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zavuch27 [327]
3 years ago
7

A 2.5 m tall steel cylinder has a cross sectional area of 1.5 mA 2 EA. At the bottom with a height of 0.5 m is liquid water and

on top of which is a 1 m high layer of gasoline (rhogasoline= 925kg/m3 ). The gasoline surface is exposed to atmospheric air at 101 kPa. Sketch the arrangement. Where is the pressure highest in the water (indicate in the sketch), and what is the value of the absolute pressure?

Physics
1 answer:
KatRina [158]3 years ago
3 0

Answer:

124 kPa

Explanation:

The highest pressure in the water is in the bottom of the cylinder. The reason is because all the weight of the water and the gasoline puts some pressure plus the atmospheric air. So, the absolute pressure would be the one exerted by the atmosphere, the gasoline and the water and is calculated as follows:

P=P_{atm}+P_{gas}+P{water}\\P=110 000Pa +1.0m*9.8m/s^2*925kg/m^3+0.5m*1000kg/m^3*9.8m/s^2\\P=123974Pa≈124 kPa

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What is a push or pull that one object excerpts on another object?
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1.) The object's Velocity

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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

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Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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