Explanation Write the expression for the energy balance equation. E in − E out = Δ E system (I) Here, the total energy entering the system is E in , the total energy leaving the system is E out , and the change in the total energy of the system is Δ E system . Substitute Q in for E in , W b , o u t for E out , and Δ U for Δ E system in Equation (I) Q i n − W b , o u t = Δ U Q i n = m ( u 3 − u 1 ) + W b , o u t (II) Here, the mass of the piston cylinder device is m , the final specific internal energy is u 3 , the initial specific internal energy is u 1 , and the work done during the process is W b , o u t . Determine the initial volume of the air. ν 1 = m R T 1 P 1 (III) Here, the mass of a piston-cylinder device is m , the initial pressure of a piston-cylinder device is P 1 , the initial temperature is T 1 , and the universal gas constant for air is R . Calculate the final volume which is doubles of initial volume. ν 3 = 2 × ν 1 (IV) Determine the final temperature of an ideal gas for polytrophic process. P 1 ν 1 T 1 = P 3 ν 3 T 3 T 3 = P 3 ν 3 P 1 ν 1 T 1 (V) Here, the initial pressure of air is P 1 , the final pressure of air is P 3 , the initial volume of air is ν 1 , and the final volume of air is ν 3 . Determine the work done during the constant pressure process. W b , o u t = ∫ 1 2 P d ν = P ( ν 3 − ν 1 ) (VI) Conclusion: From the Table A-1, “Molar mass, gas constant, and critical-point properties” to obtain the value of gas constant for air as 0.2870 kJ / kg ⋅ K . Substitute 3 kg for m , 0.2870 kJ / kg ⋅ K for R, 27 ° C for T 1 , and 200 kPa for P 1 in Equation (I). ν 1 = ( 3 kg ) ( 0.2870 kJ / kg ⋅ K ) ( 27 ° C ) 200 kPa = ( 0.861 kJ / kg ⋅ K × ( 1 kPa ⋅ m 3 1 kJ ) ) ( 27 ° C+273 ) 200 kPa = 258

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

9th Edition

ISBN: 9781260048636

Author: CENGEL

Publisher: MCG

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It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. T…

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Chapter 4.5, Problem 79P

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The work done by the air.

The heat transfer to the air during this process.

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Chapter 4.5, Problem 78P

Chapter 4.5, Problem 80P

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A spring package with two springs and an external force, 200N. The short spring has a loin of 35 mm. Constantly looking for spring for short spring so that total compression is 35 mm (d). Known values: Long spring: Short spring:C=3.98 N/mm Lo=65mmLo=87.4mmF=c·fTotal compression is same for both spring. 200 = (3.98(c1) × 35) + (c₂ × 35) 200 = 139.3 + 35c₂ 200 - 139.3 = 35c₂ 60.7 = 35c₂ c₂ = 60.7/35 Short spring (c₂) = 1.73 N/mm According to my study book, the correct answer is 4.82N/mm What is wrong with the calculating?

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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The work done by the air. The heat transfer to the air during this process.