# Solucionario cengel ed.6 cap. 1

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Review Problems 1-85 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. Assumptions 1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded. Analysis Noting that pressure is force per unit area, the pressure on the smaller piston isdetermined from F m1g P = 1 = 1 A1 πD12 / 4
=  (25 kg)(9.81 m/s 2 )  1 kN   2  1000 kg ⋅ m/s 2  π (0.10 m) / 4  

F1 25 kg

Weight 2500 kg

F2

10 cm

D2

= 31.23 kN/m 2 = 31.23 kPa From Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from
  → D 2 = 1.0 m   Discussion Note that largeweights can be raised by little effort in hydraulic lift by making use of Pascal’s principle. P1 = P2 = F2 m2 g (2500 kg)(9.81 m/s 2 )  1 kN  =  31.23 kN/m 2 = → 2  1000 kg ⋅ m/s 2 A2 πD 2 2 / 4 πD 2 / 4 

1-86 A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weightsthat will double the pressure of the gas are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis The gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield
m piston g

WEIGTHS

GAS

 (5 kg)(9.81 m/s 2 )  1 kN  = 95.66 kN/m 2 = 95.7 kPa  2 1000 kg ⋅ m/s 2  A π (0.12 m )/4   The force balance when the weights are placed is used to determine the mass of the weights (m piston + m weights ) g P = Patm + A (5 kg + m weights )(9.81 m/s 2 )   1 kN   200 kPa = 95.66 kPa + → m weights = 115.3 kg 2  1000 kg ⋅ m/s 2  π (0.12 m )/4   A large mass is needed to double the pressure. Patm = P − = 100 kPa −

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1-87 Anairplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3. Analysis The local atmospheric pressure is determined from Patm = Pplane + ρgh
 1 kN = 58 kPa + (1.15 kg/m 3 )(9.81 m/s 2 )(3000 m)  1000 kg ⋅m/s 2  The atmospheric pressure may be expressed in mmHg as
hHg =

  = 91.84 kN/m 2 = 91.8 kPa  

Patm 91.8 kPa  1000 Pa  1000 mm  =    = 688 mmHg ρg (13,600 kg/m 3 )(9.81 m/s 2 )  1 kPa  1 m 

1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight ofan 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation

 1N W = mg = (80kg)(9.807 − 3.32 × 10 −6 z m/s 2 )  1kg ⋅ m/s 2  Sea level: Denver: Mt. Ev.:

   

(z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N (z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N (z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778= 782.2 N

1-89 A man is considering buying a 12-oz steak for \$3.15, or a 320-g steak for \$2.80. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost ofeach steak is determined to be
 \$3.15   16 oz   1 lbm   = \$9.26/kg Unit Cost =       12 oz   1 lbm   0.45359 kg 

12 ounce steak: 320 gram steak:

 \$2.80   1000 g  Unit Cost =   320 g   1 kg  = \$8.75/kg      Therefore, the steak at the international market is a better buy.

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1-90 The thrust developed by the jet engine of a Boeing 777 is given to...