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sleet_krkn [62]
3 years ago
10

The two basic categories of electrical switches are manual and ??

Engineering
2 answers:
LUCKY_DIMON [66]3 years ago
8 0

Explanation:

Single Pole Single Throw Switch

Single Pole Double Throw Switch

Double Pole Single Throw Switch

Double Pole Double Throw Switch

Push Button Switch.

Toggle Switch.

Limit Switch.

Float Switches.

In the simplest case, a switch has two conductive pieces, often metal, called contacts, connected to an external circuit, that touch to complete (make) the circuit, and separate to open (break) the circuit

Ganezh [65]3 years ago
4 0

Answer:

are Manual and Automatic

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Who works alongside and assists the engineers?
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The _____ is installed in the liquid ine and is responsible for removing moisture and foreign matter.
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filter drier

Explanation:

Any refrigeration or air conditioning system needs filter driers as a crucial part. In a refrigeration or air conditioning system, a filter drier serves two crucial purposes: first, it adsorbs system impurities like water, which can produce acids, and second, it provides physical filtering. To guarantee optimal and cost-effective dryer design, each component must be evaluated. The most crucial role of a dryer is to be able to drain water from a refrigeration system. Water can originate from a variety of places, including motor windings, equipment leaks, retained air from poor drainage, and others.

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Select three possible deliverables that might be specified in a project charter.
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4 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of
Eduardwww [97]

Answer:

P_m_i_n= 356.9 KPa

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

COP=\frac{1}{\frac{T_H}{T_L}-1 }

We are given:

T_H=1.2T_L

COP=\frac{1}{1.2-1 }

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle  as:

COP_R=\frac{Q_L}{W_i_n}

Amount of heat absorbed by low temperature reservoir can be found as:

Q_L=COP_R * W_i_n

Q_L=5 * 22 KJ

Q_L=110 KJ

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

Q_H=Q_L + W_i_n

Q_H=110 KJ + 22 KJ

Q_H=132 KJ

We are given the mass of 0.96 kg. So,

q_H=\frac{Q_H}{m}

q_H=\frac{132 KJ}{0.96Kg}

q_H=137.5 KJ/Kg

Since it is a saturated liquid-vapour mixture q_H=h_f_g.

q_H=h_f_g=137.5 KJ/Kg

From Refrigerant 134-a tables T_H at h_f_g=137.5 KJ/Kg is 61.3 C. (We calculated this by interpolation)

Converting T_H from Celsius to Kelvin:

61.3^{o} C+273 = 334.3^{o} K

T_H= 334.3^{o} K

We are given:

T_H=1.2T_L

T_L=\frac{T_H}{1.2}

T_L=\frac{334.3}{1.2}

T_L=278.58^{o} K

Converting T_L from Kelvin to Celsius:

278.58^{o} K-273 = 5.58^{o} C

T_L= 5.58^{o} C

From Refrigerant 134-a tables P_m_i_n at T_L=5.58^{o} C is 356.9 KPa. (We calculated this by interpolation).

P_m_i_n= 356.9 KPa

8 0
4 years ago
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