Question

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

Answer 1

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