Example Problems (Convection)
1474./p.137 – Faires
The main trunk duct of an air-conditioning system is rectangular in cross section (16 x 30 in.) and
has air at 15 psia and 40 oF flowing through it with a velocity of 1 400 fpm. Find hi.
Given: rectangular air-conditioning duct
dimension = 16 x 30 in.
p = 15 psia
T = 40 oF
V = 40 ft/min
air
Required:
hi = ?
Solution:
Calculate first the Reynolds number to determine the flow regime. Since the shape of the duct is
rectangular, the equivalent diameter De will be used and is given as
2𝑎𝑏
𝐷𝑒 = 𝑎+𝑏
𝐷𝑒 =
2(16)(30)
16+30
𝐷𝑒 = 20.9 𝑖𝑛.
oF,
From Fig. 19/18, p. 544, Faires, at T = 40
𝑙𝑏𝑚
𝜇 = 1.25 𝑥 10−5 𝑓𝑡−𝑠𝑒𝑐
Using the ideal gas equation, the density can be determined.
𝑝
𝜌 = 𝑅𝑇
𝑙𝑏𝑓
𝜌=
144𝑖𝑛2
(15 2 )(
)
𝑖𝑛
𝑓𝑡2
𝑓𝑡−𝑙𝑏𝑓
)(40+460)°𝑅
𝑙𝑏𝑚 −°𝑅
(53.342
𝑙𝑏
𝜌 = 0.081 𝑓𝑡𝑚3
There the Reynolds number is
𝑅𝑒 =
𝑅𝑒 =
𝐷𝑒 𝑣𝜌
𝜇
1𝑓𝑡
𝑓𝑡
𝑙𝑏
1 𝑚𝑖𝑛
)(
(20.9 𝑖𝑛)(
)(1 400
)(0.081 𝑚
)
12𝑖𝑛
𝑚𝑖𝑛
𝑓𝑡3 60 𝑠𝑒𝑐
𝑙𝑏
1.25 𝑥 10−5 𝑚
𝑓𝑡−𝑠𝑒𝑐
𝑅𝑒 = 263 340 > 2 100
∴ 𝑇𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡
Using equation 19-11, p. 551, Faires,
ℎ𝑖 𝐷
𝑘𝑏
= 0.021 (
𝐷𝑣𝜌 0.8
𝜇
)
𝑏
From Table VII, p. 521, Faires, interpolate to determine the thermal conductivity k of air.
32 - 0.168
40 - k
572 - 0.312
𝐵𝑡𝑢−𝑖𝑛
𝑘 = 0.170 𝑓𝑡 2 −ℎ𝑟−°𝐹
𝐵𝑡𝑢−𝑖𝑛
∴ ℎ𝑖 = (0.021)(263 340)
ℎ𝑖 = 3.71
0.8
0.170 2
𝑓𝑡 −ℎ𝑟−°𝐹
[
]
20.9 𝑖𝑛
𝐵𝑡𝑢
𝑓𝑡 2 −ℎ𝑟−°𝐹
1476. A double-pipe, counterflow heat exchanger contains a 1.5 in. steel pipe (1.90 in. OD, 1.61
in. ID) inside a 2.5 in. steel pipe (2.88 in. OD, 2.47 in. ID). Hot oil with properties similar to those
of oil C, Fig.19/20, is flowing through the inner pipe with a velocity of 3.5 fps and a bulk
temperature of 200 oF; also, cp = 0.52 Btu/lb -oR, k = 0.96 Btu- in/ft2-hr-oF. Cold oil with
properties similar to those of oil D, Fig. 19/18, is flowing through the annular space with a
velocity of 12.5 fps and a bulk temperature of 80 oF; also, cp = 0.52 Btu/lb -oR, k = 0.94 Btu –
in/ft2–hr -oF. (a) Find the film coefficient for the inner and outer surfaces of the inner pipe. (b)
what is U for the inner pipe?
Given: double-pipe counterflow heat exchanger
2.5 in. ∅
1.5 in. ∅
oil D (cold)
oil C (hot)
inner pipe (oil C)
OD = 1.90 in.
ID = 1.61 in.
v = 3.5 fps
Tb = 200 oF
cp = 0.52 Btu/lbm -oR
k = 0.96 Btu-in/ft2-hr-oF
outer pipe (oil D)
OD = 2.88 in.
ID = 2.47 in.
v = 12.5 fps
Tb = 80 oF
cp = 0.52 Btu/lbm -oR
k = 0.94 Btu-in/ft2-hr-oF
Required:
(a) hi = ?
ho = ?
(b) U = ?
(for the inner pipe)
Solution:
For the inner pipe: (oil C inside)
From Fig. 19/20, p. 545, Faires, at Tb = 200 oF,
1
𝑙𝑏𝑚
𝜇 = 6.6 𝑐𝑝 𝑥 1 490 = 4.43 𝑥 10−3 𝑓𝑡−𝑠𝑒𝑐
To determine the density of the oil, consider the example on topic 19.33, p. 554, Faires, where
the specific gravity is calculated first using the formula:
𝑆𝐺𝑜𝑖𝑙 = 𝑆𝐺60 − 0.000 35(𝑇𝑜𝑖𝑙 − 60)
𝑆𝐺200 = 0.899 − 0.000 35(200 − 60)
𝑆𝐺200 = 0.85
𝑙𝑏
∴ 𝜌 = 0.85 (62.4 𝑓𝑡𝑚3 )
𝑙𝑏
𝜌 = 53 𝑓𝑡𝑚3
The Reynolds number can now be computed.
𝐷𝑣𝑝
𝑅𝑒 = 𝜇
𝑅𝑒 =
1 𝑓𝑡
𝑓𝑡
𝑙𝑏
)
)(3.5 )(53 𝑚
12 𝑖𝑛
𝑠𝑒𝑐
𝑓𝑡3
𝑙𝑏
𝑚
4.43 𝑥 10−3
𝑓𝑡−𝑠𝑒𝑐
(1.61 𝑖𝑛)(
𝑅𝑒 = 5 618
>
2 100 ∴ 𝑇𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡
Using equation 19-10, p. 550, Faires,
ℎ𝑖 𝐷
𝑘𝑏
𝑐𝑝 𝜇
where the term (
𝑘
𝐷𝑣𝜌 0.8 𝑐𝑝 𝜇 0.4
= 0.023 (
𝜇
)
(
𝑏
𝑘
)
𝑏
) is the Prandtl number Pr, which is another dimensionless quantity. Solving
for this value,
𝑃𝑟 =
𝐵𝑡𝑢
𝑙𝑏
)(4.43 𝑥 10−3 𝑚 )
𝑙𝑏𝑚 −°𝐹
𝑓𝑡−𝑠𝑒𝑐
𝐵𝑡𝑢−𝑖𝑛
1 𝑓𝑡
1 ℎ𝑟
(0.96 2
)(
)(
)
𝑓𝑡 −ℎ𝑟−°𝐹 12 𝑖𝑛 3 600 𝑠𝑒𝑐
(0.52
𝑃𝑟 = 103.7
The computed value is well within the limits stipulated in the conditions on the use of the
equation, which is 0.7 < Pr < 120. Solving now for the inside film coefficient of the inner pipe,
𝐵𝑡𝑢−𝑖𝑛
ℎ𝑖 = 0.023(5 618)
0.8 (103.7)0.4
0.96 2
𝑓𝑡 −ℎ𝑟−°𝐹
[ 1.61
]
𝑖𝑛
𝐵𝑡𝑢
ℎ𝑖 = 87.73
𝑓𝑡 2 −ℎ𝑟−°𝐹
For the outside film coefficient ho of the inner pipe, oil D flows within the annular space. Using
equation 19-14, p. 556, Faires,
ℎ𝐷
𝑘
𝐷𝑒 𝑣𝜌 0.8 𝑐𝑝 𝜇 𝑛
= 0.023 (
𝜇
)
(
𝑏
𝑘
)
𝑏
where n = 0.4 for heating and n = 0.3 for cooling. The equivalent diameter De = D2 – D1 is used in
the calculation of the Reynolds number.
From Fig. 19/20, p. 545, Faires, for oil D at T = 80 oF,
𝜇 = 128 𝑐𝑝 𝑥
1
1 490
= 0.086
𝑙𝑏𝑚
𝑓𝑡−𝑠𝑒𝑐
The specific gravity is
𝑆𝐺80 = 0.899 − 0.000 35(80 − 60)
𝑆𝐺80 = 0.892
𝑙𝑏
∴ 𝜌 = 0.892 (62.4 𝑓𝑡𝑚3 )
𝑙𝑏
𝜌 = 55.66 𝑓𝑡𝑚3
The equivalent diameter is
𝐷𝑒 = 𝐷2 − 𝐷1
𝐷𝑒 = 2.47 − 1.90
𝐷𝑒 = 0.57 𝑖𝑛.
Solving now for the Reynolds number,
𝑅𝑒 =
𝑅𝑒 =
𝐷𝑒 𝑣𝜌
𝜇
1 𝑓𝑡
𝑓𝑡
𝑙𝑏
(0.57 𝑖𝑛)(
)
)(12.5 )(55.66 𝑚
12 𝑖𝑛
𝑠𝑒𝑐
𝑓𝑡3
𝑙𝑏𝑚
0.086
𝑓𝑡−𝑠𝑒𝑐
𝑅𝑒 = 384
Solve likewise the Prandtl number,
𝑃𝑟 =
𝑃𝑟 =
𝑐𝑝 𝜇
𝑘
𝐵𝑡𝑢
𝑙𝑏
)(0.086 𝑚 )
𝑙𝑏𝑚 −°𝐹
𝑓𝑡−𝑠𝑒𝑐
𝐵𝑡𝑢−𝑖𝑛
1 𝑓𝑡
1 ℎ𝑟
(0.94 2
)(
)(
)
𝑓𝑡 −ℎ𝑟−°𝐹 12 𝑖𝑛 3 600 𝑠𝑒𝑐
(0.52
𝑃𝑟 = 2 055.2
Since oil D is cold, then it is used for cooling the hot oil C, use n = 0.3. Substituting the values
obtained,
𝐵𝑡𝑢−𝑖𝑛
ℎ𝑜 = 0.023(384)
0.8 (2
055.2)
0.3
0.94 2
𝑓𝑡 −ℎ𝑟−°𝐹
[ 1.90
]
𝑖𝑛
𝐵𝑡𝑢
ℎ𝑜 = 13.10 𝑓𝑡 2 −ℎ𝑟−°𝐹
(b) For the over-all coefficient of heat transfer U for the inner pipe,
1
𝑈𝑜 𝐴𝑜 = 𝑈𝑖 𝐴𝑖 =
𝐷𝑜
∑
∑
∑
1
=
𝐴ℎ
1
𝐴ℎ
1
𝑙𝑛( )
𝐷𝑖
1
+∑
𝐴ℎ
2𝜋𝑧𝑘
1
1
𝑖 𝑖
𝑜 ℎ𝑜
=𝐴ℎ +𝐴
1.61
𝐵𝑡𝑢
)
𝜋(
𝑓𝑡)(1 𝑓𝑡)(87.73 2
12
𝑓𝑡 −ℎ𝑟−°𝐹
∑
∑
1
= 0.180 507
𝐴ℎ
𝐷
𝑙𝑛( 𝑜 )
𝐷𝑖
2𝜋𝑧𝑘
∑
=
+
1
1.90
𝜋(
)(1)(13.10)
12
ℎ𝑟−°𝐹
𝐵𝑡𝑢
1.90
)
1.61
𝐵𝑡𝑢−𝑖𝑛
1 𝑓𝑡
)(
2𝜋(1 𝑓𝑡)(0.96 2
)
𝑓𝑡 −ℎ𝑟−°𝐹 12 𝑖𝑛
𝑙𝑛(
𝐷
𝑙𝑛( 𝑜 )
𝐷𝑖
2𝜋𝑧𝑘
= 0.329 490
ℎ𝑟−°𝐹
𝐵𝑡𝑢
Considering the inside area of the inner pipe as the reference area,
𝑈𝑖 =
1
ℎ𝑟−°𝐹
(0.180 507+0.329 490)
𝐵𝑡𝑢
1.61
𝜋(
𝑓𝑡)(1 𝑓𝑡)
12
𝐵𝑡𝑢
𝑈𝑖 = 4.652 𝑓𝑡 2 −ℎ𝑟−°𝐹
If the outside area is the reference area,
𝑈𝑜 =
1
ℎ𝑟−°𝐹
(0.180 507+0.329 490)
𝐵𝑡𝑢
1.90
𝜋(
𝑓𝑡)(1 𝑓𝑡)
12
𝐵𝑡𝑢
𝑈𝑜 = 3.942 𝑓𝑡 2 −ℎ𝑟−°𝐹
18.15/p. 774 – Burghardt
Water with a flow rate of 2.6 kg/s is heated from 10 oC to 24 oC as it passes through a 5 cm pipe.
The inside pipe surface temperature is 95 oC. Determine the pipe length required.
Given: a pipe with flowing water
D = 5 cm
𝑚̇ = 2.6 kg/s
T1 = 10 oC
T2 = 24 oC
Tw = 95 oC
Required:
L = ? (the pipe length)
Solution:
The bulk temperature of water is Tb = (10 + 24)/2 = 17 oC. Determine the properties at this bulk
temperature. From Table A.23, p.834, Burghardt, by interpolation:
273 K
290 K
313 K
999.3
𝜌
998.2
4 226
𝑐𝑝
4 182
0.558
𝑘
0.597
𝑘𝑔
𝜌 = 998.8
𝑚3
𝐽
𝑐𝑝 = 4 207.3 𝑘𝑔−°𝐾
𝑤
𝑘 = 0.575 𝑚−°𝐾
𝜈 = 1.456 𝑥 10−6
𝑚2
𝑠
The velocity of water is
𝑚̇
𝑣 = 𝜌𝐴
𝑣=
𝑘𝑔
𝑠
2.6
𝑘𝑔 𝜋
998.8 3 ( )(0.05 𝑚)2
𝑚 4
𝑣 = 1.33
𝑚
𝑠
Solve for the Reynolds number:
𝑅𝑒 =
𝐷𝑣𝜌
𝜇
1.789
𝜈
1.006
Since 𝜇 = 𝜈𝜌,
𝑅𝑒 =
𝑅𝑒 =
𝐷𝜐𝜌 𝐷𝑣
=
𝜈𝜌
𝜈
𝑚
𝑠
𝑚2
(0.05𝑚)(1.33 )
1.456 𝑥 10−6
𝑠
𝑅𝑒 = 45 673
>
2 100
Using equation 19-10, p. 550, Faires,
ℎ𝑖 𝐷
𝑘𝑏
𝐷𝑣𝜌 0.8 𝑐𝑝 𝜇 0.4
= 0.023 (
𝜇
)
(
𝑏
𝑘
)
𝑏
For the Prandtl number,
𝑃𝑟 =
𝑃𝑟 =
𝑐𝑝 𝜇
𝑘
=
𝑐𝑝 (𝜈𝜌)
𝑘
𝐽
𝑚2
𝑘𝑔
)(998.8 3 )
)(1.456 𝑥 10−6
𝑘𝑔−°𝐾
𝑠
𝑚
𝐽
(0.575
)
𝑠−𝑚−°𝐾
(4 207.3
𝑃𝑟 = 10.6
Therefore,
𝑊
𝑚−°𝐾
0.575
ℎ𝑖 = 0.023(45 673)0.8 (10.6)0.4 [
0.05 𝑚
]
𝑊
ℎ𝑖 = 3 633 𝑚2 −°𝐾
The heat transferred by convection is given by the equation
𝑄𝑐𝑜𝑛 = ℎ𝑖 𝐴(𝑇𝑤 − 𝑇𝑏 ) = 𝑚̇𝑐𝑝 (𝑇2 − 𝑇1 )
𝑄𝑐𝑜𝑛 = ℎ𝑖 (𝜋𝐷𝐿)(𝑇𝑤 − 𝑇𝑏 ) = 𝑚̇𝑐𝑝 (𝑇2 − 𝑇1 )
∴𝐿=
𝐿=
𝑚̇𝑐𝑝 (𝑇2 −𝑇1 )
𝜋ℎ𝑖 𝐷(𝑇𝑤 −𝑇𝑏 )
𝑘𝑔
𝐽
)(4 207.3
)(24−10)°𝐾
𝑠
𝑘𝑔−°𝐾
𝐽
𝜋(3 633
)(0.05 𝑚)(95−17)°𝐾
𝑠−𝑚2 −°𝐾
(2.6
𝐿 = 3.44 𝑚
∴ 𝑇𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡
1478./p. 138 – Faires
An oil heater, with No. 16 BWG steel tubes (0.75 in. OD, 0.620 in. ID) and effective length of 12 ft
per pass, receives oil D, at 70 oF. The oil flows through the tubes with an average velocity of 4
fps; condensing saturated steam at 15 psia surrounds the tubes. For the oil, c p = 0.5 Btu/lb-oF. If
the heater has two passes with 300 tubes per pass, find (a) the temperature of the oil at the end
of first pass, (b) the exit oil temperature, and (c) the amount of steam condensed each hour.
Given: an oil heater (with No. 16 BWG steel tubes)
OD = 0.75 in.
ID = 0.620 in.
L = 12 ft/pass
No. of tubes/pass = 300 tubes
No. of passes = 2
condensing saturated steam at 15 psia
oil D
T1 = 70 oF
v = 4 fps
cp = 0.5 Btu/lb-oF
T2
Required:
(a) T2 = ?
(b) Texit = ?
(c) 𝑚̇s = ? (in lb/hr)
Solution:
Since the temperature T2 of oil D at the end of the first pass is not given, a trial and error solution
will be presented. Assume a value of T2 = 84 oF. Evaluate the properties of oil D based on this
highest temperature. From Fig. 19/20, p. 545, Faires,
1
𝑙𝑏𝑚
𝜇 = 105 𝑐𝑝 𝑥 1 490 = 0.070 𝑓𝑡−𝑠𝑒𝑐
The specific gravity will be determined by using the equation
𝑆𝐺𝑜𝑖𝑙 = 0.899 − 0.000 35(𝑇𝑜𝑖𝑙 − 60)
𝑆𝐺77 = 0.899 − 0.000 35(77 − 60)
𝑆𝐺77 = 0.893
𝑙𝑏
∴ 𝜌 = 0.893 (62.4 𝑓𝑡𝑚3 )
𝑙𝑏
𝜌 = 55.72 𝑓𝑡𝑚3
Calculate the Reynolds number.
𝑅𝑒 =
𝐷𝑣𝜌
𝜇
𝑅𝑒 =
(
0.620
𝑓𝑡
𝑙𝑏
)
𝑓𝑡)(4 )(55.72 𝑚
12
𝑠𝑒𝑐
𝑓𝑡3
𝑙𝑏𝑚
0.070
𝑓𝑡−𝑠𝑒𝑐
𝑅𝑒 = 165
<
2 100
∴ 𝐿𝑎𝑚𝑖𝑛𝑎𝑟
Consider equation 19 – 12, p. 563, Faires,
ℎ𝑖 𝐷
𝑘𝑏
0.14 𝑚𝑐
̇ 1⁄3
𝜇
= 2.02 (𝜇 𝑏 )
𝑤
( 𝑘𝐿 )
𝑏
[𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝐿𝑖𝑞𝑢𝑖𝑑, 𝑅𝑒 < 2 100; 𝑚𝑐
̇ ⁄(𝑘𝐿) > 10]
̇
𝑚𝑐
The term ( 𝑘𝐿 ) is called the Graetz number (Gz). The mass flow rate (lb/hr) will now be
determined.
𝑚̇ = 𝜌𝑣𝐴
𝑙𝑏
𝑓𝑡
2
𝜋 0.62
𝑚̇ = (55.77 𝑓𝑡𝑚3 ) (4 𝑠𝑒𝑐) [ 4 ( 12 𝑓𝑡) ] 𝑥
3 600 𝑠𝑒𝑐
ℎ𝑟
𝑙𝑏
𝑚̇ = 1 683.7 ℎ𝑟𝑚
The value of the thermal conductivity of oil D is taken from Problem 1476. Thus,
𝑘 = 0.94
𝐵𝑡𝑢 − 𝑖𝑛.
𝑓𝑡 2 − ℎ𝑟 − °𝐹
Solving for the Graetz number,
𝐺𝑧 =
𝑙𝑏
𝐵𝑡𝑢
(1 683.7 𝑚 )(0.50
)
ℎ𝑟
𝐵𝑡𝑢−𝑖𝑛.
𝑙𝑏𝑚 −°𝐹
1𝑓𝑡
(0.94 2
)(12 𝑓𝑡)(
)
12 𝑖𝑛.
𝑓𝑡 −ℎ𝑟−°𝐹
𝐺𝑧 = 896
>
10
The dimensionless numbers satisfied the stipulated conditions. Equation 19 – 12, can be used.
Since the wall temperature inside the tube is not given, assume a 5 oF difference between the
inside and outside of the tube. At 15 psia, the saturation temperature taken from the Steam
Tables is Tsat = 213.03 oF. Thus, the inside wall temperature is 𝑇𝑤 = 213.03 − 5 = 208.03 °𝐹.
Corresponding to this temperature, 𝜇𝑤 = 5.6 𝑐𝑝. At the bulk temperature of the oil,
𝑇 +𝑇
𝑇𝑏 = 1 2 2
70+84
𝑇𝑏 = 2 = 77 °𝐹
∴ 𝜇𝑏 = 149 𝑐𝑝
149 0.14
ℎ𝑖 = 2.02 ( 5.6 )
𝐵𝑡𝑢−𝑖𝑛.
(896)1⁄3
0.94 2
𝑓𝑡 −ℎ𝑟−°𝐹
[
0.62 𝑖𝑛.
]
ℎ𝑖 = 46.74
𝐵𝑡𝑢
𝑓𝑡 2 −ℎ𝑟−°𝐹
The heat transfer by convection is determined using the equation
𝑄𝑐𝑜𝑛 = ℎ𝑖 𝐴(𝑇𝑤 − 𝑇𝑏 )
𝐵𝑡𝑢
0.62
𝑄𝑐𝑜𝑛 = (46.74 𝑓𝑡 2 −ℎ𝑟−°𝐹) [𝜋 ( 12 𝑓𝑡) (12
𝑄𝑐𝑜𝑛 = 11 926
𝑓𝑡
𝑡𝑢𝑏𝑒
)] (208 − 77)°𝐹
𝐵𝑡𝑢
ℎ𝑟−𝑡𝑢𝑏𝑒
Check whether the assumed value of T2 = 84 oF is within an acceptable range. The same amount
of heat transferred by convection is in accordance with the equation
𝑄 = 𝑚̇𝑐𝑝 (𝑇2 − 𝑇1 )
Solving for T2:
𝑇2 =
𝐵𝑡𝑢
ℎ𝑟
11 926
𝑙𝑏
𝐵𝑡𝑢
(1 683.7 𝑚 )(0.5
)
ℎ𝑟
+ 70 °𝐹
𝑙𝑏𝑚 −°𝐹
𝑇2 = 84.2 °𝐹
Therefore, the assumed value is acceptable.
Consider equation 19 – 13, as given by Sieder and Tate:
ℎ𝐷
𝐷𝑣𝜌 0.333 𝑐𝑝 𝜇 0.333 𝐷 0.333 𝜇 0.14
= 1.86 (
)
(
)
( )
( )
𝑘
𝜇
𝑘
𝐿
𝜇𝑠
Calculate the Prandtl number.
𝑃𝑟 =
𝐵𝑡𝑢
149 𝑙𝑏𝑚
)(
)
𝑙𝑏𝑚 −°𝐹 1 490 𝑓𝑡−𝑠𝑒𝑐
𝐵𝑡𝑢−𝑖𝑛.
1𝑓𝑡
1 ℎ𝑟
(0.94 2
)(
)(
)
𝑓𝑡 −ℎ𝑟−°𝐹 12 𝑖𝑛. 3 600 𝑠𝑒𝑐
(0.5
𝑃𝑟 = 2 297.9
Substituting the values,
𝐵𝑡𝑢−𝑖𝑛.
ℎ = 1.86(165)
0.333 (2
0.62 0.333 149 0.14 0.94𝑓𝑡2 −ℎ𝑟−°𝐹
297.9)0.333 ( 144 )
( 5.6 )
[ 0.62 𝑖𝑛. ]
𝐵𝑡𝑢
ℎ = 52.44 𝑓𝑡 2 −ℎ𝑟−°𝐹
The value obtained is 12 % higher than that of hi = 46.74 Btu/ft2 – hr – oF.
(b) Fort the exit oil temperature, consider a 14 oF differential in the second pass, just like in the
first pass. The exit oil temperature is then
𝑇𝑒𝑥𝑖𝑡 = 84 °𝐹 + 14 °𝐹
𝑇𝑒𝑥𝑖𝑡 = 98 °𝐹
(c) For the amount of steam condensed per hour, at p = 15 psia, hfg = 969.7 Btu/lbm (from the
Steam Tables) and since the convected heat per tube is Qcon = 11 926 Btu/hr, and there are 300
tubes/pass with 2 passes, then
𝑡𝑢𝑏𝑒
𝑄𝑐𝑜𝑛 𝑥 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑠𝑠 𝑥 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 = 𝑚̇𝑠 ℎ𝑓𝑔
𝑚̇𝑠 =
𝐵𝑡𝑢
𝑡𝑢𝑏𝑒𝑠
)(300
)(2 𝑝𝑎𝑠𝑠𝑒𝑠)
ℎ𝑟−𝑡𝑢𝑏𝑒
𝑝𝑎𝑠𝑠
𝐵𝑡𝑢
969.7
𝑙𝑏𝑚
(11 926
𝑚̇𝑠 = 7 379.2
𝑙𝑏𝑚
ℎ𝑟
1480./p. 138 – Faires
A single, 10.16 cm steel pipe, whose OD is 11.43 cm, has an outer surface temperature of 149 oC.
The horizontal pipe is located in a large room where the ambient temperature is 25.6 oC and the
barometer is standard. Determine the total heat (free convection and radiation) for 10 m of pipe
length.
Given: single horizontal steel pipe located in a large room
10.16 cm ∅
OD = 11.43 cm
Tsurf = 149 oC
Tamb = 25.6 oC
barometric pressure is standard
Required:
Qtotal = ? (free convection and radiation, for L = 10 m)
Solution:
From equation (c), p. 557, Faires, for horizon pipes and vertical pipes over 1 ft high:
∆𝑡 0.25
ℎ𝑐 = 0.27 (𝐷 )
𝑜
→
𝐵𝑡𝑢
𝑓𝑡 2 −ℎ𝑟−°𝐹
where the subscript c is used as a reminder that only convected heat is involved and that the
radiant energy should also be considered.
When converted to the SI Units,
∆𝑡 0.25
𝐵𝑡𝑢
ℎ𝑐 = 0.27 𝑓𝑡 1.75 −ℎ𝑟−°𝐹1.25 (𝐷 )
𝑜
1 055 𝐽
𝑥 (
𝐵𝑡𝑢
3.28 𝑓𝑡 1.75 1.8 °𝐹 1.25
1 ℎ𝑟
) (3 600 𝑠𝑒𝑐) (
𝑚
)
(
°𝐾
)
𝑊
∆𝑡 0.25
ℎ𝑐 = 1.32 1.75
( )
𝑚
− °𝐾 1.25 𝐷𝑜
𝑊
∴ ℎ𝑐 = 1.32 𝑚1.75 −°𝐾1.25 [
(149−25.6)°𝐾 0.25
0.114 3 𝑚
]
𝑊
ℎ𝑐 = 7.57 𝑚2 −°𝐾
The convected heat is therefore,
𝑄𝑐𝑜𝑛 = ℎ𝑐 𝐴(𝑇𝑠𝑢𝑟𝑓 − 𝑇𝑎𝑚𝑏 )
𝑊
𝑄𝑐𝑜𝑛 = (7.57 𝑚2 −°𝐾) [𝜋(0.114 3 𝑚)(10 𝑚)](149 − 25.6)°𝐾
𝑄𝑐𝑜𝑛 = 3 354.3 𝑊
For radiation, let Ɛ1 = 0.79 (for oxidized steel), F12 = 1 (case 2), and A2 is large, the total resistance
is then,
0
1−𝜖1
1
1−𝜖2
𝑅𝑡 = 𝐴 𝜖 + 𝐴 𝐹 + 𝐴 𝜖
1 1
𝑅𝑡 =
1 12
1−𝜖1
𝐴1 𝜖1
𝑅𝑡 =
2 2
1
1 (1)
+𝐴
1−𝜖1 +𝜖1
𝐴1 𝜖1
𝑅𝑡 = 𝐴
1
1 𝜖1
𝑞12 =
𝜎(𝑇1 4 −𝑇2 4 )
𝑅𝑡
= 𝐴1 𝜖1 𝜎(𝑇1 4 − 𝑇2 4 )
𝑞12 = 𝜋(0.114 3 𝑚)(10 𝑚)(0.79) (5.67 𝑥 10−8
𝑚2
𝑊
) [(422)4 − (298.6)4 ]°𝐾 4
− °𝐾 4
𝑞12 = 3 822.3 𝑊
∴ 𝑄𝑡𝑜𝑡𝑎𝑙 = 3 354.3 𝑊 + 3 822.3 𝑊
𝑄𝑡𝑜𝑡𝑎𝑙 = 7 176.6 𝑊