这不是流体力学,是热力学,解答如下,如有问题,呼我.
a)
1)open system 2) closed system
b)
i)mass flow rate=v*A*rou=105*0.05*(1/0.8)=6.5625kg/s
ii)conseveration of mass:Exit area=mass flow rate/(v*rou)=6.5625/135*1.5=0.073m^2
iii)steady state:
o=q-w+delta entropy+(v1^2-v2^2)/2----->w=-27+145-3.6=114.4KJ/Kg
power=mass flow rate *w=114.4*6.5625=750KW.
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a)Assume:I)NO HEAT LOSS.II)STEADY STATE
since pv^1.4=constant=300000*0.346^1.4=67893.p2=120000pa
power=mass flow rate*w=mass flow rate*integal-vdp=mass flow rate*67893^(1/1.4)*(1-1.4^-1)^-1*-1*(120000^0.2857-300000^0.2857)=800w
solve for mass flow rate=0.0096kg/s
b)1-->2.
v1=0.11/1.32=0.00833.v2=3*v1=0.25 since pv^2=c=p1v1^2=12*10^5*0.0833^2=8333.3,thus,p2=133333pa
w12=integal pdv=-8333.3*(0.25^-1-0.0833^-1)=66.71kJ/kg
2--->3
w23=integal pdv=p2*(v3-v2)=133333*(0.0833-0.25)=-22.23kj/kg
3--->1 no work
therefore, net work=m(w12+w23)=1.32(66.7-22.23)=1.32*44.48=58.7kj
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3
a)
idea gas law:pv=RT,v2=115mL
b)
simple
c)
1)pv=psat*relative humidity=0.75*3.1698kpa=2.377kpa
pa=100-2.377=97.62kpa
2)w=0.6219*(2.377/97.62)=0.015g/g
3)h=(ha*ma+hv*mv)/(mv+mw)
ha=cpT=1.005*(25+273)=299.5KJ/KG(dry air)
h=332.7kj/kg(dry air +vapor)
4) i) The vapor can be condensed when temperature goes down, therefore, the mass of the vapor varies as the temperature.
ii) It is the same that when the temperature rises up, more water maybe absorbed by the air.