PROBLEM 3.99
KNOWN: Dimensions and thermal conductivity of a gas turbine blade. Temperature and convection
coefficient of gas stream. Temperature of blade base and maximum allowable blade temperature.
FIND: (a) Whether blade operating conditions are acceptable, (b) Heat transfer to blade coolant.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3) Adiabatic
blade tip, (4) Negligible radiation.
ANALYSIS: Conditions in the blade are determined by Case B of Table 3.4.
(a) With the maximum temperature existing at x = L, Eq. 3.80 yields
T ( L ) − T∞
Tb − T∞
=
m
=
1
cosh mL
1/ 2
=
( hP/kA
c)
( 250W/m2 ⋅ K × 0.11m/20W/m ⋅ K × 6 ×10−4 m2 )
1/ 2
m = 47.87 m-1 and mL = 47.87 m-1 × 0.05 m = 2.39
From Table B.1, cosh mL = 5.51. Hence,
T ( L=
) 1200 C + (300 − 1200) C/5.51= 1037 C
<
and the operating conditions are acceptable.
(
(b) With M =( hPkA c )1/ 2 Θ b =250W/m 2 ⋅ K × 0.11m × 20W/m ⋅ K × 6 × 10 −4 m 2
) ( −900 C ) =−517W ,
1/ 2

Eq. 3.81 and Table B.1 yield
qf =
M tanh mL =
−517W ( 0.983) =
−508W
Hence, q b =
−q f =
508W
<
COMMENTS: Radiation losses from the blade surface and convection from the tip will contribute to
reducing the blade temperatures.
PROBLEM 3.107
KNOWN: Melting point of solder used to join two long copper rods.
FIND: Minimum power needed to solder the rods.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction along the
rods, (3) Constant properties, (4) No internal heat generation, (5) Negligible radiation
exchange with surroundings, (6) Uniform h, and (7) Infinitely long rods.
PROPERTIES: Table A-1: Copper T =
393 W/m ⋅ K.
( 250 + 30 ) / 2 C ≈ 400K: k =
ANALYSIS: The junction must be maintained at 250°C while energy is transferred by
conduction from the junction (along both rods). The minimum power is twice the fin heat rate
for an infinitely long fin,
q min
= 2q
=
f 2 ( hPkA c )
1/ 2
( Tb − T∞ ) .
Substituting numerical values,
1/2

W
W  π

2
q min =
2 10
π × 0.008m ) 393
0.008m ) 
(
(

m⋅K  4

 m2 ⋅ K

( 250 − 30 ) C.
Therefore,
q min = 31.0 W.
COMMENTS: Radiation losses from the rods may be significant, particularly near the
junction, thereby requiring a larger power input to maintain the junction at 250°C.
<
PROBLEM 5.7
KNOWN: Diameter and initial temperature of steel balls cooling in air.
FIND: Time required to cool to a prescribed temperature.
SCHEMATIC:
D = 0. 01 m
h = 25 W/m2• K
ASSUMPTIONS: (1) Negligible radiation effects, (2) Constant properties.
ANALYSIS: Applying Eq. 5.10 to a sphere (Lc = r o /3),
=
Bi
2
hLc h ( ro / 3) 25 W/m ⋅ K ( 0.005 m/3)
=
=
= 0.001.
k
k
40 W/m ⋅ K
Hence, the temperature of the steel remains approximately uniform during the cooling
process, and the lumped capacitance method may be used. From Eqs. 5.4 and 5.5,
t
=
t=
(
)
3
Ti − T∞ ρ p D / 6 cp Ti − T∞
ln
ln
=
hAs
T − T∞
T − T∞
hp D 2
ρ Vcp
7800 kg/m3 ( 0.01 m ) 600 J/kg ⋅ K
6 × 25 W/m 2 ⋅ K
ln
1150 − 325
450 − 325
=
t 589
=
s 0.164 h
COMMENTS: Due to the large value of T i , radiation effects are likely to be significant
during the early portion of the transient. The effect is to shorten the cooling time.
<
PROBLEM 5.11
KNOWN: Thickness and properties of flaked food product. Conveyor length. Initial flake
temperature. Ambient temperature and convection heat transfer coefficient. Final product temperature.
FIND: Required conveyor velocities for thick and thin flakes.
SCHEMATIC:
h = 55 W/m2·K
To = 300°C
2L = 1.2 mm
Cereal product
Lo
V
Oven
Conveyor belt
ASSUMPTIONS: (1) Constant properties. (2) Lumped capacitance behavior. (3) Negligible radiation
heat transfer. (4) Negligible moisture evaporation from product. (5) Negligible conduction between
flake and conveyor belt.
PROPERTIES: Flake: = 700 kg/m3, cp = 2400 J/kgK, and k = 0.34 W/mK.
ANALYSIS: The Biot number is
hL 55 W/m2  K  0.6  103 m
Bi 

 0.098
k
0.34 W/m  K
Hence the lumped capacitance assumption is valid. The required heating time is
t
Vc
hAs
ln
i  Lc i 700 kg/m3  0.6 103 m  2400 J/kg  K  20  300 

ln 
ln
 23 s

h

 220  300
55 W/m2  K
<
Therefore the required conveyor velocity is V = Lo/t = 3m/23s = 0.13 m/s .
If the flake thickness is reduced to 2L = 1 mm, the lumped capacitance approximation remains valid
<
and the heating time is 19 s. The associated conveyor velocity is 0.16 m/s .
COMMENTS: (1) Assuming large surroundings, a representative value of the radiation heat transfer




coefficient is hr   Ti  To  Ti2  To2  5.67  108 W/m2  K4  293  573 2932  5732 K4  20.3
W/m2K. Radiation heat transfer would be significant and would serve to increase the product heating
rate, increasing the allowable conveyor belt speed. (2) The food product is likely to enter the oven in a
moist state. Additional thermal energy would be required to remove the moisture during heating,
reducing the rate at which the product temperature increases. (3) The effects noted in Comments 1 and
2 would tend to offset each other. A detailed analysis would be required to assess the impact of
radiation and evaporation on the required conveyor velocity.
PROBLEM 5.29
KNOWN: Droplet properties, diameter, velocity and initial and final temperatures.
FIND: Travel distance and rejected thermal energy.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Negligible radiation from space.
3
PROPERTIES: Droplet (given): ρ = 885 kg/m , c = 1900 J/kg⋅K, k = 0.145 W/m⋅K, ε =
0.95.
ANALYSIS: To assess the suitability of applying the lumped capacitance method, use
Equation 1.9 to obtain the maximum radiation coefficient, which corresponds to T = T i .
h r = εσ Ti3 = 0.95 × 5.67 ×10−8 W/m 2 ⋅ K 4 ( 500 K ) = 6.73 W/m 2 ⋅ K.
3
Hence
h r ( ro / 3)
=
Bi r =
k
−3 m/3
)
( 6.73 W/m2 ⋅ K )( 0.25 ×10=
0.145 W/m ⋅ K
0.0039
and the lumped capacitance method can be used. From Equation 5.19,
(
(
)
3
L ρ c π D / 6  1
1 
=
t =
−
V 3ε π D 2 σ  T3 T3 
i 
 f
L
)
( 0.1 m/s ) 885 kg/m3 (1900 J/kg ⋅ K ) 0.5 ×10-3 m 
1
1  1


 3003 5003  K 3
18 × 0.95 × 5.67 ×10-8 W/m 2 ⋅ K 4
<
L = 2.52 m.
The amount of energy rejected by each droplet is equal to the change in its internal energy.
5 ×10−4 m )
(
885 kg/m3π
1900 J/kg ⋅ K ( 200 K )
3
Ei − =
E f ρ Vc ( Ti − T=
f)
Ei − E f =
0.022 J.
6
<
COMMENTS: Because some of the radiation emitted by a droplet will be intercepted by
other droplets in the stream, the foregoing analysis overestimates the amount of heat
dissipated by radiation to space.