Charging and discharging Rigid Vessels

S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : gunabalans@yahoo.com

Part - 2

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

3

Some familiar unsteady-flow processes are (for example)

• the charging of rigid vessels from supply lines.

• inflating tires or balloons.

• discharging a fluid from a pressurized vessel,

• cooking with an ordinary pressure cooker.

4

10 C

10 C

10 C

t1

5 C

t2

5 C

5 C

Another example.. Discharging a pressurized can.

5

Such problems can be solved by simplified model called..

Uniform State Uniform Flow model

There are 2 main assumptions in this model..

6

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= t1 Balloon at t= t2

7

10 C

10 C

10 C

t1

5 C

t2

5 C

5 C

Another example.. Discharging a pressurized can.

8

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 cross section of the inlet/exit.

• Steady means that the fluid properties do not change with time over the cross section of the inlet/exit.

P, T, v, h, u ..

P, T, v, h, u ..

P, T, v, h, u ..

t2

9

Charging Problems

Consider a process in which a gas bottle is filled from a pipeline. In the beginning the bottle contains gas of mass m, at state p1,T1,v1,h1,u1. The valve is opened and gas flows into the bottle till 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)

The conservation of energy equation for a control volume undergoing a process can be expressed as

11

Mathematical Analysis:

mmmmm systemexitin 12 D

22

2 2ei

cvi i i e e eVVQ 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

12

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

1122 umumECV D

At Time 1

13

cvee

eeii

ii EgzVhmWgzVhmQ D

22

22

In many cases, we can neglect both the KE and PE of the flowing fluid (ie at inlets/exits)

1122 umumhmhmWQ eeii

the left side

Assuming only a single inlet and a single exit stream

inlet exit 1time2time

1122 umumhmhmWQ eeii

14

2mmi

Mass balance in Charging Problems

Recall

mmmm exitin 12

Is there mass exiting? No

Is there mass at initial state?

No

For one inlet, we have:

15

1122 umumhmhmWQ eeii Recall

Energy balance in Charging Problems

Is there heat transfer? No

Is there work? No

Is there energy transferred out by mass? No

Was there an energy at time 1? No

16

22umhm ii That means the temperature in the tank is higher than the inlet temperature 2uhi

2i i iu Pv u

iuu 2

Energy balance in Charging Problems (continued..)

Therefore

But from mass balance: 2mmi

It takes energy to push the air into the tank (flow work). That energy is converted into internal energy. 풉풊 = 풖ퟐ CpTi = Cv T2 (Cp /Cv)Ti = T2 휸푻풊 = 푻ퟐ

To keep in mind 푻ퟐ = 휸푻풊

Use p for i Charging from a pipe P – pipe

푻ퟐ = 휸푻풑

17

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

1mme

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.

풉풆 = 풖ퟏ CpTe= Cv T1 (Cp /Cv)Te = T1 휸푻풆 = 푻ퟏ

18

A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300°C. 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 . 풖ퟐ− 풖풊 = 푷풊풗풊 풄풗푻ퟐ − 풄풗푻풊 = 푹푻풊 풄풗(푻ퟐ− 푻풊) = 푹푻풊

Example: Charging of a Rigid Tank by Steam

2uhi 2i i iu Pv u

푷풊푽풊 = mRT 푷풊풗풊 = RT

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

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 the valve is closed. Calculate the final temperature of the steam in the tank. (Nov.2012)

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 final mass of the air and the mass of the air left the container.

Discharging problem

Expand to atmospheric pressure p2 = 1bar V1 = V2 for a container

m2 – final mass m1-m2 = is the mass left the container , ie mass out of the system

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 by utilizing the kinetic energy of the discharged air to run a frictionless turbine. Assume reversible adiabatic expansion.(Apr. 2013)

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)

ergyInternalEn mout hout

Expand to atmospheric pressure p2 = 1bar V1 = V2 for a container

29

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. <Answer: (a) 116.1oC, (b) 0.6 kg.

Example (4-18): Cooking with a Pressure Cooker

30

(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.

31

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….

32

• 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

1122 umumhmhmWQ eeii

What conditions should you use for he?

212 hhhave

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