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Chapter 7, Solution 36.
Air is flowing over a long flat plate with a specified velocity. The distance from the
leading edge of the plate where the flow becomes turbulent, and the thickness of the
boundary layer at that location are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is
Recr = 5×105. 3 Air is an ideal gas. 4 The surface of the plate is smooth.
Properties The density and kinematic viscosity of air at 1 atm and 25°C are ρ = 1.184
kg/m3 and ν = 1.562×10–5 m2/s (Table A-15).
Analysis The distance from the leading edge of the plate where the flow becomes
turbulent is the distance xcr where the Reynolds number becomes equal to the critical
Reynolds number,
Re cr =
x cr =
Vx cr
ν
→
ν Re cr
V
=
(1.562 × 10 −5 m 2 /s)(5 × 10 5 )
= 0.976 m
8 m/s
The thickness of the boundary layer at that location
is obtained by substituting this value of x into the
laminar boundary layer thickness relation,
δx =
5x
Re 1x/ 2
→ δ cr =
5 x cr
Re 1cr/ 2
=
5(0.976 m)
(5 × 10 5 )1 / 2
V
xcr
= 0.006903 m = 0.69 cm
Discussion When the flow becomes turbulent, the boundary layer thickness starts to
increase, and the value of its thickness can be determined from the boundary layer
thickness relation for turbulent flow.
Chapter 7, Solution 37.
Water is flowing over a long flat plate with a specified velocity. The distance from the
leading edge of the plate where the flow becomes turbulent, and the thickness of the
boundary layer at that location are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is
Recr = 5×105. 3 The surface of the plate is smooth.
Properties The density and dynamic viscosity of water at 1 atm and 25°C are ρ = 997
kg/m3 and μ = 0.891×10–3 kg/m⋅s (Table A-9).
V
Analysis The distance from the leading edge of the
plate where the flow becomes turbulent is the distance
xcr where the Reynolds number becomes equal to the
xcr
critical Reynolds number,
Re cr =
x cr =
ρVx cr
→
μ
μ Re cr (0.891× 10 −3 kg/m ⋅ s)(5 × 10 5 )
=
= 0.056 m = 5.6 cm
ρV
(997 kg/m 3 )(8 m/s)
The thickness of the boundary layer at that location is obtained by substituting this value
of x into the laminar boundary layer thickness relation,
δ cr =
5x
Re 1x/ 2
→ δ cr =
5 x cr
Re 1cr/ 2
=
5(0.056 m)
(5 × 10 5 )1 / 2
= 0.00040 m = 0.4 mm
Therefore, the flow becomes turbulent after about 5 cm from the leading edge of the
plate, and the thickness of the boundary layer at that location is 0.4 mm.
Discussion When the flow becomes turbulent, the boundary layer thickness starts to
increase, and the value of its thickness can be determined from the boundary layer
thickness relation for turbulent flow.
Chapter 7, Solution 51.
The wind is blowing across the wire of a transmission line. The surface temperature of
the wire is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air
is an ideal gas with constant properties. 4 The local atmospheric pressure is 1 atm.
Properties We assume the film temperature to be
10°C. The properties of air at this temperature are
Transmission
(Table A-15)
ρ = 1.246 kg/m 3
Wind
V = 40 km/h
T∞ = 10°C
k = 0.02439 W/m.°C
ν = 1.426 × 10 -5 m 2 /s
Pr = 0.7336
Analysis The Reynolds number is
Re =
VD
ν
=
[(40 ×1000/3600) m/s](0.006 m) = 4675
1.426 ×10 −5 m 2 /s
The Nusselt number corresponding to this Reynolds number
is determined to be
hD
0.62 Re 0.5 Pr 1 / 3
Nu =
= 0.3 +
1/ 4
k
1 + (0.4 / Pr )2 / 3
[
]
⎡ ⎛ Re ⎞ 5 / 8 ⎤
⎢1 + ⎜⎜
⎟⎟ ⎥
⎢⎣ ⎝ 282,000 ⎠ ⎥⎦
4/5
5/8
0.62(4675) 0.5 (0.7336)1 / 3 ⎡ ⎛ 4675 ⎞ ⎤
⎢1 + ⎜⎜
= 0.3 +
⎟
⎟ ⎥
1/ 4
⎢⎣ ⎝ 282,000 ⎠ ⎥⎦
1 + (0.4 / 0.7336)2 / 3
[
The heat transfer coefficient is
]
4/5
= 36.0
wire, Ts
D = 0.6 cm
h=
k
0.02439 W/m.°C
Nu =
(36.0) = 146.3 W/m 2 .°C
D
0.006 m
The rate of heat generated in the electrical transmission lines per meter length is
W& = Q& = I 2 R = (50 A) 2 (0.002 Ohm) = 5.0 W
The entire heat generated in electrical transmission line has to be transferred to the
ambient air. The surface temperature of the wire then becomes
As = πDL = π (0.006 m)(1 m) = 0.01885 m 2
Q&
5W
⎯→ Ts = T∞ +
= 10°C +
= 11.8°C
Q& = hAs (Ts − T∞ ) ⎯
2
hAs
(146.3 W/m .°C)(0.01885 m 2 )
Chapter 8, Solution 1C.
Liquids are usually transported in circular pipes because pipes with a circular crosssection can withstand large pressure differences between the inside and the outside
without undergoing any distortion.
Chapter 8, Solution 2C.
Reynolds number for flow in a circular tube of diameter D is expressed as
Re =
Vavg D
where V∞ =
ν
4m&
m&
m&
=
=
2
ρAc ρ (πD / 4) ρπD 2
and ν =
μ
ρ
Substituting,
Re =
Vavg D
ν
=
4m& D
ρπD ( μ / ρ )
2
=
4m&
πDμ
m, Vavg
Chapter 8, Solution 3C.
Engine oil requires a larger pump because of its much larger density.
Chapter 8, Solution 4C.
The generally accepted value of the Reynolds number above which the flow in a smooth
pipe is turbulent is 4000.
Chapter 8, Solution 5C.
For flow through non-circular tubes, the Reynolds number as well as the Nusselt
number and the friction factor are based on the hydraulic diameter Dh defined as
Dh =
4 Ac
p
where Ac is the cross-sectional area of the tube and p is its perimeter. The
hydraulic diameter is defined such that it reduces to ordinary diameter D for circular
tubes since Dh =
4 Ac 4πD 2 / 4
=
= D.
p
πD
Chapter 8, Solution 6C.
The region from the tube inlet to the point at which the boundary layer merges at the
centerline is called the hydrodynamic entry region, and the length of this region is called
hydrodynamic entry length. The entry length is much longer in laminar flow than it is in
turbulent flow. But at very low Reynolds numbers, Lh is very small (Lh = 1.2D at Re =
20).
Chapter 8, Solution 7C.
The friction factor is highest at the tube inlet where the thickness of the boundary layer is
zero, and decreases gradually to the fully developed value. The same is true for turbulent
flow.
Chapter 8, Solution 8C.
In turbulent flow, the tubes with rough surfaces have much higher friction factors than
the tubes with smooth surfaces. In the case of laminar flow, the effect of surface
roughness on the friction factor is negligible.
Chapter 8, Solution 9C.
The friction factor f remains constant along the flow direction in the fully developed
region in both laminar and turbulent flow.
Chapter 8, Solution 10C.
The fluid viscosity is responsible for the development of the velocity boundary layer. For
the idealized inviscid fluids (fluids with zero viscosity), there will be no velocity
boundary layer.
Chapter 8, Solution 11C.
The number of transfer units NTU is a measure of the heat transfer area and
effectiveness of a heat transfer system. A small value of NTU (NTU < 5) indicates more
opportunities for heat transfer whereas a large NTU value (NTU >5) indicates that heat
transfer will not increase no matter how much we extend the length of the tube.
Chapter 8, Solution 12C.
The logarithmic mean temperature difference ΔTln is an exact representation of the
average temperature difference between the fluid and the surface for the entire tube. It
truly reflects the exponential decay of the local temperature difference. The error in using
the arithmetic mean temperature increases to undesirable levels when ΔTe differs from
ΔTi by great amounts. Therefore we should always use the logarithmic mean
temperature.
Chapter 8, Solution 13C.
The region of flow over which the thermal boundary layer develops and reaches the tube
center is called the thermal entry region, and the length of this region is called the thermal
entry length. The region in which the flow is both hydrodynamically (the velocity profile
is fully developed and remains unchanged) and thermally (the dimensionless temperature
profile remains unchanged) developed is called the fully developed region.
Chapter 8, Solution 14C.
The heat flux will be higher near the inlet because the heat transfer coefficient is highest
at the tube inlet where the thickness of thermal boundary layer is zero, and decreases
gradually to the fully developed value.
Chapter 8, Solution 15C.
The heat flux will be higher near the inlet because the heat transfer coefficient is highest
at the tube inlet where the thickness of thermal boundary layer is zero, and decreases
gradually to the fully developed value.
Chapter 8, Solution 16C.
In the fully developed region of flow in a circular tube, the velocity profile will not
change in the flow direction but the temperature profile may.
Chapter 8, Solution 17C.
The hydrodynamic and thermal entry lengths are given as Lh = 0.05 Re D and
Lt = 0.05 Re Pr D for laminar flow, and L h ≈ Lt ≈ 10 D in turbulent flow. Noting that Pr >> 1
for oils, the thermal entry length is larger than the hydrodynamic entry length in laminar
flow. In turbulent, the hydrodynamic and thermal entry lengths are independent of Re or
Pr numbers, and are comparable in magnitude.
Chapter 8, Solution 18C.
The hydrodynamic and thermal entry lengths are given as Lh = 0.05 Re D and
Lt = 0.05 Re Pr D for laminar flow, and L h ≈ Lt ≈ 10 D in turbulent flow. Noting that Pr << 1
for liquid metals, the thermal entry length is smaller than the hydrodynamic entry length
in laminar flow. In turbulent, the hydrodynamic and thermal entry lengths are
independent of Re or Pr numbers, and are comparable in magnitude.
Chapter 8, Solution 19C.
In fluid flow, it is convenient to work with an average or mean velocity Vavg and an
average or mean temperature Tm which remain constant in incompressible flow when the
cross-sectional area of the tube is constant. The Vavg and Tm represent the velocity and
temperature, respectively, at a cross section if all the particles were at the same velocity
and temperature.
Chapter 8, Solution 20C.
When the surface temperature of tube is constant, the appropriate temperature difference
for use in the Newton's law of cooling is logarithmic mean temperature difference that
can be expressed as
ΔTln =
ΔTe − ΔTi
ln(ΔTe / ΔTi )