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Republic of Iraq
Ministry of Higher Education
& Scientific Research
University of Kufa
By M.Sc. Hyder Mohammed Abdul Hussein
ABSTRACT
A theoretical study includes details flow turbulence in air-conditioned spaces with the determination of the
boundary conditions depending on the Iraqi Code of cooling is done in this research. Two kinds of two-
dimensional and three-dimensional ventilation problems have been considered:
(a) isothermal ventilation in simple rooms.
(b) non-isothermal ventilation with coupled heat or mass transfer.
The investigation has studied the flow and thermal conditions for four different diffusers (displacement, grille,
slot, and square diffusers). The dimensions of the of the physical model are (5.16×3.65 m) with (2.43 m high). The
supply condition for four diffusers are (displacement (0.0768 kg/s), grille (0.0768 kg/s), slot (0.1410 kg/s), square
(0.750 kg/s)) and temperature at supply for all types is (15.0o
C), the return considered as the type of diffusers has
been imposed zero flow pressure and temperature at (24.0o
C).
A modified version of a three-dimensional computer program (fluent 6.3.26) by using finite-volume method was
used to simulate the complex flow with buoyant inside the model room. They have been investigated numerically
by using several turbulence models and the method solution by using k-ε and k-ω models.
A THEORETICAL STUDY OF A COLD AIR DISTRIBTION
SYSTEM WITH DIFFERENT SUPPLY PATTERNS
The Iraqi Code of Cooling limited the outdoor for Baghdad and indoor conditions are
listed in Table (1) and Table (2), respectively.
Table (1) Outdoor data for Iraq
Region
DBT in summer
(ºC)
RH %
in summer
The daily
(ºC)
Altitude
(m)
Latitude
N
Longitude
E
Baghdad 47 16 18.7 34.1 33.32 44.33
Table (2) Indoor conditions
 
DBT in summer
(ºC)
RH %
in summer
Air velocity
(m/s)
Human comfort 19 - 24 40 - 60 1.8 – 2
Recommended conditions
inside the office
23 - 26 40 - 50 0.13 – 0.23
Four types of diffusers are set in three orientations all south-facing and all cases
running as constant wall temperature, but not that all walls of office are exposed to
outside. For each type of diffuser three cases are chosen, the first case just eastern and
southern walls, the second case is only the southern wall and the third case is the
southern and western wall, and the ceiling wall in all cases is included.
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Displacement Diffuser
The inlet diffuser is located near the west wall, and the exhaust opening is at the center
of the ceiling
the objects (human, computers, tables, lamps and cabinets) are simulated.
Boundary conditions:
Supply diffuser: mass flow rate of 0.0768 kg/s , turbulence intensity of 4% .
Return: The outlet is specified as pressure outlet.
Thermal conditions:
• Computer 1: 171.43 W/m2
• Computer 2: 274.6 W/m2
• Human simulators: 41.9 W/m2
• Lamps: 37.78 W/m2
Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models.
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Computation meshes: (1,480,232) cells.
Numerical schemes: discretized using the second-order upwind scheme. For the
discretization of pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The
SIMPLEC scheme is used for the pressure-velocity coupling.
Summary of boundary conditions
Table (3) summarize boundary conditions:
Cases Orientation
Air supply
ACH (kg/s)
Air velocity
(m/s)
Gross area
Agross
(m2
)
Air temp.
supply
(ºC)
Air temp.
return
(ºC)
Case 1
Eastern, southern and
ceiling walls
5.0 (0.0768) 0.35 1.1 × 0.53 15 24Case 2 Southern and ceiling wall
Case 3
Southern, western and
ceiling wall
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Simulation results
E
W
N
S
Fig. (1) Configuration of the displacement
ventilation test case.
Fig. (2) The positions of the measuring
poles for the displacement ventilation
test case.
University of Kufa
College of Engineering
Mechanical Eng. Dept.
a a
b b
Fig. (3) Distribution of calculation air temperature contours
with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (4) Distribution of calculation air temperature contours
with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (5) Distribution of calculation air temperature contours
with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (6) Distribution of calculation air temperature contours
with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
a
Fig. (7) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (8) Distribution of calculation air temperature contours
with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (9) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (10) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (11) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (12) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (14) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (15) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (16) Effect draft temperature for k-ε and
k-ω models, (a) case1, (b) case 2, (c) case
3.
a b
c
Grille Diffuser
Boundary conditions:
Supply diffuser: mass flow rate of 0.0768 kg/s , turbulence intensity of 4% .
Return: The outlet is specified as pressure outlet.
Thermal conditions:
• Computer 1: 171.43 W/m2
• Computer 2: 274.6 W/m2
• Human simulators: 41.9 W/m2
• Lamps: 37.78 W/m2
Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models.
Computation meshes: (499,952) cells.
Numerical schemes: discretized using the second-order upwind scheme. For the discretization of
pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used
for the pressure-velocity coupling.
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Summary of boundary conditions
Table (4) summarize boundary conditions:
Cases Orientation
Air supply
ACH (kg/s)
Air velocity
(m/s)
Gross area
Agross
(m2
)
Air temp.
supply
(ºC)
Air temp.
return
(ºC)
Case 1
Eastern, southern and
ceiling walls
5.0 (0.0768) 2.7 0.28 × 0.18 15 24Case 2 Southern and ceiling wall
Case 3
Southern, western and
ceiling wall
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Simulation results
Fig. (17) Configuration of grille ventilation
test case
Fig. (18) The positions of the measuring
poles for the grille ventilation test case [13].
E
W
N
S
University of Kufa
College of Engineering
Mechanical Eng. Dept.
a a
b b
Fig. (19) Distribution of calculation air temperature
contours with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (20) Distribution of calculation air temperature
contours with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (21) Distribution of calculation air temperature
contours with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (22) Distribution of calculation air temperature
contours with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (23) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (24) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (25) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (26) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (27) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (28) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (29) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (30) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a b
c
Fig. (31) Grille effect draft temperature for k-ε
and k-ω models, (a) case1, (b) case 2, (c)
case 3.
Slot Diffuser
Boundary conditions:
Supply diffuser: mass flow rate of 0.1410 kg/s , turbulence intensity of 5% .
Return: The outlet is specified as pressure outlet.
Thermal conditions:
• Computer 1: 171.43 W/m2
• Computer 2: 274.6 W/m2
• Human simulators: 41.9 W/m2
• Lamps: 37.78 W/m2
Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models.
Computation meshes: (1,071,118) cells.
Numerical schemes: discretized using the second-order upwind scheme. For the discretization of
pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used
for the pressure-velocity coupling.
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Summary of boundary conditions
Table (5) summarize boundary conditions:
Cases Orientation
Air supply
ACH (kg/s)
Air velocity
(m/s)
Gross area
Agross
(m2
)
Air temp.
supply
(ºC)
Air temp.
return
(ºC)
Case 1
Eastern, southern and
ceiling walls
9.2 (0.1410) 3.9 1.15 × 0.10 15 24Case 2 Southern and ceiling wall
Case 3
Southern, western and
ceiling wall
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Simulation results
Fig. (32) Configuration of slot ventilation
test case.
Fig. (33) The positions of the measuring
poles for the ceiling slot ventilation test
case, [13].
E
W
N
S
University of Kufa
College of Engineering
Mechanical Eng. Dept.
a a
b b
Fig. (34) Distribution of calculation air temperature
contours with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m
Fig. (35) Distribution of calculation air temperature
contours with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m
a a
b b
Fig. (36) Distribution of calculation air temperature
contours with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m
Fig. (37) Distribution of calculation air temperature
contours with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m
a a
b b
Fig. (38) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (39) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (40) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (41) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (42) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (43) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (44) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (45) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a b
c
Fig. (46) Slot effect draft temperature
for k-ε and k-ω models, (a) case1, (b)
case 2, (c) case 3.
Square Diffuser
Boundary conditions:
Supply diffuser: mass flow rate of 0.750 kg/s , turbulence intensity of 5% .
Return: The outlet is specified as pressure outlet.
Thermal conditions:
• Computer 1: 171.43 W/m2
• Computer 2: 274.6 W/m2
• Human simulators: 41.9 W/m2
• Lamps: 37.78 W/m2
Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models.
Computation meshes: (1,751,500) cells.
Numerical schemes: discretized using the second-order upwind scheme. For the discretization of
pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used
for the pressure-velocity coupling.
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Summary of boundary conditions
Table (6) summarize boundary conditions:
Cases Orientation
Air supply
ACH (kg/s)
Air velocity
(m/s)
Gross area
Agross
(m2
)
Air temp.
supply
(ºC)
Air temp.
return
(ºC)
Case 1
Eastern, southern and
ceiling walls
4.9 (0.750) 5.2 0.3 × 0.3 15 24Case 2 Southern and ceiling wall
Case 3
Southern, western and
ceiling wall
University of Kufa
College of Engineering
Mechanical Eng. Dept.
Simulation results
Fig. (47) Configuration of ceiling slot
ventilation test case.
Fig. (48) The positions of the measuring
poles for the ceiling slot ventilation test
case, [13].
E
W
N
S
Fig. (49) modeling of the square diffuser.
a a
b b
Fig. (50) Distribution of calculation air temperature
contours with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (51) Distribution of calculation air temperature
contours with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (52) Distribution of calculation air temperature
contours with k-ε , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (53) Distribution of calculation air temperature
contours with k-ω , case 1,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (54) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (55) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (56) Distribution of calculation air temperature
contours with k-ε , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (57) Distribution of calculation air temperature
contours with k-ω , case 2,
(a) plane at z=1.825m, (b) plane at z=0.4m.
a a
b b
Fig. (58) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (59) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4 m.
a a
b b
Fig. (60) Distribution of calculation air temperature
contours with k-ε , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4m.
Fig. (61) Distribution of calculation air temperature
contours with k-ω , case 3,
(a) plane at z=1.825m, (b) plane at z=0.4 m.
a b
c
Fig. (62) Slot effect draft temperature
for k-ε and k-ω models, (a) case1, (b)
case 2, (c) case 3.

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A theoretical study of a cold air distribution system with different supply patterns

  • 1. Republic of Iraq Ministry of Higher Education & Scientific Research University of Kufa By M.Sc. Hyder Mohammed Abdul Hussein ABSTRACT A theoretical study includes details flow turbulence in air-conditioned spaces with the determination of the boundary conditions depending on the Iraqi Code of cooling is done in this research. Two kinds of two- dimensional and three-dimensional ventilation problems have been considered: (a) isothermal ventilation in simple rooms. (b) non-isothermal ventilation with coupled heat or mass transfer. The investigation has studied the flow and thermal conditions for four different diffusers (displacement, grille, slot, and square diffusers). The dimensions of the of the physical model are (5.16×3.65 m) with (2.43 m high). The supply condition for four diffusers are (displacement (0.0768 kg/s), grille (0.0768 kg/s), slot (0.1410 kg/s), square (0.750 kg/s)) and temperature at supply for all types is (15.0o C), the return considered as the type of diffusers has been imposed zero flow pressure and temperature at (24.0o C). A modified version of a three-dimensional computer program (fluent 6.3.26) by using finite-volume method was used to simulate the complex flow with buoyant inside the model room. They have been investigated numerically by using several turbulence models and the method solution by using k-ε and k-ω models. A THEORETICAL STUDY OF A COLD AIR DISTRIBTION SYSTEM WITH DIFFERENT SUPPLY PATTERNS
  • 2. The Iraqi Code of Cooling limited the outdoor for Baghdad and indoor conditions are listed in Table (1) and Table (2), respectively. Table (1) Outdoor data for Iraq Region DBT in summer (ºC) RH % in summer The daily (ºC) Altitude (m) Latitude N Longitude E Baghdad 47 16 18.7 34.1 33.32 44.33 Table (2) Indoor conditions   DBT in summer (ºC) RH % in summer Air velocity (m/s) Human comfort 19 - 24 40 - 60 1.8 – 2 Recommended conditions inside the office 23 - 26 40 - 50 0.13 – 0.23 Four types of diffusers are set in three orientations all south-facing and all cases running as constant wall temperature, but not that all walls of office are exposed to outside. For each type of diffuser three cases are chosen, the first case just eastern and southern walls, the second case is only the southern wall and the third case is the southern and western wall, and the ceiling wall in all cases is included. University of Kufa College of Engineering Mechanical Eng. Dept.
  • 3. Displacement Diffuser The inlet diffuser is located near the west wall, and the exhaust opening is at the center of the ceiling the objects (human, computers, tables, lamps and cabinets) are simulated. Boundary conditions: Supply diffuser: mass flow rate of 0.0768 kg/s , turbulence intensity of 4% . Return: The outlet is specified as pressure outlet. Thermal conditions: • Computer 1: 171.43 W/m2 • Computer 2: 274.6 W/m2 • Human simulators: 41.9 W/m2 • Lamps: 37.78 W/m2 Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models. University of Kufa College of Engineering Mechanical Eng. Dept.
  • 4. Computation meshes: (1,480,232) cells. Numerical schemes: discretized using the second-order upwind scheme. For the discretization of pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used for the pressure-velocity coupling. Summary of boundary conditions Table (3) summarize boundary conditions: Cases Orientation Air supply ACH (kg/s) Air velocity (m/s) Gross area Agross (m2 ) Air temp. supply (ºC) Air temp. return (ºC) Case 1 Eastern, southern and ceiling walls 5.0 (0.0768) 0.35 1.1 × 0.53 15 24Case 2 Southern and ceiling wall Case 3 Southern, western and ceiling wall University of Kufa College of Engineering Mechanical Eng. Dept.
  • 5. Simulation results E W N S Fig. (1) Configuration of the displacement ventilation test case. Fig. (2) The positions of the measuring poles for the displacement ventilation test case. University of Kufa College of Engineering Mechanical Eng. Dept.
  • 6. a a b b Fig. (3) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (4) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 7. a a b b Fig. (5) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (6) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 8. a a b b a Fig. (7) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (8) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 9. a a b b Fig. (9) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (10) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 10. a a b b Fig. (11) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (12) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 11. a a b b Fig. (14) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (15) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 12. Fig. (16) Effect draft temperature for k-ε and k-ω models, (a) case1, (b) case 2, (c) case 3. a b c
  • 13. Grille Diffuser Boundary conditions: Supply diffuser: mass flow rate of 0.0768 kg/s , turbulence intensity of 4% . Return: The outlet is specified as pressure outlet. Thermal conditions: • Computer 1: 171.43 W/m2 • Computer 2: 274.6 W/m2 • Human simulators: 41.9 W/m2 • Lamps: 37.78 W/m2 Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models. Computation meshes: (499,952) cells. Numerical schemes: discretized using the second-order upwind scheme. For the discretization of pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used for the pressure-velocity coupling. University of Kufa College of Engineering Mechanical Eng. Dept.
  • 14. Summary of boundary conditions Table (4) summarize boundary conditions: Cases Orientation Air supply ACH (kg/s) Air velocity (m/s) Gross area Agross (m2 ) Air temp. supply (ºC) Air temp. return (ºC) Case 1 Eastern, southern and ceiling walls 5.0 (0.0768) 2.7 0.28 × 0.18 15 24Case 2 Southern and ceiling wall Case 3 Southern, western and ceiling wall University of Kufa College of Engineering Mechanical Eng. Dept.
  • 15. Simulation results Fig. (17) Configuration of grille ventilation test case Fig. (18) The positions of the measuring poles for the grille ventilation test case [13]. E W N S University of Kufa College of Engineering Mechanical Eng. Dept.
  • 16. a a b b Fig. (19) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (20) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 17. a a b b Fig. (21) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (22) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 18. a a b b Fig. (23) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (24) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 19. a a b b Fig. (25) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (26) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 20. a a b b Fig. (27) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (28) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 21. a a b b Fig. (29) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (30) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 22. a b c Fig. (31) Grille effect draft temperature for k-ε and k-ω models, (a) case1, (b) case 2, (c) case 3.
  • 23. Slot Diffuser Boundary conditions: Supply diffuser: mass flow rate of 0.1410 kg/s , turbulence intensity of 5% . Return: The outlet is specified as pressure outlet. Thermal conditions: • Computer 1: 171.43 W/m2 • Computer 2: 274.6 W/m2 • Human simulators: 41.9 W/m2 • Lamps: 37.78 W/m2 Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models. Computation meshes: (1,071,118) cells. Numerical schemes: discretized using the second-order upwind scheme. For the discretization of pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used for the pressure-velocity coupling. University of Kufa College of Engineering Mechanical Eng. Dept.
  • 24. Summary of boundary conditions Table (5) summarize boundary conditions: Cases Orientation Air supply ACH (kg/s) Air velocity (m/s) Gross area Agross (m2 ) Air temp. supply (ºC) Air temp. return (ºC) Case 1 Eastern, southern and ceiling walls 9.2 (0.1410) 3.9 1.15 × 0.10 15 24Case 2 Southern and ceiling wall Case 3 Southern, western and ceiling wall University of Kufa College of Engineering Mechanical Eng. Dept.
  • 25. Simulation results Fig. (32) Configuration of slot ventilation test case. Fig. (33) The positions of the measuring poles for the ceiling slot ventilation test case, [13]. E W N S University of Kufa College of Engineering Mechanical Eng. Dept.
  • 26. a a b b Fig. (34) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m Fig. (35) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m
  • 27. a a b b Fig. (36) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m Fig. (37) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m
  • 28. a a b b Fig. (38) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (39) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 29. a a b b Fig. (40) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (41) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 30. a a b b Fig. (42) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (43) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 31. a a b b Fig. (44) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (45) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 32. a b c Fig. (46) Slot effect draft temperature for k-ε and k-ω models, (a) case1, (b) case 2, (c) case 3.
  • 33. Square Diffuser Boundary conditions: Supply diffuser: mass flow rate of 0.750 kg/s , turbulence intensity of 5% . Return: The outlet is specified as pressure outlet. Thermal conditions: • Computer 1: 171.43 W/m2 • Computer 2: 274.6 W/m2 • Human simulators: 41.9 W/m2 • Lamps: 37.78 W/m2 Turbulence modeling: Applying that the Realizable k-ε and the SST k-ω models. Computation meshes: (1,751,500) cells. Numerical schemes: discretized using the second-order upwind scheme. For the discretization of pressure, the PRESTO! (PREssure STaggering Option) scheme is used. The SIMPLEC scheme is used for the pressure-velocity coupling. University of Kufa College of Engineering Mechanical Eng. Dept.
  • 34. Summary of boundary conditions Table (6) summarize boundary conditions: Cases Orientation Air supply ACH (kg/s) Air velocity (m/s) Gross area Agross (m2 ) Air temp. supply (ºC) Air temp. return (ºC) Case 1 Eastern, southern and ceiling walls 4.9 (0.750) 5.2 0.3 × 0.3 15 24Case 2 Southern and ceiling wall Case 3 Southern, western and ceiling wall University of Kufa College of Engineering Mechanical Eng. Dept.
  • 35. Simulation results Fig. (47) Configuration of ceiling slot ventilation test case. Fig. (48) The positions of the measuring poles for the ceiling slot ventilation test case, [13]. E W N S Fig. (49) modeling of the square diffuser.
  • 36. a a b b Fig. (50) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (51) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 37. a a b b Fig. (52) Distribution of calculation air temperature contours with k-ε , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (53) Distribution of calculation air temperature contours with k-ω , case 1, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 38. a a b b Fig. (54) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (55) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 39. a a b b Fig. (56) Distribution of calculation air temperature contours with k-ε , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (57) Distribution of calculation air temperature contours with k-ω , case 2, (a) plane at z=1.825m, (b) plane at z=0.4m.
  • 40. a a b b Fig. (58) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (59) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4 m.
  • 41. a a b b Fig. (60) Distribution of calculation air temperature contours with k-ε , case 3, (a) plane at z=1.825m, (b) plane at z=0.4m. Fig. (61) Distribution of calculation air temperature contours with k-ω , case 3, (a) plane at z=1.825m, (b) plane at z=0.4 m.
  • 42. a b c Fig. (62) Slot effect draft temperature for k-ε and k-ω models, (a) case1, (b) case 2, (c) case 3.

Editor's Notes

  1. Animated title moves behind picture (Intermediate) To reproduce the shape effects on this slide, do the following: On the Home tab, in the Slides group, click Layout, and then click Blank. On the Home tab, in the Drawing group, click Shapes, and then under Rectangles click Rectangle (first option from the left). On the slide, drag to draw a rectangle. Select the rectangle. Under Drawing Tools, on the Format tab, in the Size group, do the following: In the Shape Height box, enter 3.17”. In the Shape Width box, enter 9.5”. Drag the rectangle slightly above the middle of the slide. Select the rectangle. On the Home tab, in the Drawing group, click Arrange, point to Align, and then do the following: Click Align to Slide. Click Align Left. Under Drawing Tools, on the Format tab, in the Shape Styles group, click the next to Shape Outline, and then click No Outline. Under Drawing Tools, on the Format tab, in the Shape Styles group, click the Format Shape dialog box launcher. In the Format Shape dialog box, click Fill in the left pane, select Gradient fill in the Fill pane, and then do the following: In the Type list, select Linear. In the Angle box, enter 0. Under Gradient stops, click Add gradient stops or Remove gradient stops until two stops appear in the slider. Also under Gradient stops, customize the gradient stops as follows: Select the first stop in the slider, and then do the following: In the Position box, enter 0%. Click the button next to Color, and then under Theme Colors click Orange, Accent 6, Darker 50% (fifth row, 10th option from the left). Select the next stop in the slider, and then do the following: In the Position box, enter 100%. Click the button next to Color, and then under Theme Colors click Orange, Accent 6, Darker 25% (fourth row, 10th option from the left). To reproduce the “heading” text box on this slide, do the following: On the Insert tab, in the Text group, select Text Box. On the slide, drag to draw a text box. Enter the heading text, and then select text. On the Home tab, in the Font group, do the following: In the Font list, select Calibri. In the Font Size box, enter 38. Click Bold. Click the arrow next to Font Color, and then under Theme Colors click Orange, Accent 6, Darker 25% (fourth row, 10th option from the left). On the Home tab, in the Paragraph group, click Align Text Left. Drag the text box just above the rectangle, in the right half of the slide. To reproduce the second text box on this slide, do the following: On the Insert tab, in the Text group, click Text Box. On the slide, drag to draw a text box. Enter three lines of text with paragraph breaks, and then select the text. On the Home tab, in the Font group, do the following: In the Font list, select Calibri. In the Font Size list, select 28. Click Bold. Click the arrow next to Font Color, and then under Theme Colors click White, Background 1 (first row, first option from the left). On the Home tab, in the Paragraph group, click the Paragraph dialog box launcher. In the Paragraph dialog box, do the following: On the Indents and Spacing tab, under General, select Left in the Alignment box. Under Spacing, select 12 in the After box. Drag the second text box onto the rectangle, below the “heading” text box. To reproduce the full-color picture on this slide, do the following: On the Insert tab, in the Images group, click Picture. In the Insert Picture dialog box, select a picture, and then click Insert. On the slide, select the picture. Under Picture Tools, on the Format tab, in the Size group, click the Size and Position dialog box launcher. In the Format Picture dialog box, resize or crop the image so that the height is set to 5.08” and the width is set to 2.61”. To crop the picture, click Crop in the left pane, and in the right pane, under Crop position, enter values into the Height, Width, Left, and Top boxes. To resize the picture, click Size in the left pane, and in the right pane, under Size and rotate, enter values into the Height and Width boxes. Also in the Format Picture dialog box, click Glow and Soft Edges in the left pane, and then, in the Glow and Soft Edges pane, do the following: Under Glow, click the button next to Presets, and then click Blue, 5 pt glow Accent color 1 (first row, first option from the left). Click the button next to Color, and then under Theme Colors click White, Background 1 (first row, first option from the left). Drag the full-color picture on top of the rectangle, to the left of the text boxes. On the Home tab, in the Drawing group, click Arrange, point to Align, and then do the following: Click Align to Slide. Click Align Top. To reproduce the second picture on this slide, do the following: On the Insert tab, in the Images group, click Picture. In the Insert Picture dialog box, select the same picture, and then click Insert. On the slide, select the picture. On the slide, select the picture. Under Picture Tools, on the Format tab, in the Size group, click the Size and Position dialog box launcher. In the Format Picture dialog box, resize or crop the image so that the height is set to 2.44” and the width is set to 2.61”. To crop the picture, click Crop in the left pane, and in the right pane, under Crop position, enter values into the Height, Width, Left, and Top boxes. To resize the picture, click Size in the left pane, and in the right pane, under Size and rotate, enter values into the Height and Width boxes. On the slide, drag the new picture directly below the first one, and then, in the Format Picture dialog box, in the Crop tab, under Picture Position, adjust the Offset X and Offset Y settings to align the content of the two images so that they appear continuous. Also in the Format Picture dialog box, click Picture Corrections in the left pane, and in the Picture Corrections pane, under Brightness and Contrast, do the following: In the Brightness box, enter 70%. In the Contrast box, enter -70%. Select the smaller picture. On the Home tab, in the Drawing group, click Arrange, point to Align, and then do the following: Click Align to Slide. Click Align Bottom. Press and hold CTRL, and then select both pictures. On the Home tab, in the Drawing group, click Arrange, point to Align, and then do the following: Click Align Selected Objects. Click Align Center. To reproduce the animation effects on this slide, do the following: On the slide, select the “heading” text box. On the Animations tab, in the Advanced Animation group, click Add Animation, and then under Entrance click Fade. On the Animations tab, in the Timing group, in the Start list, select With Previous. On the Animations tab, in the Timing group, in the Duration box, enter 2. On the Animations tab, in the Timing group, in the Delay box, enter 1.5. On the slide, select the “heading” text box. On the Animations tab, in the Advanced Animation group, click Add Animation, and then under Motion Paths click Lines. On the Animations tab, in the Timing group, in the Start list, select With Previous. On the Animations tab, in the Timing group, in the Duration box, enter 2. On the Animations tab, in the Animation group, click Effect Options, and then click Left. On the Animations tab, in the Animation group, click Effect Options, and then click Reverse Path Direction. On the slide, select the motion path for the “heading” text box, point to the starting point (green arrow) of the motion path until the cursor becomes a two-headed arrow. Press and hold SHIFT, and then drag the starting point about 1.5” off the left edge of the slide. (Note: If your lines of text are longer than in the example above, you may need to further increase the length of the motion path. ) On the slide, select the second text box. On the slide, select the “heading” text box. On the Animations tab, in the Advanced Animation group, click Add Animation, and then under Entrance click Fade. On the Animations tab, in the Animation group, click the Show Additional Effect Options dialog box launcher. In the Fade dialog box, do the following: On the Effect tab, in the Animate text list, select By Letter. In the % delay between letters box, enter 5. On the Timing tab, in the Start list, select After Previous. In the Duration list, select 0.5 seconds (Very Fast). On the Text Animation tab, in the Group text list, select By 1st Level Paragraphs. To reproduce the background on this slide, do the following: One the Design tab, in the Background group, click Background Styles, and then click Format Background. In the Format Background dialog box, click Fill in the left pane, select Gradient fill in the Fill pane, and then do the following: In the Type list, select Radial. Click the button next to Direction, and then click From Center (third option from the left). Under Gradient stops, click Add gradient stops or Remove gradient stops until three stops appear in the slider. Also under Gradient stops, customize the gradient stops as follows: Select the first stop in the slider, and then do the following: In the Position box, enter 0%. Click the button next to Color, and then under Theme Colors click White, Background 1 (first row, first option from the left). In the Transparency box, enter 0%. Select the next stop in the slider, and then do the following: In the Position box, enter 40%. Click the button next to Color, and then under Theme Colors click White, Background 1 (first row, first option from the left). In the Transparency box, enter 0%. Select the next stop in the slider, and then do the following: In the Position box, enter 100%. Click the button next to Color, click More Colors, and then in the Colors dialog box, on the Custom tab, enter values for Red: 232, Green: 227, and Blue: 216.