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Solution to Assignment 3

# Solution to Assignment 3 - SOLUTION TO HOMEWORK#3 4.1 a The...

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SOLUTION TO HOMEWORK #3 4.1 a) The way to solve this problem is to begin with the system as the ball plus the water. Then, by doing an energy balance, we get: M 1 ˆ U 1 f M 2 ˆ U 2 f M 1 ˆ U 1 i M 2 ˆ U 2 i 0 Then, we can make the appropriate substitutions: M 1 C v ,1 T 1 f T 1 i M 2 C v ,2 T 2 f T 2 i 0 T f M 1 C v ,1 T 1 i M 2 C v ,2 T 2 i M 1 C v ,1 M 2 C v ,2 T f 8.31 C b) Now, we proceed to use the equation for the entropy of a solid and liquids (final and initial pressure is the same, remember): S MCp ln T 2 T 1 Therefore, for the ball we get: S 5 x 10 3 g 0.5 J gK ln 8.31 75   531.61 J K and for the water: S 12 x 10 3 g 4.2 J gK ln 8.31 5 596.22 J K So, for the system, we get: S T   S ball   S water   531.61 596.22 S T 64.61 J K Therefore, if we look at the entropy balance, for this system, we get that:

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S S gen S gen 64.61 J K 4.4 We have steam at 700 bar and 600 C being expanded to 10 bar and an unknown temperature. Therefore, by doing a mass and energy balance, we obtain: 1 2 0 dM M M dt 1 1 2 2 ˆ ˆ 0 s dU dV M H M H Q W P dt dt but, since there is no shaft work, expansion work (constant volume), and heat transferred, we get: 1 1 2 2 ˆ ˆ 0 M H M H Therefore, from the energy balance, we conclude that: ˆ H 1 ˆ H 2 (we are ignoring the change in kinetic energy). So, from the steam tables, we get the following properties: ˆ H 1 3063 kJ kg ˆ H 2 3063 kJ kg Now, since we know the final enthalpy, and that the final pressure is 10 bar, we can use the steam tables to find the final temperature, which is 308 C. Now, we find all the properties necessary from the initial and final step. This gives:
3 1 1 1 3 2 2 2 ˆ ˆ ˆ 3063 , 0.003973 , 5.522 ˆ ˆ ˆ 3063 , 0.2618 , 7.145 kJ m kJ H V S kg kg kgK kJ m kJ H V S kg kg kgK From the entropy balance, we get: 1 1 2 2 ˆ ˆ 0 gen dS Q M S M S S dt T but, we know already that there is no heat transfer, so, we get: 1 1 2 2 ˆ ˆ 0 gen M S M S S Therefore, when the above equation is solved for the generation of entropy, we get: 2 1 1 ˆ ˆ gen S S S M =(7.45-5.522) KJ/kg K 4.5 From an energy balance, we get: 2 2 1 1 f i f i s U U U U U Q W PdV but there’s not heat transfer and the system is at constant volume: U U 2 f U 2 i U 1 f U 1 i W s Doing the appropiate substitutions for a solid: W s U 2 f U 2 i U 1 f U 1 i W s MCp T 2 f T 2 i MCp T 1 f T 1 i but, we know that 1 2 f f f T T T , so:

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W s MCp 2 T f T 1 i T 2 i but, we need an expression for Tf because we don’t know it. That’s why we use an entropy balance, which is shown below: 2 2 1 1 f i f i S S S S S gen Q dt S T but, there’s no heat transfer ( 0 Q
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