Solution;
For liquid A;
mA˙=1.2lb/s=0.5443kg/s
But;
m˙=ρV˙=S.G(ρwater)V˙
Hence;
VA˙=S.GρwmA˙=1.1×10000.5443
VA˙=4.948×10−4m3/s
For liquid B;
VB˙=0.035ft3/s=9.911×10−4m3/s
Mass flow rate of liquid B is;
mB˙=S.G×ρw×VB˙
mB˙=0.9×1000×9.911×10−4
mB˙=0.8920kg/s
(a)
Time take to fill the tank;
V˙=tV
V=πr2h
V=π×42×15=753.9822in3
V=0.0123555m3
Hence;
t=(4.948+9.911)×10−40.0123555=8.3151
t=8.32s
(b)
Gage and absolute pressure:
Find the density of the liquid mixture;
Mass of mixture,mAB is;
mAB=(mA˙+mB˙)×t
mAB=(0.5443+0.8920)×8.3151
mAB=11.9429kg
Volume is ;
V=0.012355m3
Hence;
ρmixture=Vm=0.01235511.9429=966.65kg/m3
Now;
The gauge pressure;
Pg=hρg=0.381×966.65×9.81=3612.96Pa
Into psi;
Pg=0.5242psi
Absolute pressure;
Pa=Pg+Patm
Pa=0.5241+14.7
Pa=15.22psi
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