Question

2- (50 pts.) A fluid flow in a circular pipe is heated with a constant heat flux. The pipe diameter is set to D=0.2 m and the length of the pipe to L = 1.0 m. The inlet temperature is 300 K and the inlet velocity is uniform. Use the following air properties: p= 1 kg/m?, u=4x10-5 Pa-s, k=0.03 W/m-K. The Reynolds number is Rep and the Prandtl number to Pr=1. A section of the pipe is heated with a constant heat flux as it is showed in the figure. The heat flux is q"=250 W/m2. Note: Pr= и Ср = α k Rep=80 a) (5 pts.) Is the flow hydrodynamically fully developed in the heated region? b) (5 pts.) Is the flow thermally fully developed at the end of the heated region? c) (20 pts.) Which will be the mean temperature at the end of the heated region? d) (20 pts.) Which will be the maximum surface temperature? q"=250 W/m2 L=0.35 m L=0.5 m L=0.15 m

Answer 1

D m q"=250 W/m2 L=0.35 m L=0.5 m L=0.15 m

A)

Given Re = 80

FLOW IS LAMINAR!!

Hydrodynamic entrance length for Laminar Flows is given as:

$L_{hydro}=0.05\times Re\times D$

where

• Re = 80
• D = diameter of pipe = 0.2 m

$L_{hydro}=0.05\times Re\times D$

$L_{hydro}=0.05\times 80\times 0.2=>L_{hydro}=0.8\:m$

The length UPTO the end of the heated region is 0.35 + 0.5 = 0.85 m

0.8 < 0.85

YES, the fluid is hydrodynamically fully developed IN the heated region.

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B)

For laminar flows:

Thermal entrance length, x(fd,t):

2fd, 0.05 x Rex Pr D

where

• Re = 80
• Pr = 1
• D = 0.2 m

2fd, 0.05 x Rex Pr D

fdt 0.2 0.05 x 80 x 1 => Ifd,t = 0.8 m

0.8 < 0.85

YES, the fluid is thermally fully developed at the end of the heated region.

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C)

The net heat supplied to the air must be equal to the increase in internal energy of the air.

$Q=\Delta IE$

where

• Q = q" x Area = 250 x 0.31416 = 78.539 W
  • q" = Heat flux = 250 W/m2
  • Area =
    II xDx L = II x0.2 x 0.5 = 0.31416 m2
    
    • D = Dia of pipe = 0.2 m
    • L = Length of heated section = 0.5 m
• $\Delta IE=m_{air}\times C_p\times (T_{final}-T_{initial})=5.0266e-3\times 1005\times (T_{final}-300)$
  • m(air) = mass flow rate of air = Density x Volume = 1 x 5.0266e-3 = 5.0266e-3 kg/s
    • Density of air = 1 kg/m3 (given)
    • Volume = Area x Velocity = 0.31416 x 0.016 = 5.0266e-3 m3/s
      • Area = 0.31416 m2
      • Velocity = $\frac{\mu \times Re}{D\times \rho }=\frac{4e-5\times 80}{0.2\times 1}=0.016\:m/s$ ​​​​​​​
        • $\mu$ = Dynamic viscosity of air = 4e-5 Pa-s
        • Re = 80
        • D = 0.2 m
        • $\rho$ = Density of air = 1 kg/m3 (given)
  • Cp = Specific heat of air at constant pressure = 1005 J/kgK
  • T(final) = Final temp = ? (our answer)
  • T(initial) = Initial temp = 300 K

$78.539=5.0266e-3\times 1005\times (T_{final}-300)=>\mathbf{T_{final}= 315.547\:K}$

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D)

The maximum surface temperature will occur at the end of the end of the heating section. Thus,

$q"=h\times (T_{max-surface}-T_{final-fluid})$

where

• q" = Heat flux = 250 W/m2
• h = ?
• T(max-surface) = ? (our answer)
• T(final-fluid) = outlet temp of fluid from the heating section = 315.547 K

h = ?

For laminar fluid and for constant heat flux, we have,

$Nu=4.36=\frac{h\times D}{k_{air}}$

where

• D = 0.2 m
• k(air) = thermal conductivity of air = 0.03 W/mK

$4.36=\frac{h\times 0.2}{0.03}=>h = 0.654\:W/m^2K$

So,

$q"=h\times (T_{max-surface}-T_{final-fluid})$

$250=0.654\times (T_{max-surface}-315.547)=>\mathbf{T_{max-surface}=697.8099\:K}$

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