Question

Calculate the enthalpy change for:
b. Two kg mol of steam that is heated from 400 K and 100 kPa to 900 K and 100 kPa. Use three different methods and discuss your results. Answer: ΔH = 3,726 x 10 ^ 4 kJ
c. Three kg of water at 101.3 kPa and 300 K that vaporize at 15000 kPa and 800K. Answer: ΔH = 9816 kJ

Answer 1

b) enthalpy change for steam:

Moles of steam = 2kmol

Mass of steam, m = 2kmol*(18kg/kmol) = 36 kg

first method : by steam table,

initial condition :( T1 = 400K and P1 = 100 kPa)

Specific enthalpy H1 = 2720.41 kj/kg

Final condition :(T2 = 900K and P2 = 100kPa)

Specific enthalpy H2 = 3764.25 kj/kg

Change in enthalpy = m(H2 - H1)  

ΔH = 36kg*(3764.25 - 2720.41)kj/kg = 37578.24 Kj = 3.7578*10⁴ kJ

Second method: by specific heat at constant pressure Cp

Cp for superheated steam = 2.1 kJ/kg.k

Change in enthalpy ΔH = mCp(T2 - T1)  

ΔH = 36kg*2.1kj/kg.k * (900-400)k = 37800 kj = 3.78*10⁴ kJ

third method: by formula for enthalpy change ( by changing specific volume and change in internal energy)  

Initial condition :(T1 = 400 k and P1 = 100 kPa)

Specific internal energy u1 = 2547.77 kj/kg

Specific volume v1 = 1.82621 m3/kg

Final condition(T2 = 900 k and P2 = 100kPa)

Specific internal energy u2 = 3349.77 kj/kg

Specific volume v2 = 4.15204 m3/kg

Change in enthalpy,

ΔH = m(u2 - u1) + P*m(v2 - v1)

ΔH = 36kg*(3349.77 - 2547.77)kj/kg + 100kPa * 36kg*(4.15204 - 1.82621)m3/kg

ΔH = 37244.98 kj/kg = 3.7244*10⁴ KJ

Unit = 1 kPa*m3 = 1 kJ

C) 3 kg of water at P1= 101.3 kPa and T1 = 300k

P2 = 15000 kPa and T2 = 800k

Initial condition (T1 = 300K and P1 = 101.3kPa)

Specific enthalpy H1 = 112.6 kj/kg

Final condition (T2 = 800 k and P2 = 15000 kPa)

Specific enthalpy H2 = 3386.92 kj/kg

Mass of water m = 3 kg

Change in enthalpy ΔH = m(H2 - H1) = 3kg(3386.92 - 112.6)kj/kg = 9822 kj

ans: 9822 kj