Question

A proposed ocean power plant will utilize the temperature difference between surface seawater and seawater at a depth of 100 meters. Assume the surface temperature is 25° Celsius and the temperature at the 100-meter depth is 3° Celsius. 1. What is the ideal (Carnot) efficiency of the plant? 2. If the plant generates useful energy at the rate of 100 megawatts while operating with the efficiency found in part (a), at what rate is heat given off to the surroundings? 3. A nuclear power plant operates with an overall efficiency of 40 percent. At what rate must mass be converted into energy to give the same 100-megawatt output as the ocean power plant above? Express your answer in kilograms per second. The diagram below represents the Carnot cycle for a simple reversible (Carnot) engine in which a fixed amount of gas, originally at pressure po and volume V, follows the path ABCDA. Pressure A(Vo Po) Po Po D

Answer 1

Absolute temperature of the sounce. i.e. the surface beawater is 1/273 +25) - 298 K. Absolute temperature of the sink i.e. the temperature al 100-melen depth 1sT;(273+3) - 276K. .. Ideal (Carnot) efficiency of the plant is n 1 - Tz 1-276 T =0.074 on 7.4% 298 2. If Q2 is the nate of heat given off to the surroundings and S, is the nate of heat taken by the plant from then W = rate of generating eneagy = Q - Q2 Source, =) , ♡, - W+02 w O, Again, = n- 0.074 w = 0·074 -> 0.0749,- W -0.074W W92 > 0,- 0.926W --12:51 W 0.074

Here W- 100 MW Per second - 278x16'lgs' (S₂ = 1251 megawatts 3. If m m kg mass, is converted into energy then produced energy (per second) is mc² where Cz Speed of light in vaccum = 3x108m/s again, w > 100 megawatts -0.4 100x 106 75 3² x 100 m² m²5 ²x0.4 To give the same 100- megawatt output, 2. 78xjö% energy mass per second must be converted into کے 40 100 mc² =) m = 16 we can 4. Since, the given diagram represents a Cannot cycle, - say that AB is an isothermal expansion. adiabatic CD isothermal compression adiabatic and DA >> 2 BC > 7) > )

For AB, AT=0 :du o From first law of thermodynamics do= dU+P dr PdV DV - VE-VA = Vo > 0 i dg is. +ve, For BC, do=0 du - Pdv ..dv-Ve-VB>O du --ve AT DU -> 4T=-ve For CD, AT=0 do = Pov dv = Vo-Vcco id=-ve For DA, doo du-P du dv-VA-Voco i du-tve AT +ve Talat AB + o BC CD DA + of p