Manuals/calci/Examples1

Engineering Examples in z3

DESCRIPTION

• Basic Engineering examples in z3.
  
• 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
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.

* 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

/*Overall Heat transfer coefficient
U = 1/((1/hg)+(1/hw)+(x/k))
//wall very thin. So, x ≈ 0
U = 1</>((1</>hg)+(1</>hw))*/

x = 0  //wall very thin
hg = 300(W/m2.degK)    //heat transfer coefficient of water
hw = 1500(W/m2.degK)   //heat transfer coefficient of gas
cpg = 1130(J/kg.degK)  //mean specific heat capacity of gas (Cp)
cpw = 4190(J/kg.degK)  //specific heat capacity of gas
mg = 200(kg/hr)        //flow rate of gas
mw = 1400(kg/hr)       //flow rate of water
D = 75mm               //given exhaust pipe diameter

tg1 = 350degC<>degK    //temp. of gas entering 
tg2 = 100degC<>degK    //temp. of gas to be cooled to
tw1 = 10degC<>degK     //temp. of water entering

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

z3 code: Using Function

function LengthParallelContraFlow(hg,hw,cpg,cpw,mg,mw,tg1,tg2,tw1,D)
{
    
	//Overall Heat transfer coefficient
	//U = 1/((1/hg)+(1/hw)+(x/k))
	var x = 0  //wall very thin
	//U = 1</>((1</>hg)+(1</>hw))

	
        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)     //heat transfer coefficient of water
hw = 1500(W/m2.degK)    //heat transfer coefficient of gas
cpg = 1130(J/kg.degK)   //mean specific heat capacity of gas (Cp)
cpw = 4190(J/kg.degK)   //specific heat capacity of gas
mg = 200(kg/hr)         //flow rate of gas
mw = 1400(kg/hr)        //flow rate of water
D = 75mm                //given exhaust pipe diameter
tg1 = 350degC<>degK     //temp. of gas entering 
tg2 = 100degC<>degK     //temp. of gas to be cooled to
tw1 = 10degC<>degK      //temp. of water entering


LengthParallelContraFlow(hg,hw,cpg,cpw,mg,mw,tg1,tg2,tw1,D)

z3 code: USING EXAMPLE1 function SOLUTION IN EXAMPLE2 function

function HeatTransferCoefficient(hg,hw)
{
        //Overall Heat transfer coefficient
	//U = 1/((1/hg)+(1/hw)+(x/k))
	var x = 0  //wall very thin
	//U = 1</>((1</>hg)+(1</>hw))

	var U = 1</>((1</>hg)<+>(1</>hw))
							 
	return [U]
}
 

hg = 300(W/m2.degK)
hw = 1500(W/m2.degK)

HeatTransferCoefficient(hg,hw)

function LengthParallelContraFlow(tw1,mg,mw,cpg,cpw,tg1,tg2)
{
	
        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))</>(HeatTransferCoefficient(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))</>(HeatTransferCoefficient(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                 
tg1 = 350degC<>degK
tg2 = 100degC<>degK
                  
                  
LengthParallelContraFlow(tw1,mg,mw,cpg,cpw,tg1,tg2)

ExampleS2: Civil Engineering
A steel pipe 5 ft (1.5 m) in diameter and 3/5 in. (9.53 mm) thick sustains a fluid pressure of 180 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

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)

ExampleS3: Civil Engineering
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

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

z3 code: Using Function

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)

z3 code: USING EXAMPLE1 function SOLUTION IN EXAMPLE2 function

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)

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)

ExampleS4: Civil Engineering
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

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()

ExampleS5: Engineering Economics
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

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()

ExampleS6: Fluid Mechanics
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

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()