01 part6 properties pure substance more prob
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Properties of Pure Substances
S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]
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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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Problem-2
Steam initially at 0.5 Mpa, 240 is cooled at constant volume to 75 , a) state the condition of the steam, b)find the temperature at which steam become saturated vapour, c)find the quality of steam at 75 , d)what is the heat transfer per Kg of steam from 240 to 7.
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Problem-2
Given data:Steam initially at p1 = 0.5 Mpa = 5 bar, T1 = 240 is cooled at constant volume to 75 V = constantT2 = 75
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a) state the condition of the steam
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
Steam initially at p1 = 0.5 Mpa =5 bar, T1 = 240
Refer Steam table At P1 = 5 bar T sat = 151.9 < T1Since the initial temperature (T1 = 240 )is above the saturation temperature (151.9 ) ,
a) Its is a Super heated steamSo Refer Super heated steam table for further information at point 1
T2 = 75
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b) Find the temperature at which steam become saturated vapour.
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75
Steam initially at p1 = 0.5 Mpa = 5 bar, T1 = 240 Super heated steamFrom table
This is a point on saturation lineSo refer saturation table Constant For thiset The actual saturation temperatureFrom saturated steam table
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b) Find the temperature at which steam become saturated vapour.
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75
For thiset The actual saturation temperatureFrom saturated steam table
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b) Find the temperature at which steam become saturated vapour.
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52
135
140
145
150
155
Series1; 140
150f(x) = − 86.2068965517241 x + 183.879310344828R² = 1
Linear Interpolation
Vg
Tem
pera
ture
℃
Do this using calculator
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0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52135
140
145
150
155
Series1; 140
150f(x) = − 86.2068965517241 x + 183.879310344828R² = 1
Linear Interpolation
Vg
Tem
pera
ture
℃
b) Find the temperature at which steam become saturated vapour.
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75 Tsat = 143.88
= 143.88
a) Tsat = 143.88
Do this using calculator
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c)find the quality of steam at 75
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75 = 143.88
At T2 = 75 Get data from table =
Constant Volume Line
f +Substitute and find =
You can also find Terms like h, s,u,q
𝒗 𝒈𝒗 𝒇
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c)find the quality of steam at 75
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75 = 143.88
At T2 = 75 Get data from table = 0.001026
Constant Volume Line
f +Substitute and find =
You can also find Terms like h, s,u,q
𝒗 𝒈𝒗 𝒇 𝒗
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c)find the quality of steam at 75 At T2 = 75 Get data from table = 0.001026
- - 0.001026
Constant Volume Line data from super heated region ( point 1)
f + = 0.001026 + 4.133074
find = 0.112113
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d)what is the heat transfer per Kg of steam from 240 to 7.
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75 = 143.88
Magnitude and sign Heat transfer (kJ) = + w
Constant volume V1 = V2
w = p(V2-V1) = 0 =
u = = (1) – ()
For Constant volume process = =
At T2 = 75 Get data from table
h = hP sat = p2= FROM STEAM TABLE
f +
𝒗 𝒈𝒗 𝒇
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d)what is the heat transfer per Kg of steam from 240 to 7.
Magnitude and sign Heat transfer (kJ) = = (1) 240 Super heated region– () 75
For Constant volume process = =
At T2 = 75 Get data from table
h = 2321.5 KJ/KghP sat = p2= 0.38549 barf + = 0.112113 from previous calculation+ 574.17 KJ/Kg
Steam initially at p1 = 0.5 Mpa = 5 bar, T1 = 240 Super heated steamFrom table
h1 = S1 =
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d)what is the heat transfer per Kg of steam from 240 to 7.
Super heated steam
151.9
p1 =5 bar, T1 = 240
Constant Volume Line
T2 = 75 = 143.88
Magnitude and sign Heat transfer (kJ) = = (1) – ()
Substitute Q
Result: a) Super heated steam
b) Tsat =c) X =d) Q =
𝒗 𝒈𝒗 𝒇
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Enthalpy (h) – show difference between h,u,Q
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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Heat transfer
Magnitude and sign Heat transfer (kJ) = + w
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Specific heat
• The specific heat - the amount of heat required to raise a unit mass of the substance through a unit rise in temperature.
• For gases– At constant pressure
• Cp
– At constant Volume• Cv
This classic relationship between the specific heats of an ideal gas is called Mayer’s equation
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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.