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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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Component(s) that only allow(s) electrons to flow in one direction. Mark all that apply

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Answer:

A,D, and E

Explanation:

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3 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

8 0

3 years ago

A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s

KengaRu [80]

Answer:

0.00650 Ib s /ft^2

Explanation:

diameter ( D ) = 0.71 inches = 0.0591 ft

velocity = 0.90 ft/s ( V )

fluid specific gravity = 0.96 (62.4 ) ( x )

change in pressure ( P ) = 0 because pressure was constant

viscosity = (change in p - X sin∅ ) $D^{2}$ / 32 V

= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90

= - 59.904 sin (-90) * 0.0035 / 28.8

= 0.1874 / 28.8

viscosity = 0.00650 Ib s /ft^2

8 0

3 years ago

Read 2 more answers

If the load parameters are: Vln=600kV, Il=100A (resistive), calculate the source voltage and current when the line is 50Miles (s

Archy [21]

s 0Miles (short), 150 Miles(medium), and 300 Miles (long).

Explanation:

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3 years ago

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c

Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice $C_p=2.03\ KJ/kg.K$