Twoeq1

Name:	Twoeq1 - Pipe diameter calculation for specified pressure drop
Source:	Cutlip, M. B. and Shacham, M, Problem Solving in Chemical
	Engineering with Numerical Methods, Prentice Hall Inc., 1999
Reference/s
Model:	2 implicit equations, indep. variables D and fF
	Higher difficulty level
	Constraints: D>0, fF>0
	Discontinuities: The function f(fF) undefined for D<=0 and/or fF<=0
	Initial estimates: 1. D=0.04, fF = 0.001, 2. D=0.1, fF = 0.001
Solved by	Shacham, M., POLYMATH 5.1, build 19, April 1, 2001
Model Eqs.	Pipe diameter calculation for specified pressure drop |POLVER05_3
EXCEL FILE	f(D) = -dp/rho+2*fF*v*v*L/D #
TEXT FILE	f(fF) = if (Re<2100) then (fF-16/Re) else (fF-1/(4*log(Re*(fF)^(1/2))-0.4)^2) #
POLYMATH FILE	dp=103000 #
	L=100 #
	T=25+273.15 #
	Q=0.0025 #
	pi=3.1416 #
	rho=46.048+T*(9.418+T*(-0.0329+T*(4.882e-5-T*2.895e-8))) #
	vis=exp(-10.547+541.69/(T-144.53)) #
	v=Q/(pi*D^2/4) #
	kvis=vis/rho #
	Re=v*D/kvis #
	D(0)=0.04
	fF(0)=.001
Variable/function values	Variable	Value	f(x)
	D	0.04	-8.3773E+01
	fF	0.001	-4.5774E-03
	dp	103000
	L	100
	T	298.15
	Q	0.0025
	pi	3.1416
	rho	994.5715041
	vis	0.000893083
	v	1.989432136
	kvis	8.97957E-07
	Re	88620.36312
Solution	Variable	Value	f(x)
	D	0.0389653029101	-8.5265E-14
	fF	0.0045905347283	-1.7347E-18
	dp	103000
	L	100
	T	298.15
	Q	0.0025
	pi	3.1416
	rho	994.5715041241000
	vis	0.0008930825570
	v	2.0964909803150
	kvis	0.0000008979571
	Re	90973.616526310
Additional Information	If started a little farther from the solution (see 2nd initial estimate)
	most methods yield negative D or fF values leading to an execution error.
	A constrained method which keeps D and fF positive throughout the
	iterations should be used.