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I am Lyosha [343]
3 years ago
13

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.
Engineering
1 answer:
Arada [10]3 years ago
3 0

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 , $Q_H = 6.22 \times 0.434$

                                                                     $Q_H = 2.7 \ kW $

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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Answer:

Hello there, the question is not complete, but not to worry you can check the explanation section to check how you can  solve a similar question or to be be able to solve the exact question directly.

Explanation:

The flow of electrons is what is known or refer to as Current. When energy is used on a nuclei, the electrons are forced to move from one position to the other. The direction of flow of electron is from the negative terminal which then moves to the positive terminal.

Therefore, it can be said that the positive charge determines the direction of electron flow. The starting point is the negative terminal, in which it will now move in the direction in which the positive terminal is.                    

5 0
4 years ago
Need help I’m giving out brainlest whoever get it me correct
ludmilkaskok [199]

Answer:

(C) Prototype Model

Explanation:

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7 0
4 years ago
The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 40 continuous hours. During that ti
Jet001 [13]

Answer:

a) 144.000 s

b) and c)Battery voltage and power plots in attached image.

   V=-\frac{0.5}{144000} t + 1.5 V[tex]    [tex]P(t)=-(31.25X10^{-9}) t+0.0135  where D:{0<t<40} h

d) 1620 J

Explanation:  

a) The first answer is a rule of three

s=\frac{3600s * 40h}{1h} = 144000s

b) Using the line equation with initial point (0 seconds, 1.5 V)

m=\frac{1-1.5}{144000-0} = \frac{-0.5}{144000}

where m is the slope.

V-V_{1}=m(x-x_{1})

where V is voltage in V, and t is time in seconds

V=m(t-t_{1}) + V_{1} and using P and m.

V=-\frac{0.5}{144000} t + 1.5 V[tex] c) Using the equation VPOWER IS DEFINED AS:[tex] P(t) = v(t) * i(t) [tex]so.[tex] P(t) = 9mA * (-\frac{0.5}{144000} t + 1.5) [tex][tex]P(t) = - (31.25X10^{-9}) t + 0.0135

d) Having a count that.

E = \int\limits^{144000}_{0} {P(t)} \, dt  = \int\limits^{144000}_{0} {v(t)*i(t)} \, dt

E = \int\limits^{144000}_{0} {-\frac{0.5}{144000} t + 1.5*0.009} \, dt = 1620 J

4 0
3 years ago
Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time
rodikova [14]

Answer:

First we determine the tensile strength using the equation;

Tₓ (MPa) = 3.45 × HB

{ Tₓ is tensile strength, HB is Brinell hardness = 225 }

therefore

Tₓ = 3.45 × 225

Tₓ = 775 Mpa

From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10

When the percentage of cold work for steel is up to 10,the ductility is 16% EL.

And 16% EL is greater than 12% EL

Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL

7 0
3 years ago
2.24 Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same s
Gekata [30.6K]

Answer:

Explanation:

Given that:

From process 1 → 2

P_1 = 10 bar   \\  \\ V_1 = 1 m^3  \\ \\  V_2 = 4 m^3

PV^{1.5} = \ constant

\gamma = 1.5

Process 2 → 3

The volume is constant i.e V_2 =V_3 = 4m^3

P_3 = 10 \ bar

Process 3 → 1

P = constant  i.e the compression from state 1

Now, to start with 1 → 2

P_1V_1^{1.5} = P_2V_2^{1.5}

P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}

P_2 = 10 \times  (\dfrac{1}{4})^{1.5}

P_2 =1.25

The work-done for the process  1 → 2 through adiabatic expansion is:

W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1]

We know that 1 bar = 10^5 \ N/m^2

∴

W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1]

W =1000000 \ J

W_{1 \to 2} = 1000 kJ

For process 2 → 3

Since V is constant

Thus:

W = PΔV = 0

W_{2 \to 3} = 0

For process 3 → 1

W = PΔV

W _{3 \to 1} = P_3(V_1-V_3)

W _{3 \to 1} = 10 \times 10^5 (1-4)

W _{3 \to 1} = 10 \times 10^5 (-3)

W _{3 \to 1} = -3 \times 10^6 \ J

W _{3 \to 1} = -3000  \ kJ

The net work-done now  for the entire system is :

W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }

W_{net} = (1000 + 0 + (-3000)) \ kJ

W_{net} =-2000 \ kJ

The sketch of the processes on p -V coordinates can be found in the image attached below.

4 0
3 years ago
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