HW1_sol - ENGRD 221 Prof N Zabaras SOLUTIONS TO HOMEWORK 1 Problem 1 Given Oxygen in a piston-cylinder process undergoes a process where pV1.4 =

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ENGRD 221 – Prof. N. Zabaras 9/3/2007 SOLUTIONS TO HOMEWORK 1 Problem 1: Given : Oxygen in a piston-cylinder process undergoes a process where . The total work done is known. 1.4 pV = constant Find : The final volume and pressure Assumptions : (1) O 2 is a closed system (2) process is polytropic. Analysis : We start with , where W is the work done in a piston-cylinder assembly. 2 1 V V W pdV = 28 ft 3 35 lbf/in 2 -50 Btu p 1 = 35 lbf/in 2 V 1 = 28 ft 3 pV 1.4 = const. To determine V 2 , substitute the p-V relation into this equation and integrate 2 1 V -.4 -.4 21 1.4 V vv constant W dV constant v- == . 4 . But we know, constant = pV 1.4 thus the 1.4 11 constant=P V Using this expression and solving for V 2 , we get 2 -.4 -0.4 3. 4 1.4 2 3 1.4 2 0.4 W (-.4)(-50 Btu) 778 ft.lbf 1 ft V + V = (28 ft ) = 0.2928 P V (35 lbf/in )(28 ft ) 1 Btu 144 in =+ V 2 = 21.56 ft 3 Now, we use the p-V relation to get P 2 1.4 1.4 2 1 2 2 V lbf 28 P = P 35 = 50.47 lbf/in V in 21.56 ⎛⎞ = ⎜⎟ ⎝⎠ The shaded portion in the p-V diagram shows the work done. Page 1 of 6
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ENGRD 221 – Prof. N. Zabaras 9/3/2007 Problem 2 : Given : Dimensions of the swimming pool = 100 m x 50 m and depth varying from 1 m to 4m, density of water = 998.2 kg/m 3 , atmospheric pressure = 0.98 bar. Find : total force at the bottom of the pool and pressure on the floor at the center of the pool Analysis : Total pressure at the bottom of the pool = weight of water inside the pool + downward force exerted by atmosphere on the surface of the water. F total = F atm + F grav To find the weight of water, find the total mass of water inside the pool as follows: 3 3 7 kg 1 998.2 (1)(100)(50) (3)(100)(50) m m2 (998.2)(12500) 1.25x10 kg mV ρ ⎡⎤ == + Gravitational force exerted by water is given by, F grav 75 2- 2 3 m1 N1 k N mg 1.25x10 kg x 9.81 1.226x10 kN s1 k g m 0 N = F atm = 52 25 atm surface 3 10 N/m 1 kN P A (0.98 bar) (100x50)m 4.9x10 kN 1 bar 10 N Therefore, the total force acting on the bottom of the pool surface is F total = 1.226 x 10 5 kN + 4.9 x 10 5 kN = 6.126 x 10 5 kN
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This note was uploaded on 07/09/2009 for the course M&AE 326 taught by Professor Psiaki during the Spring '08 term at Cornell University (Engineering School).

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HW1_sol - ENGRD 221 Prof N Zabaras SOLUTIONS TO HOMEWORK 1 Problem 1 Given Oxygen in a piston-cylinder process undergoes a process where pV1.4 =

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