MMAN2700 Lab 1

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Oct 30, 2023

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MMAN2700 Thermodynamics - Lab 1 asdasde Waasd, z5192832 Mode of Learning: Online Allocated lab Session: 2-4pm Tuesday

4.2. Analysis (a) Describe the temperature change in each container from your observations. Idealize this process in simple terms (e.g., constant pressure, isothermal, adiabatic) for purposes of calculations. From the analysis of the results given, the temperature of Container A dropped from 24.09 (ºC) to 23.89 (ºC) showing a decrease in temperature of 0.2 (ºC). Likewise, in Container B the temperature increases from 23.8 (ºC) to 24.14 (ºC) with a similar change of 0.34 (ºC). Thus, we can idealize this process as isothermal. It could be isothermal since the change in temperature is so miniscule thus being able to be idealized. (b) Determine the mass of air initially held in each vessel. Using the ideal gas law, pV = mRT m = pV RT R = 0.287 kJ kg − 1 K − 1 Initial Mass of Container A: m = 412 × 10 − 6 ∗ 311.325 0.287 ∗ 297.24 m = 1.5 × 10 − 3 kg Initial Mass of Container B: ¿ 412 × 10 − 6 ∗ 101.325 0.287 ∗ 296.95 m = 4.9 × 10 − 4 kg (c) By considering the initial and final states of both Tank A and B together as the system boundary, determine: i. the change of the total internal energy of the complete system; ∆U = c v m∆T

Since isothermal process we can assume temperature remains constant. Total mass of system, M = 1.5 × 10 − 3 + 4.9 × 10 − 4 ¿ 1.99 × 10 − 3 kg Find final mass of Container A, m 2 a = p 2 a V RT 2 a ¿ 211.325 ∗ 412 × 10 − 6 0.287 ∗ 297.04 ¿ 1.02 × 10 − 3 kg Final mass of Container B, m 2 b = 1.99 × 10 − 3 − 1.02 × 10 − 3 ¿ 0.97 × 10 − 3 kg The change in internal energy for Container A and Container B, ∆U a = c v m 2 a T 2 a − c v m 1 a T 1 a ∆U a =( 0.718 ∗ 1.02 × 10 − 3 ∗ 297.04 )−( 0.718 ∗ 1.5 × 10 − 3 ∗ 297.24 ) ¿ − 102.6 × 10 − 3 J ( 4 sf ) ∆U b = c v m 2 b T 2 b − c v m 1 b T 1 b ∆U b =( 0.718 ∗ 0.97 × 10 − 3 ∗ 297.29 )−( 0.718 ∗ 4.9 × 10 − 4 ∗ 296.95 ) ¿ 102.6 × 10 − 3 J ( 4 sf ) Thus, the change in internal energy, ∆U system =− 102.6 × 10 − 3 + 102.6 × 10 − 3 = 0 J ii. the net work transfer; Using the isothermal work equation,

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W = P 1 V 1 ln V 2 V 1 Where V 1 = V 2 thus W = 0 . iii. the net heat transfer. Since we idealize this process as isothermal, we can assume the net heat transfer to be a value close to zero and by taking account the rounding errors in our calculation we result with ∆U = Q − W , where W = 0 thus ∆U = Q and Q = 0. 5.2. Analysis For compression process A: (a) Sketch a p-V diagram for the process. Made by using MatLab with the code. y = [101.325 121.325 141.325 161.325 181.325 201.325 221.325 241.325 261.325 281.325]

x = [0.753982237 0.698690206 0.628318531 0.552920307 0.502654825 0.462442439 0.427256601 0.397097311 0.376991118 0.356884925] plot(x,y) xlabel('Volume(m^3 x E(-3))'),ylabel('Pressure(kPa)') title('P-v diagram for compression process A ') (b) Determine the work done on the air in the glass cylinder by using the p-V diagram. By finding the area beneath the p-V diagram that determines the work done on the air in the glass cylinder. This is done with MatLab using the function work = trapz(x,y) which results with work = -68.0327 J, -0.068kJ. This is negative since work is done on the system. (c) Calculate a polytropic index n for the process from both using the initial and final state points. Why is there a difference between these two values? Rearranged from the prelab, n = ln ( p 1 p 2 ) ln ( V 2 V 1 ) n = ln ( 101.325 281.325 ) ln ( 0.35688 0.75398 ) ¿ 1.365 From prelab the equation is rearranged to get, n − 1 n = ln ( T 2 T 1 − p 2 p 1 ) n − 1 n = ln ( 296.65 296.65 ) ln ( 281.325 101.325 ) n = 1

In the calculation used to solve for the final state points where n = 1 it assumes the gas is ideal and not experimental conditions. Thus, if we use different equations for calculating polytropic index n we will get differing values. (d) Sketch the quasistatic isothermal process on the same p-V diagram, starting at the same conditions and finishing at the same volume. Since pV = constant for a quasistatic isothermal process, where constant can be calculated by using the initial pressure and volume which are 101.325 ∗ 0.753982237 = 76.39725015 . With the constant we can calculate the pressures by dividing the constant by the volumes. For compression process B: (e) Calculate a polytropic index n for this process from using the initial and final state points. n − 1 n = ln ( T 2 T 1 − p 2 p 1 )

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n = ln ( p 2 p 1 ) ln ( p 2 p 1 ) − ln ( T 2 T 1 ) n = ln ( 281.325 101.325 ) ln ( 281.325 101.325 ) − ln ( 303.05 296.65 ) n = 1.021 For both compression processes A and B: (f) Compare the experimental polytropic index for both processes with the values for the two theoretical processes (i.e., isothermal, n = 1, and frictionless adiabatic, n = γ). For the experiment, both processes A and B were not carried under the ideal conditions thus experimental errors should be present. For the polytropic index of A we calculated it to be 1.365 instead of the ideal value of 1 if it were an ideal isothermal process. For process B, it was found that n = 1.021 as opposed to n = γ where γ = c p c v = 1.4 for a frictionless adiabatic process. Process A is not completely quasistatic and for an isothermal process it normally takes time for the process to reach equilibrium where values obtained are most ideal. Process B have external factors such as the piston’s mass and friction which would affect the experimental results. (g) Comment on the direction of heat transfer, if any, during the experiment. Constant temperature is maintained in Process A, which is achieved by the continuous loss of energy in the system. On the contrary, Process B does not have any sort of heat loss due to its rapid process. Despite this, heat can enter this system as a consequence of work, where the internal energy increases thus increasing the temperature of the gas.