3 years ago

The major resisting force in gravity dam?

Engineering

2 answers:

Sloan [31]3 years ago

8 0

Explanation:

Self weight of dam.............

inna [77]3 years ago

6 0

The self weight of dam.

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A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

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Answer:

$P_2_{abs}=160\ psia$ (absolute).

Explanation:

Given that

Pressure ratio r

r=8

$r=\dfrac{P_2_{abs}}{P_1_{abs}}$

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P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure + Gauge pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5 psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

$8=\dfrac{P_2_{abs}}{20}$

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If you measured a pressure difference of 50 mm of mercury across a pitot tube placed in a wind tunnel with 200 mm diameter, what

mart [117]

Answer:

V=33.66 m/s

$Re=448.8\times 10^6$

Re>4000, The flow is turbulent flow.

Explanation:

Given that

Pressure difference = 50 mm of Hg

We know that density of Hg=136000 $Kg/m^3$

ΔP= 13.6 x 1000 x 0.05 Pa

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Property of air at 25°C

ρ=1.2 $Kg/m^3$

Dynamic viscosity

$\mu =1.8\times 10^{-8}\ Pa.s$

Velocity of fluid given as

$V=\sqrt{\dfrac{2\Delta P}{\rho_{air}}}$

$V=\sqrt{\dfrac{2\times 680}{1.2}}$

V=33.66 m/s

Reynolds number

$Re=\dfrac{\rho _{air}Vd}{\mu }$

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$Re=448.8\times 10^6$

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3 years ago

The major resisting force in gravity dam? ​

The major resisting force in gravity dam?