03 part2 charging and discharging rigid vessels
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03 part2 charging and discharging rigid VesselsTRANSCRIPT
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Charging and discharging Rigid Vessels
S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : [email protected]
Part - 2
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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-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.
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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 in this 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= t1 Balloon 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 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
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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)
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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 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
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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
1122 umumECV D
At Time 1
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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
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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:
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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
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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
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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
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
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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 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
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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 the valve 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 final mass of the air and the mass of the air left the container.
Discharging problem
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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
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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 by utilizing 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. <Answer: (a) 116.1oC, (b) 0.6 kg.
Example (4-18): Cooking with a Pressure Cooker
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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
1122 umumhmhmWQ eeii
What conditions should you use for he?
212 hhhave
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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