Difference between revisions of "Examples"

Revision as of 09:50, 6 February 2018

Engineering Examples in z3

DESCRIPTION

Basic Engineering examples in z3.
z3 codes with z3 notations are shown below.
Reflecting different domains like Engineering, Statistics, Medicine, etc.
Set of cases that are progressively complex on units are used to show the user how it goes from simple to complex cases.
Testing how we can make better solutions to the standard problems compared to other software, due to the presence of units.

Examples

ExampleS1: Chemical Engineering
<a href="http://wiki.zcubes.com/Manuals/calci/Examples1#Sutton">Sutton</a>

Difference between revisions of "Examples"

Revision as of 09:50, 6 February 2018

DESCRIPTION

Examples

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 ==Examples==
 '''ExampleS1: Chemical Engineering<br>'''
-An exhaust pipe is 75mm diameter and it is cooled by surrounding it with a water jacket. The exhaust gas enters at 350C and the water enters at 10C. The surface heat transfer coefficients for the gas and water are 300 and 1500 W/m2K respectively. The wall is thin so the temperature drop due to conduction is negligible. The gasses have a mean specific heat capacity Cp of 1130 J/kgK and they must be cooled to 100C. The specific heat capacity of the water is 4190 J/kgK. The flow rate of the gas and water is 200 and 1400 kg/h respectively. Calculate the required length of pipe for parellel flow and contra flow.<br>
+<a href="http://wiki.zcubes.com/Manuals/calci/Examples1#Sutton">Sutton</a>
- * The following example demonstrates how a problem is solved when units are used.
- * The units may be SI, Imperial or a mix of both.
- * Code shows how to assign a unit or convert a unit.
- * 3 types of calculations are shown: Normal, using Function, using Function1 in Function2.
-'''z3 code: Normal Calculation without using Function'''<br>
-<source lang="cpp">
-/*Overall Heat transfer coefficient
-U = 1/((1/hg)+(1/hw)+(x/k))
-U = 1</>((1</>hg)+(1</>hw))*/
-x = 0  //wall very thin
-hg = 300(W/m2.degK)
-hw = 1500(W/m2.degK)
-cpg = 1130(J/kg.degK)
-cpw = 4190(J/kg.degK)
-mg = 200(kg/hr)
-mw = 1400(kg/hr)
-D = 75mm
-tg1 = 350degC<>degK
-tg2 = 100degC<>degK
-tw1 = 10degC<>degK
-U = 1</>((1</>hg)<+>(1</>hw))
-delt = tg1<->tg2
-φ = mg<*>cpg<*>delt
-tw2 = tw1<+>(φ</>(mw<*>cpw))
-//Parallel flow
-delti = tg1<->tw1
-delt0 = tg2<->tw2
-A = φ<*>(log(delt0</>delti))</>(U<*>(delt0<->delti))
-L = A</>(π<*>D)
-//answer:1.48m
-//Contra Flow
-delti = tg1<->tw2
-delt0 = tg2<->tw1
-A = φ<*>(log(delt0</>delti))</>(U<*>(delt0<->delti))
-L = A</>(π<*>D)
-//answer:1.44m</source>
-'''z3 code: Using Function'''<br>
-<source lang="cpp">
-function Example1(hg,hw,cpg,cpw,mg,mw)
-{
-/*Overall Heat transfer coefficient
-U = 1/((1/hg)+(1/hw)+(x/k))
-U = 1</>((1</>hg)+(1</>hw))*/
-x = 0  //wall very thin
-var D = 75mm
-var tg1 = 350degC<>degK
-var tg2 = 100degC<>degK
-var tw1 = 10degC<>degK
-var U = 1</>((1</>hg)<+>(1</>hw))
-var delt = tg1<->tg2
-var φ = mg<*>cpg<*>delt
-var tw2 = tw1<+>(φ</>(mw<*>cpw))
-//Parallel flow
-var delti = tg1<->tw1
-var delt0 = tg2<->tw2
-var A1 = φ<*>(log(delt0</>delti))</>(U<*>(delt0<->delti))
-var LParallel = A1</>(π<*>D)
-//answer:1.48m
-//Contra Flow
-delti = tg1<->tw2
-delt0 = tg2<->tw1
-var A2 = φ<*>(log(delt0</>delti))</>(U<*>(delt0<->delti))
-var LContra = A2</>(π<*>D)
-//answer:1.44m
-return [LParallel,LContra]
-}
-hg = 300(W/m2.degK)
-hw = 1500(W/m2.degK)
-cpg = 1130(J/kg.degK)
-cpw = 4190(J/kg.degK)
-mg = 200(kg/hr)
-mw = 1400(kg/hr)
-Example1(hg,hw,cpg,cpw,mg,mw)</source>
-'''z3 code: USING EXAMPLE1 function SOLUTION IN EXAMPLE2 function'''<br>
-<source lang="cpp">
-function Example1(hg,hw)
-{
-/*Overall Heat transfer coefficient
-U = 1/((1/hg)+(1/hw)+(x/k))
-U = 1</>((1</>hg)+(1</>hw))*/
-x = 0  //wall very thin
-var U = 1</>((1</>hg)<+>(1</>hw))
-return [U]
-}
-hg = 300(W/m2.degK)
-hw = 1500(W/m2.degK)
-Example1(hg,hw)</source>
-<source lang="cpp">
-function Example2(tw1,mg,mw,cpg,cpw)
-{
-var tg1 = 350degC<>degK
-var tg2 = 100degC<>degK
-var delt = tg1<->tg2
-var φ = mg<*>cpg<*>delt
-var tw2 = tw1<+>(φ</>(mw<*>cpw))
-//Parallel flow
-var delti = tg1<->tw1
-var delt0 = tg2<->tw2
-var A1 = φ<*>(log(delt0</>delti))</>(Example1(hg,hw)<*>(delt0<->delti))
-var LParallel = A1</>(π<*>D)
-//answer:1.48m
-//Contra Flow
-delti = tg1<->tw2
-delt0 = tg2<->tw1
-var A2 = φ<*>(log(delt0</>delti))</>(Example1(hg,hw)<*>(delt0<->delti))
-var LContra = A2</>(π<*>D)
-//answer:1.44m
-return [LParallel,LContra]
-}
-cpg = 1130(J/kg.degK)
-cpw = 4190(J/kg.degK)
-mg = 200(kg/hr)
-mw = 1400(kg/hr)
-tw1 = 10degC<>degK
-Example2(tw1,mg,mw,cpg,cpw)</source>
-'''ExampleS2: Civil Engineering<br>'''
-A steel pipe 5 ft (1.5 m) in diameter and 3/5 in. (9.53 mm) thick sustains a fluid pressure of
-lb/sq.in. (1241.1 kPa). Determine the hoop stress, the longitudinal stress, and the increase
-in diameter of this pipe. Use 0.25 for Poisson’s ratio.
-'''z3 code: Using Function'''<br>
-<source lang="cpp">
-function civil1(p,D,t){
-var E = 30e+6(lb/sqin)//for steel
-var v = 0.25
-/*hoop stress
-s = pD/2t
-longitudinal stress
-s'= pD/4t
-increase in cyl diameter
-delD = D(s-vs')/E */
-var s = p<*>D</>(2<*>t)
-var sdash = p<*>D</>(4<*>t)
-var delD = D<*>(s<->v<*>sdash)</>E
-return [s,sdash,delD<>inch]
-}
-p = 180(lb/sqin)
-D = 5(ft)
-t = (3/8)<>(inch)
-civil1(p,D,t)</source>
-'''ExampleS3: Civil Engineering<br>'''
-A 1/2-in. (12.7-mm) diameter Copperweld bar consists of a steel core 3/8 in. (9.53 mm) indiameter and a copper skin 1/16 in. (1.6 mm) thick. What is the elongation of a 1-ft (0.3-m) length of this bar, and what is the internal force between the steel and copper arising from a temperature rise of 80°F (44.4°C)? Use the following values for thermal expansion coefficients: cs = 6.5*106 and cc = 9.0*106 , where the subscripts s and c refer to steel and copper, respectively. Also, Ec = 15*106 lb/sq.in. (1.03*108 kPa).
-'''z3 code: Normal Calculation without using Function'''<br>
-<source lang="cpp">
-dc = 12.7(mm)
-ds = (3/8)<>(inch)
-dcs = (1/16)<>(inch)
-Es = 30e+6(lbf/in2)
-Ec = 1.03e+11(Pa)
-cs = 6.5e-6(diffF-1)
-cc = 9e-6(diffF-1)
-L = 1(ft)
-delT = 44.4(diffC)
-//cross-sectional area
-A = π<*>(dc)^2</>4
-As = π<*>(ds)^2</>4
-Ac = A<->As
-//coeff of expansion
-//c = ((As<*>Es<*>cs)<+>(Ac<*>Ec<*>cc))</>(As<*>Es<+>Ac<*>Ec)
-a = (As<*>Es<*>cs)
-b = (Ac<*>Ec<*>cc)<>(lbf.diffF-1)
-d = (As<*>Es<+>Ac<*>Ec)
-c = (a<+>b)</>d
-//thermal expansion
-delL = c<*>L<*>delT
-//expansion w.o restraint
-delLc = cc<*>L<*>delT
-delLcs = delLc<->delL
-delLs= cs<*>L<*>delT
-delLsc = delL<->delLs
-//restraining force
-P1 = Ac<*>Ec<*>delLcs</>L
-P2 = As<*>Es<*>delLsc</>L</source>
-'''z3 code: Using Function'''<br>
-<source lang="cpp">
-function civil2(dc,ds,dcs,L){
-var Es = 30e+6(lbf/in2)
-var Ec = 1.03e+11(Pa)
-var cs = 6.5e-6(diffF-1)
-var cc = 9e-6(diffF-1)
-var delT = 44.4(diffC)
-//cross-sectional area
-var A = π<*>(dc)^2</>4
-var As = π<*>(ds)^2</>4
-var Ac = A<->As
-//coeff of expansion
-//c = ((As<*>Es<*>cs)<+>(Ac<*>Ec<*>cc))</>(As<*>Es<+>Ac<*>Ec)
-var a = (As<*>Es<*>cs)
-var b = (Ac<*>Ec<*>cc)<>(lbf.diffF-1)
-var d = (As<*>Es<+>Ac<*>Ec)
-var c = (a<+>b)</>d
-//thermal expansion
-var delL = c<*>L<*>delT
-//expansion w.o restraint
-var delLc = cc<*>L<*>delT
-var delLcs = delLc<->delL
-var delLs= cs<*>L<*>delT
-var delLsc = delL<->delLs
-//restraining force
-var P1 = Ac<*>Ec<*>delLcs</>L
-var P2 = As<*>Es<*>delLsc</>L
-return[P1,P2]
-}
-dc = 12.7(mm)
-ds = (3/8)<>(inch)
-dcs = (1/16)<>(inch)
-L = 1(ft)
-civil2(dc,ds,dcs,L)</source>
-'''z3 code: USING EXAMPLE1 function SOLUTION IN EXAMPLE2 function'''<br>
-<source lang="cpp">
-function civil2(dc,ds,dcs){
-var Es = 30e+6(lbf/in2)
-var Ec = 1.03e+11(Pa)
-var cs = 6.5e-6(diffF-1)
-var cc = 9e-6(diffF-1)
-//cross-sectional area
-var A = π<*>(dc)^2</>4
-var As = π<*>(ds)^2</>4
-var Ac = A<->As
-//coeff of expansion
-//c = ((As<*>Es<*>cs)<+>(Ac<*>Ec<*>cc))</>(As<*>Es<+>Ac<*>Ec)
-var a = (As<*>Es<*>cs)
-var b = (Ac<*>Ec<*>cc)<>(lbf.diffF-1)
-var d = (As<*>Es<+>Ac<*>Ec)
-var c = (a<+>b)</>d
-return[As,Ac,c]
-}
-dc = 12.7(mm)
-ds = (3/8)<>(inch)
-dcs = (1/16)<>(inch)
-civil2(dc,ds,dcs)</source>
-<source lang="cpp">
-function civil3(L,delT){
-var Es = 30e+6(lbf/in2)
-var Ec = 1.03e+11(Pa)
-var cs = 6.5e-6(diffF-1)
-var cc = 9e-6(diffF-1)
-//thermal expansion
-var delL = civil2(dc,ds,dcs)[2]<*>L<*>delT
-//expansion w.o restraint
-var delLc = cc<*>L<*>delT
-var delLcs = delLc<->delL
-var delLs= cs<*>L<*>delT
-var delLsc = delL<->delLs
-//restraining force
-var P1 = civil2(dc,ds,dcs)[1]<*>Ec<*>delLcs</>L
-var P2 = civil2(dc,ds,dcs)[0]<*>Es<*>delLsc</>L
-return [P1,P2]
-}
-L = 1(ft)
-delT = 44.4(diffC)
-civil3(L,delT)</source>
-'''ExampleS4: Civil Engineering<br>'''
-M1 is a 4x4, F = 5500 lb (24,464 N), and Phi = 30°. The allowable compressive
-stresses are P = 1200 lb/sq.in. (8274 kPa) and Q = 390 lb/sq.in. (2689.1 kPa). The
-projection of M1 into M2 is restricted to a vertical distance of 2.5 in. (63.5 mm).
-'''z3 code: Using Function'''<br>
-<source lang="cpp">
-function civil4(){
-var b = 3.625(inch)
-var φ = 30(deg)
-var P = 1200(lbf/sqin)
-var Q = 2689.1e+3(Pa)
-var F = 24464(N)
-var A = 13.1(sqin)
-//lengths
-var AB = b</>DSIN(φ)
-var AC = (b<*>DSIN(φ/2))</>DSIN(φ)
-var BC = (b<*>DCOS(φ/2))</>DSIN(φ)
-//stresses f1 and f2
-var f1 = (F<*>DSIN(φ))</>(A<*>DTAN(φ/2))
-var f2 = (F<*>DSIN(φ)<*>DTAN(φ/2))</>(A)
-//allowable stresses
-var N1 = P<*>Q</>((P<*>(DSIN(φ/2))^2)<+>Q<*>(DCOS(φ/2))^2)
-var N2 = P<*>Q</>((P<*>(DCOS(φ/2))^2)<+>Q<*>(DSIN(φ/2))^2)
-return[AC;BC;f1<>(lbf/sqin);f2<>(lbf/sqin);N1<>Pa;N2<>Pa]
-}
-civil4() </source>
-'''ExampleS5: Engineering Economics<br>'''
-The QRS Corp. purchased capital equipment for use in a 5-year venture. The equipment
-cost $240,000 and had zero salvage value. If the income tax rate was 52 percent and the
-annual income from the investment was $83,000 before taxes and depreciation, what was
-the average rate of earnings if the profits after taxes were invested in tax-free bonds yielding 3 percent? Compare the results obtained when depreciation is computed by the straight-line method.
-'''z3 code: Using Function'''<br>
-<source lang="cpp">
-function economics(){
-var EC = 240000
-var n = 5
-var GI = 83000
-var r = 0.52
-var i = 0.03
-//taxable income
-var DC = EC</>n
-var TI = GI<->DC
-//annual tax payment
-var TP = r<*>TI
-//net income
-var NI = GI<->TP
-//S = R(USCA)
-//SPCA = (1<+>i)^n
-var s = NI<*>(5.309)
-var sp = s</>EC
-var i = [(sp^0.2)<->1]<*>100
-return i
-}
-economics()</source>
-'''ExampleS6: Fluid Mechanics<br>'''
-A steel pipe is discharging 10 ft3/s (283.1 L/s) of water. At section 1, the pipe diameter is 12 in. (304.8 mm), the pressure is 18 lb/sq.in. (124.11 kPa), and the elevation is 140 ft(42.67 m). At section 2, farther downstream, the pipe diameter is 8 in. (203.2 mm), and the elevation is 106 ft (32.31 m). If there is a head loss of 9 ft (2.74 m) between these sections due to pipe friction, what is the pressure at section 2?
-'''z3 code: Using Function'''<br>
-<source lang="cpp">
-function fluidmechanics(){
-var d1 = 12(inch)
-var d2 = 203.2(mm)
-var p1 = 124.11e+3(Pa)
-var z1 = 140(ft)
-var z2 = 32.31(m)
-var q1 = 283.1(L/s)
-var q2 = 10(ft3/s)
-var hf = 9(ft)
-var w = (62.4/(144*12))<>(lbf/inch3)
-var g = 32.2(ft/s2)
-var a1 = π<*>(d1)^2</>4
-var a2 = π<*>(d2)^2</>4
-var v1 = q1</>a1
-var v2 = q2</>a2
-var p2 = (((v1^2<->v2^2)</>(2<*>g)<+>z1<->z2<->hf)<*>w)<+>p1
-return p2<>(lbf/in2)
-}
-fluidmechanics()</source>