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3 years ago

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The temperature at the bottom of a reservoir is TL = 280 K and the surface temperature is TH = 295 K. This temperature differenc

e is used to run a steadystate power cycle that develops a power output of 8 kW, while rejecting energy by heat transfer at the rate 14,400 kJ/min. Determine: (a) The thermal efficiency of the power cycle, in %. (b) The maximum thermal efficiency for any such power

1 answer:

Tanzania [10]3 years ago

8 0

a) For the thermal efficiency we have

$\eta_{th} = \frac{Q_{out}}{Q_{in}} = \frac{|W|}{|Q_h|}\\\eta_{th} = \frac{|W|}{|W|+|Q_2|}$

With the previously values we know that

$W=8kW$ and $Q_L = 1440/6kW$ (convert the min to sec)

Replacing the values

$\eta_{th}=\frac{8}{8+1440/6}=\frac{1}{31}\\\eta_{th}\% = 3.225\%$

b) We use the formula of carnot efficiency

$\eta_{th}=1-\frac{T_l}{T_h}\\\eta_{th}\% =(1-\frac{280}{295})*100\\\eta_{th}\%=5.085\%$

**Note that apply the formula of carnot cycle we need to consider that there is no exchange of heat, there is no friction and the reservior are completely insulated

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The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

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The complete question is:

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Ganezh [65]

Answer:

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And evaluating the integral we got:

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