ME2121/ ME2121E– Thermodynamics Chapter 3 Thermodynamic Properties of Pure Substances by KC Ng (Office: EA-07-22) Email:
[email protected] (August 2013) In Thermodynamics, matters found in the space of interest are referred to as pure substances. So, what is a pure substance and how is it defined? “A pure substance is one where its properties remain invariant (no change) when it undergoes state changes, e.g., from gaseous to liquid or to solid phase”. For example, the chemical composition of water (H2O), remains unaltered when it is in the solid (ice), gaseous (steam) or liquid states. Available in abundance, its behavior, when subjected to changes in P, T and and v, is of interest to engineers.
Other pure substances such as air (a mixture but we made an exception), methane (CH4), ammonia (NH3), carbon dioxide (CO2), etc.
State Diagrams The states of water (a condensable substance) can be plotted on a 2-dimensional state diagram using suitable axes such as P-v, T-v or P-T diagrams, diagrams, as shown below: Pcr = 221 bar
Critical point
T Vapour region
Fig. 3.1: State diagram of water
Liquid Liquid & Vapour
D
10 bar C
B A
1 bar
In Fig. 3.1, the line “A-B-C-D “A-B-C-D”” represents a constant pressure line (isobar), at atmospheric 5 pressure (1 bar =10 Pa). Assuming a certain amount of liquid (water) is initially contained in a piston-cylinder assembly and heat (dQ) could be transferred across the system boundary (shown by red arrow below). State A
State B
State B-C
State C
State D
Saturated states
V L subcooled
dQ Liquid phase
Two-phase mixture
Gaseous phase
State “A” – is termed a sub-cooled water because its temperature is below the saturation temperature, T< T sat sat . When heat is further added from the environment, its temperature rises but pressure remains constant co nstant (as the piston is free to move up). As the temperature reaches its saturation temperature, T=T sat sat , the state point is indicated by, “B”, is termed saturated liquid, and it lies on the “ f ” line (demarcates the liquid and 2-phase region). On further heating, the energy input leads to a phase transformation, i.e., boiling occurs in the water, leading to a two-phase mixture (vapour liquid phase). The state points lie on the path “B to C”, and its position in the state diagram depends on the amount of heat added. In a two-phase mixture regime, the pressure (Psat ) and temperature (T sat sat ) are coupled (no longer independent of each other). They remain unchanged from “B” to “C”. . When state point “C” is reached, boiling has ended (all liquid in cylinder
v
Chapter 3‐ME2121/E, August 2013, KC Ng
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has been evaporated). A dry-saturated vapour state is achieved. From this point, further addition of heat from surroundings causes a rise in temperature of vapor, T > Tsat, leading to a superheated state, as indicated by point “D”. The system pressure remains constant in the heating processes (as the piston is free to move), accommodating the increase in volume of steam. Similar arguments are said for other isobars (P > or < 1 bar). The highest pressure within the vapor+liquid region is called the critical pressure of water, which is at 220.9 bar. Any isobar above the critical pressure is termed the super-critical state.
where all three phases of water could co-exist - vapor, liquid and ice. The sublimation line is where solid phase changes to vapor phase directly, The melting line is where solid changes to liquid when heat is added Fig. 3.2 shows a P-v-T diagram for water - a substance that expands on freezing, (Check out the shape of state diagram when a substance that contracts on freezing!) 3-D state diagrams
In the two-phase region (denoted by region under an “envelop” region), P & T are no longer independent properties. To define a state in this region, an additional variable is required, e.g., T-v or P-v or P-x, etc. We define “ x” as the quality or dryness fraction of the mixtures, i.e., x
mass of vapour mass of liquid mass of vapour mg
, (3.1)
m g m f
where “ f” and “g” denote the saturated liquid and vapor states. In the saturated liquid region, x = 0, and in the dry or superheated vapor regions, x = 1. The following features worth mentioning:
Note the regions in the state diagram
indicating the solid, solid+liquid, solid+vapour, liquid+vapour and vapor regions, When view from the P-T plane, one sees a triple-phase line and a point
Chapter 3‐ME2121/E, August 2013, KC Ng
P-T plane The main feature on this plane is the slope of the (S+L) line, the triple-phase point, (S+V) and the (L+V) lines. For a substance that expands on freezing, such as water, the (S+L) line “leans backward” to the P axis. Thus, an increase in the pressure on the solid phase (ice) will lead to the formation of a liquid phase at the point where the pressure is exerted, as shown in Figs. 3.3 (b). However, for a substance that contracts on freezing, increasing pressure on a solid phase would remain unchanged, as seen in Fig. 3.3 (a).
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Fig. 3.3(a)
“g” refers to saturated vapour, e.g., hg is the enthalpy at dry saturated vapour conditions, “i” refers to saturated solid, e.g., for water, it would be ice at saturated conditions, “ fg” refers to the difference between the vapour to liquid properties at saturated conditions, e.g., s fg = sg - s f at saturated conditions. “u” refers to the internal energy per kg of the substance, “h” refers to the enthalpy(h=u+pv), “s” refers to the entropy (ds=dq/T),
Specific volume ( v) The specific volume is the volume per unit mass of the substance. Invoking the definition of quality(x), (equation 3.1); x
Fig. 3.3 (b)
Tables of Thermodynamic Properties of Fluids For engineering applications, one would either use a “steam table” to determine accurately the thermodynamic states, such as P, T, u (internal energy), h (enthalpy), s (entropy) of pure substances or mixtures. For example, the Tables by Rogers and Mayhew (this booklet will be used in examination) is a table for undergraduates, and one should familiar himself with the units, nomenclature, etc., found in the tables. Some of the common terminologies used are;
Chapter 3‐ME2121/E, August 2013, KC Ng
m g m f
,
the mass fraction of liquid is naturally given by (1 x)
m f m g m f
m f
(3.2)
mT
For the two-phase mixtures, the volume occupied by mixtures is the sum of the liquid and gaseous volumes, i.e., Vol of liquid
Vol , of vapour
V mixture m f v f
mg v g
and by mT gives V mixture mT
“ f ” refers to saturated liquid, e.g., u f is the internal energy per kg of water at saturated conditions,
mg
or
v
Vol of liquid
Vol , of vapour
m f
mg
mT
v f
mT
vg
v (1 x)v f xvg v fg
(3.3)
v f x(v g v f ) Page 3
Thus, small
x
Referring to the Table of saturated steam and water (pg.2), the column 1 is o the temperature of mixtures in C, and column 3 is the specific volume at dry saturated vapour. 3 At v =26.6 m /kg, we note that t of o steam/water is 34 C. Thus, we conclude that the state point sits on the dry saturated vapour line or the “g” line, shown below.
compared to v
v
v f
v fg
v v fg
(3.4)
3
For example, at low pressure, v = 1.2 m /kg 3 and v f = 0.001 m /kg (density of water is 1000 3 kg/m ), thus, v f <<< v. However, the same argument cannot be applied to the other properties such as internal energy, enthalpy or entropy. In a mixture, h f , may not significantly smaller than hg, i.e.,
Critical pt
t f-line
h (1 x) h f xhg h f xh fg
(3.5)
34oC
The internal energy and entropy of a mixture are respectively given by u (1 x )u f xu g u f xu fg
And the entropy (s) is given by s (1 x ) s f xs g s f xs fg
(3.6)
Using the linear rule, the quality ( x), can also be obtained from other given values of u, or h or s., e.g., x
s s f s fg
h h f h fg
u u f u fg
V=26 m3/kg
Critical pt
(3.7)
v
(b) At P =0.8 bar, t =100 o C, Look up page 3 of Table where column 1 is the saturated pressure, and at 0.8 bar, the corresponding temperature is o only 93.5 C. This temperature is found to be less o than 100 C (given), thus, we conclude that the steam is super-heated (that is, temperature of state higher than the saturated temperature). t 100oC
f-line
P=0.8 bar g-line Superheated state
Tsat=93.5oC
Example 3.1 Find the states of water-steam mixture for the following given conditions: o
g-line
3
(a) t = 34 C, v = 26.6 m /kg, o (b) P = 0.8 bar, t = 100 C (c) P = 2.7 bar, x = 0.5. Solutions
Chapter 3‐ME2121/E, August 2013, KC Ng
v=2.087m3/kg
(d) At P =2.7 bar, vf = 0.00107 m3/kg (pg. 10 of Table)., 3 vg =0.6686 m /kg (pg.4), we note that 2.7 bar is very much lower than Pcr =221 bar, thus, vf <<< vg., and o Tsat = 130 C.
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by definition, any other properties could
0.5
Find the specific volume of steam for o MPa, and t = 275 C.
be found; s s f h h f u u f x s fg h fg u fg u u f u fg
u u f 0.5u fg
In this region (L+V), P & T are no longer independent. P=2.7 bar
Critical pt
t f-line
g-line
130oC
X=0.5
L+V
V=26 m3/kg
v
o
t =300 C
When the states of a point fall outside the tabulated values in the Tables, an interpolation technique may be needed. For example, P2 Critical pt P t g-line P1 f-line
v2
v
v1
Sp volume to be obtained To interpolate for v at pressure (P) when values of v1 and v2 are available from the Tables at P1 and P2.
Solution The state of the steam is superheated. Use superheated tables on page 7. Values of specific volumes at P =20 o o and t= 250 C, 300 C are available and o also at 30 bars (t=250 and 300 C). It is easier to make a tabulation containing the boundary values as shown below. 3 v (m /kg) P=20 P=30 P=25 bar bar bar o t = 250 0.1115 0.0706 3 3 C m /kg m /kg o t =275 C 0.1185 Values to be 0.0759 determined =(0.1185+0.0759)/2 =0.0972
Interpolation
L+V
P =2.5
P P1 v 2 v1 v (at P) v1 P P 1 2
Sometimes, a double interpolation may be needed when values on P1 and P2 are not readily available from the tables.
Example 3.2 Chapter 3‐ME2121/E, August 2013, KC Ng
0.1255
0.0812
Gaseous Phase The state of a gas can be determined from the key properties such as pressure (P), temperature (T ), specific volume (v) and the amount of gas (mass (m) or moles (n)) in a defined volume (V ).
The equation relating these properties of a gas is called the equation of state (EOS), Experiments have been conducted to evaluate how these variables are related. For an ideal gas, the EOS is
PV R (a cons tan t ) Tm lim
(3.8)
where m is the mass in kg, R= 8314.5/ M, (8314.5) is the universal gas constant (R o) in J/kmol.K, M is the molecular mass (kg/kmol). Note: kg is the basic mass units. Equation 3.8 is also known as the ideal gas behaviour.
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If the amount of substance for a gas in a defined volume is measured in moles,
PV (3.9) Ro (universal gas cons tan t Tn lim 8.314 J / mol.K ) or 8.314 kJ / kmol.K
Note the right hand side is Ro. You recalled that a mol is the basic quantity of substance whose mass is equivalent to its molecular mass. E.g., 1 mol of carbon has a mass of 12 gm or 1 kmol of carbon has 12 kg.
Z
Pv RT
(3.12)
At low pressures (with respect tp critical pressure) and high temperatures, Z tends to 1(unity). A graphical depiction of how Z varies with T and P is shown below.
Tr = 10
Z=Pv/RT
1.0 For ideal gas at STP (standard temperature and pressure), the Avogadro’s number gives the number of gas molecules per kmol , i.e.,
Tr = 1.5
Tr = T/Tcr Tr = 1.0
26
N o = 6.02205 ×10 molecules / kmol 23
or 6.02205 ×10 molecules per mol.
5
10
Thus, the mass of a substance is related to the mols by multiplying it with the molar mass, m = n× M
(3.10)
whilst the mass of one molecule (m*) will be given by * m = M / N o. (3.11)
Working forms of Ideal Gas equation The ideal gas equation (EOS) can have the following forms:
PV = mRT basis PV = nR oT
molar basis,
molar volume v = V/n,
P v = R oT
mass
or
gravimetric
Pr = P/ Pcr
For the behaviour of real gases, several methods are available. The compressibility Z can be expanded in a power series in density (the reciprocal of specific volume v) to give Z
Pv RT
1
B v
C v
2
D v
3
(3.13)
where B,C, D… are called the second, third… virial-coefficients that are function of temperature (as well as the composition for mixtures). An alternative expression for the virial equation is a power series expansion in pressure:
R 1 B(T ) / v C (T ) / v 2 T v P
In reality, no gas behaves like an ideal gas (it is only a convenient approximation). A measure of the deviation of a real gas from the ideal gas is given by another property, called gas compressibility, Z
Chapter 3‐ME2121/E, August 2013, KC Ng
(3.13a)
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Comparing term for term for 3.13 and 3.13a, it is noted that B and C are function of temperature (T). Van der Waals Model: A more widely used equation for real gas is the correlation given by van del Waals (in 1837), where the intermolecular forces or attractions,
Thus, at the critical point, equation 3.14 can also be arranged as RoT c a Pc 2 vc b vc Differentiating,
2
( a / v ), and the presence or effect of gas molecules on (b) - the “free volume”, are both accounted for and the correlation is
RoT c 2a Pc 0 v T v c b2 v c 3 C
and
a P 2 v b RoT v
(3.14)
C
2
where a / v adds to the pressure P due to molecular interactions from per mole of
2 Pc 2 RoT c 6a 2 0 4 3 v T v c b v c
If we multiply the former by 2 and the latter by
v b and then add together, we have
molecules, and v b represents the “free volume” available to each mole of the substance. However, these empirical equations do have short comings when depicting the behaviour of condensable fluids, as shown below. P
P V T C
TC
2P
0
V 2 T C
4a
0 4
vc
3
6a ( v c b ) 4
c
6(v c b)
or b
v c
vc
3
.
Substitute back, we solve for “a”: 9 a v c Ro T c . 8 Re-writing the van der Waals equation at critical point, RoT c a 2 or Pc vc b vc
T
v
L+V
9
v
2
In the liquid+vapor (L+V) region, such a correlation does not fit well with under prediction on the “ f” -line and over predicting nearer to the g-line. At the critical point, there is an inflection point at the isotherm (Tc) where the following mathematical conditions apply:
Pc
RoT c
v c
z
Pc vc RoT c
vc 3
8
vc RoT c . vc
2
RoT c 12 9 ( ) vc 8 8
3
. 8
which is the compressibility factor (z) of the , real gas at its critical state (P c Tc).
2 P P 2 0 v T C v T C
Chapter 3‐ME2121/E, August 2013, KC Ng
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Example 3.3
v 83 RPT . Now, substitute v into a and b o
c
c
c
c
Calculate the van der Waals parameters (a and b) for the following gases with the experimental values at the critical point: Gases Pc (bar) Tc (K) Benzene 49.1 562 Water 221.2 647.3 Take R o = 8.3144 J/mol.K. We write the van der Waals equation in another form (i.e., as a cubic equation): RoT 2 a 3 ab v bv v 0 P P P And at the critical point, 3 R T 2 a ab 0 v o c b v v Pc Pc Pc
(3.15)
At the critical point, the three roots (solutions) of the cubic equation must converge, that is
v v
3
0
c
And expanding the cubic equation, we have 3
v 3v vc 3v vc vc 2
2
0
3
(3.16)
Comparing equations (3.15) and (3.16) for the roots (solution) of v, 3 ab The root of zero order: v c , (3.17) Pc
a
st
The root of 1 order : nd
The root of 2 order:
3v c , 2
Pc
RT c
2
a
From (3.18), we have
Pc (bar) or 105 Pa
Tc (K)
Benzene
49.1
562
RoT c 8 Pc
.
R T 2 b o c 27 Ro T c a 8Pc 64 P c [Jm3 / mol2]. [m3/mol] 8.314 x562 b 27 8.314 x562 2 8 x 49.1 x10
a
5
64
49.1 x10
5
-4
1.19 x10 =1.875 Water 221.2 647.3 =0.552 3.04 x10-5 Note: Parameter “a” indicates the dispersion interactions, and parameter “b” is affected by size of molecules or the occupied volume per mol.
Example 3.4 A virial equation of state of a gas is given by P R 1 B (T ) / v C (T ) / v 2 T v where B(T), C(T) etc., are functions of temperature only.
Find the expressions for the virial coefficients for gases that obey (a) the ideal gas relation, and (b) the van der Waals equation. Solution: (a) For the ideal gas, P/T = R/v, thus B(T) =0, and C(T)=0 for the equation to revert to ideal gas. (b) For the van der Waals equation, the relation is
c
v P v . b 3 3v P 3
c
Pc
Gases
2
and from (3,17) we have c
64
, and b
Now, substitute into the tables for a and b gives
P,
a 3 vc
27 RoT c
(3.18)
b 3v c (3.19)
Pc
gives
c
1
a b 2 or Pv RT 1 T v b v T v v P
R
a
2
c
c
Also, from 3.19, substitute for a, b and obtain
Chapter 3‐ME2121/E, August 2013, KC Ng
Expanding using the binomial series,
b 1 v
1
2
b 1 v v b
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Substitute and rearranging, we have
tables to compute for the missing properties as per tabulation.
R b b a 1 1 2 T v v v v RT 2
P
State
B (T ) C (T ) b2 R 1 a 1 b 2 v v RT v
Thus, the equivalent coefficients of the virial equation of van der Waal’s become
a 2 B(T) = b and C(T) = b . RT
The functional forms of the van der Waals equation can be expressed as follows:
2) 3)
a
T (°C)
v (m3/kg)
Dryness Fraction, x, (-)
1 26 226 0.05 ?? 2 ?? 300 ?? 0.8 3 15 325 ?? ?? 4 ?? 275 0.2 ?? 5 128 ?? ?? 0.6 (You steam tables to find the answer and some state may require double interpolation)
[1: x1=0.6447; 2:P=85.9 bar, v=0.0176 m3/kg; 3: v=0.1787 m3/kg, x = 1; 4:P=12.68 bar, x=1; 5: T=329.6 oC, v=0.00847 m3/kg]
Home assignment #2 (Submit to your Tutor in two weeks.)
1) P
P (bar)
R T
Q2. An industrial boiler is fitted with a sightglass device to indicate its level of liquid and the boiler contains saturated steam at 200°C.
Steam
Tsat=200oC
a
v b RoT or P o 2 , v b v v a v b T P 2 v Ro Ro T 2 a 3 ab v bv v 0 P P P 2
Using mathematics, show that the variables used in the van der Waals equation of real gas satisfying requirements of thermodynamic properties, i.e., the exactness and independency tests.
Tutorial no. 2 (Chapter 3 -2011) Q1. The table gives the properties of water under five sets of conditions. Use the steam
Chapter 3‐ME2121/E, August 2013, KC Ng
Liquid
“ blowdown” During the routine "blowdown" procedure, liquid water is purged from the bottom (muddrum) at a slow rate, and the temperature in the boiler remains constant during this process. If the cross-sectional area of the boiler is 2.5 m2 (assumed constant), and the level of the liquid drops by 150 mm, determine the mass of liquid withdrawn. (You may decouple the blowdown process – first consider the amount of water removed (vf ×dV), and then the void left behind is filled with vapour (steam). The amount of vapour is obtained from the water removed).
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[Ans: mliquid removed = (324.1-2.945) = 321.15 kg]
Q3. Water is contained in a cylinder fitted with a frictionless piston. The mass of water is 1.05 kg and the area of piston is 0.5 m2. At the initial stage, the water is at 110°C with a quality (dryness fraction) of 0.90, and the spring just touches the piston (exerts no force). Now heat is transferred to the water and the piston begins to rise. During this process, the resisting force on the spring is proportional to the distance moved, with a force (stiffness) of k = 10 N/mm. (a) Sketch this heating process on a P-v diagram for water, (b) Obtain expressions for P and V of the contents in the cylinder when the piston moves by z meters, (c) Calculate the pressure in the cylinder when the temperature reaches 200°C. (This problem coupled the piston-cylinder to a mechanical spring. The spring provides a resisting for ce on the piston). –
2
[Ans: P2 P1 = k m (v2 – v1)/A ; P =1.56 bar, v 3 = 1.391 m /kg] {ideal gas problem] Q4. A rigid vessel (A) is connected to a spherical elastic balloon (B). Both the vessel and balloon contain air at the ambient temperature of 25°C. The volume of vessel A 3 is 0.1 m and the initial pressure is 300 kPa.
Q5. One kg of air is trapped in a cylinder behind a frictionless piston. The initial temperature of air is 50°C and the density of air 3 is 1.08 kg/m . The air is compressed until the volume is 1/6 of the initial volume and the absolute pressure at the end of compression is 700 kPa. Calculate the initial volume and pressure of the air, and the final temperature of the air. If the volume is held fixed and temperature of the air reduced to 50°C by cooling, find the final pressure of the air. Indicate these processes on a P-v diagram. Assume that the molecular mass of air is 29 kg/kmol and R 0 = 8.3143 kJ/kmol.K. [Ans: P1 = 1 bar; T 2 = 376.7 K, P 3 = 6 bar]
-----------------------------------------------------Explanatory notes for Q2.: This question tests on the definition of specific 3 V m volume, i.e., v ( ). The drop in water m kg level in the boiler represents the spaces to be occupied by steam that has to be evaporated at the same pressure and temperature from the liquid. It also trains you how to read steam tables. Explanatory notes for Q3.: Spring is compressed
The initial diameter of the balloon B is 0.5 m and the pressure inside is 100 kPa. The valve connecting A and B is now opened and remains open, and the final temperature of air is uniform at 25°C.
Piston, Mass= M Z2
Z1
Determine the final pressure in the system, the final volume and the mass of air in the balloon. It may be assumed that the pressure inside the balloon is directly proportional to its diameter. [Ans: k B = 200 kN/m; P B = 200 DB,2 (kN); DB,2 = 0.683 m, V B,2 = 0.167 m3].
Chapter 3‐ME2121/E, August 2013, KC Ng
Vapor + liquid
Completely Vapor
Heat addition
Initial states
Final state 2
Area of piston ( A) = 0.5 m ., Spring stiffness, k =10 N/mm or kN/m (SI units), Mass of piston = M (kg).
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This problem involves a real fluid such as steam, and thus one must use the tables for accurate computation of fluid properties. (should you use the ideal gas equation for the equation of state, errors would be larger.) o
Initial states (1): T 1 = 110 C, m1= 1.05 kg, No force exerted by spring at the initial condition, P1 = Psat ., x1 =0.9, v1 ~ x1(vg1) – valid for low steam pressure.
conditions. As only the final temperature (T 2) is known, one has to use the superheated tables to o plot out an isotherm ( T =200 C) in a P versus v graphical plot. Using the P2 equation (as in equation (E2)) and using the same range of v2, the two lines (isotherm and the straight line of P2) to intersect for the roots (solutions, as shown below using a graphical method). P
Values taken from steam tables at 200 o C
Final states (2): T 2 = 200o C, Equation E2
Solution steps: Taking a force balance on the piston; Spring
P A Mg P A 1 o
1.56 bar
force
SF
(E1) 1.391 m3/kg
=0
V
After the addition of heat to the vapour+liquid Intersection of two lines gives the answer. mixtures, the piston expands and all liquid evaporated with increasing pressure within the cylinder (P2) and temperature (T2). Taking a Alternatively, we can approach from force balance again on the piston, noting that mechanics point of view: volume change from the initial state V V 2 V 1 A( Z 2 Z 1 ) and using the WD under PV diagram WD by spring WD against gravity definition of specific volume, i.e., ( P P1 ) 1 2 k Z Mg Po A Z 2 (V 2 V 1 ) V 2 V 1 mv 2 v1 A( Z 2 Z 1 ) , gives 2 2 And P1 A Spring force V 2 V 1 mv 2 v1 A( Z 2 Z 1 ) P A Mg P A k Z P1 A 2 o Spring force P A Mg P A . 1 o km(v 2 v1 ) From which we get P A P1 A 2 A km(v 2 v1 ) P P 1 km(v 2 v1 ) 2 2 A P P1 or (E2) 2 2 A Explanatory notes for Q4 km(v1 ) km P P1 ( v ) 2 2 2 2 A A The balloon has a certain amount of stiffness or y C mx (like a spring of stiffness k ) where it is stated Thus, we obtain a straight line relation for P2 in that the pressure in the balloon is proportional terms of P1, A, k, m and v, which are properties to the diameter: of the piston-cylinder assembly. Note that v1 3 P 100(10 ) Pa P D or k . 2(10 5 ) can be read off the steam tables at the initial D 0.5 m
Chapter 3‐ME2121/E, August 2013, KC Ng
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The total mass in the system remains unchanged before (state 1) and after (state 2) the valve was opened. Initial state
final state
m A1 m B1 mtotal m A 2 m B 2
State 2: V2 = V1/6, P2=7 bar,
where A and B denotes the vessel and the balloon, respectively, and the subscripts 1 and 2 refer to the initial and final states.
T2 = P2V2/mR
(iii) Note the ideal gas equation for mass of matter, m = PV/RT.
State 3: an isochoric process (V=constant) and T3=323.15 K. P2 /T2 = (mR/V)= P3/T3
(iv) Volume of a sphere is give by V
6
5
=(7x10 x 0.9274/6)/(287x1)=376.9 K.
5
P3 = 6 x10 Pa or 6 bar.
3
D .
Explanatory notes for Q5 Applying the Equation of state (EOS): State 1: V1 is determined. 1 bar ??
1
287
P V 1 m R
P
323.15
T
2
7 bar
V2/V1 = 1/6
T2 3 1 bar
1 T=50 C -------------------------------------------------------------------V
Chapter 3‐ME2121/E, August 2013, KC Ng
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The Steam Tables (PARTIAL LISTING ONLY) can be obtained from the booklet (recommend to buy one copy): Thermodynamic and Transport Properties of Fluids, (SI Units) by GFC Rogers & YR Mayhew, Blackwell.
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