2-m^3 rigid tank initially contains air at 100 kPa and 22degreeC. The tank is co

Question

2-m^3 rigid tank initially contains air at 100 kPa and 22degreeC. The tank is connected to a supply line through a valve. Air is flowing in the supply line at 600 kPa and 22degreeC. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77degreeC. Determine (a) the mass of air that has entered the tank and (b) the amount of heat transfer. Is it open or close system. Give the equations. Give the assumptions. Solve.

Explanation / Answer

Tank modeled as control volume since mass crosses boundary. Hence it is an open system.

Assumptions:

1 This is an unsteady process since the conditions within the device are changing during the process, but it can be solved as a uniform-flow process since the state of fluid at the inlet remains constant.

2 Air is an ideal gas with variable specific heats.

3 Kinetic and potential energies are negligible.

4There are no work interactions involved.

Solution:

The gas constant of air is 0.287 kPa.m3 /kg.K

Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and non-flowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as

Equations:

Mass balance: min mout = msystem mi = m2 m1

Energy balance: Ein - Eout= Esystem

Since the kinetic and potential energies are negligible, therefore

Qin+mihi= m2u2 – m1u1

Now, V = 2 m^3

Pi = 100 kPa

Ti = 22°C = 295.15 K

Ps = 600 kPa

Ts = 22°C = 295.15 K

Tf = 77°C = 350.15 K

(a) Acc to ideal gas law

PV = m*R/Mr*T

PiVi = mi*R/Mr*Ti

mi = PiV / R/Mr*Ti = 100*2 / 8.314/28.97 * 295.15

mi = 2.361 kg

PV = m*(R/Mr)*T

R/Mr = PfV/mTf

mf = mi*(Pf/Pi)*(Ti/Tf) = 2.361(600/100)*(295.15/350.15)

mf = 11.9409 kg

m = mf-mi = 11.9409 - 2.361

m = 9.578 kg

The heat transfer during this process is determined from

Qin+mihi= m2u2 – m1u1

Qin= -mihi + m2u2 – m1u1

=- (9.584)(295.17) + (11.941)(250.02) – (2.361)(210.49)

=-339 Kj= Qout= 339 Kj

*The values of h1, u1 and u2 are standard values of air at given temperature

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