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1.- Un gas cuyo estado se define en el diagrama P-V punto 1, pasa al estado 2 siguiendo el camino 1-C-2. al ocurrir esto al gas se le suministran 80 kj de energía en forma de calor y se obtienen de el 30 kj de trabajo. Después este mismo gas retorna al estado inicial en el proceso definido por la curva 2-a-1.
¿Qué cantidad de energia en forma de calor habra que suministrar en otro proceso 1-d-2 para obtener del gas 10 kj de trabajo, que cantidad de calor se extraen en el proceso 2-a-1, si al efectuarse este gastan en la compresión 50 kj de energia en forma de trabajo?
q=80 kj =W-30kj
q2=? W-10 kj
1-d-2 W-50kj
1+q1-W1=2+q2-w2
q1- w1 = q2-w2
80kj-30kj = q2-10kj
80kj-30k + 10kj = q2
q2 = 80kj-20 kj = 60kj
1+q1-W1=2+q2-w2
q1- w1 = q2-w2
80kj-30kj = q2-50kj
80kj-30k + 50kj = q2
q2 = 80kj+20 kj = 100kj
2.-¿Qué velocidad debe llevar una bala de plomo para que al chocar con una plancha de acero se funda totalmente? Se supone que en instante del choque la temperatura de la bala es de 27 °C. la temperatura de fusión del plomo es de 327 °C. el calor de fusión es de 20.939 kj/kg calor especifico() es de 0.1256 kj/kg °k.
V=?
T1=27°c +273.15=300.15°k
T1=327°c +273.15=600.15°k
20.939 kj/kg
Cp=0.1256 kj/kg °k
= "((125.6*300+20939.00)*2) ="(117,238.0)
v=342.40 m/s
3.-El agua con una densidad de 1000 kg/m3 fluye en una tubería de Ø 0.05 m; si en la tubería fluyen 0.5 m3 /s determine el gasto masico que fluye en kg/hr. ¿Cuál es la velocidad del fluido?
1000 kg/ m3
Ø =0.05 m.
Q=0.5 m3 /s
V=?
M=?
Q= *V*A ; m=*Q=(1000*.5)*3600= 1,800,000.0 kg/hr.
V= Q/A ; V= 0.5 / (.05*.5)2 *= 0.5/.001963= 254.65 m/s
Universidad de Quintana Roo. DCI
Ing. En sistemas de Energía.
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Enviado por: Meggatron Idioma: castellanoPaís: México