Example 1: Water flows at a velocity of 1 m/s over a plane surface 0.6 m wide and 1 m long. Calculate the total drag force acting on the surface if the transition from streamline to turbulent flow in the boundary layer occurs when the Reynolds group Rexc = 105, Solution: حفظ Us=1m/s W= 0.6m, L= Im RC-10 R 60037 R R [0·646Rx-0037R Rex-puix A total Drag friction.R A=WxL 06x10.6m² "Tu by xSul's xlrea 2)Total ρυξ فقط 1 0.037 Rex Rex 1000 ← pxus Xx → حفظ لا يعطى -0.2 + -[0.646Rex-0.037 Rexco.³] . (41) ๆ long 10s 103 × 1 × 1< Rex = 103 =106 س الزوجه الماء 10-3 R 1 2) Total pus = 0.037 × (106)-0.2 + 106 [0.646(105) 0.5-0.037(105) 0.8] R Total = 0.00214 pus R Total drag force = Total x pux Area DR pu تتخلصي من المقام رقص 8 = 0.00214 × 10³ × (1)² × 1 × 0.6 = 1.28 N Total friction Total drag force 8=2 fus طول والعرض X- A=WxL' June = 0·6×1 =0.6m² ند الرفع تقرب الاسس الاكبر Pus→ 10x (1)² = 1000 total Dra R ·x Pus x Avea Pu Example 2: Calculate the thickness of the boundary layer at a distance of 150 mm from the leading edge of a surface over which blood, of viscosity 0.05 N s/m² and density 1000 kg/m³, flows with a velocity of 0.3 m/s. What is the displacement thickness of the boundary layer? Solution: Us Rex-Pusx M exus xx 103 x 0.3 × 150 × 10-3 Rex = Laminar = 900 5. μ 0.05 hannar 5-4.64x Rex 0=0.375 × 0 -0.5 (22) Thickness of Boundary 8 = 4.64xRex -0.5 -0.5 S = 4.64 × 150 × 10¯³ × (900)¯0. = 0.0232 m = 23.2 mm 0.15 Displacement thickness 8* = 0.375 x 8 = 0.375 x 23.2 = 8.7 mm Displacement Thickness of Boundary layer
Example 1: Water flows at a velocity of 1 m/s over a plane surface 0.6 m wide and 1 m long. Calculate the total drag force acting on the surface if the transition from streamline to turbulent flow in the boundary layer occurs when the Reynolds group Rexc = 105, Solution: حفظ Us=1m/s W= 0.6m, L= Im RC-10 R 60037 R R [0·646Rx-0037R Rex-puix A total Drag friction.R A=WxL 06x10.6m² "Tu by xSul's xlrea 2)Total ρυξ فقط 1 0.037 Rex Rex 1000 ← pxus Xx → حفظ لا يعطى -0.2 + -[0.646Rex-0.037 Rexco.³] . (41) ๆ long 10s 103 × 1 × 1< Rex = 103 =106 س الزوجه الماء 10-3 R 1 2) Total pus = 0.037 × (106)-0.2 + 106 [0.646(105) 0.5-0.037(105) 0.8] R Total = 0.00214 pus R Total drag force = Total x pux Area DR pu تتخلصي من المقام رقص 8 = 0.00214 × 10³ × (1)² × 1 × 0.6 = 1.28 N Total friction Total drag force 8=2 fus طول والعرض X- A=WxL' June = 0·6×1 =0.6m² ند الرفع تقرب الاسس الاكبر Pus→ 10x (1)² = 1000 total Dra R ·x Pus x Avea Pu Example 2: Calculate the thickness of the boundary layer at a distance of 150 mm from the leading edge of a surface over which blood, of viscosity 0.05 N s/m² and density 1000 kg/m³, flows with a velocity of 0.3 m/s. What is the displacement thickness of the boundary layer? Solution: Us Rex-Pusx M exus xx 103 x 0.3 × 150 × 10-3 Rex = Laminar = 900 5. μ 0.05 hannar 5-4.64x Rex 0=0.375 × 0 -0.5 (22) Thickness of Boundary 8 = 4.64xRex -0.5 -0.5 S = 4.64 × 150 × 10¯³ × (900)¯0. = 0.0232 m = 23.2 mm 0.15 Displacement thickness 8* = 0.375 x 8 = 0.375 x 23.2 = 8.7 mm Displacement Thickness of Boundary layer
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![Example 1: Water flows at a velocity of 1 m/s over a plane surface 0.6 m wide and 1 m long.
Calculate the total drag force acting on the surface if the transition from streamline to turbulent
flow in the boundary layer occurs when the Reynolds group Rexc = 105,
Solution:
حفظ
Us=1m/s
W= 0.6m, L= Im
RC-10
R
60037 R R [0·646Rx-0037R
Rex-puix
A
total Drag friction.R
A=WxL 06x10.6m²
"Tu by xSul's xlrea
2)Total
ρυξ
فقط
1
0.037 Rex Rex
1000
←
pxus Xx
→
حفظ
لا يعطى
-0.2
+ -[0.646Rex-0.037 Rexco.³]
. (41)
ๆ
long
10s
103 × 1 × 1<
Rex =
103
=106
س الزوجه الماء
10-3
R
1
2) Total
pus
= 0.037 × (106)-0.2 +
106
[0.646(105) 0.5-0.037(105) 0.8]
R
Total
= 0.00214
pus
R
Total drag force
=
Total x pux Area
DR
pu
تتخلصي من المقام رقص 8
= 0.00214 × 10³ × (1)² × 1 × 0.6 = 1.28 N
Total friction
Total drag force
8=2
fus
طول والعرض
X-
A=WxL'
June
= 0·6×1
=0.6m²
ند الرفع تقرب الاسس
الاكبر
Pus→ 10x (1)²
= 1000
total Dra
R
·x Pus x Avea
Pu
Example 2: Calculate the thickness of the boundary layer at a distance of 150 mm from the
leading edge of a surface over which blood, of viscosity 0.05 N s/m² and density 1000 kg/m³,
flows with a velocity of 0.3 m/s. What is the displacement thickness of the boundary layer?
Solution:
Us
Rex-Pusx
M
exus xx
103 x 0.3 × 150 × 10-3
Rex =
Laminar
= 900
5.
μ
0.05
hannar
5-4.64x Rex
0=0.375 × 0
-0.5
(22)
Thickness of Boundary
8 = 4.64xRex -0.5
-0.5
S = 4.64 × 150 × 10¯³ × (900)¯0. = 0.0232 m = 23.2 mm
0.15
Displacement thickness
8* = 0.375 x 8 = 0.375 x 23.2 = 8.7 mm
Displacement Thickness of Boundary layer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd93f3317-68aa-4872-b916-bf20b3cd49f7%2Ff309f7a9-72aa-4506-a9fb-c459f2b64ad0%2Flj44w7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 1: Water flows at a velocity of 1 m/s over a plane surface 0.6 m wide and 1 m long.
Calculate the total drag force acting on the surface if the transition from streamline to turbulent
flow in the boundary layer occurs when the Reynolds group Rexc = 105,
Solution:
حفظ
Us=1m/s
W= 0.6m, L= Im
RC-10
R
60037 R R [0·646Rx-0037R
Rex-puix
A
total Drag friction.R
A=WxL 06x10.6m²
"Tu by xSul's xlrea
2)Total
ρυξ
فقط
1
0.037 Rex Rex
1000
←
pxus Xx
→
حفظ
لا يعطى
-0.2
+ -[0.646Rex-0.037 Rexco.³]
. (41)
ๆ
long
10s
103 × 1 × 1<
Rex =
103
=106
س الزوجه الماء
10-3
R
1
2) Total
pus
= 0.037 × (106)-0.2 +
106
[0.646(105) 0.5-0.037(105) 0.8]
R
Total
= 0.00214
pus
R
Total drag force
=
Total x pux Area
DR
pu
تتخلصي من المقام رقص 8
= 0.00214 × 10³ × (1)² × 1 × 0.6 = 1.28 N
Total friction
Total drag force
8=2
fus
طول والعرض
X-
A=WxL'
June
= 0·6×1
=0.6m²
ند الرفع تقرب الاسس
الاكبر
Pus→ 10x (1)²
= 1000
total Dra
R
·x Pus x Avea
Pu
Example 2: Calculate the thickness of the boundary layer at a distance of 150 mm from the
leading edge of a surface over which blood, of viscosity 0.05 N s/m² and density 1000 kg/m³,
flows with a velocity of 0.3 m/s. What is the displacement thickness of the boundary layer?
Solution:
Us
Rex-Pusx
M
exus xx
103 x 0.3 × 150 × 10-3
Rex =
Laminar
= 900
5.
μ
0.05
hannar
5-4.64x Rex
0=0.375 × 0
-0.5
(22)
Thickness of Boundary
8 = 4.64xRex -0.5
-0.5
S = 4.64 × 150 × 10¯³ × (900)¯0. = 0.0232 m = 23.2 mm
0.15
Displacement thickness
8* = 0.375 x 8 = 0.375 x 23.2 = 8.7 mm
Displacement Thickness of Boundary layer
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