First law of thermodynamics applied to a submarine


by Carbon884
Tags: thermodynamics
Carbon884
Carbon884 is offline
#1
Jun13-11, 12:49 PM
P: 5
Hallo,

I hope someone can help me with the following question:

A submarine conatins 1000m^3 of air and has a temperature and pressure of 15C 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?

V=constant=1000m^3
P=constant?
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

dQ=dU

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 ^^.
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fahim_naruto
fahim_naruto is offline
#2
Jun13-11, 02:14 PM
P: 4
R(air) = Cp - Cv
=> Cp=R(air)+Cv

so, Q=m(R(air)+Cv)(T2-T1)
nasu
nasu is offline
#3
Jun13-11, 08:55 PM
P: 1,902
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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