Question

A 380-L tank contains steam, initially at 400C, 3 bar. A valve is opened, and steam flows out of the tank at a constant mass flow rate of 0.005 kg/s. During steam removal, a heater maintains the temperature within the tank constant. Determine the time, in sec, at which 75% of the initial mass remains in the tank. Also, determine the specific volume, in m^3 /kg, and pressure, in bar, in the tank at that time.

Answer 1

Step-by-step explanation:

The given data is as follows.

Tank volume (V) = 380 L = 0.38
`m^(3)`

Initial pressure (
`P_(i)`) = 3 bar

Temperature (t) =
`400^(o)C`

Outlet mass flow rate (
`m_(o))`) = 0.005 kg/s

Final mass (
`m_(f)`) =
`0.75m_(i)`

First, we will calculate the initial mass as follows.

`m_(i) = (V)/(v_(i))`

=
`(0.38)/(1.032)`

= 0.368 kg

and,
`m_(f) = m_(i) = 0.75 * 0.368 = 0.276 kg/s`

also,
`\Delta m = 0.25m_(i) = 0.25 * 0.368 = 0.092`

As,
`\Delta m = m_(o) \Delta t`

`\Delta t = (\Delta m)/(m_(o))`

=
`(0.092)/(0.005)`

= 18.4 s

`v_(f) = (V)/(v_(f))`

=
`(0.38)/(0.276)`

= 1.376
`m^(3)/kg`

According to the table A-4, value of pressure at
`v_(f) = 1.376 m^(3)/kg` and T =
`400^(o)C` is 2.5 bar.

Therefore, value of
`-\Delta t` is 18.4 s,
`-v_(f)` is 1.376
`m^(3)/kg` and
`-P_(f)` is 2.5 bar.

Answer 2

Our values,

`V=380L = 380*10^(-3)m^3/s`

`\dot{m} = 0.005kg/s`

`P_1 = 3bar`

`T_1=T_2=400\°c`

`m_f=75\%`

We can know apply the properties of the superheated water,

`P_1 = 3Bar`

`T_1=T_2=400\°c`

So we have that

`v_1=1.032m^3/kg`

And we can find the initial mass of the tank,

`m_1 = (V)/(v)`

`m_1 = (380*10^(-3))/(1.032)`

`m_1 = 0.368kg`

How the final mass is 75%, then

`m_2 = 0.75*0.368 = 0.276 kg`

The time required is given by the change in the mass vs the rate of the mass, so

`t= (0.368-0.279)/(0.005)`

`t=17.8s`

Finally we have the specific volume in the tank, thatis

`v_2 = (V)/(m_2) = (380*10^(-3))/(0.276)`

`v_2 = 1.3768m^3/kg`

From the properties of water in the table we can find the
`P_2`, that is,

At
`v_2 = 1.378m^3/kg` and
`T_1 = T_2 =400\°c`

`P_2 = 2.8bar`