Lab 2 Thermal Brand New

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<ul><li><p>8/2/2019 Lab 2 Thermal Brand New</p><p> 1/4</p><p>Lab 2</p><p>Expansion process of an ideal gas</p><p>Aim</p><p>Objectives</p><p>1) To apply the principle of mass balance and the ideal gas equation to determine the ratio ofvolumes of the two vessels and prove that the equations hold for real gases at</p><p>approximately atmospheric pressure</p><p>2) To investigate the properties of an ideal gas, the principle of mass balance by determiningthe heat capacity ratio k = Cp/Cv for air. Also to examine other thermodynamic properties</p><p>by using P-V-T data</p><p>Objective (2)</p><p>Results</p><p>Group 1 2 3 4 5</p><p>Patm (kPa) 101.63 101.62 101.67 101.67 101.62</p><p>Tamb (C) 31.5 31.8 31.8 31.8 31.7</p><p>P1 (gauge)</p><p>(kPa) 31.37 28.33 28.55 28.63 32.8</p><p>P1 (abs) (kPa) 133 129.95 130.22 130.3 134.42</p><p>T1 (C) 31.3 31.6 31.6 31.5 31.5</p><p>P2 (gauge)</p><p>(kPa) 10.12 9.04 10 6.15 19.9</p><p>P2 (abs)(kPa) 91.51 92.58 91.67 95.52 81.72</p><p>T2 (C) 37.8 37 36.7 37.4 35.43</p><p>P3 (gauge)</p><p>(kPa) 5.45 4.92 4.92 3.96 19.29</p><p>P3 (abs) (kPa) 96.18 96.7 96.75 97.71 82.33</p><p>T3 (C) 31.6 31.7 31.8 31.7 31.6</p><p>experimental k 1.126833243 1.12391 1.151778 1.067199 1.011711</p><p>real k 1.394532034 1.39431 1.394264 1.394655 1.395212</p><p>T2(ohms) 1147 1050 1030 1203 1291</p><p>T2 (C) from</p><p>ohms 39.28205128 41.76923 42.28205 37.91111 36</p><p>T2 K 312.4320513 314.9192 315.4321 311.0611 309.15</p></li><li><p>8/2/2019 Lab 2 Thermal Brand New</p><p> 2/4</p><p>Discussions</p><p>Why is it necessary to measure the pressure and temperature of state 3?</p><p>Equation 1</p><p>()</p><p>According to the equation above, the k value can be calculated using the state 1 and state 2 values.</p><p>However, the mass of the air is unknown since the mass of the air could not be calculated</p><p>accurately. To minimise the error, it was necessary to measure state 3 values which allowed the use</p><p>of the following equation</p><p>Equation 2</p><p>See the appendix for the derivation from equation 1 to equation 2 .</p><p>The advantages and disadvantages</p><p>One of the advantages of having state 3 values is that , pressure value could be measured directly</p><p>which minimised error whereas if the mass had been calculated from states 1 and 2 using an</p><p>approximate value for the density of air and the ideal gas equation, more sources of error would</p><p>have been introduced to our experimental k value.</p><p>A disadvantage of using the state 3 pressure is that the point where thermal equilibrium is achieved</p><p>is not accurately known. For example the two temperature readings for vessels 1 and 2 were not</p><p>exactly equal. This introduces further errors into the experimental k value.</p><p>Why does the temperature change from state 1 to state 2?</p><p>We are assuming the process is adiabatic (Q=0) because the transition from state 1 to state 2</p><p>happens very rapidly. There fore the increase in temperature is a result of flow work done when</p><p>additional moles of air flow into the vessel.</p><p>Determining experimental k:</p><p>We calculated experimental k values using equation 2 above. The values for pressures were</p><p>calculated by adding (vessel 1) or subtracting (vessel 2) the gauge pressure from the atmospheric</p><p>pressure. The results are shown in the table above.</p><p>The average experimental k value can be calculated to be 1.0963.</p><p>Determining theoretical k value using the following equation;</p></li><li><p>8/2/2019 Lab 2 Thermal Brand New</p><p> 3/4</p><p>A = 3.355, B= 0.575 x 10^-3 and D = -0.016 x 10^5 as given in the lab manual . The temperature</p><p>used in the calculation is Tamb .</p><p>The average theoretical value was calculated to be 1.3946.</p><p>Conclusion</p><p>The theoretical value of k was calculated to be 1.3946 and the experimental value of k was</p><p>calculated to be 1.0963. The percentage difference is (1.3946-1.0963) / 1.3946 is 21.4 %.</p><p>The error could have sourced from</p><p>1) In the transition from state 1 to state 2 , we assumed it was adiabatic process where allthe flow work of the air was transferred to the system and none to the surrounding . (</p><p>Q= 0). In reality, Q is not zero since a small amount of heat was transferred to the</p><p>surrounding. This resulted in lower k experimental value than expected since lower</p><p>energy than expected was retained by the system. Since only partial energy being</p><p>transferred to the vessel, it means that less enthalpy carried to the vessel by air. Since</p><p>enthalpy is directly proportional to specific heat capacity at constant pressure, Cp , the</p><p>lower the enthalpy, the lower the Cp. As a result, the experimental value of k is less</p><p>than the theoretical value of k.</p><p>2) The time intervals were 1 second apart which makes it very hard to record the exactvalue of the temperature 2.</p><p>3) The assumption was made that the gas was in ideal state.Objective 1</p><p>Results</p><p>1 2 3 4 5</p><p>P1,i (abs) (kPa) 130.05 131.14 131.62 130.14 131.54</p><p>T1,i (K) 305.2 305.2 305.33 304.91 305.25</p><p>P2,i (abs) (kPa) 66.63 66.63 66.6 66.65 90.15</p><p>T2,i (K) 304.55 304.45 304.55 304.3 305.25</p><p>P1,f (abs) (kPa) 109.61 109.96 110.17 109.61 119.59</p><p>T1,f (K) 305.06 305.02 305.15 304.69 305.2</p><p>P2,f (abs) (kPa) 109.41 109.96 110.15 109.6 119.47T2,f (K) 305.15 304.95 305.15 304.8 305.25</p><p>V2/V1 0.478006 0.488143 0.492239 0.4772 0.406903</p></li><li><p>8/2/2019 Lab 2 Thermal Brand New</p><p> 4/4</p><p>Discussions</p><p>How to determine the volume of two vessels from the principle of mass balance?</p><p>By assuming the air as an ideal gas and applying it to the mass balance equation of</p><p>M1 initial M1 final = M2 final M2 initial</p><p>We can obtain equation 3 =</p><p>The long equation for objective 1</p><p>Why the density mass relationship equation is not used to determine the volume of the two</p><p>vessels?</p><p>Instead of using the equation 3, the volume of the vessel can be calculated by just simply dividingthe mass of the air in the vessel with the density of the air. However this method is not reliable as</p><p>the density of the air is not known and it is not practical to measure the mass of the air as it might</p><p>cause more error in the result.</p><p>Determining experimental volume ratio of small vessel to large vessel:</p><p>By substituting the data collected during experiment to the</p><p>Why does the temperature in vessel 1 drop while the temperature in vessel 2 increase at the final</p><p>state?</p><p>Conclusion</p><p>The theoretical ratio of small vessel to large vessel is 0.4063 and the average experimental ratio of</p><p>small vessel to large vessel is 0.4685.The percentage difference is (0.4685-0.4063) / 0.4063 is 15.3 %.</p><p>The error could have sourced from</p><p>1)</p></li></ul>