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iVinArrow [24]
3 years ago
13

A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t

op, the pipe has a diameter of 30 mm and a pressure gauge indicates a pressure of 860 kPa. At the bottom the diameter is 85 mm and a pressure gauge reading is 1 MPa. Assume the losses are negligible and determine the flov rate. Does the flow direction matter?
Engineering
1 answer:
Minchanka [31]3 years ago
4 0

Answer:

Q=7.3\times 10^{-3} m^3/s

Explanation:

Given that

At topd_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm

\rho =900\dfrac{Kg}{m^3}

We know that

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2

A_1V_1=A_2V_2

\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2

\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2

V_2=8.02V_1

Z_2=12 sin60^{\circ}

\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}

So V_1=1.30m/s

We know that flow rate Q=AV

Q=A_1V_1

By putting the values

A_1=\dfrac{\pi}{4}d^2

Q=7.3\times 10^{-3} m^3/s

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

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What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?
Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

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6 0
1 year ago
You are hitting a nail with a hammer (mass of hammer =1.8lb) the initial velocity of the hammer is 50 mph (73.33 ft/sec). The ti
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Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

Mathematically \overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}

The momentum of any body is defines as \overrightarrow{p}=mass\times \overrightarrow{v}

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}

\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds

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A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 l
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Answer:

modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048

Explanation:

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