Question

A heat engine has three quarters the thermal efficiency of a carnot engine operating between temperatures of 65°C and 435°C if the heat engine absorbs heat at a rate of 44.0 kW, at what rate is heat exhausted?

Answer 1

Answer:

The value is  $P_e = 31275.2 \ W$

Explanation:

From the question we are told that

   The efficiency of the carnot engine is  $\eta$

    The efficiency of a heat engine is $k = \frac{3}{4} * \eta$

    The operating temperatures of the carnot engine is  $T_1 = 65 ^oC =338 \ K$  to  $T_2 = 435 ^oC = 708 \ K$

    The rate at which the heat engine absorbs energy is  $P = 44.0 kW = 44.0 *10^{3} \ W$

Generally the efficiency of the carnot engine is mathematically represented as

          $\eta = [ 1 - \frac{T_1 }{T_2} ]$

=>       $\eta = [ \frac{T_2 - T_1}{T_2} ]$

=>       $\eta = 0.3856$

Generally the efficiency of the heat engine is

           $k = \frac{3}{4} * 0.3856$

=>        $k = 0.2892$

Generally the efficiency of the heat engine is also mathematically represented as

          $k = \frac{W}{P}$

Here W is the work done which is mathematically represented as

        $W = P - P_e$

Here $P_e$  is the heat exhausted

So

       $k = \frac{P - P_e}{P}$

=>    $0.2892 = \frac{44*10^{3} - P_e}{44*10^{3}}$

=>   $P_e = 31275.2 \ W$