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slavikrds [6]
3 years ago
9

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

water speed doubles and the cross-sectional area of the pipe triples, what happens to the volume flow rate of the water passing through it?
Engineering
1 answer:
Arlecino [84]3 years ago
8 0

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

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A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
2 years ago
A differential amplifier is very useful for removing common mode noise voltage that might be fed or induced in the signal cables
Reptile [31]
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I think will help
7 0
3 years ago
What are some constraints related to size, time, or materials? Remember, constraints are limitations, issues, or obstacles.
Sergeeva-Olga [200]

Answer:

too big, too small, deadlines, not enough materials

Explanation:

4 0
3 years ago
What is the real name of the town that Bud & Bugs stayed in while they were waiting for the train?
Leona [35]

Answer: Hooverville

Explanation:

Bud, not Buddy is a book about a ten years old boy who ran away from home where he stayed with his foster parents. He was treated unfairly and beaten.

He left home to seek a better living and hope that he will find his father as well. He met another orphan named Bugs and they decided to go to Hooverville so that they can get a train going to California.

6 0
3 years ago
It is desired to produce a continuous and oriented carbon fiber-reinforced epoxy having a modulus of elasticity of at least 83 G
lukranit [14]

Answer:

0.3129

Explanation:

Calculate The minimum fiber volume fraction using the relation below

E = x * E1 + ( 1 - x ) E2  ------ ( 1 )

given that :

E = 83 GPa , x = ?

E1 = 260 GPa ,  E2 = 2.4 GPA

Insert values back into equation 1

Then ; 80.6 = 257.6 x

∴ x ( volume fraction ) = 80.6 / 257.6  = 0.3129

3 0
3 years ago
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