A Carnot air conditioner takes energy from the thermal energy of a room at 63°F and transfers it as heat to the outdoors, which
1 answer:
Answer:
For the Carnot air conditioner working as a heat pump between 63 and 100°F , It would transfer 3.125 Joules of heat for each Joule of electric energy supplied.
Explanation:
The process described corresponds to a Carnot Heat Pump. A heat pump is a devices that moves heat from a low temperature source to a relative high temperature destination. <em>To accomplish this it requires to supply external work</em>.
For any heat pump, the coefficient of performance is a relationship between the heat that is moving to the work that is required to spend doing it<em>.</em>
For a Carnot Heat pump, its coefficient of performance is defined as:
Where:
- T is the temperature of each heat deposit.
- The subscript H refers to the high temperature sink(in this case the outdoors at 100°F)
- The subscript L refers to the low temperature source (the room at 63°F)
Then, for this Carnot heat pump:
So for each 3.125 Joules of heat to moved is is required to supply 1 Joule of work.
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Answer:
C = - 1.625 i + 6.06 j
Explanation:
Positive angles are measured counterclockwise.
Positive angles are measured counterclockwise. To determine the component on the x axis we use the cosine of the angle while to determine the component on the y axis we use the sine of the angle.
C = - 1.625 i + 6.06 j
Answer:
The weight of the object X is approximately 3.262 N (Acting downwards)
The weight of the object Y is approximately 8.733 N (Acting downwards)
Explanation:
The question can be answered based on the principle of equilibrium of forces
The given parameters are;
The weight of Z = 12 N (Acting downwards)
The weight of the pulleys = Negligible
From the diagram;
The tension in the in the string attached to object Z = The weight of object Z = 12 N
The tension in the in the string attached to object X = The weight of the object X
The tension in the in the string attached to object Y = The weight of the object Y
Given that the forces are in equilibrium, we have;
The sum of vertical forces acting at a point, = 0
Therefore;
Where;
= The weight of object Z = 12 N
= 12 N
= The vertical component of tension, T₂ = T₂ × sin(24°)
∴ = T₂ × sin(156°)
Similarly;
= T₃ × sin(50°)
From , and
= 12 N, we have;
12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)
Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0
Therefore at point B, we have;
T₁ₓ + T₂ₓ + T₃ₓ = 0
The tension force, T₁, only has a vertical component, therefore;
∴ T₁ₓ = 0
∴ T₂ₓ + T₃ₓ = 0
T₂ₓ = -T₃ₓ
T₂ₓ = T₂ × cos(156°)
T₃ₓ = T₃ × cos(50°)
From T₂ₓ = -T₃ₓ, we have;
T₂ × cos(156°) = - T₃ × cos(50°)...(2)
Making T₃ the subject of equation (1) and (2) gives;
Making T₃ the subject of equation in equation (1), we get;
12 = -(T₂ × sin(156°) + T₃ × sin(50°))
∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))
Making T₃ the subject of equation in equation (2), we get;
T₂ × cos(156°) = - T₃ × cos(50°)
∴ T₃ = T₂ × cos(156°)/(-cos(50°))
Equating both values of T₃ gives;
(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))
-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))
∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283
∴ T₂ ≈ -8.02 N
From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;
T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717
∴ T₃ ≈ -11.4 N
The weight of the object X = -T₂ × sin(156°)
∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N
The weight of the object X ≈ 3.262 N (Acting downwards)
The weight of the object Y = -(T₃ × sin(50°))
∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N
The weight of the object Y ≈ 8.733 N (Acting downwards)
