.The initial guess used is as followswith a positive exponent for supersonicflow:M02= (/ Frac(A/A*) + A/A*):3REMARKS:After an input of A/A* the programbegins to iterate to find M2 for futureuse. This iteration will normally takeless than one minute, but may takelonger on occasion and for extremevalues of k (1.4 is optimum) may failto converge at all.A/A* values of 1.00 are illegal inputs.M = 1 in this case.EXAMPLE 1:A pilot is flying at mach 0.93 andreads an air temperature of 15 degreesCelsius (288 K) on a thermometer thatreads stagnation temperature To. Whatis the true temperature assuming thatk = 1.38?If the pilot reads a stagnationpressure Po of 28 inches of mercury,what is the true air pressure?EXAMPLE 2:A converging, diverging passage hassupersonic flow in the diverging section.At an area ratio A/A* of 1.60, what arethe isentropic flow ratios for temperature, pressure and density? What isthe mach number? k = 1.74.


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24SOLUTION:.

1. 38 (;S818.93 X2

ST01 M2(;S831.88 f** A/A*(;S89RCL8 T/To8.86 u*288.1313 x247.35 *** T( OK)RCL58.58 *** P/Po28.88 x16.11 u* P (in. Hg)

2.1.74 (;5811.613 (;5822.11 t** M

RJ.8.27 t** p/po

RJ.18 tU P/PoRJ

13.38 *** T/To


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25User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program DD2. Enter specific heat ratio of gas k ~ ~ k + 13. If Mis known, then go to step 7. DD4. Enter area ratio. Use positive values ! A/A* ~ ~ M

for supersonic area ratios and negative I II Ivalues for subsonic area ratios DD

5. All four parameters are now in the stack I II Ior may be recalled from storage: I " IMach number (x-register) ~ [ J JI f II IX I MDensity ratio (y-register) ~ O J p/poPressure ratio (z-register) ~ O J P/PoTemperature ratio (t-register) ~ I T J T/To

6. For a new case, go to step 2 I II I7. Enter the Mach number M [ U [ 2 ]

~ I 1 I8. Calculate the area ratio G;JGJ A/A*~ O J M9. To detennine the remaininCl p a r a m e t e r ~ ao DD

to step 5. I " IDD - ~I II IDDDDDD

L ~ I rI II I


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26 PrograDl Listings91 ,LBL1 k 48 IIBS92 STD4 49 EEX83 STO? 58 CHS04 1 51 485 ST+? 52 XiY? C h a n g e ~ . 01 %?86 - 53 (;TD88? ST03 54 (;TD988 Sr:-7 55 tL8L389 2 56 RCLl18 ST+7 5? RCLJ11 + 58 x12 ST02 59 213 R/S 68 +14 ,LBL2 A/A* 61 RCL2 [ ]15 EHTt 6216 IIBS 63 ST0817 - 64 RCL?18 LSTX A/A* 1 65 'IX19 STD6 66 RCL128 EHTt 67 iX21 FRC 68 ** A/A*22 IX 69 RTH Calculate T/To23 + 78 ,L8L924 X:Y 71 225 3 72 RCLl26 x 73 RCL327 'IX ?4 x28 ST01 M ; 75 229 ,LBL8 76 +38 RCL6 7731 (;583 A/A* 78 ST0832 79 RCL833 1 88 RCL434 - 81 RCL335 . 82 -36 5 83 yx P/Po37 RCL8 84 ST0538 85 RCL539 . 86 RCL448 5 87 UX41 RCL1 88 p/po42 89 ST0643 - 98 RCLl44 91 IX t1 T45 ST+1 92 R/S *** ML ' ~ , --'Po Po To46 RCL1 ** IIPrintx ll may b inserted before IIR47 *** II Pri nt Stack ll I ay be inserted befo Nil.e IIRISI

REGISTERS0 1 M2 2 k + 1 3 k - 1 4 k 5 P/Po6 p/po 7 ( k+ 1) I (k 1)2 8 Used,T/Tn 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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27

HEAT TRANSFER THROUGHCm1POS ITE CYLI NDERS AND WALLS

F . . . . l . ~ t u b e

This program can be used to calculatethe overall heat transfer coefficientfor composite tubes and walls fromindividual section conductances andsurface coefficients.Equations:The overall heat transfer coefficientU is defined by:

q/L = U!lTor

q/A = U i fwhere is the total temperaturedifference (T 2 - T1 ), q/L is the heattransfer per unit length of pipe, andq/A is the heat transfer per unit areaof wall.For cylinders

2nU= - - - - - - - - - - ~ ~ - - - - - - - - - - - - - - - -

For wallsU= ---------'--------------

_1_ + ~ + ~ + . . . + _1_hI kl k2 hn

where

T,

h is the convective surface coeffi-cient:D is the outside diameter of thenannulus:k is the conductive coefficient;x is the thickness of a wall section.

_,+-1-"-1q

F l g u r e 2 . ~ . "

Remarks:These equations are for steady stateheat transfer through materials withconstant properties in all directions.Inputs must start with the inside convective coefficient and work out in thecase of composite cylinders.Zero is an invalid input for D, k, andh.Dimensional consistency must be maintained.


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28Example 1:A steel pipe with an inside diameter of4 inches and a thickness of 0.5 incheshas a conductivity of 25 Btu/ft-hr-oF.Two inches of asbestos (k=O.l Btu/hr-ftOF) enclose the pipe bringing the totaldiameter to 9 inches. If the insideconvective coefficient is 1000 Btu/hrft2-0F and the outside coefficient is5 Btu/hr-ft2-0F, what is the overallheat transfer coefficient? What is theheat loss for 100 feet of pipe if is115F?Example 2:A wall is composed of 1 foot of brick(k=0.4 Btu/hr-ft-OF), and 1 inch ofwood (k=0.12 Btu/hr-ft-OF). The convective coefficient on one side is23 Btu/hr-ft2-oF. The convectivecoefficient of the other side is5 Btu/hr-ft2-oF. What is the overallcoefficient? What is the heat flux ifthe temperature difference is 70F?Solutions:1.

CLRb1888.88 EHTt4.88 EHTt12.88 (convert units to feet)

b58125.88 EHTt5.88 EHTt12.88 G5828.18 EHTt9.88 EHTt12.88 .G5825.88 EHTt9.88 EHTt12.88 .

G58t8.98 *** Btu/hr-ft-OF115.88 x112.44 ***188.88 x11244.28 *** Btu/hr

2.CLRb

23.88 G5838.48 EHTt1.88 G5848.12 EHTt1.88 EHTt12.88 -G5845.88 G5838.29 *** Btu/ftLhr-oF78.88 x28.36 *** Btu/ft2-hr


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STEP

1233a

3b

3c

456a

6b

6c

7

User Instructions 29INPUT KEYS OUTPUTINSTRUCTIONS DATA/UNITS DATA/UNITS

Key in the program DDInitialize G J ~For cylinders: DDEnter the convective coefficient and the h I NIt II I l/hD or Udiameter (start from the inside and D IGSB II 1 I if outsideDD surfacework out)Enter the conductive coefficient k IENTt II 1and the diameter D IGSB II 2 I

Repeat step 3b or go to step 3a for DDoutside surface I II 1

For a new case, go to step 2 DDFor walls: DDEnter the convective coefficient (inside [ i l lJ[LJ l/h or U ifh final sur-

II faceor outside surface) I IEnter the conductive coefficient and the k IENIt ID

thickness x ~ 1 4 IRepeat step 6b or go to step 6a for final DD

surface D I IFor a new case, go to step 2 DD

I II IDDI II 1C ~ DDDDDI II I1 11 1


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30 ProgralD Listings81 tLBL182 STO? D h83 x84 Pi85 x86 fLBL387 1/,1(88 ST+889 ReL818 X=V? initial or final11 R/S entry?12 1/X13 P/S **u14 tLBL215 ReL716 Xti'17 STar18 D. 1/0.1 - 119 LN28 Xty k21 222 x2? p .24 x2526 ST-827 P/S28 fL8L429 ,l(tV38 x/k31 ST+832 R/S

**"Printx" may be inserted before"R/S".

REGISTERS0 1 2 3 4 56 7 D. 1 8 l/U 9 .0 .11 -.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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31

BLACK BODY T H E R ~ 1 A L RADIATION

Bodies with finite temperatures emitthermal radiation. The higher theabsolute temperature, the more thermalradiation emitted. Bodies which emitthe maximum possible amount of energyat every wavelength fo r a specifiedtemperature are said to be black bodies.While black bodies do not actually existin nature, many surfaces may be assumedto be black for engineering considera-tions.

Black bodymonochromaticemissive power

A. max ,

2345678Wavelength, micronsFlIP. 1.

Figure 1 is a representation of blackbody thermal emission as a function ofwavelength. Note that as temperatureincreases the area under the curves(total emissive power Eb(O-oo)) increases.Also note that the wavelengtn ofmaximum emissive power Amax shifts tothe left as temperature increases.This program can be used to calculatethe wavelength of maximum emissivepower for a given temperature, thetemperature corresponding to a parti-cular wavelength of maximum emissivepower, the total emissive power for allwavelengths and the emissive power at

a particular wavelength. It can alsobe used to calculate the emissivepower from zero to an arbitrary wavelength, the emissive power between twowavelengths or the total emissivepower.EQUATIONS:

2nCl t -T kc 2 e ~ ( t ) 3 +k=l L+ 3T + 6 1- T )2 + 6 ( T ) 3JA kc 2 i \kc2 kC 2

whereAmax is the wavelength of maximumemissivity in microns;T is the absolute temperature in

oR or K;


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32Eb (0- 00) is the tota1 emi ss i ve power

in Btu/hr-ft2 or Watts/cm 2;EbA is the emissive power at A in

B t u / h r - f t 2 - ~ m or W a t t s / c m 2 - ~ m ;Eb(O-A) is the emissive power forwavelengths less than A in

Btu/hr-ft2 or Watts/cm 2;Eb(AI- A2) is the emissive power forwavelengths between Al and A2

in Btu/hr-ft2 or Watts/cm 2.c1 = 1.8887982 X 107 B t u - ~ m 4 / h r - f t 2= 5.9544 x 10 3 W ~ m 4 / c m 2c2 = 2.58984 X 104 ~ m - o R =

1.4388 x 104 ~ m - KC3 = 5.216 X 10 3 ~ m - o R =2.8978 x 10 3 ~ m - K

a = 1.71312 X 10-9 Btu/hr-ft2-oR4=5.6693 x 10-12 W/cm2-K4

aexp = 1.731 X 10-9 Btu/hr-ft2-0R4= 5.729 x 10-12 W/cm2-K4REMARKS:A minute or more may be required to

~ b t a i n E b ( O - ~ ) or Eb(A -A ) since the1ntegrat10n 1S numerical. 2Sources differ on values for constants.This could yield small discrepanciesbetween published tables and outputs.REFERENCE:Robert Siegel and John R. Howell,Thermal Radiation Heat Transfer, Vol. 1,National Aeronautics and Space Administration, 1968.

EXAMPLE 1:What percentage of the radiant outputof a lamp is in the visible range (0.4to 0.7 microns) if the filament of thelamp is assumed to be a black body at2400 K?EXAMPLE 2:If the human eye was designed to workmost efficiently in sunlight and thevisible spectrum runs from about 0.4 to0.7 microns, what is the sun's temperature in degrees Rankine? Assume thatthe sun is a black body. Using thetemperature calculated, find the frac-tion of the sun's total emissive powerwhich falls in the visible range. Findthe percentage of the sun's radiationwhich has a wavelength less than 0.4microns.SOLUTIONS:1.

5954.48 STO!14388. 88 ST022897.88 STO:]5.669.'3-12 ST042488.88 forOS8.48 STOGB.?8ST07

I;SB44.97 ***t;SB2

J . I . constants

Eb (0 to 00)2.64 *Jlt (%)


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2. 18887982. &B sro125898.48 ST025216.88 5T03

1. 71312-89 STa48.48 HTt8.7B +2.88

}En91ish constants

8.55 u* mean valueReL31/X

9483.64 U* T, (OR)STa'58.48 STG68.785r07(;SB4

4678'556.56 H* Eb ( .4 to .7 )(;SB2138'57578.83 Ut Eb (0 to 00)

188.88 )(33.7& H* (%)8.4& STa6

(;5811168686.94 u* Eb (0 to .4 )(;5B2

188.B& x8.43 *** (%)

33


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34 User InstructionsINSTRUCTIONS INPUT KEYS OUTPUTSTEP DATA/UNITS DATA/UNITS

1 Key in the program DO2 Store constants: DO2a Engl ish units - 18887982 ] [ C J

(Btu. ft. OR) 25898.4 ] o = J5216 ISTO II 3 I

or .171312 x 10-8 ] [ L J2b SI units - 5954.4 ISTO I C L J

(w. cm. OK) 14388 ISTO " I2897.8 5!Q][LJ5.6693 x 10- 12 ISTO II 4 I3 For experimental Stefan-Boltzmann constant 1.0105 5 ! Q ] ~

instead of theoretical constant CLJD4 To calculate Amax = f(T) llihJ[LJ

T(Absol.) I t II I A m a x ( ~ m )5 To calculate T = f(A) for which A is ~ D

maximum A ( ~ m ) 0 1 I T (OAbsol.)6 To calculate total emissive power T* ][I]

~ I 2 I EbiO to 00)7 To calculate emissive power at A T* ][I]

A ISTO II 6 I~ [ L J Eb{A )8 To calculate emissive power from a to Al T* ISTO II 5 IAl 5!Q][LJ[ iJ [L] Eh(O to ). )"9 To calculate emissive power from Al to A2 T* ~ [ C J

Al ~ u l 6 I*any value of T stored previously is still A2 I TO II 7 lstored and need not be input again


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ProgralD Listings 3581 ,LBLl 58 XY,? 001%82 (;SB9 ." (;T08 yes, increment k83 R-S ***E AI) S2 RCL984 tLBL9 b(O to 53 285 8 54 x86 ST09 S5 Pi87 STOB 56 x88 tLBL8 57 RCLJ89 RCL2 58 x18 RCL5 59 RTN Eb(O to A)11 68 tLBL212 ST-8 -k c2/T 61 RCL513 3 62 414 RCL8 63 rx15 64 PCL416 peL6 65 x17 X'Z 66 **E18 67 tLBD b(O to 00)19 LSTX 1. 2 68 RCLt28 RCL6 69 221 x 78 x22 1/.'1 71 Pi23 - 72 x24 6 73 RCL62S RCL6 74 526 7S yx27 ReL8 7628 g2 77 RCL229 78 RCL638 - 7931 6 88 RCL532 RCLe 81J3 3

82 eX34 vx 83 184 -3S 8536 +37 PCL8 86 P'S **E87 tLBL4 bA38 PCLf39 88 (;SB948 eX 89 ST.8 Eb(O to AI)41 .x: 98 RCL742 ReL8 91 ST06 1.243 92 (;S8993 RC.e44 ST+9 944S RCL9 -46 95 R.-S **E47 EEX b(AI to 1.2).,48 CHS49 5 i*"*IPrintx" may be i serted before "R/S"

REGISTERS0 1 c1 2 C2 3 c3 4 a 5 T6 A 7 A? 8 - KC 2 /T 9 sum .0 used .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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36

CONSERVATION OF ENERGY

This program converts kinetic energy,potential energy, and pressure-volumework to energy. Energy is stored asa running total which may at any timebe converted to an equivalent velocity,height, pressure, or energy per unitmass. The program is useful in fluidflow problems where velocity, elevationand pressure change along the path offlow.EQUATIONS:

V212 + gZl + f.!. + J,L =p mV2

2 + gz + fz. +2 2 P mwhere:

v is the fluid velocity;z is the height above a referencedatum;P is the pressure;E is an energy term which couldrepresent inputs of work orfriction loses (negative value);g is the acceleration of gravity;p is the fluid density;m is the mass flow rate (assumed tobe unity);

subscripts 1 and 2 refer to up-stream and downstream valuesrespectively.NOTES:Downstream values should be input asnegatives. However, when an outputis called for, the calculator displaysthe relative value with no regard toupstream or downstream location.

An error will result when the totalenergy sum stored in register 8 isnegative and an attempt is made tocalculate velocity.EXAMPLE 1:A water tower is 100 feet high. Whatis the zero flow rate pressure at thebase? The density of water is62.4 lb/ft 3 If water is flowing out of the tower ata velocity of 10 ft/sec, what is thestatic pressure?What is the maximum frictionless flowvelocity which could be achieved withthe 100 foot tower?If 10000 pounds of water are pumped tothe top of the tower every hour, at avelocity of 20 ft/sec, with a frictionalpressure drop of 2 psi, how much poweris needed at the pump?EXAMPLE 2:An incompressible fluid (p = 735 kg/m )flows through the converging passageof Figure 1. At point 1 the velocityis 3 m/s and at point 2 the velocityis 15 m/s. The elevation differencebetween points 1 and 2 is 3.7 meters.Assuming frictionless flow, what is thestatic pressure difference betweenpoints 1 and 2?

3 mI.-: _ _ ------r- - -3 .7m___ __._- -_-_- ~ 1 5 - ~ .F . . . . 1.


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EXAMPLE 3:A reservoir's level is 25 meters abovethe discharge pond. Assuming 85% powergeneration efficiency, how much powercan be generated with a flow rate of20 m3 /s?

p = 1000 kg/m 3SOLUTIONS:

(1) 2S8JJ.48? STD532.17 STOf4632.48 ST0762.48 GS81188.88 GSBJ

GSB843.33 *** (psig)-18.88 GSB2GS88

42.66 *** (psig)62.48 GS81188.88 GS83GS86 (ft/sec)88.21 ***62.48 GSB!28.88 r,SB22.88 (;S84

188.8e GSB3GSB9

8.14 H* (BTU/1 b)18888.88 x1424.29 H:+: (BTU/hr)

(2 )

(3)

1.88 ST05ST079.8866S ST06735.88 GSBl

3.88 GSB23.78 f;SB3-15.88 f;SB2GSB8

-52'18.82 tt_: (Nt/m 2 )

1888.88 GSBI2S.88 GSB3GSB9

245.17 H* (joule/kg)8.85 x288.39 *** (joule/kg)28.88 E N T ~1888.88 }' (kg/s)

4167826.25 .-:* (watts)

37


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38 User InstructionsSTEP INSTRUCTIONS INPUTDATA/UNITS KEYS

OUTPUTDATA/UNITS

1. Key in the program 002. For English units: 25033.407 [ i l l ] [U32.17 [ i l l ] [L ]

4632.48 [ i l l ] [LJ 1.00For S. I. units: 9.80665 I STO II 6 I

3. Enter fluid density p GSiJ[L] 0.004 Enter tha followinJt (neCiative values are I 10

r l n w n ~ t r e ; : l m vO\lues) I II IFluid velocity v ~ ~Height from reference datum z I GSB II 3 IPressure P @Jo=JEnergy input E @JeD

5. Repeat step 4 for al l input values 006. Calculate the unknown: I II I

Fluid velocity ~ [ U vHeight from reference datum ~ 1 7 I zPressure ~ [ U PEnergy [ii][U E

7. For another case, go to step 3, or clear 00reCiister 8 and ClO to step 4 0 [ i l l ] I 8 I

00I II I00000001 II II I


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Program Listings 3981 ,L8U82 ST04 P83 884 STa8 Clear L E8S p/s86 *L8L287 EHTt88 ABS89 ::{18 211 + y2/z-12 Gras13 tL8U14 RCL6IS )(16 GTOS gz17 tL8L418 RCL?19 x28 RCL4 Pip2122 tL8LS E23 ST+824 R.'S25 tL8L626 RCL827 228 x29 .rx38 R/S *** y31 tL8L?32 RCLS33 RCL6343S R.'S *** z36 tL8L83? RCLS38 RCL?3948 RCL441 x42 R..S *** P43 tLBL944 PCL8 *** IIPrintx ll b2 inserted.45 RCLS may4647 R....S *** E

,REGISTERS

0 1 2 3 4 P 5 Used6 g 7 Used 8 E 9 .0 .1L.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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In the Hewlett-Packard tradition of supporting HP programmable calculators with quality software, the followingtitles have been carefully selected to offer useful solutions to many of the most often encountered problems in yourfield of interest. These ready-made programs are provided with convenient instructions that will allow flexibility ofuse and efficient operation. We hope that these Solutions books will save your valuable time. They provide you with atool that will multiply the power of your HP-19C or HP-29C many times over in the months or years ahead.

Mathematics SolutionsStatistics SolutionsFinancial SolutionsElectrical Engineering SolutionsSurveying Solutions

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HP-19C & 29C Solutions Mechanical Engineering 1977 B&W