Please help me with my question

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

Nataliya [291]

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

3 0

3 years ago

Explain how information is transferred from one telephone to another one when both the telephones are mobile phones

Setler79 [48]

Satellites transfer all our information from ken mobile device to another

5 0

3 years ago

Read 2 more answers

UN COLET POSTAL ESTE IMPINS DE O MASA ORIZONTALA CU O FORTA F=40N.COLETU SE DEPLASEAZA PE O DISTANTA DE 1,5M CE LUCRU MECANIC EF

Wittaler [7]

Answer:

A POSTAL PACKAGE IS PUSHED BY A HORIZONTAL TABLE WITH A FORCE F = 40N. PACKAGE MOVES ON A DISTANCE OF 1.5M WHICH MECHANICAL WORK PERFORMS THIS FORCE

PT TEST

5 0

3 years ago

An object with a resistance of 28 Ω has 76 V applied to it. How much electric current is going through this object? Answer in un

12345 [234]

Answer : The electric current of a circuit is, 2.8 A.

Explanation :

Using Ohm's law :

$V=I\times R$

Or,

$I=\frac{V}{R}$

where,

R = resistance of a circuit = 28 Ω

V = voltage of circuit = 76 volts = 76 V

I = current flowing in a circuit = ?

Now put all the given values in the above formula, we get :

$I=\frac{V}{R}$

$I=\frac{76V}{28\Omega }$

$I=2.8A$

Therefore, the electric current of a circuit is, 2.8 A

5 0

3 years ago

small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becom

Ilia_Sergeevich [38]

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

4 0

3 years ago