hp-19c & 29c solutions student engineering 1977 b&w

7/29/2019 HP-19C & 29C Solutions Student 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 problemsencou ntered.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 how to 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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TABLE OF CONTENTS

RESISTIVE/REACTIVE CIRCUIT CALCULATIONSPerforms resonance calculations. I I I I I I I IIMPEDANCE OF LADDER NETWORK I I I I I I I I Computes input impedance of arbitrary ladder network.STANDARD RESISTANCE VALUES. I I I I I I I I I IComputes nearest standard value.

1

58

EXPONENTIAL GROWTH OR DECAY I I I I I I I I . . 11Performs calculations for generalized growth and decay phenomena.EQUATIONS OF MOTION . . . . . .

Calculates an interchangeable solution for object moving underconstant acceleration.KINETIC ENERGY . . . . . . . . I I I I ICalculates an interchangeable solution for weight (mass),velocity,and kinetic energy.RPM/TORQUE/POWER. . . . . . . I I I I I I IThis program provides interchangeable solutions for RPM, torque, andpower.BLACKBODY THERMAL RADIATION . . I I I I I I I I

This program calculates the wavelength of maximum emissive power fora given temperature (or vice versa), and the emissive power (total,from Al to A2, or at A).CONSERVATION OF ENERGY. . I I I I I I I I I I This 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.MOHR CIRCLE FOR STRESS I I I I I I I I I I This program calculates the principal stresses and their orientationgiven the state of stress bn an element, and the maximum shear stressand its orientation.POLYNOMIAL EVALUATION (REAL OR COMPLEX) . I I I I I I I Evaluates polynomials with real or complex coefficients.

S I N E ~ COSINE AND EXPONENTIAL INTEGRALS. I I I I I I I IEvaluates the sine, cosine and exponential integral by means of aseries approximation.

14

17

20

23

28

32

3639


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RESISTIVE/REACTIVE CIRCUITCALCULATIONS

This program performs resonance calcu-lations fo r R-L-C circuits, calculatesthe reactance of inductive and capaci-tive branches, the equivalent value ofseries capacitors or parallel resistorsand inductors,and performs power cal-culations for resistive branches usingstraightforward manipulations of thefollowing equations:

2nILCx = 1C 2nfC

2nfLP

wherefr resonant frequency in hertzL inductance in henrysC = capacitance in faradsXc capacitive reactance inXL inductive reactance inP power in wattsI current in ampsR resistance inE voltage in volts

AI ,A 2 = the values of two parallelresistors in ohms, two parallelinductors in henrys, or twoseries capacitors in faradsA3 the resultant, equivalent resis-tance in ohms, inductance inhenrys, or capacitance in farads

NOTE: Given a resistance or capacitance,AI, the value of the circuitelement required to produce adesired resultant resistance orcapacitance may be calculated byentering Al as a negative value.EXAMPLES:

1. C = .01 jJ F, L = l60jJ h.Calculate fr2. L 2.5h, fr = 60HzCalculate C and XL at fr3. E 345v, R = 1 . 2 5 M ~Calculate P and I4. RI= l 2 0 ~ , R2 = 2 4 0 ~a. Find the equivalentresistance of these tworesistors in parallel,R 3.

b. Parallel R3 with 5 0 ~ .c. Find the resistance re -quired for a resultantresistance of 2 5 ~ .


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2

SOLUTION: ENG4168.-86 ENTt8.81-86 GSBI

125.82+8? tH f r68.8888 E N T ~2.5888 (;882

2.8145-86 t t* C68.8888 Etrrt2.5888 G5B4

942.48+88 t t t XL345.8888 E H T ~1.25+86 r,SB5

95.228-83 t t t P1.25+86 (;587276.88-86 U t I128. 88lJ8 EH'ff

; 4 ~ ~ ~ + o ~ (;889 R3t;'. ~ ~ ~ , ~ t : ' tIN58.8888 (;589

39.769+88 tUCHS25.8888 (;589

133.33+88 ***

4b4c


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3

STEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITSl . Key in the program and choose an appropriate C J C Jdisplay format C J c::=J2. Calculate f L [ENT] [ IC CGSRJ CLJ f3 r.ill rlJ lntp I nr C fr ITNfJ [_JC or L L ( l ~ Lf_J L or C

REACTANCE CALCULATIONS f_=:J L ~4 r""lrll""t". Y. f [ENT] [= Jl.. C [GSi3 -] a- - j XI"5 Calculate X f [EIT] L - _ ~ - = : JL L [-GSB] LU XI

POWER CALCULATIONS C=lc=J6a Given E' Calculate P E [ENi-l C JR [-(iSS] r-:-_J P6b Given I: Calculate P I [00_] L-:=J

I? [GSil [--5-J pGiven P,R: LJ [----]7a Calculate I P WiiJ [= :JR l!?Bl IT] I7b Calculate E P [ENT ] [ - ~ = . JR [J;SB] L a . ~ EPARALLEL RESISTANCE/SERIES CAPACITANCE L--] c::=J8. Given two circuit elements, calculate resultant Al [[NT I - ~A') rnKJ L IJ A39. Given one circuit element and the desired A, I T B ~ J n:NT I

resultant value calculate the value of the A... lGSsJ CLJ A"circuit element required. '" [ J L ~CJC.=JC J [ ][-=::J c=Jl= ] [ ][ ~ C Jc = JC Jc= JC JCJCJCJc:=JCJCJCJCJCJCJ


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4 PrograUl Listings81 t.LBLl82 )(83 rx84 GTOB85 tLBL286 x:(87 GSB888 X'i89 x:y1811 IUS L or C12 tLBL313 x14 GroBIS tLBL416 )(17 GSB818 1/X

19 XL28 tLBLS21 1/.'1,22 tLBL623 x.... ,24 X'i25 x P26 R.'S27 tLBL?28 1/X29 tLBLS38 x31 rx I or E32 R/S'33 tLBL834 23S x36 Pi7., x.II38 1.'X f or Xc39 RTH r48 tLBL941 x42 fHTt4:? fHTt44 LSTX4546 LSrX47 + A3 or A24849 R/S

REGISTERS0 1 2 3 4 56 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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IMPEDANCE OF ALADDER NETWORK

This program computes the input impedance of an arbitrary ladder network.Elements are added one at a time starting from the right. The first elementmust be in parallel. --EXAMPLE:f = 4 MHz TTT2.56/lH ~

5

Suppose we have a network whose inputadmittance is Yin' Adding a shunt R,L, or C, the input admittance becomes z ---. =.: 796 pF = ~ 2400 pF son

Ynew =

Ym +O)Y+(0 __1_)w Lp

Adding a series R, L, or C, we have

( ~ m +(Rs+jO)Y 1

( 1 )-1Ynew = Ym + (0 + w Ls )~ +(O- j _ IYUI w Cs

The program converts this admittanceto an impedance for display.NOTE: An erroneous entry may becorrected by entering thenegative of the incorrectvalue.

-SOLUTION:

FIX24.+86 :;8B158.ee GSB? IZinl58.ee tUB.ee Ut LZin2488.-12 f S P ~15.74 t.u IZi n Ix:r

"'of t " i : ~ : LZin,- tt>2.'56-e6 r;SB2GSB:: IZi n I9.f5 *:;'-1:X::""84.28 tH LZin796.-12 ;885 IZ in I97,69 Ut.. J"P IP QOt _. L. H : ~ : [Zin


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6

STEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS1 Kev in the oroaram and choose an aoorooriate c=Jc=J

rlic::nblv fnrm",t ~ c = Jv [ J C ~ J2 Initialize f ~ C O wc=J c=J3. Input a parallel element: c=J c=J

Rn []SJ [IJ IZinlIx++yl c=J LZinCo CGSBJ C:S:::] IZinl[j


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PrograUl Listings 781 *LBU f 58 GSB8 Convert Yin-+Zint32 2 51 RJ.[l7 . s ' " 52 "CLf84 p + Add admittancesI ,-',-'85 x Clear flag ..." or impedances;86 eLKG w


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8

STANDARD RESISTANCE VALUES*

For a given tolerance, a "step size"is computed which is used to determinetwo values, one below and one above thenon-standard resistance. These areconverted by a subroutine to standardvalues, and the geometric mean of thelatter is calculated. If the givennon-standard value is below the meanthen the lower standard value isselected; otherwise the larger value isselected.NOTE: Incorrect results will be obtained for tolerances other than5%, 10%, or 20%.REFERENCE: International Telephoneand Telegraph Corp.Reference Data for RadioEngineers, fourth edition,p. 78.EXAMPLES: Find the closest standardvalues for the following:

Rl = 432S"lR2 = 114 KS"lR3 = 3.5 MS"l

* Adapted from HP-65 Users Libraryprogram #00915A by Jacob Jacobs.

SOLUTION:ENG.:: . 0N1E' G S E ~ 5%

4]2.888B G S E . ~4]1]. f'9+8fl l : ~ ' R

!14.+8] GEE::'1l8.8e+8] ~ : l : : ;

J. :+@6 GEE:? J. 6888+86 * . , : ~ R

&]2.80130 GEE::'470.88+&[1 ~ : ' '

:7.5+06 GSF2

20. a880 GEL472.@1388 GEE2

478. 88+1313114. +0'! GEE::'

1ee. f'8+8J :f:: ,f

20%

R\


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9

STEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS1 Kev in the oroaram and choose an aoorooriate CJCJ

r l i ~ n l ~ v i ' n \ " m ~ + c=Jc=J.., [ lCJ2 Enter to1prance (5 10 or 20) T% [ ]S[] CIJL ~ [ _ J3. Enter non-standard resistor size and compute R,st l_GSBJ 1T1--- R ~ s tnearest standard value C - = ~ [ - ~

[ - = - ~ I . ~4. Chanae tolerance at any time by Qoing to step 2 [_- ] [ l[--] - - ~---- - - - -L ~ [ _ - ~L __ L=-]

L=-J l-='- ]l----l [- J[-- ] [---]I -][ - - j1 __ ] - = - ~ __[ - ~ _ : J l ~ ~

L - ~ L_-lL - l L=][----] [----JL--=-__ l [ lL ____ J [ -=.J[- - JL_-]

[ ~ _ ~ J L _ ~[ = ~ l L - - ' ~l---::-=J l ~

L _ ~ - l L_.=J[ ~ ~ [ - J

C = - J L ~ ]l ___ l L_ Jr-_J L .Jl __ --l ~r-=-_J l _ ~ J[ = JC JC J [=-=J[_-:J [==:J[_=--:=J C::=JL=:J C J


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10 PrograDl Listings58 x 10 EXP R781 .LBLl 51 p,,'':' *** R82 f 52 tLBL883 2 Finds standard R, . ~ : , value from multiple4 0 54 585 55 + of step size86 18)( 56 INT Round up87 ST02 10tolj120 S? 5TO(88 P.'S 289 tLBL2 ,.,iO59 618 LOG 68 iO i''-'11 EHTt 61 GT0612 !NT 62 413 5T04 63 714 - 64 RCL615 1 65 X)Y'-'16 + 66 (;T0317 18)( 6? 1 26


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11

EXPONENTIAL GROWTH OR DECAY

Many growth or decay phenomena encountered in science and engineering obeyan exponential law of the general form:- t

X = X - (X - Xo) eTt ss sswhere:

Xt = Value at any time, t , (i .e. , theinstantaneous value)X = Steady state value (i .e. , atss t =00)Xo= Initial value (i .e. , at t = 0)t = Elapsed time (time after t = 0)T Exponential time-constant forspecific phenomena

This program provides interchangeablesolutions for anyone of the fourvariables Xt , Xss ' Xo or t providedthree variables and T are known.EXAMPLE 1:Given a capacitor in series with a1 megohm resistor. 1.5 seconds afterthe circuit is completed 125 volts aremeasured across R. To what voltagewas the capacitor originally charged?Note:T = the RC time-constant, and thevo ltage at t = 0 is zero

SOLUTION:5. EHTt1" +86 ):."

ST04 T = time-constant125.88 ST01 v8.88 ST02 Xss!.58 STOJ time

bS88168.73 jU voltsEXAMPLE 2:A cobalt 60 source (half-life = 5.26years) had an activity of 3.54 curieswhen purchased 8.5 years ago. What isits present activity?Note:Activity at t = 00 will be zero,T = half-life/LN2SOLUTION:

5.26 ENTt2.8e LN.

5T04 TJ.54 STDe Xoe.885T02 Xss8.58 ST03 t

(;SB11.15 jU curies


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

1 Key in the program [ I [ [2 Store time constant T ~ [ I J T3 Store variables C]C]

Initial value Xo ~ [ O XoInstantaneous value Xt L ~ [ 1 [ XtSteady state value xc;c; ~ [ I J X"Elapsed time t I TO I ~ J t(Store any 3 of the 4 variables) I II I

4 To calculate: C]C]Xo' initial value I SB II 0 I XoXt , Instantaneous value E J ~ xtXss ' Steady state value E J ~ Xsst , elapsed time E J ~ t

5 For a new case go to step 2 [ ~ C JC]C]c ]1 IC]C]C]C]C]C]c ]1 IC]C]I II IC]C]C]C]C]C]I II II II I


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06.21824

81 tL8U82 PCL883 PCL284 -85 GSBS'8687 PCL288 +89 P'S18 tLBL211 GSBS12 EHTt13 EHTt14 PCU15 x16 PCL817 -18 x:y19 128 -2122 R.'S23 tLBL824 GSBS2S' PCU26 RCL2.,..,C,.[ -28 x29 PCL238 +31 R..S32 tL8L333 PCL834 PCL235 -36 PCUJ7 PCL238 -3948 LH41 PCL442 x4J R/S44 tLBL.545 PCU46 PCL4474849 PTH58 p.S

XoT

ProgralD Listings 13Xt

***Xss

***Xo

***t

***e -tiT

*** "Printx" may b insertedbefore "RIS".

REGISTERS1 Xt 2 Xss 3 t 4 57 8 9 .0 .1.3 .4 .5 16 1719 20 21 22 2325 26 27 28 29


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14

EQUATIONS OF MOTION

This program calculates an interchangeable solution among the variab1es:dis-placement, acceleration, initialvelocity, and time or final velocityfor an object that undergoes constantacceleration. The motion must belinear.EQUATIONS:

Final velocity v= IV02 + 2axInitial velocity Vo= Iv 2 - 2ax

v2 - vo 2Displacement x= _--=-__2av2 - vo 2Acceleration a= -"""'2=-x--

Displacement x= vot + t at2x 1Initial velocity Vo= t - 2 at

Acceleration a=

Iv 0 2 + 2ax - VoTime t=-- - - - -aREMARKS:Any consistent set of units may be used.Displacement, acceleration, and velocityshould be considered signed (vector)quantities. For example, if initial

velocity and acceleration are in opposite directions, one should be positiveand the other negative.All equations assume initial displace-ment, xO' equals zero.EXAMPLE 1:An automobile accelerates for 4 secondsfrom a speed of 35 mph and covers adistance of 264 feet. Assuming constantacceleration, what is the accelerationin ft/sec 2 ? (7.33 ft/sec 2 ) If theacceleration continues to be constant,what distance is covered in the nextsecond? (84.33 ft)SOLUTION:

254.88 STet x3S.8et ENT1'5288.88'3688.88

ST!J2 Vo4.88 BTC:: tGSB47.33 tH a

5.88 SrOJ t + 1 sec~ S B 1

348&33 U: x at t + sec264.8e x at t84.32 t* ' x (t+1) - x{t)


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EXAMPLE 2:An airplanes take off velocity is 125mph. Assume a constant accelerationof 15 ft/sec 2 How much runway lengthin feet will be used from start totake-off? (1120.37 ft.) How long willi t take for the plane to reach take-offvelocity? (12.22 sec)

SOLUTION:8.813 5T02 Va

1 5 . e ~ ST04125.80 EHTt

5 2 8 e . l 3 ~ x36813.1313 STC5 v!;SS71128.37 U t x

GSPJ12.22 t.tt t

User InstructionsSTEP INSTRUCTIONS INPUTDATA/UNITS KEYS

l . Key in the program DD2. Store variables DD

displacement x ~ Dinitial velocity Va ~ [ Utime t I STO II 3 Iacceleration a ~ I 4 Ifinal velocity v I STO I ~ J(store appropriate unknowns) I II I

3. To calculate: DDA. Displacement(a,vaand t or v known) I II I

ift is known [ i J [ l ]i fv is known [iJCLJ

B. Initial Velocity (a,x, and t or v k n o w ~ DDift is known L ~ I 2 Iif v is known ~ [ L JC Acceleration (Vn.X and t or v known) DDif t is known -- I iSB ICLJif v is known ~ G J

D Time (Vn.X and a known) ~ WE Final Velocity (Vn.X and a known) ~ I 5 I

4 For a new case 00 to steo 2 DD

15

OUTPUTDATA/UNITS

xVatav

xx

Vn

Vn

aatv


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16 ProgralD Listingsfl1 .tLEL x (t known) 51 srosfl2 ReL "'- RTN@3 PCL 53 tLPL9134 2 54 RCL4e5 .,.1 ... RCL!86 J;'!""c,


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17

KINETIC ENERGY

This program calculates an interchangeable solution among the variablesweight (or mass), velocity, and kineticenergy, for an object moving at constantvelocity. The program operates ineither English or metric units. Formetric units, any consistent set ofunits may be used; the quantity massmust be used. For English units, theenergy must be in foot-pounds, thevelocity in feet per second, and thequantity weight in pounds.K.E. = Kinetic energy

W= Weight (lb)m = Mass (kg, g)v Velocityg = Acceleration due to gravity =32.17398 ft/sec 2

EQUATIONS:English

K.E.

MetricK E = l"v 2.. 2'''ft-lb = 1.98 x 106 hp

EXAMPLE 1:The slider of a slider-crank mechanismis used to punch holes in a slab ofmetal. It is found that the workrequired to punch a hole is 775 ft-lb.If the slider weighs 5 lb . 4 oz., howfast must i t be moving when i t strikesthe metal? (97.46 ft/sec) What is therequired work in horsepower? (3.91 x10- 4 hp)

SOLUTION:b ~ B Z

64.35 itt Engl i sh Uni ts775.88 STOl88 EN7t

4.88 EHTt1f.08

+~ . 2 5 nt . w

5T02~ S B 5

9'7.4f Ut vbSB3

775.8B f** ft-l bR/53.91-84 *** hp

EXAMPLE 2:An object weighing 4.8 kg is moving withconstant velocity of 3.5 m/sec. Findits kinetic energy. (29.40 joules)SOLUTION:

~ S P l2. ee tU Metri c Units4.ee 5TD2

3 . ~ e STD3~ S B ]29.40 K.E.


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18 User InstructionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

1 Key in the program I ID2 Choose system of units DDMetric (S1) ~ o = J 2.00

or DDEnglish I SB II 2 I 64.35

3 Input two of the following variables DDKinetic energy K.E. I TO II 1 I K.E.Weight (mass) W(m) I TO II 2 I W(m)Velocity v GIQJ[LJ v

4 Compute the remaining variables: I II IKinetic energy ~ [ L J K.E.Optional: convert K.E.(ft-lb)to K.E.(hp) M JD K.E.(ho)Weight (mass) ~ [ L J W(m)Velocitv IGSB II 5 I \I

5 To change any input variable go to step 3 DD6 For a new case, go to step 2 DI IDDDD

DDI II IDDI II IDDDDDDI II I1 11 I


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PrograDl Listings 1981 .Leu t1etri c82 .,..83 STD484 rus85 .LBl2 Englishe6 687 4ae .89 ?18 411 .,..12 917 6..14 ST0415 R/S16 flBD K. E. Calc.17 ReL218 RCL319 EHTt28 .y21 x22 RCL42324 STOl ***25 RlS26 PCLl27 .L28 929 838 EX31 432 ***33 RlS34 fLBL4 W(m) Calc.35 RCLl36 RCL4"" y38 RCL339 ENTt48 x4142 5T02 ***43 R/S44 .LBLS v Calc.45 RCU 1* "Printx"46 RCL4 may be in !>erted before "RIS".47 x48 RCl24958 lX51 ST03 ***52 R/5

REGISTERS0 1 K.E. 2 W(m) 3 v 42(met) 64.3'Eno)6 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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20

RPM/TORQUE/POWER

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

RPMSI

RPMEnglishRPM

Torque nt-m ft-lbPower watts hp

EQUATIONS:RPM x Torque =Power1 hp = 745.7 watts1 ft-lb = 1.356 joules1 RPM = n/30 radians/sec

hp = 550 ft-lbsecEXAMPLE 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-m. If i t is 90%efficient, what is the power inputin both systems?

SOLUTIONS:(1)

(2)

GSB4658ft 88 ENIt

8.88 ENIt11. 88 t;SB58.89 *uR.,S12.85 ***

GSB31688.88 ENIt28.88 ENTt8.988.88 GSB5

3723.37 ***R,S4.99 ***

Torque, ft-l bTorque, nt-m

Power, wattsPower, hp


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STEP

1.

2.

3.

4.

User InstruetionsINSTRUCTIONS INPUTDATA/UNITS KEYS

Key in the program L ~ L ~Choose sytem of units: DDMetric ~ o = J

or DClEnglish I GSB II 4 IEnter variables (the unknown quantity, x, DDmust be input as zero) and compute x. I ILJRPM RPM I ENTt II ITorque Torque I ENH IDPower Power I GSB II 5 IDD(Optional) to convert torque or power to DDother system ~ D

L ~ I IDDL ~ I IDDD L ~DD[ II ID L ~I I I IC ~ DC ~ DDDL ~ I II II I

21

OUTPUTDATA/UNITS

r---1.36

5251.41

x

x converted

--1-----------


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22 Program Listings48 RCLJ81 tLBLJ 4982 3 Set up for metric 58 RCL?83 8 units 51 x84 Pi 52 R..S ** RPM85 53 tLBLl86 STO? 54 RCL487 ., 55 RCL288 4 5689 5 57 RCL?18 . 58 x11 7 59 R.S12 S105 68 RCL6 *** Torque13 1 6114 62 R/S15 3 63 tLBL8 ** Torque converted16 5 64 RCL217 6 65 RCLJ18 ST06 66 x

19 RTH 67 RCL728 tLBL4 Set up for English 6821 GSB3 units 69 R. .S *** Power22 1/X 78 RCL523 S106 7124 Xty .,'} R/S ** Power converted


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23

BLACK BODY THERMAL 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 for 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 considerations.

Black bodymonochromaticemissive power

Amax ,

2345678Wavelength, microns

Figwe 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 al lwavelengths 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:

00 [(t) 32iTCl L: -T kc2 ek=l+ 3T + 6 T 'A2 kc 2 i kC 2 + 6 k:') J

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

+


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24Eb (0- 00) is the tota1 emi ss i ve power

in Btu/hr-ft2 or Watts/cm2;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/cm2;Eb(AI- A2) is the emissive power fo rwavelengths between Al and A2

in Btu/hr-ft2 or Watts/cm2.c1 = 1.8887982 X 10 7 B t u - ~ m 4 / h r - f t 25.9544 x 10 3 W ~ m 4 / c m 2c2 = 2.58984 X 104 ~ m - o R =

1.4388 x 10 4 ~ 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

a = 1.731 X 10-9 Btu/hr-ft2-0R4exp = 5.729 x 10- 12 W/cm2-K4REMARKS:A minute or more may be required toobtain Eb(O-l) or Eb{A -A ) since theintegratlon is numerlcal. 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 fraction 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.

5954.4@ ST0114:388.13@ 8T02289 7 .88 STO.3

5.669.j-12 8T042488. 88 ST0513.48 ST06

e.?13 Sm?G884

4.9? Hlf:i;SB2

1 ie. i3g '.'

1. I . constants

2.64 ~ : : H (%)


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2.1888?9B2.@8 STC1

2'5898.48 ST025216. B88TO.?

!. 71312-89 STlJ4B.48 ENTt8. 713 +2. Be

}En91ish constants

B.'5'5 t.H mean valueReL?

1./':::24 6 73 RCLf

. . , ~ RCL6 74 '5C . ~ !26 75 yx?7 RCL8 7677 RCL228 .: 78 RCL62938 - 7931 6 88 RCL'S7" RCLe 81, - I ~ 82 eX33 3 83 134 vx 84 -3'5 8'5'36 +37 RCLe 86 R'S ~ * E87 tLBL4 bA38 RCL6 88 (;SB93948 eX 89 ST. 8 Eb(O to Ad41 )( 98 RCL742 ReL8 91 STa6 A243 92 r;SB993 RC.e44 ST+9 II 944'5 RCL9 -46 95 R'S **E47 EEX b(A1 to A2).148 CHS49 5 ~ * * " P r i n t x " may be i ~ s e r t e d before "R/S"

REGISTERS0 1 C1 2 c2 3 c3 4 (J 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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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 ) 25B?? 4B7 STCr:J2.1? STOt

4632.48 STC?62.413 t;SBl

! B ~ . B e t;SBJfSB84J.JJ H;: (psig)-lB.ee GSB2

GSB842.66 *** ( s i g)62.40 GSBlleB.BB fSBJt;SB6 (ft/sec)8B.21 *H62.4B fSBl28.eB GSB22.BB GSB4

IBB.BB GSP3GSB9

13.14 '"u (BTU/1 b)1424.29 (BTU/hr)

(2)

(3)

1.1313 ST05STC?9.813665 ST06735.88 GSS!

3.B8 GSB23.?8 GSB?

-15.813 GSB2GSB8

- 5 2 7 1 ~ . 8 2 H* (Nt/m 2 )

lel3/!.f!8 GSBl25.1313 GSBJ

GSB9245.17 *H (joule/kg)e.8: x288.39 *** (joule/kg)28.813 [NT'"leee.Be x (kg/s)

4167826.25 H;: (watts)

29


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

l . Key in the program I II I2. For English units: 25033.407 WQJ[LJ32.17 WQJ[LJ4632.48 WQJI7 I 1.00

For S.1. units: 9.80665 I STO II 6 I3. Enter fluid density p GSa] [L J 0.004 Enter the followina (neaative values are I IL J

nnwnc;:tY'Pi'lm Vi'llIIPC;:)' I II IFluid velocity v ~ ~Height from reference datum z I GSB II 3 IPressure P ~ o = JEnergy input E ~ [ D

5. Repeat step 4 for all input values DO6. Calculate the unknown: I II I

Fluid velocity I GSB ICLJ vHeight from reference datum ~ 1 7 I zPressure ~ [ L ] PEnergy ~ I 9 I E

7. For another case, go to step 3. or clear DOreCiister 8 and ClO to step 4 0 WQJI8 I

DO - -I II IDODO -OOI II II II I


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PrograUl Listings 3181 *LBL182 ST04 P83 884 STa8 Clear l: E85 R . S86 .;:LBL28? EHTt88 ABS89 x18 211 + v2 / 2-12 ~ T 0 513 tLBU14 RCL615 xf.0 ~ T a 5 gz17 tLBL418 RCL?lQ- '. x28 RCL4 Pip2122 tLBL5 E23 ST+824 I( . S25 tLBL626 RCL8"7 "... , '-28 J{29 IX38 P/S *** V31 tLBL732 RCL833 RCL63435 P/S *** z36 *LBL8r RCLS38 RCL?3948 RCL441 x42 P./S *** P43 tLBL944 RCLe *** IIPrintx ll bE! inserted before45 RCL5 may46 II RIS II 47 P/S *** E

REGISTERS0 1 2 3 4 P 5 Used6 9 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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32

MOHR CIRCLE FOR STRESS

Given the state of stress on an element,the principal stresses and their orientation can be found. The maximum shearstress and its orientation can also befound.

MeximumStr .. . State Principal Stressas Shear Stress

EQUATIONS:

ssmax . (sx sy)' + Sxy2Sx + Sy

Sl = + ssmax2Sx + Sys2 = - ssmax2

1/2 VXY )= tan -1 Sx - Sy8s = 1/2 tan- 1 - ('X - sy)2s xy

where:ssmax is the maximum shear stress;

Sl and S2 are the principalnormal stresses;8 is the angle of rotation fromthe principal axis to theoriginal axis;

8s is the angle of rotation fromthe axis of maximum shear stressto the original axis;

Sx is the stress in the x direction;Sy is the stress in the y direction;

Sxy is the shear stress on theelement .REFERENCE:Spotts, M.F., Design of MachineElements, Prentice-Hall, 1971.EXAMPLE:If Sx = 25000 psi, Sy = -5000 psi, andSxy = 4000 psi, compute the principalstresses and the maximum shear stress.

D ~ O : ' ~ - ~ I FV, - 25,000 psi s, = 5524.17psi 15524.17 psiStress Stata Principal Stressas MaximumShear Stress


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I

SOLUTION:2'5888.88 ENT1-'5888.88 ENTt4888.88 1;5812'5'524.17 H *';>"5-'5'524.17 H t

1( ...s7.47 H:*p/s-3753 tHp/s

1'5524.17 *':1/:

33

S1 (psi)S2 (psi)e (degrees)es (degrees)ssmax (psi)


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34 User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITSl . Key in the program I ID2. Enter the following: DDStress in the x direction (negative for S umD

compression) DDStress in the y direction (negative for S1 I ENTtl L ~compression) DDShear stress Sv' I IL ]" J

3 IComDute the followinc' LJI IIFirst DrinciDal stress ~ W Sl~ e c o n d principal stress I R/S II I 52

~ n g l e of rotation (principal) ~ D 8~ n g l e of rotation (shear) ~ D 8sMaximum shear stress ~ D ssmaxNOTE: Do not disturb the stack during L ~ I Istep 3 DD

4. For a new case, go to step 2. DI IDDD L ~DDI II IDDI II IDD

- C ~ D-- DD

-- I II I ----I II I


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ProgralD Listings 3581 tLBLl82 EHTt83 RJ. Sxy Sy Sx Sxy84 ST0385 RJ.86 x:y Sx87 ST0188 x:y Sy89 5T+118 - Sx - S11 2 Y12 S T ~ 11314 ST04 (s - s )/215 x Y16 5T02 ssmax17 RCLl18 +19 R/5 ** 5128 x:y 2821 RCLl22 RCL223 -24 R/5 ** S225 X:Y26 22728 R/S ** 829 RCL438 RCt::.3132 CH533 TAH-I34 23S -36 R/5 ** 8s ** IIPrintx ll may re place IIR/SII.J7 RCL238 R..'5 *** s '*** II Pri ntxll may be inserted before IIR/smax II

REGISTERS0 1 (sx+ Sy)/2 2 ssmax 3 Sxy 4 (sx-sv)/2 56 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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36

POLYNOMIAL EVALUATION--REAL OR COMPLEX

This program evaluates polynomials ofthe form:f(xo) = aO+alx+ . . . +a xnn

where the coefficientsak,k=0, . . . ( n ~ 2 8 ) and Xo are realor the coefficients and Xo arecomplex of the form

ak = Re(ak)+ i 1mZ 0 = Re (zo)+ i 1m (zo)

k = 0, 1, . . . nExample 1:

f(x) = 11-7x - 3x 2 + 5x 4 + x5for Xo = 2.5for Xo = -5

Solution:CLRG

11. 89 GSB!-7.88 IUS-3.88 P/S8.88 rus5.88 R"S1.88 R'S2.58 GSB2267.72 t**6.88 STaB-5.88 GSB2-29.89 U tf (2.5)f (-5)

Example 2:f(x) = (5-7i) - lOx + (-2+i)x 2

+ 18x 3 + (3+4i)x 4for Xo = 2 + i

Solution:1.99 fNH2.98 GSB34.99 am3.99 GSB49.88 fHTt18.88 GSB41.88 fNIt-2.88 GSB48.88 fNTt-18.88 GSB4-7.88 fNT'!'5.88 GSB5

-186.98 tt;: Re f(xo)PlS


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37

STEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS1 Kev ;n thp ornaram c=J [ IFor real polynomials: c=Jc=J[--=:J l--=J2 Initialize o:::J [Rffi]l ___ J [ ~3. Store coefficients ao LGS-Bj I LJl _ l _ ~4. Continue for i = 1 to n, n:$28 a; LR1 I __J[ _J C--=-J!) r . n m n l l t ~ of ( ll ) Xn LGSBJ ~ 2 - - - J f(xo)U L ~ ] [ ~ _ . J6. For a different Xo go to step 5 n+l [3TQ]I__Ol-- --] [ - ~- -- - - - - -For complex polynomials: l -- I l __ J7. Enter Xo 1m Xo LENftl [ - ~ J

Rp Xn rJ;SBJ [ 3__ J 0J L ~ - - . J8. Enter a . k=n n- 1. . . 1 1m al. IENTtl L = ~

1\ Re " [GSB I 4 . ~ J"L ~ _ J r - - ~ l

9 Enter an and run 1m an [NItJ L_ lRp an LGSi] [-s_l Re..fLx"l

l R 2 ~ 1 l ---1 1m f(xo)l-_-I L _ ~ _ J10. For a new case, go to step 7 l- -- j l--=-__C-l [- --l- -- - -r - ~ [----1- -- - - - - - -[ -1 [-]- - - - - -----[-- J L---=:J---- -

[----] [ - J[ - 1 [ -- -- - JL _.I L _-.JL=-] r-='J[ -j [ - = ~[ - J L Jl -] [--nul- - - - - - - - - -l _ J [----]l ] [ ~ - : ]r -- j [- J


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38

o6.21824

81 *LPU82 ISZ838m;ft4 /?/S85 POI36 tLfL2(F ENT'"38 ENTt89 EflT-tr peL;11 x12 DS:13 tLBLB1 :! F..'CL i.r1_' + , DSZ-c f;TCe1 0" , . . ~ :2e"" p . C'-'22 --:"B:"'322' 41'24 STut2: V' " V... ...26 ST02..,., {,!l.;'

r - I ! T ' ~L. .. , :29 ENTt?8 ENT'"71 pm-'.32 tLBL4J? (;SB9?-4 GTCe

tLBL3f (;8B97 7 p .. C'..... . . ".:39 p .s413 tLBL841 .:.p42 PCL 143 }:'44 g : ~ "45 RCL246 +474.':

i 7.31925

Multiply by Xo

Multiply by Xo

Divide by Xo*** f(xo,)Routines forcomplex polynomialsrePrepare for LBL 9

*** Re f(xo)*** 1m f(xo)Multiply in polarform

5852r -...54rr:;_,t5758596t!

REGISTERSr or an 2 e or al 38 9.4 .520 2126 27

* ! . . ~ L 9V-Iot.'....+-RJ+p;+

g;yP.J.V"'!{ : . ~ iPTNp'.' -{ ..:'

a2 4 ..0162228

Add real parts andimaginary parts

a 5.1172329 a28

*** "Printx" may be inserted or used to replace "RIS".


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39

SINE} COSINE} AND EXPONENTIAL INTEGRALS

Sine integral:

Si(x) f xo sin t dt- t -00 (_1)nx2n + 1n=o (2n + 1)(2n + l)!

where x is real.This routine computes successivepartial sums of the series, stopswhen two consecutive partial sumsare equal, and displays the lastpartial sum as the answer.

S,(X)

2 4 6 6 10 12Notes: When x is too large, com-puting a new term of theseries might cause an overflow. In that case, displayshows all 9 ' s and the programstops.

Si(-x) = -Si(x)Cosine integral:

C ( x) =y+1n x + f x cos - 1 d =o00 (-l)n x 2ny+ lnx+ I 2n(2n)!n=l

where x>O, and Y= 0.577215665 is theEuler's constant.

This program computes successivepartial sums of the series. Whentwo consecutive partial sums are equal,the value is used as the sum of theseries.

-to

Notes: When x is too large, computinga new term of the series mightcause an overflow. In thatcase, display shows all 9' s andthe program stops.Ci(-x) = Ci(x) - in for x>O.

Exponential integral:x etEi(x) = f t dt 00 xn= y + 1n x + I nn!n=l0 0

where x>O and y= 0.577215665 is Euler'sconstant.This program computes successive partialsums of the series. When two consecutive partial sums are equal, the valueis used as the sum of the series.

Ei(x)


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40

Note: When x is too large, computinga new term of the series mightcause an overflow. In thatcase, display shows all 9 1 sand the program stops.References: Handbook of MathematicalFunctions, Abramowitz andStegun, National Bureauof Standards, 1968.Examples:

1. Si (.69)2. Si (.98)3. Ci (1. 38)4. Ci (5)5. Ei (1. 59)6. Ei (.61)

STEP INSTRUCTIONS1 in the Droaram2. Store a.

3. For Sine IntearalFor Cosine IntegralFor Exponential Integral

Solutions:6.577215665 ST.@69 GSB10. 6 7 ~ . ' .

8.98 GSB!8,93 *-t.,1.38 GSB28 A 6 ~ ' u '5,08 GSB2-8.19 ..1.59 GSB]3.57+ . 8.61 GSB38. eg .

INPUT KEYSDATA/UNITSCJCJc = JC J

.577215665 [SIOJ ~o=J c::::=JCJCJ

x [ ] sB1ITJx [ F ~ [ I Jx CGSBJ [ 3 1~ r - - - - i

OUTPUTDATA/UNITS

a.

Si(x)Ci(x)Ei(x)


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-06.21824

81 lLBU12 STOJ93 X294 CHS85 ST01B6 j07 ST0288 Rel]eg . l91 '1& RCUi - P C ~ 21213 +l415 LSTX,.it-17 4-18 .ST02

29 PCLJ21 ..22 C'Tn723 . ~ C L 2.... ..,.;:a.r2: 4-.:::t- _ \ - i ' / ~.,.., poe:: (' :-0 R.Sr . . '29 1:11 ?..... ..... -! f .,.. ,7 ' CH.s' ...32 CT"1... : '..'J,J?34 ST03B.. ",'36 ST0277_, 1 LSTX!P U!39 RC. t:!t.'48 +4' ~ T a 842 tLBLJ4J sr!]l4445 STu?46 e47 5T0248 RCL!49 LN

Sine Integralroutine

*** Si(x)/Ci(x)Cosine Integralroutine

Exponential Integral routine

5e-' ..525?0 ;'.' t55I.e,5758596e6162'; 7.........64'-'W'661..':

R C . ~4-

UL9RCL1PCL2

4-corn.,"- : ' - ' ~RCL3

XST03RC:"2

+X ~ / "

'TO',r;:'S

REGISTERS1 used 2 used 3 used7 8 9.3 .4 .519 20 2125 26 27

*** "Pr int X" may be used to replace lORis"

41

*** Ei(x)

4 5.0 used .116 1722 2328 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 Solutions

Electrical Engineering SolutionsSurveying Solutions

GamesNavigational Solutions

Civil Engineering SolutionsMechanical Engineering Solutions

Student Engineering Solutions


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hp-19c & 29c solutions student engineering 1977 b&w

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