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3 years ago

List 3 types of heat exchangers, and classify them?

Engineering

1 answer:

frez [133]3 years ago

3 0

Answer:

Concentric tubes or double tube.

Casing and tubes.

Plates.

Explanation:

There are two classify of heat exchange:

1) Direct contact heat exchanger: the exchange heat is produced by the mixture betweenn two fluids, for example refrigeration towers.

2) Indirect contact heat exchange: the fluids never mix, and exchange the heat across a surface.

For example:

Concentric tubes or double tube: the fluid travel trough the central tube, and in the extern tube (bigger diameter) have other fluir with a different tempreature which can give o became the heat of the central fluid.

Casing and tubes are used in Industries, which have a lot of tube inside tha casing.

Plates: the fluid travel throug the metal plates.

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g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner

galben [10]

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is $88 Mpam^\frac{1}{2}$ , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = $88 Mpam^\frac{1}{2}$

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute

a = 1/π [( 88π / 1.1(300)(2/π)]²

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

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3 years ago

The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and

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Answer:

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3 years ago

A coal-burning power plant generates electrical power at a rate of 650 megawatts (MW), or 6.50 × 108 J/s. The plant has an overa

Vinvika [58]

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce 6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x 6.50 × 10⁸ J/year

Energy produce in one year = 204984 x 10¹² J/year

Energy produce in one year =20.49 x 10¹⁶ J/year

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4 years ago

A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro

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Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

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We have

= $\frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt$

$= \int\limits^a_b {(5-3t)} \, dt$

$5t - \frac{3t^2}{2} +c$

v2 = 5x2 - 3x2 + c

= 10-6+c

= 4+c

$s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c$

S2 - S1

$=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)$

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

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