Mecatronica

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Problemas



















1.-Un motor de 2HP de potencia trabaja durante 10hrs, 15min, 3/4de hr, 26hrs, 1 día 36hrs 45min y 38seg?
¿Calcular el trabajo realizado y además expresarlos en joules, Kgm, ergios y Kwh?

1HP= 75 Kgm/S 10hrs= 36,000s 3/4 hr = 2,700s1 día 36hrs 45min y 38seg= 218,7385s
2HP= 150 Kgm/S 15min= 900s 26hrs= 93,600s
T= (150kgm/s) (36,000s) = 5,400,000kgm
10hrsT= (5,400,000kgm) (9.8J) = 52,920,000 joules
T= (5,400,000kgm) (9.8×〖10〗^7erg) = 5.292×〖10〗^14erg
T= (5,400,000kgm)(2.72×〖10〗^(-6)kwh) = 14.688kwh

T= (150kgm/s) (900s) = 135,000kgm
15min T= (135,000kgm) (9.8J) = 1,323,000 joulesT= (135,000kgm) (9.8×〖10〗^7erg) = 1.323×〖10〗^13erg
T= (135,000kgm) (2.72×〖10〗^(-6)kwh) = 0.3672kwh

T= (150kgm/s) (2,700s) =405,000kgm
3/4 Hr T= (405,000kgm) (9.8J) = 3,969,000 joules
T= (405,000kgm) (9.8×〖10〗^7erg) = 3.969×〖10〗^13ergT= (405,000kgm) (2.72×〖10〗^(-6)kwh) = 1.1016kwh

T= (150kgm/s) (93,600s) = 14,040,000kgm
26hrs T= (14,040,000kgm) (9.8J) = 137,592,000joules
T= (14,040,000kgm) (9.8×〖10〗^7erg) = 1.37592×〖10〗^15erg
T= (14,040,000kgm) (2.72×〖10〗^(-6)kwh) = 38.1888kwhT= (150kgm/s) (218,732s) = 32,810,700kgm
1 día 36hrs 45min T= (32,810,700kgm) (9.8J) = 321,544,860...