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Ksivusya [100]
2 years ago
10

A heat pump designer claims to have an air-source heat pump whose coefficient of performance is 1.8 when heating a building whos

e interior temperature is 300 K and when the atmospheric air surrounding the building is at 260 K. Is this claim valid
Engineering
1 answer:
Anit [1.1K]2 years ago
3 0

Answer:

The claim is valid.

Explanation:

Let assume that heat pump is reversible. The coefficient of performance for the heat pump is:

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

COP_{HP} = \frac{300\,K}{300\,K-260\,K}

COP_{HP} = 7.5

The claim is valid as real heat pumps have lower coefficients of performance.

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Army people are good people right
Phantasy [73]

Answer: Yes army people are good people but it also depends on how you fraze that some have been in trouble before but it doesnt mean there bad people we all make mistakes

5 0
3 years ago
Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give
Lelechka [254]

Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path

Explanation:

The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction

5 0
3 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
What are the four categories of engineering materials used in manufacturing?
alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They  have low weight and are poor conductors of electricity and heat

8 0
2 years ago
Can you answer what is attached.
kirill [66]
29.4 bro I hope that helps
3 0
3 years ago
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