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Dennis_Churaev [7]
3 years ago
15

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?
Physics
1 answer:
const2013 [10]3 years ago
8 0

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

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