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natita [175]
4 years ago
10

A power cycle operates between a lake’s surface water at a temperature of 300 K and water at a depth whose temperature is 285 K.

At steady state the cycle develops a power output of 10 kW, while rejecting energy by heat transfer to the lower-temperature water at the rate 14,400 kJ/min. Determine (a) the thermal efficiency of the power cycle and (b) the maximum thermal efficiency for any such power cycle
Physics
2 answers:
nikdorinn [45]4 years ago
8 0

Answer: a) 0.04kW = 40W

b) 0.05

Explanation:

A)

Thermal efficiency of the power cycle = Input / output

Input = 10 kW + 14,400 kJ/min = 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.

Output = 10 kW

Thermal Efficiency = Output / Input = 10kW / 250kW = 0.04KW = 40W

B)

Maximum Thermal Efficiency of the power cycle = 1 - T1/T2

Where T1 = 285kelvin

And T2 = 300kelvin

Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05

yanalaym [24]4 years ago
4 0

Answer:

a) Efficiency: 4%

b) Maximum thermal efficiency: 17%

Explanation:

First we calculate the total heat that come from the hot reservoir. This heat is the work output plus the rejected energy to the cold reservoir.

Q_h=W+Q_c=10kW+14400\frac{kJ}{min}(\frac{1min}{60s})(\frac{kW}{kJ/s} ) \\\\Q_h=10kW+240kW=250kW

The efficienccy of the cycle can be calculated as the ratio between the work output and the heat input.

\eta=\frac{W}{Q_h}=\frac{10kW}{250kW}=0.04=4\%

The maximum thermal efficiency of this cycle can be determined in function of the temperatures of the reservoirs by this equation:

\eta_{max}=1-\frac{T_c}{T_h}=1-\frac{250}{300}=1-0.83=0.17=17\%

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The statements that are true are;

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<h3>What is work done?</h3>

Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.

Hence, the following are true;

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4 0
2 years ago
A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. On the same track, a 75-gram stainless steel ball is
elena-s [515]

Answer:

4.2 m/s

Explanation:

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m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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4 years ago
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lukranit [14]

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6 0
3 years ago
If a rocket transfers 2,500,000 MJ of energy when it takes off, how much work is done?
erastovalidia [21]

Answer:

2,500,000 MJ (2.5\cdot 10^6 J)

Explanation:

According to the work-energy theorem:

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In this problem, the energy transferred by the rocket is

E=2,500,000 MJ=2.5\cdot 10^6 J

Therefore, the work done must be equal to the energy transferred, therefore:

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3 years ago
A block of mass 10 kg moves from position A to position B shown in the figure above. The speed of the block is 10 m/s at A and 4
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We have that the block is moving horizontally. Hence, its potential energy due to gravity stays the same. The only change in its mechanical energy is the one due to the change of speed. This reduction of its kinetic energy, due to the conservation of energy, is equal to the work that friction does. We have that at A the kinetic energy is : K=1/2*m*u^2=10*10*10/2=500J. At B, we have that K=1/2*10*16=80J. Sine we have that the initial value is 500, the work from the friction force (opposite to the movement of the object) is 80-500=420J.
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