MMAN2700 Lab 1

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University of New South Wales *

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2700

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Mechanical Engineering

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Oct 30, 2023

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docx

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7

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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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