Genel Bilgiler General Information - Stoewer-Getriebe.de
Genel Bilgiler General Information - Stoewer-Getriebe.de
Genel Bilgiler General Information - Stoewer-Getriebe.de
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<strong>Genel</strong> <strong>Bilgiler</strong><br />
<strong>General</strong> <strong>Information</strong><br />
Einführung<br />
Eş<strong>de</strong>ğer güç hesabı Örneği<br />
Çift yönlü çalışan ham <strong>de</strong>mir had<strong>de</strong>si için aşağıdaki<br />
çalışma koşulları belirlenmiş;<br />
Veriler:<br />
Toplam bir iş çevrimi: 2 dak.<br />
1. Yük ka<strong>de</strong>mesi: 48 kNm 30sn<br />
2. Yük Ka<strong>de</strong>mesi: 32 kNm 22sn<br />
3. Yük Ka<strong>de</strong>mesi: 28 kNm 15sn<br />
4. Yük Ka<strong>de</strong>mesi: 16 kNm 10sn<br />
5. Yük Ka<strong>de</strong>mesi: 5 kNm 43sn<br />
Makina sabit <strong>de</strong>vri: 50 d/d<br />
Redüktör seçimine esas olacak eş<strong>de</strong>ğer yük<br />
aranmaktadır.<br />
Çözüm:<br />
Bir çevrimin toplam zamanı;<br />
Eş<strong>de</strong>ğer Tork;<br />
t = t + t + t + t + t = 120sn<br />
%50 eş<strong>de</strong>ğer tork;<br />
1<br />
2<br />
3<br />
1/<br />
6,<br />
6<br />
⎛ 6,<br />
6<br />
6,<br />
6 6,<br />
6<br />
30 48 .... 10 16 43 5 ⎞<br />
T ⎜<br />
⋅ + + ⋅ + ⋅<br />
⎟<br />
e<br />
=<br />
= 39,<br />
2kNm<br />
⎝<br />
120<br />
⎠<br />
0 , 5 ⋅ T<br />
e<br />
= 19,<br />
6kNm<br />
Her bir tork dilimi bu <strong>de</strong>ğerin üzerin<strong>de</strong> olmalı<br />
T < 0,<br />
5 ⋅<br />
4,<br />
5<br />
%50 torkun altındakileri çıkararak hesabı<br />
tekrarlayalım;<br />
1<br />
6,<br />
6 6,<br />
6 6,<br />
6<br />
⎜<br />
⎛<br />
6,<br />
6<br />
⎟<br />
⎞<br />
⎛ 30 48 22 32 15 28 ⎞<br />
T ⎜<br />
⋅ + ⋅ + ⋅ ⎝ ⎠<br />
⎟<br />
e<br />
=<br />
= 42.<br />
9kNm<br />
⎝<br />
67<br />
⎠<br />
Moment ve <strong>de</strong>vir <strong>de</strong>ğerlerini kullanarak eş<strong>de</strong>ğer<br />
gücümüzü hesaplayalım;<br />
Yukarıdaki Güç ve <strong>de</strong>vir <strong>de</strong>ğeri kullanılarak bu<br />
katalogda anlatılan seçim presüdürü ile redüktör<br />
seçimi yapılabilir.<br />
4<br />
T<br />
e<br />
t ' = t − t − t = 120 − 43 − 10 = 67sn<br />
P<br />
e<br />
4<br />
5<br />
Te<br />
⋅n<br />
42.<br />
9⋅1000⋅50<br />
= =<br />
≅ 225kW<br />
9550 9550<br />
5<br />
Equivalent Power Rating Sample<br />
The following data is given for a reversing blooming<br />
mill;<br />
Torque steps:<br />
Total one cycle time: 2 min.<br />
1st torque part: 48 kNm 30 s<br />
2nd torque parti: 32 kNm 22 s<br />
3th torque part: 28 kNm 15 s<br />
4th torque part: 16 kNm 10 s<br />
5th torque part: 5 kNm 43 s<br />
Machine constand speed: 50 rpm<br />
The Equivalent power which is required for gear<br />
unit selection is nee<strong>de</strong>d.<br />
Solution:<br />
Total time in a cycle;<br />
t = t + t + t + t + t = 120 s<br />
Equivalent Torque;<br />
50% of Equivalent torque;<br />
1<br />
2<br />
3<br />
1/<br />
6,<br />
6<br />
⎛ 6,<br />
6<br />
6,<br />
6 6,<br />
6<br />
30 48 .... 1016<br />
43 5 ⎞<br />
T ⎜<br />
⋅ + + ⋅ + ⋅<br />
⎟<br />
e<br />
=<br />
= 39,<br />
2kNm<br />
⎝<br />
120<br />
⎠<br />
0 , 5 ⋅ T<br />
e<br />
= 19,<br />
6kNm<br />
Every torque part must be lower then this value;<br />
T < 0,<br />
5 ⋅<br />
4,<br />
5<br />
STÖWER ANTRIEBTECHNIK GmbH<br />
We are repeating the calculation by taking out the<br />
torque parts which are below 50%;<br />
⎜⎛<br />
1<br />
6,<br />
6<br />
⎟⎞<br />
⎛ 6,<br />
6 6,<br />
6 6,<br />
6<br />
30 48 22 32 15 28 ⎞<br />
T ⎜<br />
⋅ + ⋅ + ⋅ ⎝ ⎠<br />
⎟<br />
e<br />
=<br />
= 42.<br />
9kNm<br />
⎝<br />
67<br />
⎠<br />
By using the equivalent torque and constant speed<br />
we calculate the equivalent power;<br />
Now by using the above calculated equivalent<br />
power and constant speed we can make the gearbox<br />
selection with the prosedures discribed in this<br />
catalogue.<br />
T<br />
e<br />
t ' = t − t − t = 120 − 43 − 10 = 67 s<br />
P<br />
e<br />
4<br />
5<br />
Te<br />
⋅n<br />
42.<br />
9⋅1000⋅50<br />
= =<br />
≅ 225kW<br />
9550 9550<br />
4<br />
5<br />
Åquivalenten Leistung Beispiel<br />
Die nachfolgen<strong>de</strong> Angaben sind gelten für eine<br />
Blechreversierwalze;<br />
Drehmomentstufen:<br />
Gesamte Zeit bei einem Arbeitszyclus: 2 min.<br />
Drehmomentanteil 1: 48 kNm 30 s<br />
Drehmomentanteil 2: 32 kNm 22 s<br />
Drehmomentanteil 3: 28 kNm 15 s<br />
Drehmomentanteil 4: 16 kNm 10 s<br />
Drehmomentanteil 5: 5 kNm 43 s<br />
Maschinen konstante Drehzahl: 50 upm<br />
Gesucht wird die äquivalente Leistung die für<br />
die <strong>Getriebe</strong>auslegung notigwendig ist;<br />
Lösung:<br />
Gesamte Zeit bei einem Arbeitszyclus;<br />
t = t + t + t + t + t = 120 s<br />
1<br />
Åquivalentes Drehmoment;<br />
2<br />
50% von äquivalenten Drehmoment;<br />
3<br />
1/<br />
6,<br />
6<br />
⎛ 6,<br />
6<br />
6,<br />
6 6,<br />
6<br />
30 48 .... 10 16 43 5 ⎞<br />
T ⎜<br />
⋅ + + ⋅ + ⋅<br />
⎟<br />
e<br />
=<br />
= 39,<br />
2kNm<br />
⎝<br />
120<br />
⎠<br />
0 , 5 ⋅ T<br />
e<br />
= 19,<br />
6kNm<br />
Drehmomentanteile müssen größer als dieser Wert;<br />
Wir wie<strong>de</strong>rholen die Berechnung nochmal ohne<br />
die kleine Drehmomentanteile;<br />
Mit <strong>de</strong>r Hilfe <strong>de</strong>s äquivalenten Drehmoments und<br />
konstanter Drehzahl berechnet man die<br />
äquivalenten Leistung;<br />
4<br />
T < 0,<br />
5 ⋅<br />
4,<br />
5<br />
⎜⎛<br />
1<br />
6,<br />
6<br />
⎟⎞<br />
⎛ 6,<br />
6 6,<br />
6 6,<br />
6<br />
30 48 22 32 15 28 ⎞<br />
T ⎜<br />
⋅ + ⋅ + ⋅ ⎝ ⎠<br />
⎟<br />
e<br />
=<br />
= 42.<br />
9kNm<br />
⎝<br />
67<br />
⎠<br />
T<br />
e<br />
t ' = t − t − t = 120 − 43 − 10 = 67s<br />
P<br />
e<br />
4<br />
5<br />
Te<br />
⋅n<br />
42.<br />
9⋅1000⋅50<br />
= =<br />
≅ 225kW<br />
9550 9550<br />
Nach Bestimmung <strong>de</strong>r äquivalenten Leistung und<br />
konstanter Drehzahl, erfogt die <strong>Getriebe</strong>auslegung<br />
wie vorher beschrieben.<br />
5<br />
23