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evablogger [386]
3 years ago
8

When removing a diesel engine from a truck, Technician A says it is OK to disconnect an air con­ditioning hose, but the refriger

ant should be vented to the outside of the shop. Technician B says do not loosen any air conditioning hoses until the refrigerant can be recovered. Who is right? *
Engineering
1 answer:
agasfer [191]3 years ago
5 0

Answer:

Technician B is right.

Explanation:

Air conditioning refrigerant contains Freon R22 and R410a, which have been linked to environmental damages, including ozone depletion, global warming, and energy-inefficiency.  For environmentally-savvy entities and individuals, there is the modern move to a more environment-friendly refrigerant, known as R-32.   Technician A's advice to vent the refrigerant outside the shop is in bad taste.  He does not seem to be aware of the environmental footprint of such an action.  Venting gas outside, in addition to the environmental damages, is also a waste of resources, and therefore, costly.  This is why Technician B's advice should be preferred.

You might be interested in
Can you carry 1 m3 of liquid water? Why or why not? (provide the weight to support your answer)
djverab [1.8K]

Answer:

No we cannot carry 1 cubic meter of liquid water.

Explanation:

As we know that density of water is 1000 kilograms per cubic meter of water hence we infer that 1 cubic meter of water will have a weight of 1000 kilograms of 1 metric tonnes which is beyond the lifting capability of strongest man on earth let alone a normal human being who can just lift a weight of 100 kilograms thus we conclude that we cannot lift 1 cubic meter of liquid water.

5 0
3 years ago
Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)
meriva

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

E(t)=E_o[1-a\frac{T}{T_m}]

where,

E(t) is the modulus of elasticity at any temperature 'T'

E_o is the modulus of elasticity at absolute zero.

T_{m} is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

6 0
2 years ago
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF
Naddik [55]

Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

Thus,

Their equivalent becomes:

1/Ceq = (1/C₁ + 1/C₂ +1/C₃

1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]

1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)

1/Ceq =[(0.1724)/(0.01034)]

1/Ceq = [(16.67)]

1/Ceq =(1/16.67) = 0.059μf

Ceq = 0.059μf ≈ 0.060μf

Therefore the equivalent capacitance of the three series capacitors is 0.060μf

4 0
3 years ago
#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing idea
otez555 [7]

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}}  \right )^{\dfrac{\gamma -1}{\gamma }}

Therefore, we have;

T_{2} = 294.2611 \times \left (\dfrac{5}{1}  \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K

The p·dV work is given as follows;

p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)

Therefore, we have;

Taking air as a diatomic gas, we have;

C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)

The molar mass of air = 28.97 g/mol

Therefore, we have

c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)

The work done per unit mass of gas is therefore;

p \cdot dV =W =   1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.

8 0
2 years ago
It is illegal to improperly dispose of antifreeze.<br> True<br> False
Reil [10]
Ehheem✔️

Explanation:
✔️✖️✔️✖️✔️✖️
6 0
2 years ago
Read 2 more answers
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