|Jun13-11, 12:49 PM||#1|
First law of thermodynamics applied to a submarine
I hope someone can help me with the following question:
A submarine conatins 1000m^3 of air and has a temperature and pressure of 15°C and 0.1MPa respectively. Due to the cold seawater a heatflow of 60 MJ/h occurs. The machines on the otherhand adds disspiative Work of 21 kW to the system. The specific heat capacity of air is c(p,air) = 1.005 kJ/Kg*K.
what average Temperature will the air have after one hour of diving?
T1=288.15K => T2=?
Firstly i found the density and mass of the gas:
R(air)= 286.9 J/K*Kg => density= P/T*R(air)=1.209 Kg/m^3 => m= V*density= 1209.63 Kg
Secondly i found the total Energy:
Q/h=Wdiss - Q/h = 15600 KJ/h
Then i used the first law:
dW=0 because V=constant
Q=mcp(T2-T1)= m(R(air)-cv)*(T2-T1)= 288.19K .... which can't possibly be right because that is essentially my T1 temerature. So what did I do wrong?
Thanks for your help in advance ^^.
|Jun13-11, 02:14 PM||#2|
R(air) = Cp - Cv
|Jun13-11, 08:55 PM||#3|
The specific heat at constant pressure is given in the problem.
If you assume a constant-volume process, you can calculate cv=cp-R.
Either way the temperature increase is of the order of 10K (around 14K if you consider constant volume).
Review your last step.
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