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marta [7]
4 years ago
9

What are the basic types of heat exchangers?

Engineering
1 answer:
liraira [26]4 years ago
5 0

Answer:

The two large divisions of heat exchangers are direct contact between fluids and indirect contact between fluids.

Explanation:

A heat exchanger is one of the most used equipment at the level of thermal installations, both at the building, tertiary and industrial levels. A heat exchanger is a device designed to transfer heat between two fluids. These two fluids (liquids, gases) can be in contact or separated by a solid barrier. Its use is basic in all types of air conditioning or refrigeration, air conditioning, energy transfer or chemical processes. Heat transmission occurs through convection and conduction.

Classifying heat exchange systems can be carried out using many different criteria. When classifying different types of heat exchangers, different criteria can be taken into account. Taking into account the degree of contact between the fluids, they are grouped into two different types:

Direct Contact Heat Exchanger:

In direct contact exchangers, heat transfer occurs through a physical mixture of the fluids involved in the process. An example of this type of exchangers are the cooling towers. In this case, direct contact occurs between a stream of hot water (fluid to be cooled) using dry and colder air.

Indirect Contact Heat Exchanger :

In a direct type exchanger there is no direct contact between the fluids and they never mix. The fluids are separated by a solid barrier and may also not coincide at the same time.

Indirect contact heat exchangers can be of various types, being the most used, according to their constructive typology:

  • Concentric tubes or double tube .
  • Shell and tubes .
  • Of plates .
  • Compact heat exchangers .
  • Regenerators .

The concentric tube equipments are the simplest that exist since they are composed of two concentric tubes of different diameter so that one of the fluids circulates inside the smaller one and the other does it through the annular space between both tubes.

Shell and tube exchangers are widely used at an industrial level and use a housing with a multitude of tubes inside.

The equipment of plates are formed by a succession of sheets of metal, armed in a frame and separated by joints, which are fixed with a steel shell. The fluid circulates between these sheets.

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Question 1
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A. from zero at the center to maximum at the circumference.

Explanation:

The shear stress describes the process where a body deforms as a result of a part sliding over the other.

The value of shear stress which is induced in the shaft due to the applied couple varies from zero at the center to maximum at the circumference.

The shear stress in a shaft is given by this formula;

τ = Tr/I

where;

τ = shear stress (MPa).

T = applied torque (Nmm).

r = distance from center to stressed surface in the given position (mm).

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3 0
3 years ago
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has
trasher [3.6K]

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0
3 years ago
An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat
IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

b)

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature T_L = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature T_H = 99° F = ( 99 + 460 )R = 559 R

Cooling Load Q_L = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

a)

Coefficient of performance of the air conditioner;

COP_{air-condition = Cooling Load Q_L  / power P

we substitute

COP_{air-condition = 30000 Btu/h / 8396.619 Btu/h

COP_{air-condition = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

b)

- Power input required ( in hp )

Q_L / P_{required = T_L / ( T_H - T_L )

we substitute

30000 Btu/h / P_{required = 530 R / ( 559 R - 530 R )

30000 Btu/h / P_{required = 530 R / 29 R

we solve for P_{required

P_{required  = ( 30000 Btu/h × 29 R ) / 530 R

P_{required  = ( 870000 Btu/h / 530 )

P_{required  = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

P_{required  = ( 1641.5094 / 2544.43 ) hp

P_{required  = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

COP_{rev-air-condition = T_L / ( T_H - T_L )

we substitute

COP_{rev-air-condition = 530 R / ( 559 R - 530 R )

COP_{rev-air-condition = 530 R / 29 R

COP_{rev-air-condition = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0
3 years ago
State two main consumers of electricity transmitted at low frequency Calculate the primary voltage of a transformer if the prima
bagirrra123 [75]

Answer:

1) a. Customers requiring AC electric power transmission for powering remote devices which may include a subsea transmission system where power is distributed to subsea devices

b. Customers that utilize commutator-type motors

2) The primary voltage is 150 volts

Explanation:

The parameters given are;

Number of primary winding, N_p = 50 turns

Number of secondary winding, N_s = 150 turns

Voltage in secondary winding, V_s = 450 volts

Voltage in primary winding = V_p

The relation between N_p, N_s, V_s and V_p is as follows;

\dfrac{V_p}{V_s}  = \dfrac{N_p}{N_s}

Which gives;

V_p  = \dfrac{N_p \times V_s}{N_s}

From which we have;

V_p  = \dfrac{50\times 450}{150} = 150 \ volts

The primary voltage = 150 volts.

7 0
4 years ago
Read 2 more answers
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