Solution to Assignment 2

Solution to Assignment 2 - SOLUTION TO HOMEWORK #2 3.2...

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SOLUTION TO HOMEWORK #2 3.2 Starting with the energy balance, we get: from problem statement, we can eliminate the heat, shaft work, and expansion work. Also, since we are at steady state, all the d/dt terms cancel out. In addition, because the temperature of the fluid does not change, the internal energy of the fluid stays the same (it is a liquid), so we can say 0 kk k MU . Finally, the fluid is in contact with the atmosphere so the pressures at the beginning (top of the fall) and the end (bottom) are the same. Density is also the same (it is liquid), so we can say: ˆ 0 kkk k M PV Therefore, we are left with 2 2 v gh At the bottom of the fall, the water is going to have two components (the x and y components), so we need to take them into consideration. Now, v is given by: 2 2 2 i ix iy v v v  Therefore:
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    2 2 2 2 2 2 2 1 1 x y x y v v v v v However 21 xx vv Then , we get: gh v v y y 2 ) ( 2 1 2 2 but 1 0 y v (the fluid is traveling horizontally first). Then s m m s m v v v y y y 8 . 32 ) 55 . 0 )( 81 . 9 ( 2 2 2 2 2 2 2 2 Now, in order to find the final velocity, we do: hr km hr s m s m hr v v v v y x / 53 . 118 ) 1 3600 )( 1000 1 )( 8 . 32 ( ) 8 ( 2 2 2 2 2 2 2 2
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3.3 a) Assume it is a closed system with constant pressure and no shaft work, so using energy balance we get: so we are left with (after multiplying by dt the left and right side of the energy balance equation, and looking at the system as a before and after type system): 2 1 2 2 1 1 2 1 2 1 ˆˆ () ˆ ˆ ˆ ˆ ( ) ( ) ˆ ˆ ˆ ˆ ( ) ( ) V V d MU Qdt MPdV Integrating MU M U M U Q MP dV or M U U Q MP V V  By rearranging the equation to get enthalpy, we get:
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Q H H M Q V P U M V P U M V V MP Q U U M ) ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ ( 1 2 1 1 2 2 1 2 1 2 We knew this from theory, right? Therefore, we can calculate the enthalpy before and after by using the steam tables. So the work done is:
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    2 1 21 33 2 () 1 (1 )(3.6322 1.8161 ) 100000 1 )( )(3.6322 1.8161 ) 181.6 V V W P dV P V V mm W Kg bar kg kg N m Kg bar bar kg kg W kJ          So in the end, our answers are W=-181.6kJ, Q=793.2kJ, H = 793.2kJ, U = 609.2kJ, T = 514.5C b) Constant volume, we get that the energy balance written above: ˆˆ d MU Q MPdV reduces to: Q U U M ) ˆ ˆ ( 1 2 Since we double the pressure, we get that for the final state, P=2x1.013bar = 2.026bar and since the volume is constant, kg m V V 3 1 2 8161 . 1 . So, by linear interpolation again, we get:
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so, the amount of heat added is: kJ Q kg Q 6 . 611 ) 8 . 2544 4 . 3156 ( 1 and the amount of work is zero (constant volume). In addition, H = U + (PV) In the end, W= 0, Q=611.6kJ , ∆H = 795.57kJ , ∆U = 611.6kJ, T = 515C. c) doing the same as part a) but assuming ideal gas behavior, we can utilize the ideal gas law to come up with the value of the temperature:
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Now, utilizing the results from part a, we get that the heat is: 21 () N H H Q  so, the change in enthalpy can be calculated by using the heat capacity at constant pressure
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Solution to Assignment 2 - SOLUTION TO HOMEWORK #2 3.2...

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