Page 57 - BASIC CONCEPTUAL OF THERMOFLUID

Points to note :

The average velocity of flow of fluid through a turbine
is normally high, and the fluid passes quickly through
the turbine. It may be assumed that, because of this,
heat energy does not have time to flow into or out of
the fluid during its passage through the turbine, and
hence Q = 0 .
Although velocities are high the difference between
them is not large, and the term representing the
change in kinetic energy may be neglected.

Potential energy is generally small enough to be
neglected.

W is the amount of external work energy produced
per second.

Turbine equation
2
2
ܳ − ܹ = ṁ[(ℎ − ℎ ) + ( ܥ − ܥ 1 ) + ( ܼ݃ − ܼ݃ )]
2
2 1 2 1
2

ܹ = ṁ(ℎ − ℎ 2
1
EXAMPLE 2

A fluid flows through a turbine at the rate of 45 kg/min. Across the turbine the specific enthalpy drop
of the fluid is 580 kJ/kg and the turbine loss 2100 kJ/min in the form of heat energy. Determine the
power produced by the turbine, assuming that changes in kinetic and potential energy may be
neglected.

Nozzle

Points to note :

The average velocity of flow through a nozzle is
high, hence the fluid spends only a short time in
the nozzle. For this reason, it may be assumed that
there is insufficient time for heat energy to flow
into or out of the fluid during its passage through
the nozzle, i.e. Q = 0.
ܥ2 = √2(ℎ1 − ℎ2
Since a nozzle has no moving parts, no work energy
will be transferred to or from the fluid as it passes
through the nozzle, i.e. W = 0.

Potential energy is generally small enough to be
neglected.

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