chemical engineering thermodynamics
1. Ask students to use their knowledge of chemical thermodynamics to choose at least two topics from the following topics for discussion: (total score 100)
1. There are many equations of state in the textbook. Please choose a system and a state equation according to my work or life, analyze the calculation accuracy of PVT relationship, and put forward the direction and suggestions for improvement.
PVT state analysis of propylene
Recently, I was trained in the methanol workshop of a coking plant in Wujiaqu, Xinjiang. Propylene is the most used refrigerant in methanol purification. While studying the propylene compression section, I also have a deep understanding of the physical and chemical properties of propylene.
Physical and chemical properties of propylene: propylene is a colorless and slightly sweet flammable gas, with molecular formula of CH3CH=CH2, molecular weight of 42.08, boiling point of -47.7℃, melting point of-185.25℃ and density of air.
1.46 times, critical temperature 9 1.8℃, critical pressure 4.6Mpa, explosion limit 2.0 ~1%(by volume), flash point-108℃. (Therefore, propylene should be carefully stored. If it leaks, because it is heavier than air, it will accumulate in low-lying places and ditches. If it meets Mars in the process of flowing, it will easily cause an explosion and lead to serious consequences. )
The calculation accuracy of PVT relationship of liquid propylene is analyzed by using R-K equation of state. The critical data of propylene is Tc=364.9K, which comes from Chemical Thermodynamics edited by Chen Guangjin et al. pc=46.0* 10- 1MPa，
The following is the propylene performance data given by Shanghai Coking Plant.
For the convenience of calculation, the new data obtained through excel conversion and simple calculation are as follows:
temperature
-40-30-20-10010 20 30 40 pressure
Atm)1.4012.097 3.023 4.257 5.772 7.6851.04612.916.307 turnover.
mL/g) 12966 6404 4639 3423 2569 1957 10 15 10 10 1 177 50 20.299 922(℃)(
temperature
-40 -30 -20 - 10 0 10 20 30 40
(℃)
Temperature (k) 233 243 253 263 273 283 293 303 313 pressure p.
1.4 196 2. 1248 3.063 1 4.3 134 5.8485 7.7868 10. 179 1 13.082 1 16.523 1
(1 * 10- 1 MPa)
Molar volume v
54560.928 26948.032 19520.9 12 14403.984 108 10.352 8235.056 6354.080 6354.080 4952.8 16
(1*10-5m3/mol)
R-K equation: p? radiogram
v？ b？ a
T0.5vv？ b
0.42748R2T2.52.5
Answer? c？ 0.42748? 8.3 146? 364.9? 16.3409? m6？ Dad? K0.5
p.6？ 106? mol？ 2?
Complement fourth component deficiency
b？ 0.08664RTc？ 0.08664? 8.3 146? 364.9? 5.7 145? 10? five
p.6？ 106? m3？ mol？ 1?
Complement fourth component deficiency
The molar volume v is also known from the above table, so according to the R-K equation, the temperature can be calculated by excel.
Pressure value P 1 degree:
temperature
(℃)-40-30-20-1001020 30 40 pressure p
1 * 10- 1 MPa) 1.4 196 2. 1248 3.0654438+0 4.365435.
p- 1 1.0288 2. 1706 3. 1 182 4.3903 6.0679 8.2505 1 1.0602 1 1.444 12 15.654438
1( 1 * 10 MPa)
Comparing the data calculated by R-K equation of state with the given value, we can get the following data diagram: 50 323 20.5680 3879.776 50 (
Comparing the calculation with the data in the above figure, it can be concluded that the pressure value P 1 calculated by substituting the V value into the R-K equation has little deviation from the given P value. That is to say, for gaseous propylene, it is very reliable to calculate its PVT relationship with R-K equation of state.
Brief introduction of refrigeration process:
The propylene gas from the recovered alcohol is mixed with the gas discharged from the shell side of the propylene subcooler at a pressure of 0. 13Mpa and a temperature of -40℃. After gas saturation measurement, it enters at a pressure of 0. 12Mpa and a temperature of -40℃. In the first stage of propylene compressor, the pressure of propylene flash steam from flash tank is 0.525Mpa and the temperature is -5.5℃. Entering the middle section of propylene compressor, the two gases are compressed to 1.9Mpa and discharged at 102℃. Compressed gas is condensed by propylene condenser, and liquid propylene enters propylene storage tank under pressure.
The temperature of 1.85Mpa is 45℃. In order to prevent the propylene compressor from surging, the makeup gas at the gas outlet of the propylene compressor is insufficient and returns to the loop pipeline of the inlet separator. Liquid propylene from propylene storage tank enters the flash tank, and flash steam enters the middle section of propylene machine. The first pipeline from the outlet of propylene compressor with the temperature of℃ is the anti-surge secondary loop. Liquid propylene has a pressure of 0.525Mpa and a temperature of -5.5, and is led out from the bottom of the flash tank. All the way into the compressor inlet separator, adjust the liquid level to supplement the imported propylene airflow. The other path enters the propylene refrigerator, and the propylene gas in the shell side flashes to cool down. The liquid from the tube side is further adjusted to -20℃ with a pressure of 0.485Mpa, and then leaves the system and enters the recovery liquid. In order to prevent the trace moisture in propylene from freezing during flash evaporation, it is necessary to inject a small amount of methanol into the system. The spraying of methanol is completed by adjusting the spraying and mixing of methanol at the outlet of metering pump and liquid propylene from propylene storage tank to the flash tank.
Because propylene is flammable and explosive, N2 should be used to replace propylene before and after equipment maintenance, and then N2 should be replaced by air. When driving, replace air with N2 first, and then replace N2 with propylene.
2. According to the principle of work-heat conversion, select a system or working condition to analyze the energy-saving process. Detailed calculation steps and process analysis are required.
Principle of air conditioning refrigeration
Air conditioning can be seen everywhere in daily life. The refrigeration process and principle are analyzed by temperature entropy relation and pressure enthalpy relation respectively. The refrigeration principle of air conditioner involves the first law of thermodynamics and the second law of thermodynamics, which is an inverse Carnot cycle. The following is the analysis and calculation of theoretical refrigeration cycle.
1. Inverse Carnot cycle-functional calculation of ideal refrigeration cycle;
Figure 1 temperature difference diagram
It includes two isothermal processes and two adiabatic processes. Assuming that the temperature of the low-temperature heat source (i.e. the cooled medium) is T0 and the temperature of the high-temperature heat source (i.e. the environment) is Tk, the temperature of the working substance in the heat absorption process is T0 and the temperature of the working substance in the heat release process is Tk, that is, there is no temperature difference between the working substance and the cold source and the high-temperature heat source in the heat absorption and release process, that is, the heat transfer is carried out at an isothermal temperature, and the compression and expansion processes are carried out without any loss. The circulation process is as follows:
The working medium firstly absorbs heat q0 from the cold source (i.e. the cooled medium) at time T0, and performs isothermal expansion of 4- 1, then increases its temperature from T0 to the temperature of the environmental medium through adiabatic compression of 1-2, and then performs isothermal compression of 2-3 at time Tk, releasing heat qk to the environmental medium, and finally performs adiabatic expansion of 3-4, so that.
For the inverse Carnot cycle, as can be seen from the figure:
q0=T0(S 1-S4)
qk=Tk(S2-S3)=Tk(S 1-S4)
w0 = qk-Q0 = Tk(s 1-S4)-T0(s 1-S4)=(Tk-T0)(s 1-S4)
The refrigeration coefficient εk of the inverse Carnot cycle is: εk = w0/ qk=(Tk-T0)/Tk.
As can be seen from the above formula, the refrigeration coefficient of the inverse Carnot cycle has nothing to do with the nature of the working medium, but only depends on the cold source (i.e.
The temperature T0 of the cooled object and the temperature tk of the heat source (i.e. the environmental medium); Reducing Tk and increasing T0 can improve the refrigeration coefficient. In addition, the second law of thermodynamics can be used to prove that "the refrigeration coefficient of inverse Carnot cycle is the highest in a given temperature range of cold and heat sources". The refrigeration coefficient of any actual refrigeration cycle is less than that of the inverse Carnot cycle.
Four components of refrigeration system and the changing process of refrigerant;
The vapor compression refrigeration cycle system is mainly composed of four parts: compressor, condenser, throttling element and evaporator, which are connected in series with pipes with different diameters to form a closed system that can make refrigerant circulate. The refrigeration compressor is driven by the motor and other prime movers, and continuously sucks the refrigerant vapor in the evaporator, compresses it into high pressure (pk) and superheated vapor, and sends it to the condenser after being discharged. It is precisely because of this high pressure that the refrigerant vapor releases heat in the condenser and transfers it to the surrounding environmental medium, so that the refrigerant vapor condenses into liquid. Of course, the temperature at which the refrigerant vapor condenses must be higher than the temperature of the surrounding medium. The condensed liquid is still under high pressure and flows through the throttling element into the evaporator. In the throttling element, the refrigerant is reduced from high pressure pk at the inlet end to low pressure p0 and from high temperature tk to t0, and a small amount of liquid is evaporated into steam.
2. Process Calculation of Inverse Carnot Cycle-Ideal Refrigeration Cycle
According to the hypothesis of theoretical cycle, the working process of theoretical cycle of single-stage vapor compression refrigeration is shown in figure 2 on the pressure enthalpy diagram.
Fig. 2 pressure enthalpy diagram of theoretical refrigeration cycle
1) The refrigeration compressor sucks saturated refrigerant vapor with evaporation pressure p0 from the evaporator (state point 1) and compresses it to condensation pressure pk along the isentropic line (state point 2), and the compression process is completed.
2) The high-temperature and high-pressure refrigerant vapor at state point 2 enters the condenser, exchanges heat with ambient medium air or water through the condenser, releases heat qk, and then cools to saturated steam state point 2 along isobar pk? And then condensed to the saturated liquid point 3, and the condensation process is completed. During the cooling process (2-2? ) refrigerant and environment interface
There is a temperature difference (2? -3) There is no temperature difference between the refrigerant and the ambient medium.
3) After the saturated refrigerant liquid at the state point 3 is throttled and depressurized by the throttling element, it drops from the condensation pressure pk to the evaporation pressure p0 along the isenthalpic line (the enthalpy is unchanged during throttling) and reaches the wet steam state point 4, and the expansion process is completed.
4) The wet refrigerant vapor at the state point 4 enters the evaporator, absorbs the heat of the cooled medium, and evaporates along the isobar p0 to reach the saturated vapor state point 1, and the evaporation process is completed. There is no temperature difference between the evaporation temperatures of the refrigerant and the cooled medium.
Calculation method of theoretical period:
1, refrigerating capacity per unit mass The refrigerating capacity produced by the cooled medium per cycle 1kg of refrigerant of the refrigeration compressor is called refrigerating capacity per unit mass and expressed by q0.
Q0 = h 1-H4 = r0( 1-x4)( 1- 1)
Where q0 unit mass cooling capacity (kJ/kg);
H 1 specific enthalpy (kj/kg) corresponding to inspiratory state;
H4 specific enthalpy of wet steam after throttling (kJ/kg);
Latent heat of evaporation of refrigerant at r0 evaporation temperature (kJ/kg);
Dryness of gas-liquid two-phase refrigerant after x4 throttling.
The refrigerating capacity per unit mass q0 is equivalent to the projection of hydrograph 1-4 on the H axis on the pressure enthalpy diagram (see figure 1-2).
2. The refrigerating capacity per unit volume The refrigerating capacity obtained from the cooled medium after each inhalation of 1m3 refrigerant vapor by the refrigeration compressor is called the refrigerating capacity per unit volume, which is denoted by qv.
qv？ q0h 1？ h4？ v 1v 1 ( 1-2)
Where qv is the refrigerating capacity per unit volume (kJ/m3);
V 1 specific volume of refrigerant in suction state (m3/kg).
3. The work consumed by the theoretical specific work refrigeration compressor for compressing and conveying 1kg refrigerant vapor each time is called theoretical specific work, which is expressed by w0.
w0=h2-h 1 ( 1-3)
Where the theoretical specific work W0 (kJ/kg);
Specific enthalpy (kJ/kg) of refrigerant in the exhaust state of h2 compressor;
H 1 specific enthalpy of refrigerant in suction state of compressor (kJ/kg).
4. Unit condensing heat load The heat released by the refrigeration compressor per kloc-0/kg refrigerant in the condenser is called unit condensing heat load, which is expressed by qk.
qk=(h2-h2？ )+(h2？ -h3)=h2-h3 ( 1-4)
Where qk unit condensation heat load (kj/kg);
h2？ Specific enthalpy of dry saturated steam corresponding to condensation pressure (kj/kg); H3 corresponds to the specific enthalpy (kJ/kg) of saturated liquid at condensation pressure;
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In the pressure enthalpy diagram, qk is equivalent to isobaric cooling condensation process line 2-2? The projection of -3 on the H axis (see Figure 2).
Comparing equations (1- 1), (1-3), (1-4) and h4=h3, we can see that the theoretical cycle of single-stage vapor compression refrigeration has the following relationship.
qk = q0 +w0 ( 1-5)
5. Refrigeration coefficient The ratio of refrigerating capacity per unit mass to theoretical specific work, that is, the ratio of profit to cost of theoretical cycle, is called theoretical cycle refrigeration coefficient. 0 means, that is
q0h 1？ h4？ w0h2？ h 1 ( 1-6)？ 0?
According to the above performance indicators, we can further get the data of refrigerant circulation, heat release of condenser, theoretical power required by compressor and so on.
3. Choose a suitable phase equilibrium calculation method for the treatment of phenol-containing aqueous solution, give the detailed calculation process and steps, and analyze and discuss the results.
A mixed integer nonlinear programming model for phase equilibrium calculation of mixed electrolyte solution is established, and a genetic algorithm for solving this model is proposed. Firstly, based on the principle of Gibbs free energy minimization, the phase equilibrium calculation model of electrolyte system is established by coding the salts precipitated in liquid phase and solid phase, and the phase equilibrium calculation problem is transformed into a constrained optimization problem. Secondly, the genetic algorithm is used to solve the problem, and the dynamic boundary feasible region coding method and sequential convergence technology are used to optimize the variables to ensure the effective implementation of the algorithm, so as to realize the solid-liquid equilibrium calculation and get the number of precipitated crystals, salts, solid content and liquid phase composition; Finally, the calculation results of various systems show the feasibility and effectiveness of this method.
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