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Natural convection from the horizontal tube surface

# Natural convection from the horizontal tube surface

Horizontal steel tube, filled with the flowing hot water is surrounded by the still air.

**Problem type:**

Plane problem of the steady-state heat transfer.

**Given:**

inner diameter of the tube *D1* = 120 mm,

outer diameter of the tube *D2* = 140 mm,

temperature of the water *T*_{water} = 90°C,

temperature of the ambient air *T*_{air} = 20°C,

thermal conductivity of the tube λ_{tube} = 40 W/(m·K),

thermal conductivity of the ambient air λ_{air}=0.027 W/(m·K).

**Task:**

Calculate the temperature of the tube outer surface and the heat flux per meter of the tube length.

**Solution:**

Flowing through the tube water heats its internal surface up to the water temperature. Heat from the tube surface dissipates as a result of convection. Convection coefficient depends on many parameters including temperature of the agent, shape and orientation of the surface. For the horizontal cylinder an average convection coefficient could be calculated using the similarity theory of heat transfer. The model is the thermal conductivity problem in the tube with the boundary conditions of the fixed temperature on the inner surface and the natural convection on the outer surface. The procedure of the solution is:

- Prandtl number (
*Pr*), Grashof number (*Gr*), Relay number (*Ra*) and Nusselt number (*Nu*) are calculated by the known formulae^{*}:

*Pr* = μ·*C* / *λ*_{air}, where μ - dynamical viscosity of the air, *C* - specific heat of the air;

*Gr* = *L*^{3}ρ^{2}*g*·Δ*T*β / μ^{2}, where *L* -
characteristic size, in our case - *D*2, ρ - air density, *g* - acceleration of gravity, Δ*T* - temperature difference, β - coefficient of the air thermal expansion;

*Ra* = *Gr*·*Pr*;

*Nu* = [0.60 + 0.387·*Ra*^{1/6} / (1 + (0.559/*Pr*)^{9/16})^{8/27}]^{2}.
Parameters of the air are calculated for the average temperature T_{avg} = (T_{water}+T_{air})/2 = (90+20)/2=55(°C)^{**}.

Numerical values of the parameters:

ΔT = 90-20 = 70°C;

ρ = 1.1 kg/m^{3};

*C* = 1000 J/(kg·K);

μ = 1.87·10^{-5} N·s/m^{2};

β = 0.003047 1/K;

*Pr* = 0.7;

*Gr* = 1.99·10^{7};

*Ra*=1.38·10^{7};

*Nu*=31;

Calculations are automated using the spreadsheet prepared by Harlan Bengtson and Lamar Stonecypher and downloaded from the website http://www.brighthubengineering.com/hvac/92660-natural-convection-heat-transfer-coefficient-estimation-calculations/;
Spreadsheet for the horizontal cylinder convection is attached.

- Using the numerical values, we have

α = *Nu*·λ_{air} / *d*_{2}

After the substitution of the values, numerical value could be obtained

α = 5.97 W/(m^{2}·K).
- QuickField problem is solved and the resulting temperature of the tube surface and the heat flux through the surface are calculated

**Result:**

Tube surface temperature: *T* = 363.04 K (89.89 °C).

Heat flux per meter of length: *F* = 183.5 W.

View simulation report in PDF.

Download simulation files and parameter calculation spreadsheet.

^{*}CIBSE Guide C: Reference Data (2010). Butterworth-Heinemann, ISBN: 0750653604. Table 3.5.

^{**}Remark: since the natural convection from the different parts of the curved cylindrical surface actually differs, the formula above gives the average value of the Nusselt number along surface. This approach is very approximate by its nature, so there is no reason to make iterations to calculate the accurate Nusselt number related to actual surface temperature.