Question

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.

Answer 1

Solution;

For liquid A;

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

But;

$\dot{m}=\rho\dot{V}=S.G(\rho_{water})\dot{V}$

Hence;

$\dot{V_A}=\frac{\dot{m_A}}{S.G\rho _w}=\frac{0.5443}{1.1×1000}$

$\dot{V_A}=4.948×10^{-4}m^3/s$

For liquid B;

$\dot{V_B}=0.035ft^3/s=9.911×10^{-4}m^3/s$

Mass flow rate of liquid B is;

$\dot{m_B}=S.G×\rho_w×\dot{V_B}$

$\dot{m_B}=0.9×1000×9.911×10^{-4}$

$\dot{m_B}=0.8920kg/s$

(a)

Time take to fill the tank;

$\dot{V}=\frac Vt$

$V=πr^2h$

$V=π×4^2×15=753.9822in^3$

$V=0.0123555m^3$

Hence;

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

$t=8.32s$

(b)

Gage and absolute pressure:

Find the density of the liquid mixture;

Mass of mixture,m_AB is;

$m_{AB}=(\dot{m_A}+\dot{m_B})×t$

$m_{AB}=(0.5443+0.8920)×8.3151$

$m_{AB}=11.9429kg$

Volume is ;

$V=0.012355m^3$

Hence;

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

Now;

The gauge pressure;

$P_g=h\rho g=0.381×966.65×9.81=3612.96Pa$

Into psi;

$P_g=0.5242psi$

Absolute pressure;

$P_a=P_g+P_{atm}$

$P_a=0.5241+14.7$

$P_a=15.22psi$