Heat Transfer –
Practical Lecture 7 (Solved Problems)
58. The temperature profile in the thermal boundary layer of a flow over a heated flat surface is given by the expression below, where y is the normal distance to the surface, the Prandtl number is 0.7, T∞ = 400 K, Ts = 300 K, andu∞/ν = 5000 m−1. Determine the convection coefficient.
T −Ts
T∞−Ts = 1−exp
−P ru∞y ν
(1)
Solution:
The local surface heat flux – heat flux calculated at a surface exposed to bulk fluid motion in which the surface and free stream temperatures (Ts and T∞) differ – is computed according to the Fourier’s law – see Equation (2).
qs00 =−kf ∂T
∂y y=0
(2) The Fourier equation is suitable for the calculation of the surface heat flux because at the surface y= 0 the fluid velocity is zero and heat is transferred exclusively by diffusion (conduction). In Equation (2), kf is the fluid thermal conductivity and y in the coordinate direction normal to the surface.
Similarly, the Newton’s law of cooling can be applied to calculate the local surface heat flux – see Equation (3).
qs00=h(Ts−T∞) (3)
The local convection heat transfer coefficient,h, can be calculated equating the right-hand sides of Equations (2) and (3) as follows – see Equation (4).
h= −kf ∂T /∂y|y=0
(Ts−T∞) (4)
The wall temperature gradient (temperature gradient at the surface), ∂T /∂y|y=0 can be evalu- ated considering the provided temperature distribution in the thermal boundary layer (Equation (1)). Equation (1) can be written as follows:
T (y) = (T∞−Ts) h
1−exp
−P ru∞y ν
i
+Ts (5)
Taking into account Equation (5), the wall temperature gradient is calculated in accordance to Equation (6).
∂T
∂y y=0
=−(T∞−Ts)
−P ru∞
ν
exp
−P ru∞y ν
y=0
⇔
⇔ ∂T
∂y y=0
=P ru∞
ν (T∞−Ts)
(6)
Finally, replacing Equation (6) in Equation (4), the expression to compute the local convection heat transfer coefficient is obtained – see Equation (7).
h= −kf ∂T /∂y|y=0
(Ts−T∞) ⇔h= −kfP r(u∞/ν) (T∞−Ts) (Ts−T∞) ⇔
⇔ h=kfP ru∞
ν
(7)
61. A bearing operates at 3600 rpm and is lubricated with oil with the following properties: ρ = 800 kg m−3 , ν = 10−5m s−2 and k = 0.13 W m−1K−1. The journal diameter is 75 mm and the spacing between the journal and the bearing is 0.25 mm.
(a) Determine the temperature distribution in the oil assuming that there is no heat transfer to the journal and that the lubricated surface of the bearing is maintained at 75◦C.
Solution:
The following figure presents a schematic representation of the journal bearing – shaft (journal) rotating in a hole. The bear clearance is filled by an oil film layer of thickness h. The right inset of this figure shows the coordinate system (and axis origin) considered for the solution procedure. End (border) effects are neglected in the current analysis – the same is to say that the bearing sleeve is very long, and consequently, velocity and temperature gradients in the oil film are negligible along the perpendicular direction to the 2D (x, y) coordinate system.
The boundary layer equations allow the determination of the velocity and temperature fields in the velocity and thermal, respectively, boundary layers. Knowing the velocity and temperature fields in the boundary layers the surface shear stress (τs =µ ∂u/∂y|y=0) and the surface heat flux (qs00=−kf ∂T /∂y|y=0) can be computed.
For a steady-state, two-dimensional flow of an incompressible fluid with constant properties (ρ, µ, k, and cp), the laminar boundary layer equations – derived from the convection transfer equations (developed through the application of the laws of mass and energy conservation and Newton’s second law of motion to a differential control volume located in the flow) after applying the corresponding simplifications (boundary layer approximations) – are written as follows. Equations (8), (9), and (10) are the continuity, momentum, and thermal energy equations, respectively.
∂u
∂x +∂v
∂y = 0 (8)
u∂u
∂x +v∂u
∂y =−1 ρ
dp
dx +ν∂2u
∂y2 (9)
u∂T
∂x +v∂T
∂y =α∂2T
∂y2 + ν cp
∂u
∂y 2
(10) Since the fluid properties are constant (do not depend on temperature), the solution for the velocity components (u(x, y) and v(x, y)) can be computed through a decoupled fashion in relation to the thermal energy equation – the velocity components are determined exclu- sively through the solution of the continuity and momentum equations (Equations (8) and (9)), i.e, without considering the thermal energy equation. The temperature field (T(x, y)) is calculated through the solution of the thermal energy equation taking into account the velocity field previously determined.
Assuming a fully developed flow in the bearing clearance, the velocity and temperature gra- dients along thex-coordinate direction are negligible and neglecting the pressure gradient, Equations (8) to (10) are simplified as follows.
dv
dy = 0 (11)
vdu
dy =νd2u
dy2 (12)
vdT
dy =αd2T dy2 + ν
cp du
dy 2
(13) Equations (11) to (13) are subjected to the boundary conditions given by Equations (14) to (18) – see the coordinate system in the previous figure. Equations (14) and (15) are imposed by the no-slip condition at the static housing (bearing) surface. U corresponds to the (linear) velocity of the rotating shaft (journal) surface and T0 to the bearing surface temperature.
u(y= 0) = 0 (14)
v(y = 0) = 0 (15)
T(y= 0) =T0 (16)
u(y=h) = U (17)
dT dy y=h
= 0 (18)
Integrating Equation (11) and considering Equation (15) to obtain the integration constant, it follow that:
v(y) = 0 (19)
Considering v = 0 (Equation (19)), Equation (12) can be simplified in accordance to Equation (20).
d2u
dy2 = 0 (20)
Integrating twice Equation (20) results in the general solution forugiven by Equation (21).
u(y) =C1y+C2 (21)
The integration constants (C1 and C2) are obtained by applying the corresponding bound- ary conditions – Equations (14) and (17) – and the (particular) solution for u is given by Equation (22).
u(y) =y h
U (22)
Substituting Equations (19) and (22) in Equation (13) and considering α=k/(ρcp) (ther- mal diffusivity) and ν=µ/ρ (kinematic viscosity) the following equation is obtained – see Equation (23).
d2T
dy2 =−µ k
U h
2
(23) The integration of Equation (23) provides the general solution for the temperature distri- bution in the the thermal boundary layer – see Equation (24).
T(y) =−µ k
U h
2
y2
2 +C1y+C2 (24)
The integration constants are evaluated through the application of the boundary conditions for the thermal energy equation – Equations (16) and (18). The integration constants C1 and C2 are given by Equations (25) and (26), respectively.
C1 = µ k
U2
h (25)
C2 =T0 (26)
Considering the evaluated integration constants (Equations (25) and (26)) in the general solution (Equation (24)), the particular solution for the temperature field in the boundary layer is obtained – see Equation (27).
T(y) = µ kU2
y h − 1
2 y
h 2
+T0 (27)
The (linear) velocity at the journal surface, U[m s−1], can be computed with Equation (28), whereω[rad s−1] andD[m] corresponds to the journal angular velocity and diameter, respectively.
U =ω D
2
(28) The journal angular velocity, ω[rad s−1], is computed with the number of shaft revolution per minute (N[rpm]) according to Equation (29).
ω = 2πN
60 (29)
Finally, Equation (30) is obtained by considering Equations (28) and (29) in Equation (27).
T(y) = µ k
πDN 60
2 y h −1
2 y
h 2
+T0 (30)
The following figure presents the solution for the velocity x-component distribution along the oil film thickness (u(y), 0≤y≤h) – solution for the velocity boundary layer equations (Equations (11) and (12)) – and the solution for the oil temperature distribution (T(y), 0≤ y ≤ h) – solution for the velocity and thermal boundary layer equations (Equations (11) to (13)). Consequently, the left (right) inset of the figure was computed applying Equation (22) (Equation (30)) with the geometrical and thermophysical properties provided by the problem statement.
Note that the boundary conditions for the velocity x-component (given by Equations (14) and (17)) as well as the boundary conditions for the temperature distribution (given by Equations (16) and (18)) are clearly satisfied in the previous figure.
(b) What is the heat transfer rate for the bearing and what power is required to keep the bearing running?
Solution:
• Heat transfer rate for the bearing
The heat transfer rate from the bearing surface to the oil (in the positive y-direction) per unit sleeve length is computed with the Fourier’s law applied at the surface of the bearing where the heat transfer is driven exclusively by diffusion/conduction (no macroscopic/bulk fluid motion at the bearing surface due to the no-slip boundary condition) – see Equation (31). The temperature gradient at the surface of the bearing is computed with the temperature distribution in the oil determined in the previous question – see Equation (30).
q
L =−kA L
dT dy y=0
⇔ q
L =−k[2π(D/2 +h)] dT dy y=0
⇔
⇔ q
L =−[2π(D/2 +h)]µ
ω D
2 2
1 h ⇔ q
L =−[2π×(0.075/2 + 0.00025)]× 10−5×800
×
(3600×2π/60)×
0.075 2
2
× 1 0.00025 ⇔
⇔ q
L ≈ −1516.951 W m−1
(31)
The negative value for the evaluated heat transfer rate per unit sleeve length (Equa- tion (31)) implies that the heat is being transferred from the oil to the bearing surface and not from the bearing surface to the oil. Consequently, the heat transfer rate for the bearing is given by Equation (32).
q
L ≈1516.951 W m−1 (32)
• Power required to keep the bearing running
The power per unit sleeve length that is required to keep the bearing running is computed with Equation (33). In this equation, T corresponds to the torque that is equal to the product between the tangential force (surface shear stress times surface area) and the shaft radius.
P L = T
Lω⇔ P L =τ
A L
Rω ⇔ P
L =µ du dy y=h
2πRL L
R
2πN 60
⇔
⇔ P
L =ρνU h
(2πR)2N
60 ⇔ P
L =ρνωR h
(2πR)2N
60 ⇔
⇔ P
L =ρν2πN R 60h
(2πR)2N
60 ⇔ P
L = ρν
3600hN2(πD)3 ⇔
⇔ P
L = 800×10−5
3600×0.00025 ×36002×(π×0.075)3 ⇔ P
L ≈1506.905 W m−1
(33)
