5.3 Simulation of Planar Interface Evaporation
5.3.2 Simulation of Two-Phase Stefan Problems
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76 Engineering Mathematics
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‘Cross-multiplying’ gives:30(1+100α)=35(1+50α) 30+3000α=35+1750α 3000α−1750α=35−30
1250α=5
i.e. α= 5
1250= 1
250 or 0.004 Substitutingα= 1
250 into equation (1) gives:
30=R0
1+(50) 1
250
30=R0(1.2) R0= 30
1.2 =25 Checking, substitutingα= 1
250and R0=25 in equation (2) gives:
RHS=25
1+(100) 1
250
=25(1.4)=35=LHS
Thus the solution isα=0.004/◦C and R0=25.
Problem 16. The molar heat capacity of a solid compound is given by the equation c=a+bT, where a and b are constants. When c=52, T=100 and when c=172, T=400. Determine the values of a and b
When c=52, T=100, hence
52=a+100b (1)
When c=172, T=400, hence
172=a+400b (2)
Equation (2) – equation (1) gives:
120=300b from which, b=120
300=0.4
Substituting b=0.4 in equation (1) gives:
52=a+100(0.4)
a=52−40=12 Hence a=12 and b=0.4
Now try the following exercise
Exercise 36 Further practical problems
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Chapter 10
Transposition of formulae
10.1 Introduction to transposition of formulae
When a symbol other than the subject is required to be calculated it is usual to rearrange the formula to make a new subject. This rearranging process is called transposing the formula or transposition.
The rules used for transposition of formulae are the same as those used for the solution of simple equa-tions (see Chapter 8)—basically, that the equality of an equation must be maintained.
10.2 Worked problems on transposition of formulae
Problem 1. Transpose p=q+r+s to make r the subject
The aim is to obtain r on its own on the left-hand side (LHS) of the equation. Changing the equation around so that r is on the LHS gives:
q+r+s=p (1)
Substracting (q+s) from both sides of the equation gives:
q+r+s−(q+s)=p−(q+s) Thus q+r+s−q−s=p−q−s
i.e. r=p−q−s (2)
It is shown with simple equations, that a quantity can be moved from one side of an equation to the other with an appropriate change of sign. Thus equation (2) follows immediately from equation (1) above.
Problem 2. If a+b=w−x+y, express x as the subject
Rearranging gives:
w−x+y=a+b and −x=a+b−w−y Multiplying both sides by−1 gives:
(−1)(−x)=(−1)(a+b−w−y) i.e. x= −a−b+w+y
The result of multiplying each side of the equation by−1 is to change all the signs in the equation.
It is conventional to express answers with positive quantities first. Hence rather than x= −a−b+w+y, x=w+y−a−b, since the order of terms connected by+and−signs is immaterial.
Problem 3. Transposev=fλto makeλthe subject
Rearranging gives: fλ=v Dividing both sides by f gives: fλ
f = v f
i.e. λ= v
f Problem 4. When a body falls freely through a height h, the velocityvis given byv2=2gh.
Express this formula with h as the subject Rearranging gives: 2gh=v2 Dividing both sides by 2g gives: 2gh
2g = v2 2g
i.e. h= v2
2g Problem 5. If I=V
R, rearrange to make V the subject
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78 Engineering Mathematics
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Rearranging gives: V R=I Multiplying both sides by R gives:R V
R
=R(I)
Hence V=IR
Problem 6. Transpose: a=F m for m Rearranging gives: F
m=a
Multiplying both sides by m gives:
m F
m
=m(a) i.e. F=ma
Rearranging gives: ma=F
Dividing both sides by a gives: ma a =F
a
i.e. m=F
a Problem 7. Rearrange the formula: R=ρl
a to make (i) a the subject, and (ii) l the subject (i) Rearranging gives:ρl
a =R Multiplying both sides by a gives:
a ρl
a
=a(R) i.e. ρl=aR Rearranging gives: aR=ρl
Dividing both sides by R gives:
aR R = ρl
R
i.e. a= ρl
R (ii) Multiplying both sides ofρl
a =R by a gives:
ρl=aR
Dividing both sides byρgives: ρl
ρ =aR
ρ
i.e. l=aR
ρ
Now try the following exercise Exercise 37 Further problems on
transposition of formulae Make the symbol indicated the subject of each of the formulae shown and express each in its simplest form.
1. a+b=c−d−e (d) [d=c−a−b]
2. x+3y=t (y)
y=1
3(t−x)
3. c=2πr (r)
r= c
2π
4. y=mx+c (x)
x=y−c m
5. I=PRT (T )
T= I
PR
6. I=E
R (R)
R=E
I
7. S= a
1−r (r)
⎡
⎢⎣R=S−a S or 1−a
S
⎤
⎥⎦
8. F=9
5C+32 (C)
C=5
9(F−32)
10.3 Further worked problems on transposition of formulae
Problem 8. Transpose the formula:v=u+ft mto make f the subject
Rearranging gives: u+ft
m=v and ft
m=v−u Multiplying each side by m gives:
m ft
m
=m(v−u) i.e. ft=m(v−u) Dividing both sides by t gives:
ft t = m
t (v−u) i.e. f =m t (v−u)
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Transposition of formulae 79
Sec tion 1
Problem 9. The final length, l2of a piece of wire heated throughθ◦C is given by the formula l2=l1(1+αθ). Make the coefficient of expansion, α, the subject
Rearranging gives: l1(1+αθ)=l2
Removing the bracket gives: l1+l1αθ=l2
Rearranging gives: l1αθ=l2−l1
Dividing both sides by l1θgives:
l1αθ
l1θ =l2−l1
l1θ i.e. α=l2−l1
l1θ
Problem 10. A formula for the distance moved by a body is given by: s=1
2(v+u)t. Rearrange the formula to make u the subject
Rearranging gives: 1
2(v+u)t=s Multiplying both sides by 2 gives: (v+u)t=2s Dividing both sides by t gives:
(v+u)t t = 2s
t
i.e. v+u= 2s
t Hence u=2s
t −v or u= 2s−vt t Problem 11. A formula for kinetic energy is k=1
2mv2. Transpose the formula to makevthe subject
Rearranging gives: 1 2mv2=k
Whenever the prospective new subject is a squared term, that term is isolated on the LHS, and then the square root of both sides of the equation is taken.
Multiplying both sides by 2 gives: mv2=2k Dividing both sides by m gives: mv2
m =2k m
i.e. v2=2k
m
Taking the square root of both sides gives:
√v2= 2k
m
i.e. v=
2k m
Problem 12. In a right-angled triangle having sides x, y and hypotenuse z, Pythagoras’ theorem states z2=x2+y2. Transpose the formula to find x
Rearranging gives: x2+y2=z2
and x2=z2−y2
Taking the square root of both sides gives:
x= z2−y2
Problem 13. Given t=2π
l
g find g in terms of t, l andπ
Whenever the prospective new subject is within a square root sign, it is best to isolate that term on the LHS and then to square both sides of the equation.
Rearranging gives: 2π
l g=t Dividing both sides by 2πgives:
l g= t
2π Squaring both sides gives: l
g= t
2π 2
= t2 4π2 Cross-multiplying, i.e. multiplying each term by 4π2g, gives:
4π2l=gt2
or gt2=4π2l
Dividing both sides by t2gives: gt2
t2 =4π2l t2
i.e. g=4π2l
t2 Problem 14. The impedance of an a.c. circuit is given by Z=√
R2+X2.Make the reactance, X, the subject
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80 Engineering Mathematics
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Rearranging gives:R2+X2=Z Squaring both sides gives: R2+X2=Z2
Rearranging gives: X2=Z2−R2
Taking the square root of both sides gives:
X= Z2−R2
Problem 15. The volume V of a hemisphere is given by V=2
3πr3. Find r in terms of V
Rearranging gives: 2
3πr3=V Multiplying both sides by 3 gives: 2πr3=3V Dividing both sides by 2πgives:
2πr3 2π = 3V
2π i.e. r3= 3V 2π Taking the cube root of both sides gives:
√3
r3= 3 3V
2π i.e. r= 3 3V
2π Now try the following exercise
Exercise 38 Further problems on