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8 days 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

2 answers:

inna [2.9K]8 days ago

8 0

Response: a) 0.04 kW = 40 W

b) 0.05

Explanation:

A)

The thermal efficiency of the power cycle is calculated as Input / Output

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

Output equals 10 kW

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

B)

Maximum Thermal Efficiency of the power cycle is defined as 1 - T1/T2

where T1 = 285 Kelvin

and T2 = 300 Kelvin

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

Sav [3K]8 days ago

4 0

Response:

a) Efficiency: 4%

b) Maximum thermal efficiency: 17%

Explanation:

First, we compute the total heat from the hot reservoir. This heat consists of work output plus the energy expelled 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 efficiency of the cycle is determined as the ratio between work output and heat input.

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

The maximum thermal efficiency for this cycle can be established based on the reservoir temperatures using the following equation:

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

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Scenario (2)

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The final energy when the plate separation transitions to d₂ can be written as:

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From equation (2),

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Thus, in this particular scenario,

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Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

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