I need to find the specific Enthalpy and Volume change of an in compressible sys
ID: 3163829 • Letter: I
Question
I need to find the specific Enthalpy and Volume change of an in compressible system of water, given its change in Temperature and pressure., and the constant specific heat. --------------------------------------- the temperature and pressure inside a segment of a water solar collector, initially at 25 C and 14.7psi is increased to 100 C and 17 psi due to heating by the sun. Determine the specific volume change and then the specific enthalpy change. Assume the water is in compressible at its initial volume 4.185 kJ/kg/K. I used NIST for part A) I found delta h=315.06 kJ/kg and delta v = 4.05 * 10^-5 m^3/kg
Explanation / Answer
The flow is incompressible that means it is an isochoric process.
from the first law of thermodynamics: dU=dQ-PdV but in isochoric process dV=0 then du=dQ
that means volume remain constant
P1=14.7psi=14.7*6894.76Pa=101352.97Pa
T1=25+273K=298K
P2= 17psi=17*6894.76Pa=117210.92Pa
T2=100+273K=373K and R=8.314J/K/mol
specific volume at P1 and T1 is given by v1=nRT1/P1=8.314*298/101352.97n=.02444n
specific volume at P2 and T2 is v2=nRT2/P2=8.314*373/117210.92n=.0264n
but we need to calculate n which is given by n=PV/RT using Ideal gas law
then using V1=4.185*1000=4185J/Kg/K
we get n=101352.97*4185/8.314*298=171200.7
then change in specific volume deltav=n(.0264-.02444)=3.35*10^-5m^3/Kg
now enthalpy H=U+PV
then change in enthalphy deltaH=mCv(T2-T1)+(P2V2-P1V1)
taking Cv=3/2R
we get deltaH=mCv(T2-T1)+(P2V2-P1V1)=(3.35*10^-5/4185)*3/2*8.314*(373-298)+(117210.92* 4185-101352.97*4185)=663.65Kj/kg