Question

Ammonia at 70 F with a quality of 50% and a total mass of 4.5 lbm is in a rigid tank with an outlet valve at the bottom. How much liquid mass can be removed through the valve, assuming the temperature stays constant

Answer 1

Answer:

0.10865 killograms

Step-by-step explanation:

calculating the liquid mass of ammonia removed through the bottom value from a rigid tank at constant temperature.

Given:

temperature: $T=70 F$

quality : 50% = 0.5

initial mass: $m1= 4.5 lbm$

to find the removed liquid mass first we have to find total volume from which we can find remaining mass. as the tang is rigid the temperature and volume remains constant.

by taking the difference of mass we can determine the mass of liquid removed.

 we have two phases at temperature $T= 70 F$ with specific volume for liquid  $vf=0.02631 ft^3/lbm$  and specific volume for vapor is  $vg=2.3098 ft^3/lbm$ .

The Volume in the initial state is given by, (Using definition of specific volume)

                                          $V= m1v1$

 using $v1=x(vf+vg)$

                                    $V= m1x(vf+vg)$

substituting $m1= 4.5 lbm$ , $vf= 0.02631 ft^3/lbm$ , $vg=2.3098 ft^3/lbm$

we get

        $V= (4.5 lbm)(0.5)(0.02631 ft^3/lbm +2.3098 ft^3/lbm)$  

finally          $V=5.2625 ft^3$  

we know the formula to find liquid mass is

$mass =density *volume$

density of ammonia is  $0.73 kg/m^3$

inserting the values into the formula we get the value for liquid mass removed through the valve.

$m = (0.73 kg/m^3)*(5.25625 ft^3)$

the final answer is

                           $m= 0.10865 kg$