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RUDIKE [14]
2 years ago
9

Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to

the bend is 1 ½ in. inside diameter, and the exit is 1 in. inside diameter. Ignoring losses, calculate the force (magnitude and direction) exerted by the fluid on the bend relative to the direction of the entering stream. Express your answer in units of Newtons. Hint. Use Bernoulli’s equation between the inlet and outlet to determine the upstream pressure P1.
Physics
1 answer:
Nataliya [291]2 years ago
3 0

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1}  =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}

As we know, P1 must be equal to \frac{F_{1} }{A_{1}}, so, replacing P1 in the equation, we have:

P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})

And

F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as Q = V A, where Q is flow, V is velocity and A is area. As the same, we can write this: Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}. In the last two equations, let's clear Velocities.

V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

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You would need to freeze it in a freezer. Hope this helps if it does could I have brainlist thanks
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Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene
sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0
3 years ago
How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?
saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

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2 years ago
Wood is an example of a translucent material.<br> True<br> False
Svetradugi [14.3K]
False, wood is a solid structure that is not see through
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2 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
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