01 part5 properties pure substance

Properties of Pure Substances

S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]

Part - 2

Properties of Pure Substances

fg – latent heat of vaporization

f – Fluid (Saturated liquid)

g – gas (saturated vapour- steam)

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.

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.

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

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

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)

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

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)

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).

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 )

FormulaSpecific volume

f +

Specific enthalpy

Specific entropy

FormulaSpecific volume

f +

Specific enthalpy

Specific entropy

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).

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.

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).

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).

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.

Given data:

Vessel volumeV= 0.04 m3 mixture of saturated water and saturated steam at T= 250 0C.

mf = 9 kg.

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

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

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.

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.