In: Other
Heat volumetric flow V = 100 SCM
SCM - standard cubic meter
At Standard condition 1 mol having 22.4 L or
1kmol having volume of ideal gas = 22.4 m3
1 m3 having = 1/22.4 kmol
100 m3 having = 100/22.4 kmol = 4.464 kmol
Nitrogen moles n= 4.464 kmol
T1 = 26.85 C = 273 +26.85 = 299.85 K
T2 = 126.85 C = 273 + 126.85 = 399.85 K
Specific heat of Nitrogen,
Cp(kj/kmol.k) = a + bT + cT^2 + dT^3
By table,
a = 28.90
b = -0.1571 *10^-2
c = -0.8081*10^-5
d = -2.8713*10^-9
Change in enthalpy ,
dH = nCpdT
dH = n(a + bT + cT^2 + dT^3)dT
Integrate both sides, from temperature T1 to T2
ΔH = n[ a(T2 - T1) + b/2 * (T2^2 - T1^2) + c/3 * (T2^3 - T1^3) + d/4 * (T2^4 - T1^4)
ΔH = 4.464 kmol*[ 28.90*(399.85 - 299.85) - 0.1571*10^-2/2 *(399.85^2 - 299.85^2) - 0.8081*10^-5/3 *(399.85^3 -299.85^3) - 2.8713*10^-9/4 *(399.85^4 - 299.85^4)]
ΔH =4.464*( 2890 -54.96 -99.58-12.546)
ΔH = 12155 kJ
Heat required for heating in this case,
Q = ΔH = 12155 kJ