Question

A solar collector receiving solar radiation at the rate of $0.6{\text{kWm}}^{-2}$ transforms it into the internal energy of a fluid at an overall efficiency of $50\%$. The fluid heated to $350\text{K}$ is used to run a heat engine which rejects heat at $313\text{K}$. If the heat engine is to deliver $2.5\text{kW}$ power, the minimum area of the solar collector required would be

• A. $83.33{\text{m}}^2$
• B. $16.66{\text{m}}^2$
• C. $39.68{\text{m}}^2$
• D. $79.36{\text{m}}^2$

Answer 1

The correct option is D $79.36{\text{m}}^2$

Energy flow diagram:


Overall efficiency of heating of the fluid system,
${\eta }_f=50\%=0.50$

${\eta }_f=\frac{q}{0.6}=0.5$
$\Rightarrow q=0.50\times 0.6=0.3{\text{kWm}}^{-2}$

Let, $A=$ Miniumum area of the solar collector.

So, the total heat transfer to the fluid,
$Q=qA=0.3A\text{kW}$

The fluid at $350\text{K}$ acts as the source for the heat engine which rejects heat to a sink at $313\text{K}$.
$\therefore T_1=350\text{K}$
$T_2=313\text{K}$
& $W=2.5\text{kW}$ (power delivered)

Minimum energy input required (minimum area of solar collector) is when the heat engine is reversible.

Thermal efficiency of the reversible heat engine,

$\eta =1-\frac{T_2}{T_1}=\frac{W}{Q_1}$

where $Q_1$ is the heat input to the engine per unit time.

Substituting the values from diagram,

$\eta =1-\frac{313}{350}=0.105$

$\Rightarrow 0.105=\frac{2.5}{Q_1}=\frac{2.5}{0.3A}$

[since heat extracted from the liquid = increase in internal energy]

$\therefore A=79.36{\text{m}}^2$ is the minimum area of solar collector.

Hence, option (d) is correct answer.