Page 29 - Design and Operation of Heat Exchangers and their Networks

16 Design and operation of heat exchangers and their networks

2.1.1.1 Fully developed laminar flow in straight circular tubes
In a thermally and hydrodynamically developed laminar flow through a
straight circular tube, the Nusselt number is a constant. For constant wall
temperature,

Nu T ¼ 3:66 Re < 2200, Pr > 0:6, RePrd h =L < 10ð Þ (2.7)
For constant heat flux,

Nu H ¼ 4:36 Re < 2200, Pr > 0:6, RePrd h =L < 10ð Þ (2.8)

Example 2.1 Sizing an electrically heated tube
The compressed air at 1.5bar with a normal volumetric flow rate of
3
1.2Nm /h shall be heated from 20°Cto80°C by the heating wire
uniformly wrapped around the tube as it flows through the tube. The
tube outside diameter is 25mm, tube wall thickness is 2mm, and thermal
conductivity of the tube material is 15W/mK. The tube temperature
shall not exceed 200°C. Determine the length of the tube heating section.
Solution
The mean temperature of air is
t m ¼ t in + t out Þ=2 ¼ 50°C
ð
The properties of air are calculated with RefProp. At normal pressure
3
p N ¼1.01325bar and normal t N ¼0°C, the density ρ N ¼1.293kg/m ,
resulting in the mass flow rate as

_ m ¼ ρ V N ¼ 1:293 1:2=3600 ¼ 4:310 10 4 kg=s
N
The mass velocity is
_ m 4:310 10 4 2
G ¼ 2 ¼ 2 ¼ 1:244 kg=m s
πd =4 π 0:021 =4
i
At the mean temperature, we have μ¼1.964 10 5 sPa,
λ¼0.02810W/mK, c p ¼1008J/kgK, and Pr¼0.7047. Then, we have
the Reynolds number

Gd i 1:244 0:021
Re ¼ ¼ ¼ 1331 < 2200
μ 1:964 10 5
The heat power transferred through the tube wall to the air flow is
evaluated as
Q ¼ _mc p t out t in Þ ¼ 4:310 10 4 1008 80 20Þ ¼ 26:07W
ð
ð
Because the uniform electric heating implies the constant heat flux
boundary condition, we chose at first Eq. (2.8) for the Nusselt number,
which yields