01 part5 properties pure substance
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Transcript of 01 part5 properties pure substance
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Properties of Pure Substances
S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]
Part - 2
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Properties of Pure Substances
fg – latent heat of vaporization
f – Fluid (Saturated liquid)
g – gas (saturated vapour- steam)
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Enthalpy (h)
Enthalpy is a measure of the total energy of a thermodynamic system.
Enthalpy is a thermodynamic potential. It is a state function and an extensive quantity.
It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.
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Sensible heat (hf – Enthalpy of saturated fluid)
• Sensible heat is heat exchanged by a body or thermodynamic system that has as its sole effect a change of temperature
specific enthalpy (kJ/kg) =• This is given in the steam table for different
pressure and temperature.
The specific heat - the amount of heat required to raise a unit mass of the substance through a unit rise in temperature.
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Latent heat (hfg – Latent heat of vaporisation)
• The latent heat (hfg) is the amount of heat transfer required to cause a phase change in unit mass of a substance at a constant pressure and temperature
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Enthalpy of steam (h)
• Wet steam
f – Fluid (hf – enthalpy of Saturated liquid)fg – latent heat or enthalpy of vaporizationX – Dryness fraction
– mass of water particle in the steam – mass of dry steam
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Enthalpy of Dry and saturated steam(hg)
• Dry and saturated steam ( x = 1)
f – Fluid (hf – enthalpy of Saturated liquid)fg – latent heat or enthalpy of vaporizationX – Dryness fraction
g – gas (saturated vapour- steam)
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Mollier Diagram• Plots the total heat against entropy, describing the enthalpy of a
thermodynamic system.• It shows enthalpy h in terms of
• internal energy u, • pressure P and • volume v
using the relationship • chart covers a
• pressure range of 0.01 - 1000 bar,• Temperatures up to 800 degrees Celsius
Enthalpy–entropy chart
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Problem -1
A vessel of volume 0.04 m3 containing a mixture of saturated water and saturated steam at a temperature of 250 0C. The mass of the liquid present is 9 kg. find the pressure , the mass, specific volume, enthalpy, entropy and internal energy. (Apr/may 2011)
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Problem -1Given data:volume (v)= 0.04 m3 mixture of saturated water and saturated steam at a Temperature (T)= 250 0C. Mass of liquid present(mf) = 9 kg.
find 1. the pressure (p), 2. the mass (m) (mass of liquid (mf) + mass of vapour (mg)),
3. specific volume (), 4. Enthalpy (h), 5. Entropy (s) and 6. internal energy (u).
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Data From steam table
1. Temperature (t)2. Absolute pressure (p)
1. Specific volume (f, )
2. Specific enthalpy (h - hf,+ hfg ,= hg )
3. Specific entropy (s - sf, + sfg, = sg )
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FormulaSpecific volume
f +
Specific enthalpy
Specific entropy
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FormulaSpecific volume
f +
Specific enthalpy
Specific entropy
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Problem -1Given data:v= 0.04 m3 mixture of saturated water and saturated steam at a T= 250 0C. mf = 9 kg.
find 1. the pressure (p), 2. the mass (m) (mass of liquid + mass of vapour), 3. specific volume (), 4. Enthalpy (h), 5. Entropy (s) and 6. internal energy (u).
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Collect data from steam table
From page 6 , for Temperature t = 250 0CSteam Tables
Absolute pressure p = Specific volume ()1. f =
Specific enthalpy (h )2. hf =
3. hfg =
4. hg =
Specific entropy (s )5. sf =
6. Sfg =
7. Sg =
Given data:v= 0.04 m3 mixture of saturated water and saturated steam at T= 250 0C. mf = 9 kg.
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Collect data from steam table
From page 6 , for Temperature t = 250 0CSteam Tables
Absolute pressure p = Specific volume ()1. f =
Specific enthalpy (h )2. hf =
3. hfg =
4. hg =
Specific entropy (s )5. sf =
6. Sfg =
7. Sg =
find 1. the pressure (p) – direct from
steam table, 2. the mass (m) (mass of liquid +
mass of vapour), 3. specific volume (), 4. Enthalpy (h), 5. Entropy (s) and 6. internal energy (u).
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Collect data from steam table
From page 6 , for Temperature t = 250 0CSteam Tables
Absolute pressure p = 39.73 barSpecific volume ()• f = 0.001251 m3/kg
Specific enthalpy (h )1. hf = 1085.3 KJ/Kg
2. hfg = 1714.6 KJ/Kg
3. hg = 2802 KJ/Kg
Specific entropy (s )4. sf = 2.793 KJ/KgK
5. Sfg = 3.277 KJ/KgK
6. Sg =
6.073 KJ/KgK
find 1. the pressure (p) – direct from
steam table, 2. the mass (m) (mass of liquid +
mass of vapour), 3. specific volume (), 4. Enthalpy (h), 5. Entropy (s) and 6. internal energy (u).
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2. mass (m) m=mass of liquid + mass of vapour
To find mg
From steam table at T= 250 0C
• f = 0.001251 m3/kg
Given data
mf = 9 kg
Substitute in
=
Given data:
Vessel volumeV= 0.04 m3 mixture of saturated water and saturated steam at T= 250 0C.
mf = 9 kg.
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Given data:
Vessel volumeV= 0.04 m3 mixture of saturated water and saturated steam at T= 250 0C.
mf = 9 kg.
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3. Specific volume ()
f +
• f = 0.001251 m3/kg
= Or get directly from table
X – Dryness fraction
mf = 9 kg
= calculatedSubstitute in f +
Given data:
Vessel volumeV= 0.04 m3 mixture of saturated water and saturated steam at T= 250 0C.
mf = 9 kg. Specific volume ( – unit m3/kg)f = 0.001251 m3/kg
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1. Enthalpy (h), 2. Entropy (s) and 3. internal energy (u).
Specific enthalpy
Specific entropy
Specific internal Energy
From steam tableT= 250 0C.hf = 1085.8 KJ/Kg hfg = 1714.6 KJ/Kg
sf = 2.794 KJ/KgK sfg = 3.277 KJ/KgK
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Exercise - 1
A vessel of volume 0.06 m3 containing a mixture of saturated water and saturated steam at a temperature of 230 0C. The mass of the liquid present is 8 kg. find the pressure , the mass, specific volume, enthalpy, entropy and internal energy.
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Reference• Rajput, R. K. 2010. Engineering thermodynamics. Jones and Bartlett
Publishers, Sudbury, Mass.• Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill,
New Delhi.