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

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,

$m_1 = \frac{V}{v}$

$m_1 = \frac{380*10^{-3}}{1.032}$

$m_1 = 0.368kg$

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

$t= \frac{0.368-0.279}{0.005}$

$t=17.8s$

Finally we have the specific volume in the tank, thatis

$v_2 = \frac{V}{m_2} = \frac{380*10^{-3}}{0.276}$

$v_2 = 1.3768m^3/kg$

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

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

$P_2 = 2.8bar$

Answer 2

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.

          $m_{i} = \frac{V}{v_{i}}$

                       = $\frac{0.38}{1.032}$

                       = 0.368 kg

and,     $m_{f} = m_{i} = 0.75 \times 0.368 = 0.276 kg/s$

also,      $\Delta m = 0.25m_{i} = 0.25 \times 0.368 = 0.092$

As,      $\Delta m = m_{o} \Delta t$

        $\Delta t = \frac{\Delta m}{m_{o}}$

                      = $\frac{0.092}{0.005}$

                      = 18.4 s

         $v_{f} = \frac{V}{v_{f}}$

                   = $\frac{0.38}{0.276}$

                   = 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.