A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to the outside air. At steady‐state, the AC cycle requires 0.434kW and has a coefficient of performance (COP) of 6.22. Determine the rate at which the energy is removed from the room air, in kW. If electricity is valued at $0.10/kw-hr, determine the cost of operating the unit for 24hrs.

Answers

Answer 1

Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by  = 6.22

Therefore he heat removed is given by , [tex]$Q_H = 6.22 \times 0.434$[/tex]

                                                                     [tex]$Q_H = 2.7 \ kW $[/tex]

Now if the electricity is valued at  0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

                                                                                            = 6.48

Therefore the operating cost = $ 6.48 for 24 hours.


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Earth completes one full ____ on its axis every 24 hours

Answers

Answer:

rotation

Explanation:

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rotation. i hope this helped:)

If Nick's average stride length is 2.7 feet, how many strides will it take him to walk to school?

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To understand the concept of moment of a force and how to calculate it using a scalar formulation.
The magnitude of the moment of a force with a magnitude F around a point O is defined as follows:
MO=Fd
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A stool at a restaurant is anchored to the floor. When a customer is in the process of sitting down, a horizontal force with magnitude F1 is exerted at the top of the stool support. When the customer is seated, a vertical force with magnitude F2 is exerted on the stool support. If the maximum moment magnitude that the stool support can sustain about point A is MA = 160 Nm , what is the maximum height d1 that the stool can have if the magnitudes of the two forces are F1 = 300 N and F2 = 720 N . Assume that moments acting counterclockwise about point A are positive whereas moments acting clockwise about A are negative.

Answers

Answer:

When analyzing forces in a structure or machine, it is conventional to classify forces as external forces;

constraint forces or internal forces.

External forces arise from interaction between the system of interest and its surroundings.

Examples of external forces include gravitational forces; lift or drag forces arising from wind loading;

electrostatic and electromagnetic forces; and buoyancy forces; among others. Force laws governing these

effects are listed later in this section.

Constraint forces are exerted by one part of a structure on another, through joints, connections or contacts

between components. Constraint forces are very complex, and will be discussed in detail in Section 8.

Internal forces are forces that act inside a solid part of a structure or component. For example, a stretched

rope has a tension force acting inside it, holding the rope together. Most solid objects contain very

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Explanation:

What is the difference between digital instruments and decimal scaled instruments to measure

Answers

Answer Digital measuring instruments are self-contained devices that automatically present the value of the measured quantity on a digital display. And Decimal Scaled Instruments: Record all digits that you can certainly determine from the scale markings and estimate one more digit. I hope this Helped I´m new to this.

Explanation:

A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.

Answers

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

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e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

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