Isentropic Expansion of Ideal Gas - Coursework Example

ISENTROPIC EXPANSION OF IDEAL GAS H. Alshekh, ME Butt, ME Brown, ME ME 2160 Fluids Laboratory Laboratory Experiment: 30 Jan 2015 Submission Date: 30 April 2015 Submitted To: Dr. Jonathan Naughton Contents LIST OF SYMBOLS 2 LIST OF TABLES 3 LIST OF FIGURES 3 ABSTRACT 3 INTRODUCTION 4 THEORITICAL BACKGROUND 5 APPARATUS AND MATERIALS 6 Fig. 1: An air tank fitted with inlet/outlet valves and pipes. A pressure gauge fitted to outlet pipe to measure pressure at different states 6 EXPERIMENTAL PROCEDURE 7 RESULTS AND DISCUSSION 7 GRAPHS 9 Conclusions 10 References 11 APPENDIX 12 Appendix I: Derivation of Formula to Calculate the Specific Heat (Cp): 12 Appendix II: Derivation of Formula to Calculate the Temperature (T2) 14 Appendix III: Derivation of Formulae to Calculate Entropy (S) 14 LIST OF SYMBOLS SYMBOL DEFINITION P Pressure (Pa) T Temperature (K) Cp Molar heat capacity (kJ/kg. K) V Volume (m3) U Internal Energy (kJ) R Gas Constant for air (287 J/kg .K) W Work Done (kJ) Q Heat transferred (kJ) γ Adiabatic Index [γ air = 1.4] LIST OF TABLES Table 1: measured values Table 2: calculated values LIST OF FIGURES Figure 1: apparatus set up. ABSTRACT The aim of the experiment was to determine the specific heat capacity of air at room temperature and pressure, using an isentropic expansion followed by an isochoric process on air contained in a tank. By recording and using relevant pressure measurements with the help of the suitable pressure gauges. Value of the specific heat capacity of air could be calculated. A relatively low percentage error in the readings and average value very close to the expected value (precision value) led to the conclusion that the experiment and the simulated adiabatic expansions were quite precise and reliable. INTRODUCTION The objective for this experiment is to determine the specific heat capacity of air at room temperature by allowing the air in a pressurized tank to expand adiabatically, during a quick opening and closing action of a valve attached to the tank. This ensured that the expansion could be considered as adiabatic. The specific heat (C) is critical because it determines how quickly a substance will heat up or cool down. It is a quantity of heat that is required to change the temperature of 1 kilogram of a substance by 1 degree Celsius. Specific heat capacity is a quantity central to thermodynamic analysis of a substance. In this experiment, pressure was the best parameter to monitor and evaluate, and thus an equation relating the specific heat and the two pressure readings had to be derived. The derivation required the application of the first law of thermodynamics to the adiabatic expansion process and using the ideal gas law, making an assumption that the air behaves like an ideal gas. An average value for the specific heat was determined using the results from several trials and percentage error was analyzed to verify the experiment reliability. THEORITICAL BACKGROUND In thermodynamics, an isentropic process is a process that takes place from initiation to completion without an increment or decrement in the entropy of the system. In other words, the entropy of the system remains constant. Entropy is a form of energy. If a process is both reversible and adiabatic, then it is an isentropic process. An isentropic process is an idealization of an actual process, and serves as a limiting case for an actual process. For an adiabatic process, there is no transfer of heat energy. Specific heat is a quantity of heat that is required to change the temperature of 1 kilogram of a substance by 1 degree Celsius. The specific heat is denoted by Cp. Using Gibb’s equation: (kJ/kg) Gay- Lussac law: P α T The relation for the specific heat can be derived from the above. It must be noted that the expansion is perfectly isentropic and that air behaves like an ideal gas. To get entropy we make use of the first law of thermodynamics and the nature of system work, this can be written as Where S is entropy and other parameters as described earlier. A and B are replacing 1 and 2. APPARATUS AND MATERIALS Fig. 1: An air tank fitted with inlet/outlet valves and pipes. A pressure gauge fitted to outlet pipe to measure pressure at different states The system shown in figure 1 above was used for the experiment. It consists of a vessel and the associated tubing, valves and a pressure gauge for measuring the pressure in the vessel. Other apparatus required were Barometer: for taking pressure measurements Mercury in glass thermometer: for taking temperature measurements EXPERIMENTAL PROCEDURE Before the experiment, the atmospheric pressure, P2 was measured using the barometer so as to determine the absolute pressures of the cylinder. In a similar was, the room temperature T1 was measured using the mercury in glass thermometer. The vessel was gradually pressurized to 5psig and the air supply turned off followed by the closing of the valve in order to isolate the compressed air inside the tank. It was then allowed a few minutes to come down to thermal equilibrium (room temperature). This condition was termed as equilibrium condition 1. The valve was slowly opened allowing the vessel to be exhausted to atmospheric pressure. The valve was closed after the discharge had stopped. A lower temperature T2 than the starting temperature was observed inside the cylinder during the process. This equilibrium state was termed as equilibrium condition 2. Upon closing the valve, the cylinder was left undisturbed and the remaining contents of the vessel warmed up, with a subsequent increase in pressure p3. The temperature T3 was measured and it was equal to initial room temperature, T1. The corresponding final equilibrium state was termed as equilibrium condition 3. The above procedure was repeated four times to get four sets of p1 and p3 readings. RESULTS AND DISCUSSION The following is the data obtained while undertaking the experiment. Different values of pressure and temperature at different experimental stages were measured and recorded. The values of pressure is Atmosphere (atm) while that of temperature was in Kelvin (K). The data is tabulated as shown below Table 1: measured experimental values Pressure (kPa) P1 P2 P3 80.40 87.30 Temperature (K) T1 T2 T3 293 293 Table two: calculated values No P1(kPa) P2(kPa) T1(K) T2 (K) T3 (K) V1(m3) V2 (m3) V3 (m3) CP(kJ/KgK) 1 114.18 80.40 293 269.86 293 0.74 0.96 0.31 1.22 2 114.53 80.40 293 269.86 293 0.73 0.96 0.31 1.23 3 114.12 80.40 293 269.86 293 0.75 0.96 0.31 1.16 The experimental data collected from each trial has been laid out in the observation table above. The relevant absolute pressure P1 and P3 have been tabulated. They were used in the calculation of the specific heat, Cp. the formula used for this calculation has been derived in the appendix. Cp was calculated from the average of all the trials. The mean value of the specific heat has been given below as Cp (mean) = 1.20 kJ/kg K Cp (known at 293 K) = 1.005 kJ/kg. K CV (mean) = 0.92 kJ/kg. K CV (known at 293 K) = 0.718 kJ/kg. K Work Done (W): W1-2 = [P1V1 – P2V2] / [γ - 1] [Process 1-2 is isentropic expansion] W1-2= [(114.18×0.74) – (80.40×0.96)]/[1.4 - 1] W1-2 = 18.27 kJ Also, W2-3 = 0 [Isochoric process] GRAPHS PV diagram T-S diagram For a process to be adiabatic, heat transfer should not take place and the experiment uses the fact that heat transfer is very slow to make the expansion precisely adiabatic. This is explained through the reasons below. A large bore valve was opened for a very short lapse of time (quick snap action). This meant that a relatively large amount of air was allowed out in a very short time and therefore, the rate of expansion was quite high. Hence, heat transfer would not have had the time to occur during such an expansion. The reading of the pressure p1 was taken immediately after the valve was closed and the expansion had stopped by considering the minimum value to which the pressure had dropped. This ensured that the value of pressure obtained was not affected by any heat transfer and meant that the expansion could therefore be regarded as adiabatic. Conclusions The expected specific heat of air at room temperature is around 1.005 and considering the theoretical value of 1.20 obtained from the experimental analysis, it can be said that the results are accurate and reliable. The values obtained for each trial consistent. There were error sources in the experiment that led to the variations in the experimental values. Firstly, the response time of pressure sensors or limitations in human sight could have been an issue when taking pressure readings. In fact, this measurement required accuracy of the observer to be able to see the minimum value of pressure while the pressure sensors and the console might not have had the time to detect and show the effective minimum value reached. Secondly, the time taken to open and close the valve for the expansion process varied, since it was done manually. If it was opened for a little longer in one particular trial, this might have allowed some heat transfer to occur, hence making the expansion non-adiabatic and affecting the values of pressure. Considering the average values for the specific heat obtained from the experiment is appropriate accuracy to the expected value and that the percentage error in the results is about 19.4% and the readings for specific heat obtained in each trials were precise. Thus it can be considered that the experiment was reliable and the objective was achieved. The specific heat value Cp was found, and the general basics of isentropic expansion of an ideal gas understood. This beyond doubt, is a proof that the experiment was successful, and its object achieved. References Cengel, Y.A. and Boles, M.A., Thermodynamics, an Engineering Approach, 6th Ed., McGraw-Hill, Inc., 2008, Sections 7-3, 7-5, 7-7, 7-9  R. Turns, Stephen, Thermodynamics: Concepts and Applications, Cambridge University Press, 2006 A. Rock Peter, Chemical Thermodynamics, McGraw-Hill, 1983 APPENDIX Appendix I: Derivation of Formula to Calculate the Specific Heat (Cp): The differential form of the conservation of energy equation for a closed stationary system (a fixed mass) containing a simple compressible substance can be expressed for an internally reversible process as: But, And, Thus, (kJ) Or, (kJ/kg) [1] Also, Differentiating both sides, we get: [2] From 1 and 2, we get: Now, Integrating both sides from 1 to 2, we get: Since, the process is adiabatic in nature, entropy change will be zero. Appendix II: Derivation of Formula to Calculate the Temperature (T2) During the process 2-3, the gas warms up at constant volume. , We can thus calculate the temperature T2, with the equation below, since we know P2, P3 and T3. Where it may be assumed that: T3 = T1. (From the results) Appendix III: Derivation of Formulae to Calculate Entropy (S) From the first law of thermodynamics and the nature of system work, this can be written The parameters are as explained in the theory section. Read More