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igomit [66]
3 years ago
12

A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

n average temperature of 25o C. (a) If the house loses heat to the atmosphere at the rate of 60,000 kJ/h, determine the minimum power supplied to the heat pump (in kW) that can maintain the interior temperature of the house at 25o C. (b) Suppose the heat pump absorbs energy from
Engineering
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

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Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are
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Answer:

Please check explanation for answer

Explanation:

Here, we are concerned with stating the advantages and disadvantages  of using a 6 tube passes instead of a 2 tube passes of the same diameter:

<u>Advantages</u>

* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface

* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too

            Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.

<u>Disadvantages</u>

* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.

* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of  manufacturing costs

5 0
3 years ago
What is hardness and how is it generally tested?
drek231 [11]

Answer:

Hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

Explanation:

Hardness of a material is understood as the resistance that the material opposes to its permanent surface plastic deformation by scratching or penetration. It is always true that the hardness of a material is inversely proportional to the footprint that remains on its surface when a force is applied.

In this sense, the hardness of a material can also be defined as that property of the surface layer of the material to resist any elastic deformation, plastic or destruction due to the action of local contact forces caused by another body (called indenter or penetrator), harder, of certain shape and dimensions, which does not suffer residual deformations during contact.

That is, hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

The following conclusions can be drawn from the previous definition of hardness:  

  1) hardness, by definition, is a property of the surface layer of the material, and is not a property of the material itself;  

  2) the methods of hardness by indentation presuppose the presence of contact efforts, and therefore, the hardness can be quantified within a scale;

  3) In any case, the indenter or penetrator must not undergo residual deformations during the test of hardness measurement of the body being tested.

To determine the hardness of the materials, durometers with different types of tips and ranges of loads are used on the various materials. Below are the most commonly used tests to determine the hardness of the materials.

   Rockwell hardness :

It refers to the Rockwell hardness test, a method with which the hardness or resistance of a material to be penetrated is calculated. It is characterized by being a fast and simple method that can be applied to all types of materials. An optical reader is not required.

    Brinell hardness :

Brinell hardness is a scale that is used to determine the hardness of a material through the indentation method, which consists of penetrating with a hardened steel ball tip into the hard material, a load and for a certain time.  

This test is not very precise but easy to apply. It is one of the oldest and was proposed in 1900 by Johan August Brinell, a Swedish engineer.

    Vickers hardness:

Vickers hardness is a test that is used in all types of solid and thin or soft materials. In this test, a square-shaped pyramid-shaped diamond and a   136° vertex angle are placed on the penetrating equipment.

In this test the hardness measurement is performed by calculating the diagonal penetration lengths.

However, its result is not read directly on the equipment used, therefore, the following formula must be applied to determine the hardness of the material: HV = 1.8544 · F / (dv2).

3 0
2 years ago
For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a domin
Ad libitum [116K]

Answer:The awnser is 5

Explanation:Just divide all of it

4 0
2 years ago
To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr
alukav5142 [94]

<u></u>\ T_{c} has greater effect.

<u>Explanation</u>:

\eta_{\max }=1-\frac{T_{c}}{T_{A}}

T_{c}\\ = Temperature of cold reservoir

T_{H} = Temperature of hot reservoir

when T_{c} is decreased by 't',

$\eta_{\text {incre }}$ = 1-\frac{\left(\tau_{c}-t\right)}{T_{H}}

=n \ + \frac{t}{T_{n}}      -(i)

when {T_{H}} is increased by 'T'

\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)

\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}

7 0
3 years ago
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Answer:

4.17x10^-3 kW/K

Explanation:

Detailed explanation and calculation is shown in the image below

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3 years ago
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