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natita [175]
3 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]3 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]3 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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Answer:3.77 T/s

Explanation:

Given

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N=no of turns=61

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Answer:

v = 17.71 m / s

Explanation:

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sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

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The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

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(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

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\boxed{v_0 = 28.58m/s.}

(b).

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v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

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\boxed{v =33.3m/s.}

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Answer:

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