01 part6 properties pure substance more prob
19
Properties of Pure Substances S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : [email protected]
-
Upload
gunabalan-sellan -
Category
Education
-
view
195 -
download
0
description
properties pure substance more prob
Transcript of 01 part6 properties pure substance more prob
Properties of Pure Substances
S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]
Properties of Pure Substances
fg – latent heat of vaporization
f – Fluid (Saturated liquid)
g – gas (saturated vapour- steam)
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.
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
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
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
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
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
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
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
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
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
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
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 =
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
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.
Heat transfer
Magnitude and sign Heat transfer (kJ) = + w
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
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.
