Example of change of enthalpy:

Q 0.5kg/s of a fluid flows in steady state process. The properties of fluid at the entrance are measured as p₁ = 1.4bar, density = 2.5kg/m³, u₁ = 920Kj/kg while at exit properties are p₂ = 5.6 bar, density = 5 kg/m³, u₂ = 720Kj/kg. The velocity at entrance is 200m/sec, while at exit it is 180m/sec. It rejects 60kw of heat and rises through 60m during the flow. Find change of enthalpy and rate of work done.     

Sol: Given that:

m_f  = 0.5kg/s

P₁  = 1.4bar, density = 2.5kg/m³,

u₁  = 920Kj/kg

P₂  = 5.6 bar, density = 5 kg/m3,

u₂  = 720Kj/kg. V₁  = 200m/sec

V₂  = 180m/sec Q = - 60kw

Z₂  - Z₁  = 60m h = ?

W_S  = ?

Since           h₂ - h₁  =   U +   P

h₂ - h₁  = [U₂  - U₁    + (P₂/  2  - P₁/   1)]

= [(720 -920) × 103  +   (5.6/5 - 1.4/2.5) × 105]

= [-200 × 103  + 0.56 × 105] = - 144KJ/kg

H =  mf  × (h₂ - h₁) = 0.5 × (-144) Kj/kg = -72KJ/sec                    .......ANS

By Applying SFEE

60 × 103  - W_S  = 0.5[ - 144 × 103  + (180²  - 100²)/2 + 9.81 × 60]

WS  = 13605.7 W = 136.1KW                    .......ANS

Posted Date: 10/16/2012 6:33:39 AM | Location : United States

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