MIE 210 Thermodynamics Lab 5

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MIE210: Thermodynamics Lab 5 Report Group PRA0104-B1 Table of Contents 1. Introduction 2 2. Theory and Scientific Background 3 3. Experiments and Methodology 4 4.0 Results and Discussion 6 4.1 Reversible or Irreversible Gas Expansion? 6 4.2 Entropy Generation 16 4.3 Steady State Flow Test 21

5. Conclusion 25 6. References 25 1. Introduction Entropy can be defined as measure of thermal energy of a system per unit temperature that is unavailable for doing useful work.[1] It is often interpreted as the measure of disorder or randomness of a system. The primary objective of this experiment is to understand changes in entropy. Through three procedures, different aspects of the concept of entropy will be investigated. First part of the experiment deals with reversibility, or irreversibility of expansion of a gas. A reversible process in thermodynamics for a system is defined as a process that a system undergoes which can be restored to initial conditions along the same path on a P-V diagram where every point on the diagram is in equilibrium.[2] Otherwise, the process is deemed irreversible. In this part, irreversibility of the expansion of air is investigated. The second part of the lab examines the entropy generation of a system due to heat transfer. Heat is added to a pressurized tank using heaters whose energy consumption is carefully monitored. Using the second law of thermodynamics, the rate of entropy generation of the system and its surroundings is calculated. The third part of the lab considers the rate of heat loss and generation of entropy for an open system at steady-state conditions. In this experiment, no mass is accumulated by creating a control volume system. The temperature is kept steady using the heater, while pressure and mass are kept constant by allowing inflow rate to equal the outflow rate. 1

2. Theory and Scientific Background In the first part of the experiment, rapid expansion of air from a system to its surroundings is investigated. When a gas undergoes expansion, it follows the relation PV n = constant . This process is called a polytropic process. If n is constant, then the equation for two different states becomes V 1 V 2 ¿ n P 2 P 1 = ¿ . When this relationship is combined with the ideal gas law, this equation becomes P 2 P 1 ¿ n − 1 n T 2 T 1 = ¿ , where n is determined by the specific heat transfer process that occurs during the expansion. For the purposes of this experiment, n for isothermal and isentropic processes is given. For isothermal processes, n = 1 is used. For isentropic processes, n = k = C p C v , which for air is 1.4 . In the second part of the experiment, the change in entropy is investigated. The change in entropy can be mathematically written as Δ ´ S = ´ Q T 0 , where Δ ´ S is the rate of change in entropy, ´ Q is the rate of heat transfer and T 0 . The rate of entropy generated can be written as ´ Δ S gen = ´ Δ S sink + ´ Δ S source . 2

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For the third part of the experiment, a steady-state control volume system needs to be analyzed for entropy generation. For this system, the equation ´ Δ S gen = ´ Δ S sink + ´ Δ S source − ´ Q T provides the entropy generated during the experiment. 3. Experiments and Methodology This lab consisted of three parts. Each was done separately and centered around a different aspect of the concept of entropy. The first section of this lab was focused on reversible and irreversible gas expansion. It was started by filling the left tank up to 5 psig. This was done by opening the A2 valve and the left solenoid and setting the flow rate to 50 g/min. Once the pressure reached 5 psig, the valve and the solenoid were closed. Next, the right tank was vacuumed to -2.5 psig by opening valves A1 and A5 as well as the right solenoid and setting the flow rate to 50 g/min. Again, the valves and solenoid were closed once the desired pressure was reached. Afterwards, the gas was expanded rapidly into the atmosphere by opening the A4 valve and both the center and right solenoids one after the other. Once the pressured reached the ambient pressure in approximately 10 seconds, both solenoids and the A4 valve were closed and the pressure was allowed to stabilize. This series of steps was then repeated for 10 psig and -2.5 psig, 25 psig and -5 psig, 80 psig and -10 psig in the left and right tanks respectively. Before this series of steps was taken for each set of pressures, the B1 valve was opened to depressurize the tanks. The second section of this lab was centered around entropy generation. It was started by filling the left 3

tank to 40 psig. This was done by opening valve A2 and the left solenoid and setting the flow rate to 50 g/min. Once this pressure was reached, the valve and solenoid were closed. Next, the air was heated to 40 by turning the heaters on and allowing the temperature to stabilize. The tanks were kept in this ℃ condition for 2-3 minutes and data was collected. This process was then repeated with the temperature being held at 60 instead. Finally, the left tank was cooled down by letting air run through it by ℃ opening the left solenoid and C1 and C2 valves until the temperature returned to the ambient temperature. The B1 valve was then opened to depressurize the left tank in preparation for the next section of the lab. The third and final section of this lab was focused on analyzing steady state flow. It was started by opening valves A2, A5, and B2 as well as the left and right solenoids. Next, the needle valve was opened for five turns and the flow rate was set to 50 g/min. Once the tank pressure reached 40 psig, the flow rate was readjusted to 3 g/min and the air temperature was set to 30 and allowed to stabilize. At this ℃ point, the needle valve was used to keep the pressure stable at 40 psig and the number and direction of turns needed to do this was carefully noted. Afterwards the temperature on both sides of the needle valve was monitored until it stabilized and became constant and data was then collected for 2-3 minutes. The heaters were then turned off. It should be noted that this section of the lab could only be completed if the temperature of the air in the left tank was 30 or lower. ℃ For each of these sections, data recording took place through the LabVIEW software and was started at the beginning of each experiment or trial and concluded once enough data had been gathered. 4.0 Results and Discussion The ambient temperature recorded was 21.7 °C = 294.7 K, and the ambient pressure was 29.59 inHg = 14.53 pisa = 100.2 kPa. 4

4.1 Reversible or Irreversible Gas Expansion? Graph 4.1.1. Temperature and Pressure of Two Tanks for Part A. Graph 4.1.2. Needle Valve Inlet and Outlet Temperature for Part A. 5

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Graph 4.1.3. Mass Flow Rate and Heater Energy for Part A. Graph 4.1.4. Temperature and Pressure of Two Tanks for Part B. 6

Graph 4.1.5. Needle Valve Inlet and Outlet Temperature for Part B. Graph 4.1.6. Mass Flow Rate and Heater Energy for Part B. 7

Graph 4.1.7. Temperature and Pressure of Two Tanks for Part C. Graph 4.1.8. Needle Valve Inlet and Outlet Temperature for Part C. 8

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Graph 4.1.9. Mass Flow Rate and Heater Energy for Part C. Graph 4.1.10. Temperature and Pressure of Two Tanks for Part D. 9

Graph 4.1.11. Needle Valve Inlet and Outlet Temperature for Part D. Graph 4.1.12. Mass Flow Rate and Heater Energy for Part D. 1. In a polytropic gas expansion process, the relationship of the properties of an ideal gas is 10

PV = constant. By rearranging and through substitution, we get that P 2 P 1 ¿ n − 1 n T 2 T 1 = ¿ . The values of temperature and pressure will be read from the graphs, and are listed below in tables. After isolating and solving for n, the above expression becomes n = 1 1 − ln T 2 T ! ln P 2 P 1 . The results for all four sections are shown in the table below, using only data from the left tank. Table 4.1.1. Values of n Based on Pressure and Temperature Readings Part T 2 [K] T 1 [K] P 2 [psia] P 1 [psia] n A 22 + 273 25.5 + 273 2.5 + 14.53 6 + 14.53 1.067 B 20 + 273 25 + 273 5 + 14.53 10 + 14.53 1.080 C 17 + 273 25.5 + 273 14 + 14.53 25 + 14.53 1.097 D 10 + 273 25.5 + 273 47 + 14.53 79 + 14.53 1.146 The calculated values of n are all significantly lower than the k value for air, which is 1.4. Thus, it can be said that these expansion processes cannot be considered isentropic, and thus irreversible. The trial which was the closest to the reversibility conditions would be the one which had the highest n value. Thus, trial D was the closest to being a reversible expansion process, though the other three trials were not much worse in that regard. 2. The change in entropy in an ideal gas, through a relationship of temperature and pressure, can be 11

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calculated as Δ s = ∫ 1 2 c p ( T ) dT T − Rln P 2 P 1 . If one were to assume an ideal gas with constant specific heats, then Δ s = c p ln T 2 T 1 − Rln P 2 P 1 . Considering that the experiment is run at about 25 °C, we take a constant c p = 1.005 kJ/kg K, and R = 0.287 kJ/kg K. As a function of temperature, the specific heat at ⋅ ⋅ constant pressure for air is c p = 0.969 +( 6.783 × 10 − 5 ) T +( 1.656 × 10 − 7 ) T 2 +(− 6.779 × 10 − 11 ) T 3 . Using this method, it would be necessary to integrate as a function of T, namely Δ s = 0.969ln T 2 T 1 +( 6.783 × 10 − 5 )( T 2 − T 1 )+ 0.5 ( 1.656 × 10 − 7 )( T 2 2 − T 1 2 )+ 1 3 (− 6.779 × 10 − 11 )( T 2 3 − T 1 3 ) −( 0.287 kJ / kg ⋅ K ) ln P 2 P 1 ❑ Table 4.1.2. Calculation of Entropy based on Constant and Varying Specific Heats Part T 2 [K] T 1 [K] P 2 [psia] P 1 [psia] Δs constant c p Δs varying c p A 22 + 273 25.5 + 273 2.5 + 14.53 6 + 14.53 0.04179 0.04182 B 20 + 273 25 + 273 5 + 14.53 10 + 14.53 0.04841 0.04847 C 17 + 273 25.5 + 273 14 + 14.53 25 + 14.53 0.06456 0.06466 D 10 + 273 25.5 + 273 47 + 14.53 79 + 14.53 0.06659 0.06680 Comparatively, the entropy calculated with a constant c p was very similar to the one where it was assumed to be a function of temperature. This is due to the fact that the temperature change on an absolute scale was not too large, so an assumption of a constant value was fairly accurate. However, it can be seen that as the temperature difference in the trial increases (ie. from A to D) the difference in the calculated values for Δs is steadily increasing. 12

3. In order to plot the graph, it is necessary to calculate the specific volume of air at each of the states. This can be easily calculated using the ideal gas law, taking R = 0.287 kJ/kg K. Rearranging ideal gas law, ⋅ the specific volume is calculated as v = RT P . Table. 4.1.3. Calculations of Specific Volume. Trial State Pressure [kPa] Temperature [K] Specific volume [m 3 /kg] A 1 141.5 298.5 0.6054 2 117.4 295 0.7212 3 117.4 297.5 0.7273 B 1 169.1 298 0.5058 2 134.7 293 0.6243 3 134.7 297.5 0.6339 C 1 272.5 298.5 0.3144 2 196.7 290 0.4231 3 196.7 297 0.4333 D 1 644.9 298.5 0.1328 2 424.2 283 0.1915 3 424.2 296 0.2003 13

Graph 4.1.13. P-v diagram for the Four Trials If the process were instead a slow one, then it would happen isothermally since there is enough time for heat to transfer to the environment. Thus pressure as a function of specific volume would be P = C/v, where C = RT = constant, and so these would look like rational functions. The think black lines have been drawn as accurately as possible to represent what isothermal processes would look like on a P-v diagram. 4. In a perfect vacuum, no mass will be inside the tank. Thus, heat transfer by conduction and convection will be impossible. The wall is in contact with the surroundings, so the temperature there will be detected as the same as the ambient temperature. However, inside the tank, there is no way for heat to be transferred inside the tank, and so the sensor will read a temperature of 0 Kelvin. 4.2 Entropy Generation 14

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Graph 4.2.1. Temperature and Pressure of Two Tanks for Part A. Graph 4.2.2. Needle Inlet and Outlet Valve Temperature for Part A. 15

Graph 4.2.3. Mass Flow Rate and Heater Energy for Part A. Graph 4.2.4. Temperature and Pressure of Two Tanks for Part A. 16

Graph 4.2.5. Needle Inlet and Outlet Valve Temperature for Part A. Graph 4.2.6. Mass Flow Rate and Heater Energy for Part A. 17

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1. In order to calculate the rate of heat transfer, the the total heat transfer will be divided by the total amount of time the heat transfer took place in order to calculate the average rate of heat transfer. Table 4.2.1. Rate of Heat Transfer. Part Heater Energy [kJ] Time Start [s] Time Stop [s] Rate of Heat Transfer [kW] A 60 57 276.4 0.2735 B 106 42 276.3 0.4524 The rate of entropy generation of the tank is simply the change in entropy from the initial state to the final state. It is know that for an ideal gas, the change in entropy is Δ s = c p ln T 2 T 1 − Rln P 2 P 1 . Let us consider a constant rate of entropy generation, taken from the beginning of the heating period. For the rate of entropy change in the surroundings, since there is no mass transfer, this is equal to Δ S = Q T o , where T o is the temperature of the heat source (40 °C). The total rate of entropy generation is the summation of the two rates of entropy generation. Table 4.2.2. Rate of Total Entropy Generation. Part Mass [kg] Time Elapsed [s] Rate of Entropy Generation in the Tank [W/K] Rate of Entropy Generation in the Surroundings [W/K] Total Rate of Entropy Generation [W/K] A 0.02417 219.5 0.74 0.8738 1.6138 B 0.0225 234.3 1.0158 1.4454 2.4612 Positive entropy changes are in agreement with the second law of thermodynamics. Additionally, it can be seen that the rate of entropy change in Part B is greater than that of Part A, which is correct conceptually since Part B is performed at a higher temperature, which would result in a higher rate of entropy generation. 18

4.3 Steady State Flow Test Graph 4.3.1. Temperature and Pressure of the Two Tanks. 20

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Graph 4.3.2. Needle Inlet and Outlet Temperature. Graph 4.3.3. Mass Flow Rate and Heater Energy. 1. Clearly, the mass flow rate at the inlet of the left tank is equal to the mass flow rate which can be read 21

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from the graph, namely 3 g/min or ṁ = 0.05 g/s. But considering the fact that the pressure, temperature and volume stay constant, we can deduce that the type of flow is a steady-state flow. This means that at both inlets and outlets of both tanks, the mass flow rate must be 0.05 g/s. Taking the ideal gas law, PV = mRT and dividing through by time, we get P ´ V = ´ m RT , so ´ V = ´ m RT P . It is known that ´ V = Av , where A is the cross-sectional area and v is the velocity. Therefore, velocity at each valve can be calculated. Table 4.3.1. Calculations of Velocity at Inlets and Outlets. Valve Volumetric Flow Rate [m 3 /kg] Cross Sectional Area [m 2 ] Velocity [m/s] Left, Inlet 1.1564 × 10 -5 7.854 × 10 -5 0.1472 Left, Outlet 1.1411 × 10 -5 1.590 × 10 -5 0.7177 Right, Inlet 3.7260 × 10 -5 1.590 × 10 -5 2.343 Right, Outlet 3.7386 × 10 -5 7.854 × 10 -5 0.476 2. Reading from the chart, c v is taken as 0.002 since the number of turns needed for the needle valve to maintain steady state was 6.25 turns. Knowing also, that T 1 = 26 °C = 299 K, and that P 1 = 40 psig = 54.53 psia = 376 kPa, we calculate Q = 0.471 ( 6816 )( 0.002 )( 376 kPa ) √ ❑ . It is known that ρ = m V = ´ m ´ V = 0.05 × 10 − 3 kg / s 0.002327 m 3 / s = 0.0215 kg / m 3 . 3. In the left tank, the rate of heat loss can be calculated by comparing the rate of heat transfer to the 22

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change in internal energy. By the first law of thermodynamics, E ¿ − E out = Δ E system ⟶ Q ¿ − Q out = U 2 − U 1 . But knowing that for an ideal gas, the internal energy is directly proportional to the temperature, and knowing that temperature is constant, we can conclude that Q ¿ − Q out = 0 , or ´ Q ¿ = ´ Q out . The total heater energy is 9.8 kJ, over a time 377 seconds, so the average rate of heater energy is 25.99 W. This means the rate of heat loss in the left tank is 25.99 W. Now considering the rate of heat loss in the needle valve, the first law of thermodynamics must be used again. After equating and substituting for the incoming energies, exiting energies, and the change in energy of the system, ´ m ( h ¿ ❑ + v ¿ 2 2 )= ´ m ( h out + v out 2 2 )+ Q out . Enthalpies are directly related to temperature, through the relation h = c p T . Taking values for velocities from Table 4.3.1., where v ¿ = 0.7177 m / s , and v out = 2.343 m / s , and taking T ¿ = 26 + 273 = 299 K and T out = 23 + 273 = 296 K , we get that Q out = 0.0264 W . 4. As this is a steady state problem, Δ ´ S ¿ − Δ ´ S out + ´ S gen = 0 since there is no change in state internally. Both the entropy generated in the tank and the needle are positive values. The entropy generated in the tank is Q out T o = 25.99 W 303 K = 0.08578 W / K . For the entropy generated in the needle valve, this is equal to Q out T o = 0.0264 W 296 K = 8.9189 × 10 − 5 W / K . Thus ´ S gen = Δ ´ S out − Δ ´ S ¿ = 25.99 W 303 K + 0.0264 W 296 K = 0.08586 W / K . 23

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5. Conclusion In conclusion, this lab provided a practical insight into the concept of entropy through consideration of the concepts of reversible and irreversible gas expansion, entropy generation, and steady state flow. The first part of this lab pointed to the idea that the rapid expansion of air in this experiment is not an isentropic process. It also showed that the process of gas expansion seemed to become more reversible as the pressure difference increases. The second part of this lab showed that higher entropy generation takes place at higher temperatures, validating theoretical ideas on the concept of entropy generation and reaffirming the second law of thermodynamics. The third and final part of this lab helped connect the ideas of heat loss and entropy generation and showed the relationship between the two concepts. The conclusions obtained in this lab connect the theoretical concepts and relationships in thermodynamics with practical results that can be recorded through simple experiments. These conclusions show that, while generally correct, the laws of thermodynamics are very idealistic with the simplifications used and cannot be made use of in the real world without taking into account numerous systematic and random errors that take place in the real world, as well as unwanted heat, work and mass transfers. Results like these show the importance of combining experimentation with theoretical frameworks to obtain a better understanding of phenomena in the real world. 6. References [1] "entropy | Definition and Equation", Encyclopedia Britannica , 2018. [Online]. Available: https://www.britannica.com/science/entropy-physics. [Accessed: 03- Apr- 2018]. 24

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