Question

for the following system we have a balloon that is required to be inflated when the valve opens

the balloon has the following

p2= 1.195 bar

m2 = 71.77 g

v2 = 0.0515 m3

T2 = 25.58 celcius

REQUIREMENTS:

2 analysis

1-transient analysis for balloon filling

2-transient analysis for tank discharging

in order to find P1,m1,v1,T1

assume the process occurs rapidly

3- prove that there will be no beat transfer

Balloon Valve air Tank Find mm P2=1.195 bar my=71.779 V2=0.0515 m T2=25.58c° Pipe mi Vi

Answer 1

For transient analysis of balloon filling, we consider the ballon as a system.

Sum of inflow energies = Energy stored in the system + Sum of outflow energies

.........1

SQ + dEj = dEgys + SW + de

In the above case sum of outflows is zero as the system gas enters doesn't leave the system Therefore dE2 = 0.

dEsys = 0, as the balloon is completely empty at the initial stage.

The Sum of inflow energies includes the heat transfer (if any) and the specific energy transfer associated with the mass entering the balloon.

The specific energy entering includes enthalpy, Kinectic energy, and potential energy.

The specific energy transfer is given by (h + C2/2 + gZ)

The energy entering the balloon depends upon the mass transfer rate (dm/dt)

So rearranging the above equation we get,

$\frac{\delta Q}{dt}-\frac{\delta W}{dt} =\frac{dm}{dt}(h+\frac{C^{2}}{2}+gz)$

Integrating the above equation with respect to time gives the transient analysis of filling the balloon

$\int_{t_{1}}^{t_{2}}(\frac{\delta Q}{dt})dt-\int_{t_{1}}^{t_{2}}(\frac{\delta W}{dt})dt =\int_{t_{1}}^{t_{2}}(\frac{dm}{dt}(h+\frac{C^{2}}{2}+gz))dt$

When the balloon expands there is a work done on the system for expanding it which is given by

$P\Delta V$

P - is the pressure

$\Delta V$ - is the change in volume (V2 - V1) - (Final Volume - Initial Volume)

As mentioned in the question as this is an adiabatic process the work equation can be substituted to simplify the equation.

When considering the case of tank discharge we consider the tank as a system with a fixed volume. In this case, we can use equation (1) for our analysis.

Where

dE2 and dEsys are not zero

Therefore the final equation includes integral of these terms with respect to time, given as

$\int_{t_{1}}^{t_{2}}(\frac{\delta Q}{dt})dt-\int_{t_{1}}^{t_{2}}(\frac{\delta W}{dt})dt =\int_{t_{1}}^{t_{2}}(\frac{dE_{sys}}{dt})dt+\int_{t_{1}}^{t_{2}}(\frac{dm_{2}}{dt}(h_{2}+\frac{C_{2}^{2}}{2}+g_{2}z))dt-\int_{t_{1}}^{t_{2}}(\frac{dm_{1}}{dt}(h_{1}+\frac{C_{1}^{2}}{2}+g_{1}z))dt$

Where C is the respective velocities.

In this case of the tank, only mass leaves the system. Therefore mass entering the system will be zero (dm1/dt = 0)

Wok done is also zero as no shaft work is involved. So the final equation becomes

$\int_{t_{1}}^{t_{2}}(\frac{\delta Q}{dt})dt=\int_{t_{1}}^{t_{2}}(\frac{dE_{sys}}{dt})dt+\int_{t_{1}}^{t_{2}}(\frac{dm_{2}}{dt}(h_{2}+\frac{C_{2}^{2}}{2}+g_{2}z))dt$

And if the process takes place adiabatically we can neglect heat transfer as well.

As enthalpy h = u + PV

where u is the internal energy

P is the pressure

V is the volume

We can substitute in the above integral equation to find out the variables.

When we assume the process occurs rapidly the time interval tends to zero. ie we can assume the process took place within no time.

The basic heat transfer equation is

$Q=mC\Delta t$

Where,

Q - Heat transfer

m - mass

C - constant of the process (Cp - for constant pressure and Cv - for constant volume process)

$\Delta t$ - time interval (t2 - t1)

When the process takes place quickly, $\Delta t\approx 0$ .

Therefore the heat transfer is equal to zero.

So this process can be considered as an adiabatic process.

Usually, in cases where tanks are insulated or process takes place quickly, those processes are considered to be adiabatic processes because the heat transfer is nil or negligible.