Answer to Question #259814 in Mechanical Engineering for Jool

Question #259814

Two liquid streams are pouring in a container, 8 in. diameter and 15in. high. The mass flow rate of stream A is 1.2


lb/sec and the volume flow rate of B is 0.035 ft3/s. If the specific gravities of A and B are 1.1 and 0.9, respectively, in how many seconds will the container be filled? What is the gage and absolute pressure at the bottom of the filled tank in psia.

1
Expert's answer
2021-11-04T00:41:46-0400

Solution;

For liquid A;

mA˙=1.2lb/s=0.5443kg/s\dot{m_A}=1.2lb/s=0.5443kg/s

But;

m˙=ρV˙=S.G(ρwater)V˙\dot{m}=\rho\dot{V}=S.G(\rho_{water})\dot{V}

Hence;

VA˙=mA˙S.Gρw=0.54431.1×1000\dot{V_A}=\frac{\dot{m_A}}{S.G\rho _w}=\frac{0.5443}{1.1×1000}

VA˙=4.948×104m3/s\dot{V_A}=4.948×10^{-4}m^3/s

For liquid B;

VB˙=0.035ft3/s=9.911×104m3/s\dot{V_B}=0.035ft^3/s=9.911×10^{-4}m^3/s

Mass flow rate of liquid B is;

mB˙=S.G×ρw×VB˙\dot{m_B}=S.G×\rho_w×\dot{V_B}

mB˙=0.9×1000×9.911×104\dot{m_B}=0.9×1000×9.911×10^{-4}

mB˙=0.8920kg/s\dot{m_B}=0.8920kg/s

(a)

Time take to fill the tank;

V˙=Vt\dot{V}=\frac Vt

V=πr2hV=πr^2h

V=π×42×15=753.9822in3V=π×4^2×15=753.9822in^3

V=0.0123555m3V=0.0123555m^3

Hence;

t=0.0123555(4.948+9.911)×104=8.3151t=\frac{0.0123555}{(4.948+9.911)×10^{-4}}=8.3151

t=8.32st=8.32s

(b)

Gage and absolute pressure:

Find the density of the liquid mixture;

Mass of mixture,mAB is;

mAB=(mA˙+mB˙)×tm_{AB}=(\dot{m_A}+\dot{m_B})×t

mAB=(0.5443+0.8920)×8.3151m_{AB}=(0.5443+0.8920)×8.3151

mAB=11.9429kgm_{AB}=11.9429kg

Volume is ;

V=0.012355m3V=0.012355m^3

Hence;

ρmixture=mV=11.94290.012355=966.65kg/m3\rho_{mixture}=\frac mV=\frac{11.9429}{0.012355}=966.65kg/m^3

Now;

The gauge pressure;

Pg=hρg=0.381×966.65×9.81=3612.96PaP_g=h\rho g=0.381×966.65×9.81=3612.96Pa

Into psi;

Pg=0.5242psiP_g=0.5242psi

Absolute pressure;

Pa=Pg+PatmP_a=P_g+P_{atm}

Pa=0.5241+14.7P_a=0.5241+14.7

Pa=15.22psiP_a=15.22psi


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