Question

A 10 kg steam initially at 2 MPa saturated vapor is inside an insulated piston cylinder. The cylinder has a bottom valve. As this valve opens, steam at 800 kPa and 200C exits while the steam inside the tank has 400kPa saturated vapor. Find the exit mass of steam from the cylinder in kg.

Answer 1

Ans -

Initial mass in the cylinder, m1 = 10 kg

Let the exit mass be 'me' and final mass into the cylinder be 'm2'

Then from the first law of thermodynamics for unsteady flow, we have

where, = m2u2 - m1u1

= change in the internal energy of the system

Q = heat transfer into the system

subscript 'i' represents the inlet condition of the gas

subscript 'e' represents the exit condition of the gas

W = work done by the gas

Now, the piston-cylinder is insulated, therefore, Q = 0

no work is done by the gas, therefore, W = 0

no mass enters into the cylinder, therefore, mi = 0

exit mass, me = m1 - m2

So, we have

.................................. (1)

Now,

u1 = 2600 kJ/kg (at 2 MPa, saturated vapour, from steam table)

u2 = 2554 kJ/kg (at 400 kPa, saturated vapour, from steam table)

he = 2839.8 kJ/kg (at 800 kPa and 200 oC , from steam table)

Putting all the values in equation (1), we have

  

Therefore, exit mass of steam, me = m1 - m2

= 10 - 8.39

= 1.61 kg (ans)