(b) During a certain machining experiment at the UCSI workshop, it is observed that temperature at the tool workpiece interface is 1200 °C at a cutting speed of 300 mm/min with a feed rate of 0.002 mm/rev. (1) Analyse how the temperature will be affected if the cutting speed is increased by 100 %. (ii) Determine the cutting speed necessary to achieve a maximum cutting temperature of 900 °C."
(b) During a certain machining experiment at the UCSI workshop, it is observed that temperature at the tool workpiece interface is 1200 °C at a cutting speed of 300 mm/min with a feed rate of 0.002 mm/rev. (1) Analyse how the temperature will be affected if the cutting speed is increased by 100 %. (ii) Determine the cutting speed necessary to achieve a maximum cutting temperature of 900 °C."
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Please use formulas attached

Transcribed Image Text:(b)
During a certain machining experiment at the UCSI workshop, it is observed
that temperature at the tool workpiece interface is 1200 °C at a cutting speed of
300 mm/min with a feed rate of 0.002 mm/rev.
(1)
Analyse how the temperature will be affected if the cutting speed is
increased by 100 %.
(ii)
Detemine the cutting speed necessary to achieve a maximum cutting
temperature of 900 °C.
![Formula Sheet
A = 2RD14 + DL
F, = F. coso - F, sing
y= tan(6- a) - cot ø
V= TD*L/4
tan ¢ = r cos a/[1-r sin a]
r= (4. - ANA,
Trs = C.(VIA)?.
d= t, - tf
N= waD, F. = F. sino + F.cos
I= VIRR
RAR = vfd
27 D
ADL
I=7 duax = µ R
nc = vwC
A = r Dh +
Tm
RMR = Vwwd
I = VDd HP.
V = nD*/4
33,000
Min no of passes = (t, - ta/duax
ø = 45+.
a B
F,
S =
tow,Vo = tW¡V1 =t,w2V2
A,
T= CIDF
A, =
sinø
Hm = UmV, V= TD*d/4
H=FRt
T= (C/v)'n
U/n
v = TDN
A = 1 D² +2r D² /4
Tm = V/MRR
A, – A
d = D, - D
A.
={(D – 0.9382 p)²
A.
o = Yje = 7, In
Af
-Y,(1+
%3D
tanga
= In
F = Aja4 = A,V; (1+Je In
A = T(R1 + R2){ + (R1 - R),05
tan a.
A
hị +V*/2g = b; +V2g
N= F. cosa - F, sina
=(» - )
0.9743
s = (V - V)/Vr
A = a(R, + R;){h* + (R; – R3);"
M.R.R = wdv
%3D
-
7,-
D. - D,
2 sin o
tan ¢ = r cos a[1-r sin )
t,w,Vo = TWNE
V=A.d:
(1 + s)v, = Vf
s = (V - V,)/v,
1+n
E = Inrx
A,
A.
In-
Af
F = V/Nn
D
$ = 0.88 + 0.12-
Af
= Y; In rx
G= Y/(1 + p/tan a)e(n A/4)
F = F. sina + F, cosa
F.=F,cos (B - a)/cos(o +B - a)
F.=F,sin (B- a)/cos(o - B-a)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F33488049-1422-455e-b50d-4d67f5321948%2F08b48845-3307-40ca-b9bd-9f76e31df780%2Fux2vcqs_processed.png&w=3840&q=75)
Transcribed Image Text:Formula Sheet
A = 2RD14 + DL
F, = F. coso - F, sing
y= tan(6- a) - cot ø
V= TD*L/4
tan ¢ = r cos a/[1-r sin a]
r= (4. - ANA,
Trs = C.(VIA)?.
d= t, - tf
N= waD, F. = F. sino + F.cos
I= VIRR
RAR = vfd
27 D
ADL
I=7 duax = µ R
nc = vwC
A = r Dh +
Tm
RMR = Vwwd
I = VDd HP.
V = nD*/4
33,000
Min no of passes = (t, - ta/duax
ø = 45+.
a B
F,
S =
tow,Vo = tW¡V1 =t,w2V2
A,
T= CIDF
A, =
sinø
Hm = UmV, V= TD*d/4
H=FRt
T= (C/v)'n
U/n
v = TDN
A = 1 D² +2r D² /4
Tm = V/MRR
A, – A
d = D, - D
A.
={(D – 0.9382 p)²
A.
o = Yje = 7, In
Af
-Y,(1+
%3D
tanga
= In
F = Aja4 = A,V; (1+Je In
A = T(R1 + R2){ + (R1 - R),05
tan a.
A
hị +V*/2g = b; +V2g
N= F. cosa - F, sina
=(» - )
0.9743
s = (V - V)/Vr
A = a(R, + R;){h* + (R; – R3);"
M.R.R = wdv
%3D
-
7,-
D. - D,
2 sin o
tan ¢ = r cos a[1-r sin )
t,w,Vo = TWNE
V=A.d:
(1 + s)v, = Vf
s = (V - V,)/v,
1+n
E = Inrx
A,
A.
In-
Af
F = V/Nn
D
$ = 0.88 + 0.12-
Af
= Y; In rx
G= Y/(1 + p/tan a)e(n A/4)
F = F. sina + F, cosa
F.=F,cos (B - a)/cos(o +B - a)
F.=F,sin (B- a)/cos(o - B-a)
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