ahmedmoon
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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
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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 ??
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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
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Thanks so much
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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 ??
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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 ..
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¨(v) change in enthalpy.Cp (T2 - T1 )1.005(806 - 40)=769.83
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for a sould we use
T1V12^()=T2V2^()
and what the unit of 15 ??
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the heat transferred = Cv(T2 - T1 )
0.71005(806 - 40)= 54.860 -
Hi
already I posted my question and my answer could please check my answer for each part
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thanks so much .
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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
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¨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.Kmy answerp1v1=mRT4.5 X 100 X 0.4 = m X 0.28 X 403m= 1595 kgnowi) p2v2 = mRT9 X 100 X 0.4 = 1.1595 X 0.28 X TT = 806 Kii ) the heat transferredno heat transferred because the work = 0¨(iii) change in internal energyCv(T2 - T1 )0.71005(806 - 40)= 54.86¨ (iv) work done = 0¨(v) change in enthalpy.Cp (T2 - T1 )1.005(806 - 40)=769.830 -
Carnot efficiency = w/Qh
20/40 = 0.5
then what ?
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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
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Hi
Can please give me the idea for solving these two questions
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A Carnot engine operates between reservoirs at 20 and 200°C. If 10 kW of power is produced, the rejected heat rate is nearest
in Homework Help
Posted
Total energy = work + change in heat .
Please help me ...