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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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Answer:

37 psi

Explanation:

For ideal gases this equation applies:

p1*V1/T1 = p2*V2/T2

Since we are assuming volume remains constant:

V2 = V1

p1/T1 = p2/T2

p2 = p1*T2/T1

The temperatures must be in absolute scale.

T1 = 15 + 273  = 288 K

T2 = 60 + 273 = 333 K

Then:

p2 = 32 * 333 / 288 = 37 psi

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Which of the following elements would make a good insulator?
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Answer:

Glass

Explanation:

Please mark me the brilliant

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2 years ago
The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is
nexus9112 [7]

Answer:

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4 0
3 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
3 years ago
A developer is having a single-lane raceway constructed with a 100 mph design speed. A curveon the raceway has a radius of 1000-
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Answer:

(a) 36+45.00

(b) 24+65.00  

(c) 6+517.500

(d) 12+324.800

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3 years ago
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