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Mechanical engineering Questions and solutions set - 6

Mechanical engineering Questions and solutions set - 6
Question:-
Option (A)
3.17kN
Option (B)
4.17kN
Option (C)
5.17kN
Option (D)
8.17kN
Correct Option:
B
Question Solution:
Reynolds similarity low (Rc)m=((VmLn)/Vm) = (Re)p= ((VpLp)/Vp)
Vr = (Vr/Lr)
If vr=I ( i.e Pm = Pp, mm = mp), then Vr (1/Lr)
Vm =(Vp/Lr)=60 x 6 = 360 Km/hr = 100 m/sec
Force ratio = (fm/fp)=( m2r/pr) If Pr=1 & mr =1 and (Fm/Fp) = 1.0 then fp= 250 N
Power to overcome drag in the prototype Pp = Fp Vp = 250 x (60 x 103)/(3600)= 4167 W = 4.17 kw
Question:-
Option (A)
p2ET/L2
Option (B)
2p2EI/L2
Option (C)
p2EI/4L2
Option (D)
4p2EI/L2
Correct Option:
A
Question Solution:
Standard
Question:-
Option (A)
P will be in tension and Q in compression
Option (B)
Both P&Q will be compression
Option (C)
P will be in compression and Q in tension
Option (D)
Both P&Q will be in tension
Correct Option:
C
Question Solution:
As P will expand more than Q . therefore P will be in compression and Q in tension
Question:-
Option (A)
0.21 pas
Option (B)
0.23 pas
Option (C)
0.29 pas
Option (D)
0.32 pas
Correct Option:
B
Question Solution:
The torque is transmitted through the fluid layer to the outer cylinder and the gap between the cylinder is small
Þ tangential velocity of inner cylinder rw=(0.12m)(.2 p rad/sec )
The small space between cylinder the velocity gradient radius can be used then
(dv/dy) = (0.767)/(0.128-9.122) = 127.851
Torque applied = tongue resisting
0.881 = t (area)(arm)
= t (2p x 0.125 x 0.305 )(0.125)
t= 29.4 pas
m = (t/(dv/dy))=(29.4/127.8)= 0.230 pas
Question:-
Option (A)
37.7%
Option (B)
47.7%
Option (C)
57.7%
Option (D)
67.7%
Correct Option:
C
Question Solution:
Compressor :-
(T1’/T1)=(P2/P1)(r-1/r) =(4)(0.4/1.9) = 1.486
T1 = (273+25) x 1.486 = 442.8

T12-T1=144.8k
hc = 0.75 = (T12-T1) / (T2-T1)=(144.8)/(T2T1) =T2= 491.KK
T2-T1=193.1K
Turbine :-
For zero % cycle efficiency net work % is zero turbine work = compressor work
T3-T4= T2-T1
hT(T3-T4)=(T12-T11/hc)
(T3/T4)=(P3/P4)(r-1/r)=1.486
T14= ((273+750)/1.486) = 688.4K
ht(1013-688.4)=193.1
ht=(193.1/334.6)=0.577 or 57.7 %
Question:-
Option (A)
94.73, 103.27mm
Option (B)
92.73, 105.27mm
Option (C)
94.27, 103.73mm
Option (D)
None of the above
Correct Option:
A
Question Solution:
Given 2x = 30° or x= 15°
Pn=0.35 N/mm2, b=(R/3) , m=0.15
P=22.5 KW = 22.5 X 103W, N = 2000rpm , w=(2p x 2000)/(60) = 209.5 red/sec
r1= outer radius of the contact syfoce in mm
r2= inner radius of the contact syfoce in mm
R= mean with radius of the contact syfoce in mm
B= face with radius of the contact syfoce in mm = (R/3)
T= torque transmitted by the clutch N-m
Power transmitted = 22.5 x 103 = T.W =T x 209.5
T=107.4 N-m= 107.4 x 103 N-mm
Toque transmitted = 107.4 x 103 = 2 p m PnR2b
=2p x 0.15 x 0.35 x R2 x (R/3) = 0.11R3
R3 = (107.4 x103)/(0.11)=976 x 103
TR= 99 red /s
Dimensions of the constant surfaces
R1-r2 = b sinx =(R/3) sinx =(99/3) sin15° = 8.54 mm
Or r1+r2=2R=2 x 99 = 198 mm
R1= 103.27,
R2 = 94.73 mm