2 years ago

Which of these are not referenced in an assembly?

Engineering

1 answer:

Otrada [13]2 years ago

7 0

E. Parts they don’t resemble

You might be interested in

You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______

stich3 [128]

I believe it’s D) 20 feet

3 0

2 years ago

Explain by Research how a basic generator works ? using diagram<br>

natulia [17]

Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry

8 0

2 years ago

Outline the process used to test the following hypothesis: Titanium cages are stronger than steel cages for hockey goalie masks.

son4ous [18]

Answer:

true

Explanation:

8 0

2 years ago

What kind of abilities are needed to run a business

nordsb [41]

Answer:

Essential business skills:

1. Financial management. Being able to effectively manage your finances is critical. ...

2. Marketing, sales and customer service. ...

3. Communication and negotiation. ...

4. Leadership. ...

5. Project management and planning. ...

6. Delegation and time management. ...

7. Problem solving. ...

8. Networking.

8 0

2 years ago

Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige

Lera25 [3.4K]

Answer:

A) $COP = \frac{16.97}{9.8} = 1.731$

B) $P_{IN} = 0.4763$

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

$T_1 = 28$ degree celcius

$T_2 = 12$ degree celcius

$\dot m = 1140 kg/h$

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

$A) Q = \dot m Cp \Delta T$

$= 1140 \times 3.35\times (28-12) = 61,104 kJ/h$

$Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec$

$COP = \frac{16.97}{9.8} = 1.731$

$COP ∝ \frac{1}{P_{in}}$

$P_{in}$ wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

$COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625$

we know that

$COP = \frac{heat\ obsorbed}{P_{in}}$

$P_{IN} = \frac{16.97}{35.62} = 0.4763$

c) second law efficiency

$\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%$

exergy destruction os given as $X = W_{IN} - X_{Q2}$

= 9.8 - 0.473 = 9.3237 kW

8 0

3 years ago