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Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per

maksim [4K]

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

$V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}$

V = 7.07 m/s

from bernou;ii's theorem we have

$\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l$

as we know pipe is horizontal and with constant velocity so we have

$\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}$

$P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma$

$135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81$

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have $\epsilon = 0.15 mm$

$\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025$

reynold number is

$Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}$

Re = 378750

from moody chart

$For Re = 378750 and \frac{\epsilon}{d} = 0.0025$

$f_{new} = 0.025$

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have $\frac{\epsilon}{d} =0.006$

$\epsilon = 0.006 \times 60$

$\epsilon = 0.36 mm$

thus pipe is old one with increased roughness

5 0

3 years ago

What are the complex structures and the advantages and disadvantages

Svetllana [295]

Answer:

A complex system is a system composed of many components which may interact with each other.

ADVANTAGES

Structs are marginally faster at runtime than classes, due to optimisations done by the compiler. You can enforce full immutability. If you declare a struct instance as let, you will not be able to change its properties.

DISADVANTAGES

A complex corporate structure makes communication more difficult. For instance, when workers must interact with several supervisors, the various directives might work at cross purposes. Also, messages might get lost in the shuffle if there is no simple way to communicate within the organization.

4 0

3 years ago

Five Kilograms of continuous boron fibers are introduced in a unidirectional orientation into of an 8kg aluminum matrix. Calcula

Lunna [17]

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

7 0

3 years ago

A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the

Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

$I_{load}$ = 0.75 < 25.84°

attached below is the remaining part of the solution

B) Find the input current on the primary side in real units

load current in primary = 31.38 < 25.84 A

C) find the input power factor

power factor = 0.9323 leading

attached below is the detailed solution

8 0

3 years ago

A beam spans 40 feet and carries a uniformly distributed dead load equal to 2.2 klf (not including beam self-weight) and a live

salantis [7]

Answer:

A beam is a structural member that is subjected primarily to transverse loads ... stress equal to σy, then the section Moment - Curvature (M-φ) response for .... Example 2.2 Design a simply supported beam subjected to uniformly distributed dead ... (live load). • Step I. Calculate the factored design loads (without self-weight).

Explanation:

6 0

3 years ago