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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 coefficient:η1=1 ■ Geaar reducer transmission coefficient:η2=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 |
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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 |
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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 |
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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) |
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Model Selected | According to the model-reduction ratio reference table,model 32.1HP、reduction ratio1/60 are selected. |
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