Thermodynamics Lab

Mark powers Period 2 C physics March 17 Thermodynamics testing ground I. The take of this experiment is to keep a certain amount of pee as hot as possible for as long as possible. We also accomplish to use whatever significants we want as long as our product is or is smaller than 20cm. II. For my project I use Styrofoam, tinfoil, insulation, silicone, and a cork. I used the Styrofoam as a recess to put the can in and the Styrofoam also acts as a thermal insulation. I used the tin foil because it serves as an insulator and keeps the radiant agitate inside the box.Next I used insulation, which does what its name is, it keeps whatever inside of it insulated or prevents convection. I used silicone to block eat up the holes from the Styrofoam to make a good box for the screw up to stay in. finally I used a cork on the top of my can to consec appreciate the smallest hole for the heat to escape it. III. For this experiment I used Styrofoam, tinfoil, insulation, silicone, and a cor k. I also need a thermometer to measure the change in temperature of the water during the course of the experiment.IV. First I cut out Styrofoam to the appropriate lengths and made sure it wasnt over the 20 cm limit. I then cover the Styrofoam with tin foil and sealed off all the cracks on the inside with silicone. After that I put insulation on the whole inside of the box with enough room to put the can in on the inside. I finally added a cork on the top of the can to make the smallest interruption so that the heat had little room to escape to conclude my project. Time Temperature (C) 0 minutes 20 minutes V. Data VI. 1. I think the around important type of the 3 types of energy transfer that we had to minimize was convection because convection is heat transfer via heating surrounding fluid and then the fluid moves. 2. The sun doesnt heat us through convection or conduction because for heat to heat through conduction it need to be twining it and we dont touch the sun, and for co nduction is heat transfer via heating surrounding fluid and then the fluid moves but us humans are solids, not liquids. . Mammals that live in cold climates and cold water insulate themselves with their fur and being cold blooded. The mammals that dont live in water use their fur as an insulator to keep their warm air inside them and keep the cold air outside, and the cold-water mammals usually are cold blooded so that their temperature doesnt unfeignedly change with the cold conditions that they are around all the time. VII.Thermodynamics Lab uprise This report will show the acquired understanding of the refrigeration one shot by development first and second laws of thermodynamics. In order to analyze this system several assumptions where made such like an isentropic physical process at the compressor an isenthalpic expansion in the throttling valve. Diagrams will be provided to depict these thermodynamic processes in addition to computing the heat transferred to the system and the work input to the compressor. Table of Contents Abstract . Table List . 3 excogitation 4 Descriptions. 5 Theory Calculation 8 Discussion. 11 References. 15 Appendix. 16 ObjectivesThe Refrigeration lab was conducted to gain a better understanding of the refrigeration cycle, Carnot cycle and to compare the ideal cycle. Also, determent the heat transfers by using the mass and energy balance. Introduction The term refrigeration may be defined as the process of removing heat from a substance under controlled conditions. It also includes the process of reducing and maintaining the temperature of a body to a lower place the general temperature of its surroundings. In other words, the refrigeration means a continued extraction of heat from a body whose temperature is already below temperature of its surroundings.In this lab 1,1,1,2-Tetrafluoroethane (R-134a) was used as cold. Since energy cannot be sunk according to first law of thermodynamics. In a refrigerator, heat is virtually pumped from a lower temperature to a higher temperature. According to Clausius Statement of Second Law of Thermodynamics states that heat will not pass cold to hotter region without the aid of an away agency. Thus, process can only be performed with the aid of some external work. It is thus obvious that supply of power is regularly required to drive a refrigerator.Theoretically, a refrigerator is a reversed heat engine or a heat pump which pumps heat from a cold body and delivers it to a hot body. The substance, which works in a pump to extract heat from a cold body and to deliver it to a hot body, is known as refrigerating. Description of Refrigeration cycle description Most commonly used refrigeration/heat pump cycle and involves the same four processes as a heat engine cycle but in the reverse order (i. e. evaporation compression condensation expansion/throttling). flesh 1 gives a schematic theatrical performance of the four essential mechanical components in this cycle. light out 3 2 Condenser Throttle Work in Valve Evaporator Compressor 4 1 Heat in date 1 Refrigeration Cycle components Compressor The compressor in a refrigeration system is essentially a pump. It is used to pump heat uphill from the cold side to the hot side of the system Condenser The condenser or cooler consists of coils of pipe in which the high pressure and temperature vapour refrigerant is cooled and condensed.Expansion valve As the high-pressure high temperature liquid refrigerant passes the throttling valve seating, its pressure and temperature drop to that of the evaporator. The drop in temperature is changed liquid phase to a mixture of unaggressive and temperature liquid and vapour enter the evaporator without any change in enthalpy Evaporator The purpose of the evaporator is to remove unwanted heat from the product, via the liquid refrigerant 3 2 4 1 Superheating Sub-cooling Critical Point Liquid Pressure Enthalpy 3 2 4 1 Superheating Sub-cooling Critical PointLiquid Pressure Enthalpy Figure 2 Refrigeration cycle on pressure enthalpy diagram. In cycle the refrigerant vapour is soused to a higher temperature and pressure (12). The compressed vapour is then condensed isobarically which results in heat rejection to the surroundings (23). The next step is the adiabatic throttling of the refrigerant to the low temperature and pressure (34). The final step is where the refrigerant is evaporated at low temperature and pressure, which results in the absorption of heat from its surroundings (41). Theory /AnalysisFigure 3 According to first law of thermodynamics energy (heat, work), cannot destroyed but energy changing from one form to another. ?H+? Ke+? Pe=? Q+? W (1) (Balmer, 2011) Where H=enthalpy, K=kinetic energy, P= potential energy, Q= heat and W= work. In order to simplify the calculation for the energy balance, the following assumptions were made Kinetic and potential energy changes are negligible throughout the system So, ?H=? Q+? W In a con stant pressure for the condenser (work=0). The heat transfer of the condenser is the heat loss, in other words Q2-3 and is given byQ2-3= m(h3-h2) In a constant pressure for evaporator (work=0), he heat gain by the evaporator is equal to Q1-4 and is given by Q4-1= m(h1-h4) The work done by the compressor is the same as the work in the cycle and is given by Q1-2= (h2-h1) The power drawn by the compressor defined by the following equation Power= Voltage ? latest ? Power Factor Coefficient of performance is defined as savvy=Q4-1? Q2-3? -Q4-1 (2) (Balmer, 2011) -The heat transfer water can be mensural from the water feed in place and temperature changes using the following formula where Cp is specific heat capacityQ= mCp ? T Convective heat transfer occurs between the air and the refrigerant and this is can be auspicated using the following Qconv= hA (T? -TS) (3) (Balmer, 2011) Where to (h) is convective heat transfer coefficient =25, A= area Sample Calculations mr=76mlmin ? 1L molar concentrationml? 1m31000ml? 1 min60s= 0. 001505kgs mw, cond=11. 7Lhr? 1m31000L? 1 hr3600s =0. 00322kgs mw, evap=27. 5Lhr ? 1m31000L ? 1hr3600s=0. 00764kgs Q23, cond= 0. 001505kgs ? 83-300kJkg? 1000=-324. 5w Q14, evap= 0. 001505kgs? 260-83kJkg? 1000=266. w Qab, cond=0. 00322kgs? 4. 2kJkg. K ? 34. 9-16. 2K? 1000=253w Qcd, evap=0. 00764kgs? 4. 2kJkg. K ? 8. 4-16. 2K? 1000=-250. 2w W12 =-0. 001505kgs? 300-260kJkg? 1000= -60. 13w COPc=260-83kJkg300-260kJkg=4. 425 Pelectric compressor=228. 2V ? 0. 79A? 0. 75=135W Results Heat transfer refrigerant Heat transfer water Q2-3 (W) -326. 54 Q2-3(W) 253. 07 Q3-4 0 Q1-4 (W) -250. 25 Q4-1 (W) 266. 34 W1-2 -60. 129 COP 4. 425 Area (m2) 0. 162 Q convective 2-3(w) 87. 062 Q connective 4-1(w) -41. 904 conv HTC of air (m2*k) 25 Total energy balance (w) -12. 20984307Carnot Refrigeration Cycle between to two pressure Figure 4 A reversed Carnot cycle is shown on p-h diagrams in Figures respectively. The processes of the cycle are as follows The re frigerant is compressed isentropically as shown by the curve 1-2 and 3-4 on p-h diagrams. During this process the T2 temperature of refrigerant decrees from. We know that during isentropic compression, no heat is absorbed or rejected. The refrigerant is now (compressed/ evaporated) isobarically (P2 = P3) and (P4=P3) as shown by the p-h. We know that the heat rejected in 2 to 3 and heat absorbed from 4 to 1.Result Carnot cycle kj/kg h1 278 h2 300 h3 118 h4 83 COP 8. 86 paragon cycle Fgure 5 A reversed ideal cycle is shown on p-h diagrams in Figures respectively. The processes of the cycle are as follows Ideal cycle h1 250 h2 275 h3 100 h4 100 COP 6 The refrigerant is compressed isentropically as shown by the curve 1-2 . in the exit of the condenser the refrigerant is in a saturated liquid in curve 3-4 on p-h diagrams. After that refrigerant exit the evaporator in saturated vapour. In a isobarically process 4-1 as shown by the p-h ResultDiscussion It was seen in the refrigeration cyc le Carnot cycle and the Ideal cycle there are different coefficients of performance on refrigerant. If we look at COP formula the low heat is carve up by the work in in this topic where the refrigerant entered the compressor could significantly hinder the performance of this component by change (superheating). Also the refrigerant deviation the condenser will effect the COP (sup-cooling) . If we compare Carnot cycle to the actual cycle, in Carnot refrigerant super- heating has increased, which will increase the COP.In the other hand, in Ideal cycle the sub-cooling has decreased which the sub-cooling decrease the liquid during expansion to saturated liquid-vapour and increases the refrigerating performance, but by decreases the sub-heating in the evaporator exit from vapour to saturated vapour. Which in this case the COP decreased but it will be higher that the actual cycle. The heat transfer occur in the basic refrigeration cycle, by calculate the energy balance in the cycle. The sum of the energy will not be even close to zero and this violates the first law of thermodynamics.So, there is heat lost in the cycle according to the second law of thermodynamics, heat always flows from a material at a high temperature to a material at a low temperature. For heat to transfer there has to be a temperature difference between the two materials. This heat can be careful by Convection is the transfer of heat from one place to a different location by circulating it with a fan or natural movement. Figure 6Temperature vs. distance condenser Figure 7 Temperature vs. distance evaporator In figure 6 and 7 shows the heat exchange, in figure 6 the cold-water flow co-current with the hot refrigerant.With both entering the same end. This application will be utilitarian to if we want a limited outlet temperature. Its also useful to drop more quickly temperature because the large difference in the temperature. Where in figure 7 the hot-water flow countercurrent with the cold ref rigerant the temperature is achievable by the cold stream can never exceed that of the hot steam. (Ulrich, 2004) Conclusion This experiment was carried out in order to observe the properties and functions of a refrigeration cycle. With the properties of the refrigerantR134a at all the stages of the refrigeration cycle known, the heat transfer at the condenser and evaporator were calculated. The heat absorbed by the system (evaporator) was determined to be 16 w. The heat given off by the condenser was calculated to being -73. 46 w. The work done by the compressor was also calculated to being -60. 19 W. With this, the refrigeration performance of the cycle was calculated and was equal to 4. 42. Works Cited Balmer, R. T. (2011). Modren Engineering Themodynamics. Burlington, MA, USA Acadimic Press. Richard M Felder, Ronald W. Rousseau. (2005). Elementary Preincipeles of Chemical Processes (Third ed. . Raleigh, NC John Wiley & Sons, Inc. Ulrich, G. D. (2004). Chemical Engineering Process protrude and Economics. Durham, New Hampshire, USA Process Publishing. Appendix bar temp C l/hr kg/s h (kj/kg) Tatm 18 High P 9. 1 Low P 3. 25 3. 25 T2 62. 1 300 T3 22. 7 83 T1 11. 7 260 T4 4 wet Tin 16. 2 Water Cond T out 34. 9 Water Evap T out 8. 4 Ref Flow rate 76 0. 0015048 Water Con flow rate 11. 6 0. 003222222 Water evap flow rate 27. 5 0. 007638889 Voltage