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# 2 - Prob 2.1 PROBLEM 2.1 PROBLEM STATEMENT Liquid water...

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Prob 2.1 3/3/2005 PROBLEM 2.1 PROBLEM STATEMENT: Liquid water flows at a velocity of 2 m/s from the pipe inlet (point A) through a horizontal pipe of 1 cm inside diameter (ID). The horizontal part of the pipe is 2 m above the floor. The pipe then turns rises through a vertical distance of 20 m, and again turns to the horizontal at the new height, where the pipe ID then changes to 3 cm (point B). If the velocity of water is inversely proportional to the cross-sectional area of the pipe, (a) What is the change in specific potential energy (kJ/kg) of the water between points A and B? (b) What is the change in the specific kinetic energy (kJ/kg) between points A and B? (c) If the water temperature entering the pipe is 27 ° C, where its specific internal energy is 113.12 kJ/kg, what is the specific energy e in (kJ/kg) of the water at point A, and what percentage of the specific energy is composed of potential, kinetic and internal energy? DIAGRAM DEFINING SYSTEM AND PROCESS: D B =3 cm v A =2m/s D A =1 cm 20m 2m GIVEN: Point A: v A =2 m/s, D A =1 cm, z A =2m, u A =113.12 kJ/kg, T A =27 ° C Point B: D B =3 cm, z B = z A +20m v =Const/A pipe =Const/D 2 FIND: (a) pe 2 -pe 1 (b) ke 2 -ke 1 (c) e A and % pe and ke ASSUMPTIONS: Constant gravitational acceleration, g = 9.81 m/s 2 GOVERNING RELATION: 2 c c gz = and ke = g 2 v pe g QUANTITATIVE SOLUTION: Using the governing equation, the change in potential energy is: ( ) ( )( ) 2 2 2 1 2 1 c g(z - z ) pe -pe = = 9.81 m/s 22 2 m 1(kg m s N) g =196(N m/kg) = 0.196kJ/kg ×

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Prob 2.1 3/3/2005 The velocity at B can be calculated from the given velocity-area relationship = = = = = = A A B B 2 2 A B A A B A 2 B K / A and K / A D 1cm (A / A ) 2m/s 0.22m/s 3cm D v v v v v And the change in kinetic energy is 2 2 2 2 2 2 2 1 c -3 1 m ke -ke = = (0.22 2 ) 1(kg m s N) 2 s 2g = -1.976 J/kg = -1.976 10 kJ /kg × 2 1 v v Specific energy e A =u A +ke A +pe A 2 A A A A A A A c c 2 2 2 2 2 v gz e u ke pe u 2g g 4m /s 9.81m/s 2m 113.12kJ/kg 2 1kg m/s N 1kg m/s N (113.12 0.002 0.0196)kJ/kg 113.14kJ/kg = + + = + + = + + = + + = u represents 113.12/113.14 = 99.9% of the total specific energy in this case, and pe and ke together account for less than 0.1% DISCUSSION OF RESULTS: For many of the problems we deal with in thermodynamics, changes in potential and kinetic energy will be very small relative to changes in internal energy, and hence these terms will often be assumed negligible. In this problem, the velocity and the height would have to be much larger for ke and pe to be of significance.
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