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Example Of Selection

 ■ Total wright of cargo:W=300kgTotal wright of cargo:W=300kg
 ■ Carrying speed:V=9.5m / m
 ■ Chain pulley transmission coefficientη1=1
 ■ Geaar reducer transmission coefficientη2=0.9
 ■ Operation time:2hour / day
 ■ Start frepuency:1 time / min,medium shock
 ■ power:3 phase 220V,60Hz
Ratio Noticws Caloulation Example Load Condition Example
   Redueion Ratio Notices Calculation Example Ten reduction ratio is based
 on input / output shaft revolutions.
 1.Find the revolution of conveyer pulley(N1) first N1=carrying
    apeed / (pulley D×π)
 2.Find the output shaft revolution of gear reducar(N2)=N2=N1×(chain
    pulleyspeed / gear mumder of reducer)
 3.Caiculate reduction ratio (τ)based on 30,60Hz motor τ=output
    shaft ravlution / input shaft revolution(motor rpm N)
 1.N1=V / (D×π)
    =9.5 / (0.2×3.14)=15r / min
 2.N2=N1 / 1=15(2 / 1)=30r / min
 3.τ =output shaft revolution / input
     shaft ravlution
     =30 / (1800 / Motor rpm N)=1 / 60
Torque Noticws Caloulation Example Load Condition Example
   After reduction ratio is decided,calculate the torpue by the condition of the
 machine used.
 1.Find the torque of conveyer(T1)T1=(μ×Xliad×pulley radius) / η1
 2.Find the torque meeded from the output shaft of reduce(T2)T2=(T1×
    reduction ratio of chain pulley) / η2
 1.T1=W(D / 2)×(1 / η1)
        =300×(0.2/2)=30kg-m
 2.T2=T1×1/2×1 / η2
        =30×1 / 2×1 / 0.9
        =16.7Kg-m
Loa conditions Noticws Caloulation Example Load Condition Example
   1.Find corrective torque(T3) according to operation condition
    T3=T2×opertion comditition(coefficient K)
 T3=T2×K
     =16.7
×1=16.7kgf-m
Horse Power Noticws Caloulation Example Load Condition Example
   1.Find house power(HP) HP=(T×N) / 716.2 Hp=(T / N) / 716.2
     =(16.7×30) / 716.2=0.69<(HP)
Model Selected  According to the model-reduction ratio reference table,model 32.1HP、reduction ratio1/60 are selected.
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