- #1
Carbon884
- 5
- 0
First law of thermodynamics
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 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?
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 ^^.
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 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?
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 ^^.
Last edited: