COE_cv05
Air flows through a nozzle with an inlet diameter of 200 mm, velocity of 400 m/s, pressure of 7 kPa (abs),
and temperature of 420 °C. The nozzle exit diameter is adjusted such that the exiting velocity is 700 m/s.
Determine:
a. the exit temperature, and
b. the mass flow rate through the nozzle
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COE_cv05
SOLUTION:
Vi = 400 m/s
di = 200 mm
pi = 7 kPa
Ti = 420 °C
Vo = 700 m/s
Apply the First Law to the CV shown assuming 1D, steady flow, no heat transfer, and no “other” work
done on the CV besides inlet/outlet pressure work,
d
2
+ W!other,
(1)
into
∫ eρ dV + CS∫ h + 12 V + gz ( ρurel ⋅ dA ) = Q!CV
dt CV
on CV
where,
d
(2)
ò er dV = 0 (steady state)
dt CV
(
∫ (h +
CS
1
2
)
)
(
V 2 + gz ( ρ u rel ⋅ dA ) = m! ho + 12 Vo2 − hi − 12 Vi 2
)
(3)
(The change in potential energies is assumed negligible and, since the flow is steady and 1D, the
mass flow rate is the same at the inlet and outlet.)
(4)
Q! into = 0 (Assume little heat transfer occurs over the nozzle surface area.)
CV
W!other, = 0 (No work is performed other than inlet/outlet pressure work.)
(5)
on CV
Substitute and simplify,
(
)
(
2
2
m! ho + 12 Vo2 − hi − 12 Vi 2 = 0 Þ ho = hi + 12 Vi - Vo
)
(6)
For the given conditions,
hi = 705.75 kJ/kg (from thermodynamics tables for air, assumed to be an ideal gas, at Ti = 420 °C)
Vi = 400 m/s
Vo = 700 m/s
Þ ho = 540.75 kJ/kg Þ To = 264 °C (from thermo tables assuming air is an ideal gas)
The mass flow rate is,
p
m! = ρiVi Ai = i Vi π4 di2
RTi
(7)
Using the given parameters:
pi = 7 kPa (abs),
R = 287 J/(kg×K),
Ti = 420 °C = 693 K,
=> ρi = 0.03624 kg/m3,
Vi = 400 m/s,
di = 0.2 m => Ai = 0.03142 m2,
Þ m! = 0.442 kg/s
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