A cast iron flywheel is requires to absorb 25000 N-m of energy as speed is increased from 120 to 125 rpm. If the wheel is to be solid disc having a diameter 8 times its thickness. Determine its diameter. Density of $C.I.=7200 kg/m^3$.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 8 Marks

Year: Dec 2015

ΔE=20000Nm

$w_1 =120 =2ΠN_1 /60=12.56 rad/sec \\ w_2 =125=2ΠN_2 /60=13.08 rad/sec \\ ρ=.7200g/m^3$

b=5t [assumed not given in the problem]

d=8t

to find d=?

m=vol.*density

=ΠDA*ρ

=ΠD(bt)ρ

=ΠD(5tt)7200

$Cs=[w_1-w_2]/w \\ N=(N_1+N_2)/2=(120+125)/2 \\ =245/2=122.5 rpm \\ W=2ΠN/60 \\ =12.82 rad/sec$

Cs=0.0408

$ΔE=m * v_2 * Cs \\ 25000=Π*D*5t2*7200*0.04*[Π*8t*122.5/60]2 \\ 25000=Π*8t*5t2*7200*0.04*[Π264*t2*15006.25/3600] \\ =Π3*40*t5*7200*64*15006.25/3600 \\ =2382274250*t5 \\ t5=1.04*10-5 \\ t=.100 M \\ t=100mm \\ D=8t \\ =8*100 \\ D=800mm$