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3 years ago

How many gallons of water is needed to fill a pool.

Engineering

2 answers:

zalisa [80]3 years ago

6 0

Answer:

depends on the size

Explanation:

Length x width x depth x 7.5 = volume (in gallons)

Length times width gives the surface area of the pool. Multiplying that by the depth gives the volume in cubic feet. Since there are 7.5 gallons in each cubic foot, multiply the cubic feet of the pool by 7.5 to arrive at the volume of the pool, expressed in gallons.

kotykmax [81]3 years ago

6 0

18,000-20,000 gallons.
The average swimming pool takes 18,000-20,000 gallons of water to fill.
I WANT POINTS PLEASE

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A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

2 years ago

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$ = 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

4 0

3 years ago

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