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1 month ago

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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 [3.1K]1 month 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 [3.1K]1 month 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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2 months ago

Read 2 more answers

A 10 cm wide box is held between two springs in a 1 m gap on a frictionless surface. The left spring has a natural length of 80

Maru [3345]

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The distance measures $x =0.291 \ m$

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According to the problem statement,

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The first spring's natural length is $y = 80 \ cm = \frac{80 }{100} = 0.8 \ m$

The spring constant for the first spring is $k_1 = 200 N/m$

The second spring has a natural length of $z = 90 \ cm = \frac{90}{100} = 0.9 \ m$

The second spring's spring constant is $k_2 = 350 \ N/m$

We denote the distance from the center of the box to the left edge as x.

At equilibrium,

The force exerted by the first spring is

$F_1 = k_1 * (0.8 -x)$

while the force from the second spring is

$F_2 = k_2 * [ 0.9 - (0.9 -x)]$

Thus, at equilibrium,

$F_1 = F_2$

Substituting values gives us

$k_1 * (0.8 -x) = k_2 * [ 0.9 - (0.9 -x)]$

which leads to

$200 * (0.8 -x) = 350 * [ 0.9 - (0.9 -x)]$

resulting in

$160 -200x) = 350x$

and finally,

$160 =550x$

this simplifies to

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