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1-116E A water pipe is connected to a double-U manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Assumptions 1 All the liquids are incompressible. 2 The solubility of the liquids in each other is negligible. Properties The specific gravities of mercury and oil aregiven to be 13.6 and 0.80, respectively. We take the density of water to be w = 62.4 lbm/ft3. Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the gh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives
Pwater pipewater gh water oil ghoil Hg gh Hg oil ghoil

1-49
Oil

Oil

35 in 60 in

40 in

15 in
Water

Patm SG oil hoil )

Mercury

Solving for Pwater pipe, Pwater pipe Substituting,
Pwater pipe

Patm

water g ( h water

SG oil hoil

SG Hg hHg

14.2 psia (62.4 lbm/ft 3 )(32.2 ft/s 2 )[(35/12 ft) 0.8(60/12 ft) 13.6(15/12 ft) 0.8(40/12 ft)] 22.3 psia 1 lbf 32.2 lbm ft/s 2 1 ft 2144 in 2

Therefore, the absolute pressure in the water pipe is 22.3 psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.
 


 

Analysis Atmospheric pressures at the location of the plane and the ground level are Pplane ( g h) plane 1N 1 kPa (13,600 kg/m 3 )(9.81 m/s2 )(0.690 m) 2 1 kg m/s 1000 N/m 2 92.06 kPa
Pground ( g h) ground (13,600 kg/m )(9.81 m/s )(0.753 m) 100.46 kPa
3 2

1N 1 kg m/s
2


  1 kPa
  1000 N/m 2

h


  Taking an air column between the airplane and the ground and writing a force balance per unit base area, we obtain Wair / A Pground
  Pplane

0 Sea level

( g h ) air (1.20 kg/m 3 )(9.81 m/s 2 )(h) 1N 1 kgm/s
2

Pground
  Pplane (100.46
  92.06) kPa

 
 
 

1 kPa 1000 N/m 2

It yields h = 714 m which is also the altitude of the airplane.


 

1-106 A 10-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Properties The density of water is given to be =1000 kg/m3. The specific gravity of oil is given to be 0.85. Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water,
SG
H 2O

Oil SG = 0.85 h = 10 m Water

(0.85)(1000 kg/m ) 850 kg/m

3

3

The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, Ptotal PoilPwater ( gh) oil ( gh) water
(850 kg/m 3 )(9.81 m/s 2 )(5 m) (1000 kg/m 3 )(9.81 m/s 2 )(5 m) 90.7 kPa

 
 
 
 
 
 

1 kPa 1000 N/m 2

 


 
 
 
 
 
 
4-92

ke pe negligible.

4-142 A piston-cylinder device contains an ideal gas. An external shaft connected to the piston exerts a force. For an isothermal process of the ideal gas, theamount of heat transfer, the final pressure, and the distance that the piston is displaced are to be determined. W Assumptions 1 The kinetic and potential energy changes are negligible,
0 . 2 The friction between the piston and the cylinder is

Analysis (a) We take the ideal gas in the cylinder to be the system. This is a closed system since no mass crosses the system boundary. The energy balancefor this stationary closed system can be expressed as
Ein Eout Qout Wb,in Qout Wb,in Esystem
Change in internal, kinetic, potential, etc. energies Net energy transfer by heat, work, and mass

GAS 1 bar 24 C

Q

Wb,in

U ideal gas Qout

mcv (T2 T1 )ideal gas )

0 (since T2

T1 and KE

PE

0)

Thus, the amount of heat transfer is equal to the boundary work input
0.1 kJ

(b)...
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