## Example of change of enthalpy, Mechanical Engineering

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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 p= 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:

mf  = 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 = ?

WS  = ?

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

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

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