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hw2_solutions

# hw2_solutions - ES330 Assignment 2 Solutions Chapter 3 Due...

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ES330 Assignment 2 Solutions Chapter 3 Due Date: Thursday September 9, 2010 1 Similar to Problem 3-43 in Text The 500- kg load on the hydraulic lift shown in the figure is to be raised by pouring mercury ( ρ = 13560 kg/m 3 ) into a thin tube. Determine how high h should be in order to begin to raise the weight. Solution In order to solve the problem, we need to balance the hydrostatic force of the fluid with the weight of the load. It is also important to note that atmospheric pressure acts of both sides of the lift, so we don’t need to include it in our calculations. The force balance becomes: F weight = F fluid (1) m load g = P fluid A = ρ fluid ghA (2) Solving for h and assuming that the lift is a cylinder: h = m load πr 2 ρ fluid = 500 kg (1 / 4) π (1 . 2 m 2 )(13560 kg/m 3 ) = 0 . 0326 m (3) h=3.26 cm 1

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2 Similar to Problem 3-48 in Text Consider a double-fluid manometer attached to an air pipe shown in the figure. If the specific gravity of one fluid is 13.55, determine the specific gravity of the other fluid for the indicated absolute pressure of air. Take the atmospheric pressure to be 100 kPa . Solution To solve the problem, we can start with the pressure of the air. If the change in height of a particular fluid is negative as we move towards the outlet, then we add it to the pressure of the air. If it is positive, then we subtract it. Therefore: P Air + ρ 1 gh 1 - ρ 2 gh 2 = P atm (4) P atm - P Air = SG 1 ρ water gh 1 - SG 2 ρ water gh 2 (5) Solving for SG 2 we get: SG 2 = P atm - P Air water - SG 1 h 1 - h 2 = SG 1 h 1 h 2 + P Air - P atm water h 2 (6) SG 2 = (13 . 55) 0 . 22 m 0 . 8 m + 76 , 000 Pa - 100 , 000 Pa (9 . 81 m/s 2 )(1000 kg/m 3 )(0 . 8 m ) (7) SG 2 = 0 . 668 2
3 Similar to Problem 3-62 in Text Consider an 8 m long, 8 m wide, and 2.0 m

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hw2_solutions - ES330 Assignment 2 Solutions Chapter 3 Due...

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