ahmedmoon

Total energy = work + change in heat . Please help me ...

A Carnot engine operates between reservoirs at 20 and 200°C. If 10 kW of power is produced, the rejected heat rate is nearest 26.3 kJ/s 20.2 kJ/s 16.3 kJ/s 12.0 kJ/s Above is full question please could help me

T teacher said the answer will be kj per second How is that .. I will try n = 1 - tmin / tmax So 1 - 293/473 = 0.38 THen power = work divide time What should I do next ??

Hi Can please give me the idea for solving this quetion A Carnot engine operates between reservoirs at 20 and 200°C. If 10 kW of power is produced, the rejected heat rate is nearest

Thanks so much

Hi I have this ideal gas rule Pv =nRT If we have p in kpa and m in mole Is sould to convert mole to kilo mole ??

Hi Here we have ischoric process because the volume is constant Now I will solve the question agine I Found T2 = 806 By using p2/p1 = T2/T1 The heat transferred Q = cv X delta T 0.71005(806 -403 ) = 228kj Is Correct ..

¨(v) change in enthalpy. Cp (T2 - T1 ) 1.005(806 - 40) =769.83

for a sould we use T1V12^()=T2V2^() and what the unit of 15 ??

the heat transferred = Cv(T2 - T1 ) 0.71005(806 - 40) = 54.86

Hi already I posted my question and my answer could please check my answer for each part

thanks so much .

see Carnot efficiency = 83.3 % efficiency of inventor's machine = w/Qh = 50 % Carnot efficiency is more than efficiency of inventor's machine so claim is possible

¨A certain Gas of volume 0.4 m3, pressure of 4.5 bar and temperature of 1300 C is heated to in a cylinder to 9 bar when the volume remains constant. Calculate ¨ (i) Temperature at the end of the process, ¨ (ii) the heat transferred ¨(iii) change in internal energy ¨ (iv) work done ¨(v) change in enthalpy. ¨Assume Cp = 1.005 kJ/kg.K and Cv= 0.71005 kJ/kg.K my answer p1v1=mRT 4.5 X 100 X 0.4 = m X 0.28 X 403 m= 1595 kg now i) p2v2 = mRT 9 X 100 X 0.4 = 1.1595 X 0.28 X T T = 806 K ii ) the heat transferred no heat transferred because the work = 0 ¨(iii) change in internal energy Cv(T2 - T1 ) 0.71005(806 - 40) = 54.86 ¨ (iv) work done = 0 ¨(v) change in enthalpy. Cp (T2 - T1 ) 1.005(806 - 40) =769.83

Carnot efficiency = w/Qh 20/40 = 0.5 then what ?

Yes .. for first one I have efficiency = 1 - Tlow /T high so 1 - 200/1200 efficiency = 5/6 then what should I do and please can give me the idea for solving both questions thanks Molecule

Hi Can please give me the idea for solving these two questions