6-44Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.

Answers

Answer 1

Answer:

The correct answer will be "8640 kWh".

Explanation:

The given values are:

Rate (Ql)

= 5040 kJ/h

As we know,

⇒  [tex]COP = \frac{Cooling}{heat \ required}[/tex]

On putting the values, we get

⇒           [tex]=\frac{5040 \ kJ/h}{1 \ kW}[/tex]

⇒           [tex]=1.4[/tex]

Now,

⇒  [tex]Q = Ql + W[/tex]

        [tex]= 1 + 1.4[/tex]

        [tex]=2.4 \ kW[/tex]

        [tex]=2.4\times 3600[/tex]

        [tex]=8640 \ kWh[/tex]


Related Questions

At 7:00PM the temperature was 40 degrees F. If the temperature dropped steadily at a rate of 6 degrees per hour, what was the temperature at 11:00 PM? Do you know what it is

Answers

Answer:

Temperature at 11:00 PM = 16 degree

Explanation:

Given:

Temperature at 7:00 PM = 40 degree

Decrease rate = 6 degree per hour

Find:

Temperature at 11:00 PM

Computation:

Total time = 11 - 7 = 4 hour

Total decrease in temperature = 4 x 6 = 24 degree

Temperature at 11:00 PM = 40 - 24

Temperature at 11:00 PM = 16 degree

A heat exchanger is to heat water (cp = 4.18 kJ/kg·°C) from 25°C to 60°C at a rate of 0.5 kg/s. The heating is to be accomplished by geothermal water (cp = 4.31 kJ/kg·°C) available at 140°C at a mass flow rate of 0.3 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit tempera

Answers

Answer:

73.15 kW, 196.6°C

Explanation:

Energy in - Energy out = change in energy

[tex]E_{in}-E_{out}=\Delta E\\\\\Delta E=0\\\\E_{in}-E_{out}=0\\\\E_{in}=E_{out}\\\\Q_{in}+\dot mh_1=\dot mh_2\\\\Q_{in}=\dot mh_2-\dot mh_1\\\\Q_{in}=\dot m c_p(T_2-T_1)[/tex]

The rate of heat transfer to cold water is given as:

[tex]Q_{in}=\dot m c_p(T_2-T_1)\\\\\dot m =0.5\ kg/s,c_p=4.18\ kJ/kg.^oC, T_2=60^oC,T_1=25^oC\\\\Q_{in}=(0.5 *4.18)(60-25)=73.15\ kW[/tex]

For the geothermal water:

[tex]Q_{in}=\dot m c_p(T_2-T_1)\\\\\dot m =0.3\ kg/s,c_p=4.31\ kJ/kg.^oC, ,T_1=140^oC\\\\T_2=\frac{Q}{\dot m c_p} +T_1=\frac{73.15}{0.3*4.31}+140=196.6\\ \\T_2=196.6^oC[/tex]

Service entrance conductors must be kept a minimum of ____ feet from the sides and bottom of windows that can be opened

Answers

Minimum 2-1/2 inches

What type of engineer constructs the infrastructure necessary for roads airfields and water ports

Answers

the answer is civil engineer

Answer:

Civil Engineer

Explanation:

Structural engineers and Civil engineers are the ones who design, build, and maintain the foundation for our modern society.

Like our roads and bridges, drinking water and energy systems, sea ports and airports, and the infrastructure for a cleaner environment and many more.

In a two dimensional flow, the component of the velocity along the X-axis and the Y-axis are u = ax2 + by + cy2 and v = cxy. What should be the condition for the flow field to be incompressible

Answers

Answer:

The condition for the flow field to be incompressible is independent of a and c

Explanation:

We are given the component of the velocity along the X-axis and the Y-axis as;

u = ax² + by + cy²

v = cxy

Now, the condition for the flow to be incompressible is;

du/dx + dv/dy = 0

Now,

du/dx = 2ax

dv/dy = cx

Thus;

2ax + cx = 0

(2a + c)x = 0

Thus,the condition for the flow field to be incompressible is independent of a and c

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