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3.7) The following table lists temperatures and specific volumes of water vapor at two pressures:
Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate:
(a) the specific volume at T = 240 oC, p = 1.25 MPa, in m^3 /kg.
(b) the temperature at p = 1.5 MPa, v = 0.1555 m^3 /kg, in oC.
(c) the specific volume at T = 220 oC, p = 1.4 MPa, in m^3 /kg.
(a) To determine the temperature we use linear interpolation. From the table at 240 oC we have:
v(P=1.0 MPa) = 0.2275 m^3 /kg v(p = 1.5 MPa) = 0.1483 m^3 /kg. Therefore:
𝑣[1.5 𝑀𝑃𝑎] − 𝑣[1.0 𝑀𝑃𝑎] 𝑚^3 /𝑘𝑔
(𝑝 1 − 𝑝 2 )𝑀𝑃𝑎 =^
𝑣 − 𝑣[1.0 𝑀𝑃𝑎] 𝑚^3 /𝑘𝑔
0.1483 − 0.2275 𝑚^3 /𝑘𝑔
𝑣 − 0.2275 𝑚^3 /𝑘𝑔
−0.0792 𝑚^3 /𝑘𝑔
𝑣 − 0.2275 𝑚^3 /𝑘𝑔
𝑣 − 0.2275 𝑚^3
3 𝑘𝑔 0.5 𝑀𝑃𝑎 0.25^ 𝑀𝑃𝑎^ =^ −0.^
𝑚^3
𝑚^3
𝑚^3
(b) Finding bounding values on the table at 1.5 MPa we find v(T=240 oC) = 0.1482 and v(T=280 oC) = 0.1627. Therefore:
𝑣[240 ℃] − 𝑣[280℃] 𝑚^3 /𝑘𝑔
(𝑇 1 − 𝑇 2 )℃ =^
𝑣 − 𝑣[280℃] 𝑚^3 /𝑘𝑔
0.1482 − 0.1627 𝑚^3 /𝑘𝑔
(240 − 280)℃ =^
0.1555 − 0.1627 𝑚^3 /𝑘𝑔
−0.0145 𝑚^3 /𝑘𝑔
−0.0072 𝑚^3 /𝑘𝑔
3 𝑘𝑔 0.0145 𝑚
3 𝑘𝑔
(c) Since neither of our given values appear on the table, to solve for these we will need to interpolate three times. Order will not matter on this, but interpolating one way and then another may be easier computationally. For this case and the way the table is set up, interpolating first on temperature and then on pressure appears easiest. Therefore, we’ll start with a pressure of 1.0 MPa:
𝑣[240 ℃] − 𝑣[200℃] 𝑚^3 /𝑘𝑔
𝑣 − 𝑣[200℃] 𝑚^3 /𝑘𝑔
0.2275 − 0.2060 𝑚^3 /𝑘𝑔
(240 − 200)℃ =^
v − 0.2060 𝑚^3 /𝑘𝑔 (220 − 200)℃
0.0215 𝑚^3 /𝑘𝑔 40 ℃ =^
v − 0.2060 𝑚^3 /𝑘𝑔 20 ℃
v − 0.2060 𝑚^3 /𝑘𝑔 =
0.0215 𝑚^3 /𝑘𝑔
20 ℃ = 0.01075 𝑚^3 /𝑘𝑔
𝑣(1.0 𝑀𝑃𝑎, 220℃) = 0.01075 𝑚^3 /𝑘𝑔 + 0.2060 𝑚^3 /𝑘𝑔 = 0.21675 𝑚^3 /𝑘𝑔
This is now the specific volume at P=1.0 MPa and T = 220 oC. Now we repeat this interpolating on the 1.5 MPa table
𝑣[240 ℃] − 𝑣[200℃] 𝑚^3 /𝑘𝑔
𝑣 − 𝑣[200℃] 𝑚^3 /𝑘𝑔
0.1483 − 0.1325 𝑚^3 /𝑘𝑔
v − 0.1325 𝑚^3 /𝑘𝑔 (220 − 200)℃
3.39) Determine the values of the specified properties at each of the following conditions.
(a) For Refrigerant 134a at T = 60 oC and v = 0.0762 m^3 /kg, determine p in kPa and h in kJ/kg.
(b) For ammonia at p = 8 bar and v = 0.005 m^3 /kg, determine T in oC and u in kJ/kg.
(c) For Refrigerant 22 at T = -10oC and u = 200 kJ/kg, determine p in bar and v in m^3 /kg.
a) From Table A-12 in the appendix we have:
For Refrigerant-134a at this temperature and specific volume, it is a superheated vapor. This is determined by looking at Table A-10 and noting that at 60 oC the specific volume of the saturate vapor is 0.0114 m^3 /kg. The specific volume of our state is larger than this, indicating a superheated vapor state. Proceeding, we use Table A-12 and look for two values that boarder
our specific volume. By inspection of the table one finds: v(P=3.2 bar)=0.08106 m^3 /kg and v(P=4.0 bar) = 0.06405 m^3 /kg. Interpolating off of these values to find our pressure we find:
𝑣[3.2 𝑀𝑃𝑎] − 𝑣[4.0 𝑀𝑃𝑎] 𝑚^3 /𝑘𝑔
(𝑝 1 − 𝑝 2 )𝑀𝑃𝑎 =^
𝑣 − 𝑣[4.0 𝑀𝑃𝑎] 𝑚^3 /𝑘𝑔
0.08106 − 0.06405 𝑚^3 /𝑘𝑔
(3.2 − 4.0)𝑀𝑃𝑎 =^
0.072 − 0.06405 𝑚^3 /𝑘𝑔
0.01701 𝑚^3 /𝑘𝑔
0.00795𝑚^3 /𝑘𝑔
0.00795𝑚^3
3 𝑘𝑔
Repeating this procedure for the enthalpy we find: h(3.2 MPa)=302.72 kJ/kg and h(4.0 MPa)=301. kJ/kg. Therefore:
[3.2 𝑀𝑃𝑎] − [4.0 𝑀𝑃𝑎] 𝑘𝐽/𝑘𝑔
(𝑝 1 − 𝑝 2 )𝑀𝑃𝑎 =^
− [4.0 𝑀𝑃𝑎] 𝑘𝐽/𝑘𝑔
(3.2 − 4.0)𝑀𝑃𝑎 =^
−0.8 𝑀𝑃𝑎 =^
−0.8 𝑀𝑃𝑎 −0.374^ 𝑀𝑃𝑎^ = 0.5657𝑘𝐽/𝑘𝑔
Using this quality we can solve for the internal energy. Looking at the table again we find: uf = 262.64 kJ/kg and ug = 1330.64 kJ/kg. Combining all of this information:
𝑢 = 1 − 𝑥 𝑢𝑓 + 𝑥𝑢𝑔
𝑘𝑔 +^ 0.02^ 1330.^
c) From Table A-7 we have:
Since we are given the specific internal energy we first investigate the saturated liquid and vapor tables to see if we are in a two phase region. By inspection of the table above at a temperature of -10 oC we find: ug = 232.02 kJ/kg and uf =33.27 kJ/kg. Since our given specific internal energy of 200 kJ/kg lies between these two values we are in a two phase region. Therefore there is only one pressure at this temperature:
𝑝 = 3.5485 𝑏𝑎𝑟𝑠
To find the specific volume we need to find the quality of the mixture as follows:
3.56) As shown in Fig. P3.56, 0.1 kg of propane is contained within a piston-cylinder assembly at a constant pressure of 0.2 MPa. Energy transfer by heat occurs slowly to the propane, and the volume of the propane increases from 0.0277 m^3 to 0.0307 m^3. Friction between the piston and cylinder is negligible. The local atmospheric pressure and acceleration of gravity are 100 kPa and 9.81 m/s^2 , respectively. The propane experiences no significant kinetic and potential energy effects. For the propane, determine (a) the initial and final temperatures, in oC, (b) the work, in kJ, and (c) the heat transfer, in kJ.
(a) Since we are given the pressure and volume of the propane for the initial and final states, we can use the tables to determine the initial and final temperatures. We first need to calculate the specific volumes for use in the tables by dividing the total volume by the mass of the propane. Doing this yields:
0.0277 𝑚^3
= 0.277 𝑚^3 /𝑘𝑔
0.0307 𝑚^3
0.1 𝑘𝑔 = 0.307^ 𝑚
Converting our pressure from MPa to bars yields a pressure of 2 bars. Looking at the tables for propane for this pressure and looking at the saturated liquid and saturated vapor specific volumes shows vg(2 bars) = 0.2192 m^3 /kg and vf(2 bars) = 0.001781 m^3 /kg. Since both of our values for specific volume are above those for the saturated vapor we use the superheated vapor tables. Looking at these tables for 2 bars we find v(T=30oC) = 0.277 m^3 /kg and v(T = 60 oC) = 0.307 m^3 /kg. Since our specific volumes are found directly on the tables:
𝑇 1 = 30 ℃
(b) Since the propane is kept at a constant pressure, we can find the work by:
𝑉 2 𝑉 1 𝑊 = 𝑝(𝑉 2 − 𝑉 1 )
𝑊 = 0.2 𝑀𝑃𝑎 (0.0307 𝑚^3 − 0.0277 𝑚^3 )
𝑊 = 0.2 𝑀𝑃𝑎 0.003𝑚^3 = 0.0006 𝑀𝑃𝑎 𝑚^3 𝑥
1 𝑁/𝑚^2
(c) To find the heat input we can use the energy balance. Since there are no significant changes in kinetic or potential energies, to find the total change in energy we need to find the internal energy at states 1 and 2. Using the superheated vapor tables for our two temperatures we find:
𝑢 1 = 476.3 𝑘𝐽/𝑘𝑔
𝑢 2 = 524.3 𝑘𝐽/𝑘𝑔
𝑈 1 = (𝑢 1 )𝑚 = 476.
𝑘𝑔 0.1^ 𝑘𝑔^ = 47.63^ 𝑘𝐽
Since we now have the change in energy and the work of the system, we can use our energy balance equation to solve for the heat of the system:
𝑑𝐸 𝑑𝑡
Therefore our reduced temperature and pressure are:
𝑇𝐶^ =^
647.3𝐾 = 1.148^ ≅^ 1.
𝑃𝐶^ =^
Using the generalized compressibility chart (Figure A-1) we have:
b) To solve using the steam tables we need to use the analytical form for the compressibility:
𝑍 =
Since we are given p and T we can use the steam tables to find the specific volume (v) and look up the value of R for water. From Table 31. in the text (page 111) we have:
𝑅𝑤𝑎𝑡𝑒𝑟 = 0.
Looking up the temperature of a saturated mixture of water at 200 bar (the temperature will be constant anywhere in the two phase dome, from saturated liquid to saturated vapor) we find a temperature of 365.5 oC. Since our given temperature is greater than this we have a superheated vapor. Looking up values for our given pressure of 200 bars in the superheated water vapor tables we find: v(T=440 oC) = 0.01222 m3/kg and v(T=480 oC) = 0.01399 m^3 /kg. Since our given temperature of 47609 oC does not appear on the tables, we need to interpolate between these two values. Doing so yields:
𝑣[440 ℃] − 𝑣[480℃] 𝑚^3 /𝑘𝑔
𝑣 − 𝑣[480℃] 𝑚^3 /𝑘𝑔
0.01222 − 0.01399 𝑚^3 /𝑘𝑔
(440 − 480)℃ =^
v − 0.01399 𝑚^3 /𝑘𝑔 (470 − 480)℃
−0.00177 𝑚^3 /𝑘𝑔 − 40 ℃ =^
v − 0.01399 𝑚^3 /𝑘𝑔 − 10 ℃
v − 0.01399 𝑚^3 /𝑘𝑔 =
0.00177 𝑚^3 /𝑘𝑔
− 10 ℃ = −0.0004425 𝑚^3 /𝑘𝑔
𝑣 = −0.0004425 𝑚^3 /𝑘𝑔 + 0.01399 𝑚^3 /𝑘𝑔 = 0.0135 𝑚^3 /𝑘𝑔
Plugging all of these values into our compressibility formula yields:
𝑍 =
200 𝑏𝑎𝑟 (0.0135^ 𝑚
3 𝑘𝑔 ) 0.4614 (^) 𝑘𝑔 ∙ 𝐾𝑘𝐽 (470℃)
Converting units and solving:
0.0135 𝑚^3
3.119) As shown in Fig. P3.119, a piston-cylinder assembly whose piston is resting on a set of stops contains 0.5 kg of helium gas, initially at 100 kPa and 25 oC. The mass of the piston and the effect of the atmospheric pressure acting on the piston are such that a pressure of 500 kPa is required to raise it. How much energy must be transferred by heat to the helium, in kJ, before the piston starts rising? Assume ideal gas behavior for the helium.
To solve for the heat needed to be input into the system we first set up an energy balance: 𝑑𝐸 𝑑𝑡
Neglecting any kinetic or potential energy terms and noting that since the volume is fixed no work is done, we have:
0 + 0 + ∆𝑈 = 𝑄 − 0
∆𝑈 = 𝑄
Since we are given the initial and final pressures, and the initial temperature, we would normally be able to look up some of these values (the internal energy) in a table and solve this equation. In the book there are no tabulated data for Helium, therefore another approach is needed.
Since we are treating the air as an ideal gas we know the equation of state:
𝑃𝑉 = 𝑛𝑅𝑇
Also the system is closed and the volume doesn’t change, thus n is a constant and V 1 = V 2. We may therefore solve for the final temperature of the system:
𝑉 =
Solving for T 2 we find:
𝑇 2 =
We now need an equation relating our new temperatures and the internal energy at a constant volume. Since we are treating the Helium as an ideal gas the specific internal energy depends only on temperature and is related by the specific heat capacity of the gas:
Since we want the total internal energy, not the specific internal energy, we need to multiply this by the mass of the gas (which we are given). Therefore:
∆𝑈 = 𝑚∆𝑢 = 𝑚∆𝑇 𝑐𝑣 𝑇 = ∆𝑇 𝑐𝑣 (𝑇)
The book does not give values for the specific heat capacity at constant volume for Helium. The data for the specific heat capacity of helium (and all the monatomic noble ideal gases) at a constant pressure can be found. This can be seen from Fig. 3.13 on page 120 of the text. The relationship is:
𝑐𝑝 =
2 8.^
𝑘𝑚𝑜𝑙 ∙ 𝐾 = 20.^
Supplemental Question 1) In the Virial equation of state, what type of molecular interactions do the first three terms taken into consideration?
A sample virial equation of state (where virial stems from the Latin word for force) giving the compressibility factor as an infinite series expansion in pressure is:
𝑍 = 1 + 𝐵 𝑇 𝑝 + 𝐶 𝑇 𝑝^2 + 𝐷 (𝑇)𝑝^3 + ⋯
Each of the terms in the expansion takes into account interactions among progressively more molecules, where the interactions take place at the same time. The first term (1) takes into account interactions of the molecule with itself, the next term takes into account the interaction of the molecule with one other molecule, and the third term takes into account the simultaneous interaction of the molecule with two other molecules. The expansion continues up to however many simultaneous molecular interactions one wishes to consider (thought the terms get progressively smaller as the likelihood of these interactions decreases).
Supplemental Question 2) For a system whose state coincides with the two phase region of a P-v diagram, is that system a simple system?
A simple system is one that is composed of a pure substance or a uniform mixture of non-reacting gases. A pure substance is one that has a uniform and invariable chemical composition. Since a system that is in the two phase region of a P-v diagram has two states in it (either: solid- liquid, liquid-gas, or gas-solid), it does not have a uniform chemical composition and as such is NOT a simple system.
