TITLE.PM5

FIRST LAW OF THERMODYNAMICS 165

dharm

/M-therm/Th4-4.pm5

Solution. Flow of fluid = 10 kg/min

Properties of fluid at the inlet :

Pressure, p 1 = 1.5 bar = 1.5 × 10^5 N/m^2

Density, ρ 1 = 26 kg/m^3

Velocity, C 1 = 110 m/s

Internal energy, u 1 = 910 kJ/kg

Properties of the fluid at the exit :

Pressure, p 2 = 5.5 bar = 5.5 × 10^5 N/m^2

Density, ρ 2 = 5.5 kg/m^3

Velocity, C 2 = 190 m/s

Internal energy, u 2 = 710 kJ/kg

Heat rejected by the fluid,

Q = 55 kJ/s

Rise is elevation of fluid = 55 m.

(i) The change in enthalpy,

∆h = ∆u + ∆(pv) ...(i)

∆(pv) = pv^221 −pv^11

=−=

pp 2 × − ×

2

1

1

51055

5

510

ρρ 26

.5

.5

1.

= 1 × 10^5 – 0.0577 × 10^5

= 10^5 × 0.9423 Nm or J = 94.23 kJ

∆u = u 2 – u 1 = (710 – 910) = – 200 kJ/kg

Substituting the value in eqn. (i), we get

∆h = – 200 + 94.23 = – 105.77 kJ/kg. (Ans.)

55 m

Q

1

2

Fluid in

Fluid out

Boundary

Fig. 4.44

(ii) The steady flow equation for unit mass flow can be written as

Q = ∆ KE + ∆ PE + ∆ h + W

where Q is the heat transfer per kg of fluid