03 part2 charging and discharging rigid

7/28/2019 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]

Part - 2


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2

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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3

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

03 part2 charging and discharging rigid

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