Approximations - Memorial University of Newfoundland · 2010-09-26 · • The ideal gas law is an equation of state. • There are many equations of state which are utilized for

Faculty of Engineering and Applied Science Memorial University of Newfoundland

St. John’s, Newfoundland, Canada

33

Approximations •  For sub-cooled states, the tables are not broad

in coverage. The following approximations are useful for calculations

•  Tables show that specific volume and specific internal energy change little in these regions with changes in pressure for fixed temperature

•  We can approximate properties as:

€

v(T, p) ≈ v f (T)u(T, p) ≈ uf (T)



34

Approximations •  The specific enthalpy may be approximated

as:

•  or since we can also write that:

•  combining, we obtain:

•  This is useful for finding enthalpy due to pump work on a fluid.

€

h(T, p) ≈ uf (T) + pv f (T)

€

hf (T) = uf (T) + psatv f (T)

€

h(T, p) ≈ hf (T) + (p − psat )v f (T)



35

Example - 6 •  Compare values of v, u, and h for water

at 40 C and 10 MPa obtained from the compressed liquid tables with those obtained by assuming that the water is incompressible.



36

Final Notes on Tables •  For sub-cooled and superheated properties,

only two properties are required to establish the state.

•  In the two phase region (vapor dome), pressure and temperature are not independent.

•  Always start with the saturated tables first and apply the tests given earlier.

•  Also make sure you have the correct table!



37

Specific Heat •  Two additional properties that we frequently

require are termed specific heats. – constant volume specific heat, cv

– constant pressure specific heat, cp

•  They are mathematically related to u(T,p) and h(T,p) as:

€

cv =∂u∂T v

cp =∂h∂T p



38

Specific Heat

cp for water vapor in SI and EES units



39

Specific Heat •  We define their ratio as k:

•  They are also referred to as heat capacities in some instances, because they relate the temperature change of a mass to the energy added from heat transfer.

•  For gases cp and cv are different, but for incompressible substances (dv = 0) and they are equal.

€

k =cpcv



40

Specific Heat •  For incompressible substances the specific

internal energy is only a function of T:

•  But specific enthalpy is a function of both p and T for an incompressible substance:

•  Taking the derivative of h(T,p) wrt T gives:

€

cv (T) =dudT

€

h(T, p) = u(T) + pv

€

∂h∂T p

=dudT

→ cp = cv = c



41

Specific Heat •  In general we find (for incompressible

substances):

•  The last term is often small as v ~ small, but it also depends on dp!

€

u2 − u1 = c(T)T1

T2

∫ dT

u2 − u1 = c(T2 −T1)

€

h2 − h1 = u2 − u1 + v p2 − p1( )h2 − h1 = c T2 −T( )1 + v p2 − p1( )

often−small



42

Ideal Gas Law •  All students are familiar with the ideal gas law

in the form of:

–  where is the universal gas constant. •  In thermodynamics we often work with

alternate forms of the ideal gas law, one of which is:

–  where R is the gas constant for a particular gas (not universal). We will address other forms later.

€

pV = nR T

€

R

€

pv = RT



43

Ideal Gas Law



44

Ideal Gas Law •  The ideal gas model in thermodynamics is more

appropriately summarized as:

–  at this point we should also state that it is only valid for pressures that are small relative to the critical pressure of the gas. In actual fact, u and h depend on pressure under many conditions.

•  Compressibility also effects ideal gas behaviour.

€

pv = RTu = u(T)

h = h(T) = u(T) + RT



45

Ideal Gas Law •  Other forms of the ideal gas law can be written

using the following definitions*:

–  where m is mass, M is the molecular weight of the gas, and n is the number of mols of gas. Also note that: €

v =Vm

v =v M

v = Vn

R =R M

€

R =8.314⋅ KJ /kmol⋅ K1.986⋅ Btu / lbmol⋅ R1545⋅ ftlbf / lbmol⋅ R

⎧

⎨ ⎪

⎩ ⎪

€

pV = nR T pv = RT pV = mRT pv = R T

* Properties denoted by a bar are / mol (per mol)



46

Ideal Gas Law •  Introducing the compressibility factor:

–  as Z approaches unity, we get ideal gas behaviour. •  We can create a generalized compressibility chart

which demonstrates the departure of a gas from ideal gas behaviour.

•  It is plotted according to reduced pressure and reduced temperature. That is we normalize with respect to the critical pressure and temperature.

€

Z =pvRT

=p

ρRT



47

Ideal Gas Law

Generalized compressibility chart for gases.

€

pR =ppc

andTR =

TTc

Tc and Pc are the critical values Figs. A1-A3 give More detail.



48

Equations of State •  Analytical representations of the p-v-T relationship

of gases are called equations of state. •  The ideal gas law is an equation of state. •  There are many equations of state which are

utilized for extending the applicability of the ideal gas law to higher pressures: –  Van der Waals Equation –  Redlich-Kwong Equation –  Virial Equation –  Benedict-Webb-Rubin Equation –  Beattie-Bridgeman Equation



49

Equations of State •  Redlich-Kwong Equation

– coefficients in Table A24

•  Van der Waals Equation – coefficients in Table A24 €

p =R T

v − b−

av (v + b)T

12

€

p =R T

v − b−

av 2

*Note: the a and b are different constants in each equation, see Table A24.



50

Ideal Gas Law and u and h •  From the definitions given earlier for specific

heat we find that to evaluate the change in specific internal energy and specific enthalpy we have:

•  We also saw earlier that:

€

u2 − u1 = cv (T)T1

T2

∫ dT

h2 − h1 = cp (T)T1

T2

∫ dT

€

h(T) = u(T) + RT



51

Ideal Gas Law •  This leads to (for gases):

•  Finally, we also obtain:

€

cp (T) = cv (T) + Ror

R = cp (T) − cv (T)

€

cp (T) =kRk −1

cv (T) =Rk −1



52

Ideal Gas Law and u and h •  Since we must evaluate integrals of

temperature varying specific heats it is convenient to use equation fits:

– coefficients tabulated in Table A21 •  We may also use the ideal gas tables

given in the Appendix A22 and A23 €

c p (T)R

= α + βT + γT 2 +δT 3 +εT 4



53

Example - 7 •  A storage vessel has a volume of 2000

m3. If the tank is initially filled with air at 20 C and atmospheric pressure of 101.3 kPa, determine the mass of air held inside. If the mass is increased 10 fold, for the same temperature, what is the new pressure inside the tank. Assume the vessel is rigid.



54

Example - 8 •  A worker pressurizes a section of rigid

pipe 20 m long with an inside diameter of 30 mm to check for leaks. Initially the temperature of air is 35 C and its pressure is 205 kPa. 24 hours later the worker returns and finds the pressure has dropped to 183 kPa while the temperature has dropped to 21 C. Has the pipe leaked. If so how much?



55

Example - 9 •  Compressed air is used to power a

pneumatic tool. Air enters the tool at 560 kPa and 350 K. Air exits the tool at atmospheric pressure 101.3 kPa and 290 K. Calculate the change in u and h using: –  i) constant average specific heats –  ii) the ideal gas tables



56

Example - 10 •  Determine the specific volume of water

vapor at the following two states: –  i) P = 40 MPa, T = 500 C –  ii) P = 60 MPa, T = 400 C Use the ideal gas equation of state and compare with values found in the superheated tables.



57

Example - 11 •  Methane is at reservoir conditions of

32.48 MPa and 109 C. What is the gas density? –  i) assume ideal gas behavior –  ii) find the compressibility factor and re-calculate



58

Polytropic Processes •  A polytropic process in a closed system

is one in which:

– where n is the polytropic constant.

€

pV n = Cor

pvn = C



59

Polytropic Processes •  Polytropic constant n for a number of special

process is summarized below:



60

Polytropic Processes •  Using the expression for simple compression

work:

•  We can work in either the v or V forms, just as long as we know the difference.

•  We can also introduce the ideal gas law as an equation of state in order to develop further expressions.

€

W12 = pdV1

2

∫ or w12 = pdv1

2

∫



61

Polytropic Processes •  If we integrate the polytropic equation

valid for all n except n = 1. •  When n = 1 we obtain: €

w12 = pdv1

2

∫ = Cv −ndv1

2

∫ =p2v2 − p1v11− n

€

w12 = pdv1

2

∫ = C dvv1

2

∫ = C ln v2v1

= p1v1 lnv2v1

= p2v2 lnv2v1



62

Polytropic Processes •  We can also introduce the ideal gas law

as an equation of state to get:

€

T2T1

=p2p1

⎛

⎝ ⎜

⎞

⎠ ⎟

n−1n

=v1v2

⎛

⎝ ⎜

⎞

⎠ ⎟

n−1

ideal − gas

w12 = pdv =R(T2 −T1)1− n1

2

∫ n ≠1

w12 = pdv = RT⋅ ln v2v21

2

∫ n =1



63

Example - 12 •  Air in a piston assembly is compressed

from an initial state of 100 kPa and 25 C, to a final state of 1000 kPa isothermally. What is the work done by the piston and how much heat is transferred in this process?



64

Example - 13 •  Air in a piston assembly is compressed

from an initial state of 100 kPa and 25 C, to a final state of 1000 kPa adiabatically. What is the work done by the piston and what is the final temperature of the gas?

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Approximations - Memorial University of Newfoundland · 2010-09-26 · • The ideal gas law is an equation of state. • There are many equations of state which are utilized for