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TEMAT: PROJEKT MECHANIZMU JAZDY |
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WCIĄGARKI . |
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1. ZAŁOŻENIA DO ZADANIA |
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siła cieżkości udżwigu |
Q |
63 |
kN |
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względny czas pracy silnika |
e |
25 |
% |
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siła ciężkości zblocza |
Qo |
1100 |
N |
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prędkość jazdy wózka |
V`j |
19,2 |
m/min |
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sila ciężkości wózka |
Gw |
30 |
kN
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prędkość obrotowa silnika |
ns |
1000 |
obr/min |
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a) . Przyjeto rownomierny nacisk na wszystkie koła wózka |
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2.NACISK NA KOłA JEZDNE WÓZKA (Pmax) |
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Q +Qo+Gw |
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94100 |
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Pmax = |
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23525,00 |
N |
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4 |
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4 |
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3. DOBÓR KOŁA JEZDNEGO |
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a) dobór szyny jezdnej - przyjęto |
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szerokość szyny |
K |
50 |
mm |
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wysokość szyny |
H |
25 |
mm |
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promień zaok. główki |
r1 |
3 |
mm |
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b) dobór średnicy nominalnej koła jezdnego Dk |
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max. nacisk na koło jezdne ------- |
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Pmax = |
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23525
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dop. nacisk powierz.koła z szyną --- |
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Pdop = |
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750
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(wg tabeli 6.4 ) |
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czynna szer. styku koła i szyny --- |
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bc = |
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44
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{ bc =K - 2r1} |
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6000 |
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Dk = |
0.2 *(Pmax / bc)*{(600/Pdop)^2 } |
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68,44
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c) dobór zestawów kołowych wg |
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ZESTAW KOŁOWY NAPĘDZANY -------------- |
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250-A002DNk |
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ZESTAW KOŁOWY NIENAPĘDZANY --------- |
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250-A004DNi |
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c 1 )parametry zestawów |
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średnica nominalna koła |
Dk |
250 |
mm |
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średnica zewnętrzna |
D1 |
280 |
mm |
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średnia średnica łożyska obliczona |
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wg wymiarów z katalogu łożysk |
d1 |
95 |
mm |
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szerokość bieżni |
b |
65 |
mm |
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średnica wału wyjściowego |
dw |
50 |
mm |
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długości czopa wału zestawu kół jezdn. |
l |
110 |
mm |
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nr łożyska tocznego |
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22312 |
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c 2 ) sprawdzenie luzu obustronnego między obrzeżami koła jezdnego a szyną . |
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{ b - K } > { (b - K)min } |
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wartość ( b - K )min podana jest w |
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(wg PN - 75 / M -84601) |
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( b - K ) = |
15 |
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( b - K)min = |
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10 |
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WARUNEK SPEŁNIONY |
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4 . OPORY JAZDY |
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a ) ustrój obciążony (W) |
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wsp . tarcia koła o szynę |
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b |
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1 |
wg tabeli 6.5 (skrypt) |
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wsp . tarcia potoczystego |
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f |
mm |
0,3 |
wg tabeli 6.6 (skrypt) |
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wsp . tarcia w łożyskach |
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m |
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0,015 |
wg tabeli 6.7 (skrypt) |
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G = Q+Qo+Gw |
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G |
kN |
94,1 |
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śr . nominalna koła jezdnego |
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Dk |
mm |
250 |
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śr . czopa |
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d |
mm |
95 |
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2f + md |
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W = ( 1 + b ) * G * |
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N |
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Dk |
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b ) ustrój nieobciąźony (Wo) |
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2f +md |
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Wo =( 1 + b )* (Qo+Gw )* |
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N |
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Dk |
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5 . WSTĘPNY DOBÓR SILNIKA |
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5 a . ) wymagane obroty koła jezdnego ( nk ) |
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Vj |
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nk = = |
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24,46
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p * Dk |
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5 b .) przełożenie reduktora |
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ns |
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i = =
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nk |
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PRZYJĘTO PRZEKŁADNIĘ |
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5 c . ) sprawność napędu obciążonego |
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wg tabeli 6.2 (skrypt) |
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h= h1 * h2 |
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sprawność jednego stopnia |
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0,98 |
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h = |
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0,960 |
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5 d . ) moment oporu zredukowany na wał silnika ( ustrój obciążony ) |
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W * Dk |
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M` =
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4,85
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2 * i * h |
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5 e . ) sprawność napędu nieobciążonego (w zależności od wsp . k) |
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Qo + Gw |
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k = |
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0,3305 |
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Q + Qo + Gw |
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DLA k = |
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0,3305 |
ODCZYTANO WARTOŚĆ Z WYKRESU |
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SPRAWNOŚCI ( skrypt tablica 6 . 1 ) |
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ho= |
0,82 |
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5 d . ) moment oporu zredukowany na wał silnika ( ustrój nieobciążony ) |
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Wo * Dk |
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Mo` =
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Nm |
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2 * i * ho |
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5 e . ) moment średni |
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Mśr = |
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= |
3,68
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2 |
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5 f . ) moc średnia |
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Pśr = 0,1047 * Mśr * ns * 10 |
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-3 |
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prędkość obrotowa silnika |
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Pśr = |
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0,385
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1000 |
obr/min |
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0,385 |
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Dla mocy śr. Pśr = |
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0,385 |
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silnik asynchroniczny pierścieniowy |
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parametry |
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typ silnika |
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SUDf 100 L-6A |
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moc znamioowa |
P |
1,1 |
kW |
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względny czas pracy |
e |
25 |
% |
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wsp. przeciążenia momentem |
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mr |
1,6 |
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prędkość obrotowa |
n |
850 |
obr/min |
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moment bezwł. silnika |
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Is |
0,011 |
kgm
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śr. czopa końc. wału sil. |
d |
28 |
mm |
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6 . DOBÓR PRZEKŁADNI ZĘBATEJ |
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Przełożenie reduktora |
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wg katalogu PRZEKŁADNIE ZĘBATE dobrano przekładnię |
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40,89 |
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zębatą o parametrach |
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; |
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przełażenie ip |
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40,2 |
- |
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moc Pt |
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2,3 |
kW |
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śred. czopa szybkoobr. |
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25 |
mm |
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śred. czopa wolnoobr. |
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45 |
mm |
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wielkość przekładni |
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250 |
mm |
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(odległość między wałkami skrajnymi, a) |
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6 . a ) poprawione momenty oporu |
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- ustrój obciążony |
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- ustrój nieobciążony |
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i |
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i |
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M = M` * |
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Mo = M`o * |
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ip |
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ip |
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M= |
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4,94
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Mo = |
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1,91
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6 . b ) poprawiona prędkość jazdy wózka |
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- prędkość obrotowa kół |
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- prędkość wózka |
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n |
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nk =
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Vj = p *Dk * nk |
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ip |
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nk = |
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21,14
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Vj = |
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16,60
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#REF! |
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7 . DOBÓR HAMULCA I SPRZĘGŁA SILNIKOWEGO |
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7 a . ) moment oporu na wale silnika przy hamowaniu przyjęto |
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ustrój obciążonny M = |
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4,94 |
Nm |
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ustrój nieobciążonny Mo = |
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1,91 |
Nm |
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7 b . ) wstępny dobór hamulca i bębna hamulcowego |
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-moment hamowania |
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Mh > M = |
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4,94 |
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Nm |
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Dla Mh dobiera się hamulec szczękowy wg |
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o oznaczeniu HAMULEC SZCZĘKOWY |
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120-EB 120-40 C120.500V AC/50Hz-261AHH |
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ze zwalniakiem |
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EB 120-40 C120 |
( typ ) |
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Parametry hamulca |
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szerokość szczęki |
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40 |
mm |
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śr . bębna hamulcowego |
Dh |
120 |
mm |
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moment znamion. hamul. Mhk |
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50 |
Nm |
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Ze względu na przyjęty hamulec i średnicę czopa wału końcowego silika d = |
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28
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dobrano bęben ham. wg |
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120-50-60-28/60-120BH |
( OZNACZENIE ) |
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Parametry bębna hamulcowego |
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moment bezwł . |
It |
0,004 |
kgm
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śr . bębna hamulcowego |
Dh |
120 |
mm |
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szerokość bębna |
B |
50 |
mm |
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7c ) dobór sprzęgła silnikowego zębatego |
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moment obliczeniowy sprzęgła |
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Mnom > M * k1 * k2 * k3 = Msp |
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wsp . dynamiczny |
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k1 |
1,5 |
( tabela 2 normy ) |
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k2 |
1 |
( tabela 3 normy ) |
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Mnom > |
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7,9 |
Nm |
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k3 |
1,07 |
( tabela 4 normy ) |
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Dla Msp = |
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7,9 |
Nm dobiera się wg ZN -81/1232 - 32356 |
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silnik |
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przekładnia |
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śr. czopa końc. wału sil. |
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= |
28
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śred. czopa szybkoobr. |
= |
25
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SPRZĘGłO ZĘBATE DWUSTRONNE |
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wyk. 8/5 M-001 Asz |
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o parametrach |
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śr. nominalna tarczy sprzęgła |
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D |
mm |
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160 |
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moment bezwładności |
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Isp |
kgm |
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0,03 |
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moment nominalny |
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Mnom |
Nm |
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710 |
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Mnom = |
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710 |
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Nm > M sp = |
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7,9 |
Nm |
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WARUNEK SPEŁNIONY |
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8 . SPRAWDZENIE HAMULCA NA CZAS ZAHAMOWANIA |
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I SPRAWDZENIE CIEPLNE |
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a . ) redukcja momentów bezwładności na wał silnika przy hamowaniu |
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-- ustrój obciążony |
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( Q + Qo + Gw )* Dk |
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Izrh = d * ( Is + Isp + It ) + |
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4 * g * ip |
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wsp . uwzgl . pozostałe masy wirujące ( 1,06 -1,20 ) |
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d |
1,12 |
--- |
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przyspieszenie ziemskie |
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g |
9,81 |
m/s |
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z wykresu sprawności hh = h1h *...hnh |
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hh |
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0,87 |
--- |
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Izrh = |
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0,131 |
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0,0504 |
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0,080687784055062 |
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94100 |
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-- ustrój nieobciążony |
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Dla k = |
0,330 |
odczytano wartość h oh --------- |
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0,885 |
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( Qo + Gw )* Dk |
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0,027127050269822 |
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Izroh = d * ( Is + Isp + It ) + |
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* hoh |
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0,077527050269822 |
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4 * g * ip |
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Izroh = |
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0,078 |
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b . ) dopuszczalne maksymalne opóźnienia przy hamowaniu |
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algorytm mechanizmu jazdy, skrypt nr 1124 - strona 64; skrypt nr 1553 - strona 70 |
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przyjęto wsp . tarcia koła o szynę m1 = |
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0,2 |
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g |
4f + md |
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ahmax < |
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* m1 + |
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ahmax < |
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1,033 |
m / s
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2 |
Dk |
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2,625 |
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0,0105 |
0,2105 |
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c . ) dopuszczalny maksymalny moment hamujący ( bez poślizgu kół ) |
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2 * ahmax * ip * Izroh |
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Mo |
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MHmax = |
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- ho * |
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25,10 |
Nm |
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Dk |
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1 + ß |
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25,7430744279433 |
0,642307835820895 |
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SPRAWDZENIE |
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POWINNO BYĆ ; MHmax > Mhk |
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JEST ; |
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MHmax = |
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25,1
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Mhk = |
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50,0
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HAMULEC NALEŻY WYREGULOWAĆ Z MOMENTU Mhk = |
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50
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na Mhrz < |
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25
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d . ) czas zahamowania |
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2,27620175373134 |
27,4 |
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-- ustrój obciążony |
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Izrh |
n |
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th = |
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*
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0,426 |
s |
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9,55 |
h * M |
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Mhrz + |
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1 + ß |
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-- ustrój nieobciążony |
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0,642307835820895 |
25,7 |
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Izroh |
n |
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toh = |
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*
|
|
0,268 |
s |
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9,55 |
ho * Mo |
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Mhrz + |
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1 + ß |
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|
--- granice czasu zahamowania |
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Vj |
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thmin = |
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= |
0,268
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ahmax |
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Sprawdzenie |
|
thmax = |
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2 ÷ 5 s |
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5 |
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toh |
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0,268
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thmin |
|
< < thmax
|
0,27 |
s |
< < |
2 ÷ 5 s |
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th |
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0,426
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|
e . ) sprawdzenie cieplne hamulca dla Vj < 60 m/min |
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p * V < ( p * V ) dop = ( 1,5 ÷ 3 ) MPa *m/s |
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--prędkość obwodowa bębna hamulcowego V |
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p * DH * n |
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V = |
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V = |
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5,34
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60 |
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-- jednostkowy nacisk powierzchniowy p |
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2 *MHrz |
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1 |
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p = |
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*
|
* 10 |
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m * DH |
|
DH |
a |
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4 * b sin |
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45 |
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2 |
2 |
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0,785 |
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wsp . tarcia (hamulec / bęben) |
|
|
m |
0,3 |
--- |
( skrypt tablica 6 . 8 ) |
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|
kąt opasania bębna (80 lub 90°) |
|
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a |
90 |
° |
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p = |
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0,19
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p * V = |
|
0,99 |
MPa*m/s < ( pV ) dop = ( 1,5 ÷ 3 ) MPa * m/s |
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|
HAMULEC SPEŁNIA WARUNEK CIEPLNY |
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9 . SPRAWDZENIE WSTĘPNIE PRZYJĘTEGO SILNIKA NA CZAS |
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ROZRUCHU I GRZANIE - METODĄ MOMENTÓW ZASTĘPCZYCH |
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a . ) sprawdzenie silnika na czas rozruchu |
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--- moment rozruchowy średni |
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przyjęto spadek napięcia w sieci D U = |
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10% |
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30 * P |
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Mrśr = 0,85 Mmax = 0,85 * Mn * mr * ( 0,9 ) = 0,85 * * 10 * 2,4 * 0,81 |
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p * n |
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Mrśr = |
|
13,6 |
Nm |
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--- zredukowany na wał silnika moment bezwładności ustroju obciążonego |
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( Q + Qo + Gw )* Dk |
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Izr = d * ( Is + Isp + It ) + |
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4 * g * ip * h |
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0,096568699889249 |
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Izr = |
|
0,147
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|
--- czas rozruchu ustroju obciążonego |
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Izr n |
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tr = *
|
|
|
1,506
|
|
2,61619675195521 |
48,8225253261009 |
|
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|
|
|
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|
9,55 Mrśr - M |
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7,55 |
53,76 |
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|
t rmin < t r < t rmax |
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0,27 |
5 |
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0,268 |
< |
1,506 |
< |
5
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1,50618903792544 |
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|
WARUNEK SPEŁNIONY |
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13,6203821656051 |
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|
--- zredukowany na wał silnika moment bezwładności ustroju nieobciążonego |
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1,50618903792544 |
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( Qo + Gw )* Dk |
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|
Izro = d * ( Is + Isp + It ) + |
|
|
|
0,09
|
0,037380529516084 |
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|
4 * g * ip * hoh |
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|
--- czas rozruchu ustroju nieobciążonego |
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Izro n |
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tor = *
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|
|
0,67
|
|
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|
9,55 Mrśr - Mo |
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|
b . ) sprawdzenie silnika na grzanie |
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Mn > Mz = |
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T c * e |
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|
|
|
--- moment nominalny i maksymalny |
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30 * P |
|
|
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|
|
M n =
|
|
12,36
|
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|
p * n |
|
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M max = |
|
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|
|
16,02
|
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|
|
|
|
--- średni czas trwania cyklu |
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47,5 |
44,6 |
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|
Tc * e - tr - t or - t h - toh |
|
|
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|
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|
t śr = |
|
|
|
22,32
|
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2 |
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|
gdzie T c = |
|
|
190 |
-- czas trwania cyklu pracy |
|
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|
Stąd |
|
|
|
|
558,058494893534 |
625,070341282185 |
1183,12883617572 |
0,25 |
|
|
|
|
|
|
|
|
|
|
|
|
4,99078906571257 |
|
24,9079754984362 |
|
|
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|
|
|
|
|
|
M z = |
|
4,99
|
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|
M z = |
|
4,99 |
< M n = |
|
12,36
|
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|
SILNIK SPEŁNIA WARUNEK NA GRZANIE |
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|
10. DOBÓR SPRZĘGIEŁ ZĘBATYCH NA WALE WOLNOOBROTOWYM wg |
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|
|
ZN - 79 /1232-32341 |
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|
a . moment obrotowy na wale w ruchu ustalonym przy ustroju obciążonym |
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W * Dk |
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|
|
Mw = =
|
190,55 |
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2G |
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|
b . moment dynamiczny |
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Izr n |
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|
|
Mdyn = * * ip * h
|
|
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9,55 tr |
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335,30 |
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c . moment nominalny sprzęgła |
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k1 |
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Mspnom > ( Mw + Mdyn )k2 * k3
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k2 |
1 |
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g2 |
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k3 |
1,07 |
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335,30 |
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1,506 |
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Mspnom > |
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281,3 |
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DLA |
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średnicy czopa wału wolnoobrotowego reduktora d2m7 = |
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45 |
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średnicy czopa wału zestawu kół jezdnych dm7 = |
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50 |
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długości czopa wału zestawu kół jezdnych lw = |
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110 |
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długości czopa wału l2 = |
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110
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średnica wału pędnego dw = |
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50 |
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DOBRANO z normy |
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sprzęgła zębate |
jednostronne |
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o oznaczeniach : |
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od strony reduktora |
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wyk. 22/26 M-002 Asg |
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Moment nominalny |
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1400 |
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od strony zestawu kołowego |
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wyk. 26/26 M-002 Asg |
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Moment nominalny |
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1400 |
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sprzęgła zębate |
dwustronne |
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o oznaczeniu : |
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wyk. 22/26 M-003 Asz |
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Moment nominalny |
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3150 |
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