Thermodynamics and Heat Transfer
ECE 309 Tutorial # 4
First Law of Thermodynamics: Control Volumes
_______________________________________________________________
Problem 1: Air enters an adiabatic nozzle steadily at 300 kPa, 200°C, and 30 m/s and
leaves at 100 kPa and 180 m/s. The inlet area of the nozzle is 80 cm2. Determine
(a) the mass flow rate through the nozzle,
(b) the exit temperature of the air, and
(c) the exit area of the nozzle.
3-D view of a nozzle
Cross-sectional view of a nozzle
Solution:
Step -1: Schematic Diagram
P1=300 kPa
T1=200 °C
V1=30 m/s
A1=80 cm2
P1=100 kPa
V2=180 m/s
T2=?
A2=?
AIR
m& =?
Step-2: Solve for
(a) The mass flow rate through the nozzle,
(b) The exit temperature T2 of the air,
(c) The exit area A2 of the nozzle.
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Step-3: Make table of values for the substance or fluid given
Initial Condition
Final Condition
Pressure
(kPa)
300
100
Temperature
(K)
200+273=473
Unknown
Area
(m2)
0.008
Unknown
Velocity
(m/s)
30
180
Step-4: Analysis
The region within the nozzle is selected as the system, and its boundaries are indicated by
the dashed lines in the schematic diagram. Mass is crossing the boundaries, thus it is a
control volume. And since there is no observable change within the control volume with
time, it is a steady-flow system. As the specified conditions, the air can be treated as an
ideal gas since it is at a high temperature and low pressure relative to its critical values
(Tcr= – 147 °C and Pcr=3390 kPa for nitrogen, the main constituent of air).
(a) To determine the mass flow rate, we need to find the specific volume of the air
first. This is determined from the ideal-gas relation at the inlet conditions:
Specific volume at inlet: υ1 =
Mass flow rate: m& =
1
υ1
(
RT1
0.287 kPa ⋅ m 3 /kg ⋅ K
=
P1
300 kPa
A1V1 =
(
1
0.008 m 2
3
0.4525 m /kg
) (473 K ) = 0.4525 m /kg
3
) (30 m/s ) = 0.5304 kg/s
Since the flow is steady, the mass flow rate through the entire nozzle will remain constant
at this value.
(b) A nozzle normally involves
(i) no shaft or electrical work (w=0),
(ii) negligible heat transfer (q ≈ 0), and
(iii) a small (if any) elevation change between the inlet and exit ( ∆ pe ≈ o).
Then the conservation of energy relation on a unit mass basis for this single-stream
steady-flow system reduces to
q − w = ∆h + ∆ke + ∆pe
0 − 0 = ∆h + ∆ke + 0
V − V1
0 = h2 − h1 + 2
2
2
0 = C p (T2 − T1) +
2
2
V22 − V12
2
Using specific heats at the anticipated average temperature of 450K (Cp=1.020 kJ/kg⋅K,
Cv=0.733 kJ/kg⋅K, and R=Cp–Cv=0.287 kJ/kg⋅K; see Table A-2), the exit temperature of
air is determined to be
T2 = T1 −
V 22 − V12
180 2 m 2 /s 2 − 30 2 m 2 /s 2
= 200 o C −
2C p
2 × (1020 J/kg ⋅ o C )
= 200 C −
o
180 2 m 2 /s 2 − 30 2 m 2 /s 2
2 × (1020 m /s ⋅ C )
2
2 o
= 184.6 o C

kg ⋅ m m
J
N⋅m
m2
 Note :1
=1
=1 2
=1 2
kg
kg
kg
s
s




This shows that the temperature of the air is decreased by about 15°C. The temperature
drop of the air is mainly due to the conversion of internal energy to the kinetic energy.
(c) To determine the exit area, we need to find the specific volume of the exit air
from the ideal- gas relation.
(
)
RT2
0.287 kPa ⋅ m 3 /kg ⋅ K (184.6 + 273)K
υ2 =
=
= 1.313 m 3 /kg
P2
100 kPa
Since the mass flow rate of the air is constant, exit area can be found from the mass flow
rate equation.
m& =
1
υ2
A2V 2 → 0.5304 kg/s =
1
A2 (180 m/s )
1.313 m 3 /kg
A2 = 0.00387 m 2 = 38.7 cm 2
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Problem2: A hair dryer is basically a duct in which a few layers of electric resistors are
placed. A small fan pulls the air in and forces it through the resistors where it is heated.
Air enters a 1200 watt hair dryer at 100 kPa and 22°C and leaves at 47°C. The crosssectional area of the hair dryer at the exit is 60cm2. The power consumed by the fan is 2
watt and the heat losses through the walls of the hair dryer is 5 watt, determine
(a) the volume flow rate of the air at inlet
(b) the velocity of the air at the exit
3-D view of hair dryer
Cross-section of hair dryer showing heating
element and fan
Solution:
Step -1: Schematic Diagram
Fan
Heating element
Q = -5 watt
P2=?
T2=47°C
υ2=?
A2=60cm2
P1=100 kPa
T1=22°C
υ1=?
A1=?
Wf = -2 watt
Wh = -1200 watt
Step-2: Solve for
&
(a) the volume flow rate of the air at inlet, ∀
in
(b) the velocity of air at the exit, V2
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Step-3: Make table of values for the substance or fluid given
Pressure
(kPa)
100
?
Inlet condition
Exit condition
Temperature
(K)
22+273=295
47+273=320
Area
(m2)
?
0.006
Velocity
(m/s)
?
?
Step-4: Analysis
The region within the hair dryer is selected as the system, and its boundaries are indicated
by the dashed lines in the schematic diagram. Mass is crossing the boundaries, thus it is a
control volume. And since there is no observable change within the control volume with
time, it is a steady-flow system. As the specified conditions, the air can be treated as an
ideal gas since it is at a high temperature and low pressure relative to its critical values
(Tcr= – 147 °C and Pcr=3390 kPa for nitrogen, the main constituent of air).
(a) The conservation of energy relation in the rate form for this single-stream steadyflow system is
Q& − W& = m& [∆h + ∆KE + ∆PE ]


 V22 − V12 
&
&
 + g ( z 2 − z1 )
⇒ Q − W = m& (h2 − h1 ) + 
2




Assumptions:
(a) fixed elevation; that is, negligible potential energy (∆PE≈0)
(b) change in kinetic energy is small (∆KE≈0)
(c) inlet and exit sections of the hair dryer are exposed to atmosphere; that is, P1=P2
Therefore, the reduced form of the energy equation is
Q& − W& = m& (h − h ) = m& C (T − T )
2
1
p
2
1
⇒ ( −5 watt ) − ( −1200 watt − 2 watt ) = m& (1.005 kJ/kg ⋅ K ) ( 320 K − 295 K )
1197 watt
1197 J/s
⇒ m& =
=
= 0.047641 kg/s
(1.005 kJ/kg ⋅ K ) ( 25 K ) (1005 J/kg ⋅ K ) ( 25 K )
Note: In the above calculation specific heats are calculated at the anticipated average
temperature of 300K (Cp=1.005 kJ/kg⋅K, Cv=0.718 kJ/kg⋅K, and
R=Cp–Cv=0.287 kJ/kg⋅K; see Table A-2).
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Now, using the equation of state: Pυ = RT
(
)
RT1
0.287 kPa ⋅ m 3 /kg ⋅ K ( 295 K )
Specific volume at inlet: υ 1 =
=
= 0.84665 m 3 /kg
100 kPa
P1
Using the mass conservation equation
Mass flow rate: m& = ρ 1 V1 A1 =
V1 A1
υ1
=
&
∀
1
υ1
& is the volume flow rate at inlet.
where, ∀
1
& = υ m& = ( 0.84665 m 3 /kg) ( 0.047641 kg/s) = 0.04033 m 3 /s
Volume flow rate in inlet: ∀
1
1
(b) mass flow rate must be constant at the inlet and outlet, therefore
m& = ρ 2 V 2 A2 =
V 2 A2
υ2
⇒ V2 =
m& υ 2
A2
The only unknown in above equation is υ2 and it is calculated from
υ2 =
(
)
RT2
0.287 kPa ⋅ m 3 /kg ⋅ K (320 K )
=
= 0.9184 m 3 /kg
100 kPa
P2
Now the velocity at the exit is
V2 =
m& υ 2 ( 0.047641 kg/s) (0.9184 m 3 /kg)
=
= 7.2922 m/s
A2
( 0.006 m 2 )
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