03 part2 charging and discharging rigid
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Charging and discharging Rigid
VesselsS.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]
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Unlike steady-flow processes, unsteady-flow
process start and end over some finite time period
instead of continuing indefinitely.
Therefore, we deal with change that occur over
some time interval Dt instead of the rate of
change.
Unsteady flow problems
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Some familiar unsteady-flowprocesses are (for
example)
the charging of rigidvessels from supplylines.
inflating tiresor balloons.
discharging afluid from apressurizedvessel,
cookingwith an
ordinarypressurecooker.
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10 C
10 C
10 C
t1
5 C
t2
5 C
5 C
Another example..
Discharging a pressurized can.
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Such problems can be solved by
simplified model called..
Uniform State Uniform
Flow model
There are 2 main assumptions inthis model..
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Assumption 1: Uniform State over
the CV The state of the mass within the CV may change with time
but in a uniform manner.
Uniform means that the fluid properties do not change
over the control volume.
Balloon at t= t1Balloon at t= t2
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10 C
10 C
10 C
t1
5 C
t2
5 C
5 C
Another example..
Discharging a pressurized can.
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Assumption 2: Uniform Flow over
inlets and exits The fluid flow at inlets/exits is assumed to be uniform and steady.
Uniform means that the fluid properties do not change over the crosssection of the inlet/exit.
Steady means that the fluid properties do not change with time over thecross section of the inlet/exit.
P, T, v, h, u ..
P, T, v, h, u ..
P, T, v, h, u ..
t2
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Charging Problems
Consider a process in which a gas bottle is filledfrom a pipeline. In the beginning the bottlecontains gas of mass m, at state p1,T1,v1,h1,u1.The valve is opened and gas flows into the bottletill the mass of gas in the bottle is m2
at state p2, t2, v2, h2 and u2.The supply to the pipeline is very large so that the
state of gas in the pipeline is constant at pp, tp, vp,hp, up, and Vp.
1. Derive Energy balance equation for charging and discharging of a tank (Nov. 2010)
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The conservation of energy equation for a control
volume undergoing a process can be expressed as
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Mathematical Analysis:
mmmmm systemexitin 12 D
22
2 2
eicvi i i e e e
VVQ W m h gz m h gz E
D
Recall the general mass balance equation
Recall also the general energy balance equation
Let us analyze the right side first then the left
side of this equation
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2 1 2 2 1 1CVE U U U m u m uD D
2E
1E
At time 2
PEKEUECV
DDDDKE of the CV PE of the CV
Usually, they both equal 0, therefore
1122umumE
CVD
At Time 1
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cvee
eeii
ii EgzVhmWgzVhmQ D
2222
In many cases, we can neglect both the KE and PE of theflowingfluid (ie at inlets/exits)
1122umumhmhmWQ
eeii
the left side
Assuming only a single inlet and a single exit stream
inlet exit 1time2time
1122umumhmhmWQ
eeii
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2mmi
Mass balance in Charging Problems
Recallmmmm
exitin 12
Is there mass exiting? No
Is there mass at initial
state?
No
For one inlet, we have:
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1122umumhmhmWQ eeii
Recall
Energy balance in Charging Problems
Is there heat transfer? NoIs there work? No
Is there energy transferred out by mass? No
Was there an energy at time 1? No
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22umhm
ii
That means the temperature in the tank is
higher than the inlet temperature2uhi
2i i iu P v u
iuu 2
Energy balance in Charging Problems
(continued..)Therefore
But from mass balance:2
mmi
It takes energy to push the air into the
tank (flow work). That energy is
converted into internal energy. =
CpT i = Cv T2(C p /C v)T i = T2 =
To keep in mind =
Use p for i
Charging from a pipe
P pipe
=
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11umhm
ee
That means the temperature in the tank is
lower than the exit temperature
1uhe
1uvPu
eee
euu 1
Mass and Energy balance in
discharging Problems
1mm
e
It takes energy to push the air out of the can
(flow work).
That energy comes from the energy of
the air that remains in the can.
= CpTe= Cv T1(Cp /Cv)Te = T1 =
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A rigid, insulated tank that is initially evacuated is connected
through a valve to a supply line that carries steam at 1 MPa and
300C. Now the valve is opened, and steam is allowed to flow
slowly into the tank until the pressure reaches 1 MPa, at which
point the valve is closed. Determine the final temperature of the
steam in the tank .
=
= ( ) =
Example: Charging of a Rigid Tank by Steam
2uh
i
2i i iu P v u
= mRT = RT
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Gas Constant R
Gas Molar Weight ( M)Kg/Kmol Gas Constant (R )KJ/KgK
Air 28.97 0.287
Nitrogen 28.01 0.297
Oxygen 32 0.260
Hydrogen 2.016 4.124
Helium 4.004 2.077
Carbon dioxide 44.01 0.189
Steam 18.02 0.461
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2. An insulated rigid tank, initially at zero
pressure is connected through a valve to an
infinite source supply line of steam at 600KPa
and 250C. The valve is slowly opened to allow
the steam to flow in to the tank until the
pressure reaches 600 KPa and at which point thevalve is closed. Calculate the final temperature
of the steam in the tank. (Nov.2012)
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2. An air container of volume 4 m3 contain air at
14 bar and 40C. A valve is opened to the
atmosphere. Then the valve is closed and
pressure is dropped to 10 bar. Calculate the finalmass of the air and the mass of the air left the
container.
Discharging problem
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Expand to atmospheric pressure p2 = 1barV1 = V2 for a container
m2 final mass
m1-m2 = is the mass left the container , ie mass out of the system
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Discharging problem
A 1.6 m3 tank is filled with air at a pressure of 5
bar and temperature of 100 C. the air is let off to
the atmosphere through the valve. Assume no
heat transfer determine the work obtained byutilizing the kinetic energy of the discharged air
to run a frictionless turbine. Assume reversible
adiabatic expansion.(Apr. 2013)
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Discharging problem
V1 = 1.6 m3
P1 = 5 bar
T1 = 100 C.
the air
Assume no heat transfer
determine the work obtained by utilizing the kinetic
energy of the discharged air
Assume reversible adiabatic expansion.(Apr. 2013)
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ergyInternalEn mout hout
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Expand to atmospheric pressure p2 = 1bar
V1 = V2 for a container
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A certain pressure cooker has a volume of 6 L
and operating pressure of 175 kPa gage.Assuming an atmospheric pressure of 100
kPa. Initially, it contains 1 kg of water. Heat is
supplied to the pressure cooker at a rate of
500 W for 30 min after the operating pressure
is reached. Determine:(a) the temperature at which cooking takes
place and
(b) the amount of water left in the pressure
cooker at the end of the process.
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(2) As long as there is a
liquid, the phase will be sat.
mixture and the
temperature will be sat.temperature at the cooking
pressure.
(3) The steam
leaving the
cooker will be
saturated vapor.
Solution:(1) This is a discharging problem.
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Time-dependent inlet/exit conditions
We assumed that the flow into or out of the
CV is steady.
How would we handle inlet or exit conditions
that change with time? The best we can do at this point is to take the
time average.
to see this approximation, see next slide.
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Consider air coming out of a can. It gets colder
with time.
That means the exit conditions are not
constant since Te is decreasing with time.
So, how can we deal with the 1st low
1122umumhmhmWQ
eeii
What conditions should you
use for he?
2
12hh
have
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Reference
Rajput, R. K. 2010. Engineering thermodynamics. Jones and Bartlett
Publishers, Sudbury, Mass.
http://ocw.kfupm.edu.sa/ocw_courses/user061/ME203/Lecture%20Notes
/Ch04c%201st%20law%20OS.ppt