Thermo mini project

Thermodynamics Mini Project Title: Evaporator Group members: KA20068 1. Pui Wen Jie KA 20074 2. Choong Wai Kit KA 20075 3. Tan Wei Hang KA 20084 4. Chan Guo Quan Group Member Contribution: 1. Pui Wen Jie - 100 % 2. Choong Wai Kit – 100% 3. Tan Wei Hang – 100% 4. Chan Guo Quan – 100%

Abstract Evaporator is a system that transfers heat to an aqueous mixture like salted solution and to concentrate the solution with a higher concentration of solute by boiling off the solvent. The concentration process is done by supplying steam through calandria to transfer heat to the incoming feed. Separation of the feed occurs and the condensed liquid leaving from the evaporator is the product with higher concentration compared to feed before processed. Main findings of our project is to calculate the enthalpy, heat and entropy for the evaporation process. As a conclusion, we found that the enthalpy of feed Hf= 42.022kJ/kg, enthalpy of product, Hp1=83.91 kJ/kg, enthalpy of vapor released, Hv1= 2537.4 kJ/kg, enthalpy of saturated vapor in calandria, Hvs= 2675.6 kJ/kg, enthalpy of condensed steam in calandria, Hcs= 419.17kJ/kg. For the heat transfer between calandria and feed, Q= 15675 kJ/h and the entropy generated Sgen= 11.530 kJ/K•h.

Introduction Evaporation is a concept that transfers heat to a boiling liquid and to concentrate a non- volatile solute from a solvent by boiling off the solvent. As cited by Vanitha Dunna and P.S Kishore (2015), evaporation is an operation used to remove a solvent as vapour from a solution. Thus, this is a thermal concentration process and the most common solvent in the evaporation systems is water. The process of evaporation is usually stopped before the non-volatile solute begins to precipitate. If the solute is precipitated, it will be better considered as a crystallization process. Leena M. Borkar (2013) mentioned the difference between evaporation and crystallization processes. In evaporation, the main function is to vaporize the solvent but crystallization focuses on the crystal formed as the main function in the process. As stated by Paul E. Minton (1986), there are three principal elements in a evaporator design, which are heat transfer, vapor-liquid separation, and efficient utilization of energy. First, condensing steam supplied heat for the process. Large amounts of heat are transferred across a calandria to achieve the separation of liquid and vapor from the feed. Inadequate separation of the mixture may happen due to factors such as fouling of the equipment downstream, pollution and corrosion of the downstream equipment. This will result in the inefficient operation or pumping problem of the evaporator. Performance of the evaporator is rated as the ability of solvent evaporated per pound of steam used. A higher performance evaporator makes more efficient use of the energy. Energy efficiency could be increased by exchanging heat between the feed and the condensate. Since the evaporators deal with the aqueous solution and there are non-volatile solutes like salt dissolve in the solvents. Therefore, the effect of boiling point elevation must be accounted for. Angelico Obille (2017) related that when solutes dissolved in pure solvents, colligative properties exist as a change in physical properties. The solute is considered as impurities which will bring effects of freezing point depression (FPD) and boiling point elevation (BPE) on the solution. Hence, most of the substances are considered mixtures because it is difficult to maintain its substance purity.

There are two different types of evaporators: single-effect evaporator and multi-effect evaporator. In a single-effect evaporator, the vapor is contaminated and cannot be reused. So this is normally used when the throughput is low and a cheap supply of steam is available. In a single-effect evaporator, more than 1 kg of steam is needed in order to evaporate 1 kg of water. In a multi-effect evaporator, there is a higher efficiency in using energy and less heat loss if compared to a single-effect evaporator. This is because the vapor formed will be used to heat the second evaporator. Therefore, normally 1 kg of steam is used to evaporate 1kg of water in Mv, Hv1 Mf, Hf Ms, Hvs Ms, Hcs Mp, Hp1 a multi-effect evaporator. (Rucha G. Bansod, Isha C. Deshpande, Gandhar Y. Chaudhari & Ravi W. Tapre, 2018) Figure 1: Single-effect evaporator Figure 2: Multi effect evaporator

In sugar concentration of ethanol production by single-effect evaporator and multi- effect evaporator, the single-effect evaporator achieved the desired sugar concentration while the multi-effect evaporator exhibited a product with lower sugar concentration. Although the product from multi-effect systems is diluted compared to single-effect systems, this residue achieves the human’s need. The multi-effect evaporators also shows it has a greater ability to concentrate the feed which could reduce the energy consumption. This means less energy is required to produce the same amount of vapour from the feed. Therefore, the multi-effect evaporator has a greater economic performance than the single-effect evaporator. (Cristina A. Pradoa, Ozair Souzaa, Noeli Sellina & Cintia Marangonib, 2015)

Theoretical framework Figure 3: Input and output of the single effect evaporator The diagram shown above is an example of a single-effect evaporator. In our report, we are assuming that the evaporator used is a single-effect evaporator. The mechanism of a single-effect evaporator involves the principles of thermodynamics such as mass balance, energy balance and entropy balance. The process of a single-effect evaporator begins when the feed enters the evaporator with a flow of Mf, which is the mass flow rate of the feed at a temperature of Tf, which is 10C. The feed contains the aqueous 20% table salt mixture and settles inside the evaporator. At the same time, a supply of saturated steam enters the evaporator as well with set conditions of pressure which is assumed to be 101.42 kPa in this case with a flow of Ms, which is the mass flow rate of steam through another inlet. The saturated steam will then pass through the evaporator and exit through an outlet as condensate The temperature of the evaporator is set to be 20ºC according to the question. After some time, the solution evaporates and leaves the evaporator. The evaporated solution leaves as a saturated vapour with a flow of Mv, which is the mass flow rate of saturated vapour. The remaining solution is the concentrated product and leaves at a flow of Mp, which is the mass flow rate of the concentrated product. According to this, the mass balance of the evaporator system can be deduced as, Mf = Mv + Mp as well as Mvs=Mcs where: Mf= mass flow rate of feed Mvs= Mass flow rate of saturated steam Mv= mass flow rate of vapour Mcs= Mass flow rate of condensate Mp= mass flow rate of concentrated product

The energy balance of the evaporator system can also be deduced as, MfHf + MsHvs = MvHv1 + MpHp1 + MsHcs where: Mf= Mass flow rate of dilute feed Ms= Mass flow rate of steam in calandria Mv= Mass flow rate of vapour out of the evaporator Mp= Mass flow rate of concentrated product out of evaporator Hf= Enthalpy of feed Hvs= Enthalpy of saturated vapour in the calandria at temperature Ts Hcs= Enthalpy of condensate out of the calandria at temperature Ts Hp1= Enthalpy of concentrated product at temperature T1 Hv1= Enthalpy of saturated vapour out of the evaporator While on the other hand, the entropy balance of the system is as follows, .. . (ΔSsys= 0) Sin - Sout + Sgen = ΔSsys Sgen = Sout- Sin Sin= Mf•Sf + Ms•Svs Sout= Ms•Scs + Mp•Sp1 + Mv•Sv1 Sgen= Ms(Scs- Svs) + Mp•Sp1 + MvSv1 - Mf•Sf Sgen= MvSv1 + MpSp1 - MfSf where: Sgen= Entropy generation Mv= Mass flow rate of saturated vapour out of evaporator Sv1= Entropy of saturated vapour out of evaporator at temperature T1 Mp= Mass flow rate of concentrated product Sp1= Entropy of concentrated product at temperature T1 Mf= Mass flow rate of feed Sf= Entropy of feed Svs = Entropy of saturated vapor in calandria at temperature Ts Scs = Entropy of condensate in calandria at temperature Ts

Discussion The assumption that we made are mixture is well mixed, the evaporator works under lower pressure than atmospheric pressure, the temperature of steam that flow in the calandria is equal to the temperature of condensed vapor that exits the calandria (100C at 101.42 kPa), there is no elevation of boiling point for the mixture, mass of feed, mass of vapor and mass of concentrated product is equal to their respective mass flow rates(kg/h), no heat loss to the surrounding, heat of mixing is zero. First, we calculate the density of the solution and we get the density of solution is 1229.6 kg/m^3. After that we use the mass = density/volume to find out the mass of the aqueous table salt mixture and we get the answer 24.592kg. Then, we can calculate the mass of NaCl in the 20L of aqueous salt mixture and we get 4.9184kg. From the internet, we found that the maximum mass of table salt that can dissolve in 100g of water is 35.89g. By using this, we can get this we could calculate for the solution to exceed supersaturation limit, the maximum of 4.1984kg of NaCl can dissolve in 13.703kg of water. Since we know that the mass of water in 20L of aqueous salted mixture is 19.6736kg, we could determine the mass of water to be evaporated, V by subtracting the mass of water in the mixture and the mass of water that can dissolve 4.9184 kg of NaCl, so V=5.9696 kg Since, we assume that all the masses for feed in, vapor out and concentrated product out are equal to their respective mass flow rates. Mass flow rate of vapor flowing out of evaporator, mv= 5.9696 kg/h. Mass flow rate of feed, mf= 24.592 kg/h. By using the mass balance, we could determine the mass flow rate for the concentrated product, mp=18.6224kg/h. Then, we calculate the mass flow rate of the steam. First, we determine for the enthalpy of feed Hf= 42.022kJ/kg, enthalpy of product, Hp1=83.915 kJ/kg, enthalpy of vapor released, Hv1= 2537.4 kJ/kg, enthalpy of saturated vapor in calandria, Hvs= 2675.6 kJ/kg, enthalpy of condensed steam in calandria, Hcs= 419.17 kJ/kg. By using the energy balance and substituting the values of enthalpies, we could determine the mass flow rate of the steam, which is 6.947kg/h. By using this mass flow rate of the steam, we can further determine the rate of heat transfer from the steam supplied from the calandria to the feed and we get the value Q= 15675 kJ/h. To calculate the entropy generation of the evaporator by using entropy balance, we find those entropy values for the entropy of feed, Sf= 0.1511 kJ/kg•K. Entropy of product, Sp1= 0.2965 kJ/kg•K. Entropy of vapor released, Sv1= 8.6661 kJ/kg•K. entropy of saturated vapor

in calandria, Svs= 7.3542 kJ/kg•K. Entropy condensed steam in calandria, Scs= 1.3072 kJ/kg•K. After this, we substitute back these value into the entropy balance and we get the answer for entropy generation, Sgen= 11.530 kJ/K•h.

Conclusion Evaporation is a common process applied for different purposes in industry works such as food processing, chemical, kraft paper, and also pharmaceutical. The evaporator used in different industries will be a bit different in design depending upon the characteristic and the type of feed. In the food processing industry, evaporator is used to concentrate the fresh product like juice. By reducing the water concentration in the food, there will be a lower water activity for bacteria and hence having a longer preservation period. Since the product is concentrated, it also reduces the transportation costs and storage costs. In chemical and kraft paper industries, evaporator is applied to recover valuable chemicals and then refined for reuse in production. In the pharmaceutical industry, evaporator is normally used to produce a concentration solution. The concentrated solution is then dried up and produces the product in powdered form. In conclusion, the enthalpy of feed, Hf= 42.022kJ/kg; the enthalpy of vapor released, Hv1 = 2537.4 kJ/kg; the enthalpy of concentrated product, Hp1= 83.915 kJ/kg; the enthalpy of saturated vapor, Hvs= 2675.6 kJ/kg; the enthalpy of condensed steam, Hcs= 419.17 kJ/kg. The rate of heat transfer from calandria to the feed, Q= 15675 kJ/h. The rate of entropy generated, Sgen= 11.530 kJ/K•h.

Reference 1. Dunna, V., & Srinivas Kishore, P. (2015, August). Thermal Analysis on Triple Effect Falling Film Evaporator. Retrieved January 08, 2021, from https://www.researchgate.net/profile/Pisipaty_Srinivas_Kishore/publication/338533829_Ther mal_Analysis_on_Triple_Effect_Falling_Film_Evaporator/links/5e19d36c4585159aa4c8aca0 /Thermal-Analysis-on-Triple-Effect-Falling-Film-Evaporator.pdf 2. L. Earle., R. (1983). UNIT OPERATIONS IN FOOD PROCESSING. Retrieved January 08, 2021, from https://nzifst.org.nz/resources/unitoperations/evaporation1.htm 3. Rps8tube. (2015, September 04). Single Effect Evaporator - Mass and Enthalpy Balance. Retrieved January 08, 2021, from https://www.youtube.com/watch?v=yUisQIRkMnA&ab_channel=R.PaulSingh 4. R.M. Price. (2003). Evaporation. Retrieved January 08, 2021, from http://facstaff.cbu.edu/rprice/lectures/evap1.html 5. Leena M. Borkar. (2013, June). SOFTWARE FOR SINGLE-EFFECT EVAPORATOR. Retrieved from https://www.technicaljournalsonline.com/ijaers/VOL%20II/IJAERS%20VOL%20II%20ISS UE%20III%20APRIL%20JUNE%202013/314.pdf 6. Paul. E. Minton. (1986). Handbook of Evaporation Technology. Retrieved from http://www.iqytechnicalcollege.com/Handbook%20of%20Evaporation%20Technology.pdf 7. Rucha G. Bansod, Gandhar Y. Chaudhari, Isha C. Deshpande, & Ravi W. Tapre. (2018, October). A REVIEW ON STUDY OF DIFFERENT TYPES OF EVAPORATORS AND THEIR SOFTWARE. Retrieved from https://www.irjet.net/archives/V5/i10/IRJET- V5I1053.pdf 8. Cristina A. Prado, Ozair Souza, Noeli Sellin, & Cintia Marangoni. (2015). Comparison between Single and Multi-Effect Evaporators for Sugar Concentration in Ethanol Production. Retrieved from https://www.aidic.it/cet/15/43/091.pdf 9. The Essentials of Continuous Evaporation. (2018, May 17). Retrieved January 08, 2021, from https://www.aiche.org/resources/publications/cep/2018/may/essentials-continuous- evaporation