Example 1 – Natural Convective Heat Transfer

A fluid flows over a plane surface 1 m by 1 m with a bulk temperature of 50°C. The temperature of the surface is 20°C. The convective heat transfer coefficient is 2000 W/m²°C.

Answer:

The equation for heat transfer by convection is: Q = hc . A . dT

Where;

Q = Heat transfer by convection (Watts) or (kW)

hc = Heat transfer coefficient or film coefficient (W/m²degC) or (kW/m²degC)

A = Surface area (m²)

dT = Temperature difference, fluid to surface or surface to fluid (degC) or (K)

Therefore Q = 2000 x 1 x 1 (50 - 20)

Q = 60,000 Watts or 60 kW.

Example 2 – Forced Convective Heat Transfer

Calculate the heat transfer coefficient for water flowing through 25mm diameter pipe when the water velocity is 3.06 m/s.

The mean water temperature is 40^oC. The dynamic viscosity is 0.000651 kg m/s.

Answer:

The Dittus-Boelter equation can be used for smooth pipes if the flow is turbulent.

To check for turbulent flow use;

Re = ( r . v . d ) / m

where;

r = density (kg/m³), v = bulk velocity (m/s), d = diameter (m), m = dynamic viscosity (kg m/s)

Re = ( 1000 . 3.06 . 0.025 ) / 0.000651 = 117,512

Therefore the flow is turbulent i.e. Re is above 4000.

The Dittus-Boelter eqn. is: Nu _d = 0.023 Re_d ^0.8 Pr ⁿ

Where;

Nu = Nusselts number

Re = Reynolds number

Pr = Prandtl number

n = 0.4 for heating

n = 0.3 for cooling.

The suffix d refers to diameter of tube.

Nu _d = 0.023 x 117,512 ^0.8 Pr ^{n = 0.4}

Prandtl number Pr = Cp ( m / k )

Where;

hc = Surface heat transfer coefficient for convection (kW/m²K)

L = Characteristic dimension

k = Thermal conductivity of fluid (kW/mK) from Steam tables page 10, k_f = 632 x 10^-6 kW/m K @40 ^oC

m = dynamic viscosity (kgm/s) ….0.001 kgm/s at 20^oC for water.

Cp = Specific heat capacity (kJ/kg K) 4.179 @ 40^oC

Pr = 4.179 ( 0.000651 / 632 x 10^-6 ) = 4.30

Therefore: Nu _d = 0.023 x 117,512 ^0.8 4.30 ^0.4

Nu _d = 0.023 x 11,378 x 1.792

Nu _d = 469.0

Also Nusselt number Nu = h_c . L / k

Therefore; h_c = Nu . k / L

h_c = 469 x 632 x 10^-6 / 0.025

h_c = 11.86 kW/m² K

Example 3 – Forced Convective Heat Transfer

A copper tube is heated externally to 100^oC.

Air is heated by passing it through the 40mm diameter tube.

The air enters the tube at 20^oC and leaves at 88^oC.

The average air velocity is 8 m/s.

Calculate the heat transfer coefficient.

Dittus Boelter : Nu _d = 0.023 Re_d ^0.8 Pr ⁿ

Where;

Nu = Nusselts number

Re = Reynolds number

Pr = Prandtl number

n = 0.4 for heating

n = 0.3 for cooling.

The suffix d refers to diameter of tube.

Answer:

The mean film temperature is; T _{mean film} = ( T_w + T _b ) / 2

where;

T_w = wall or surface temperature (K)

T_b = mean bulk fluid temperature (K).

T _{mean film} = (100 + ( (20 + 88) / 2 ) ) 2

T _{mean film} = 77^oC (350 K)

From steam tables page 16 for air; density at 350 K = 1.009 kg/m³.

Dynamic viscosity of air from steam tables page 16 at 350 K = 2.075 x 10^-5 kg/ms.

Calculate Reynolds number from; Re = ( r . v . d ) / m

where;

r = density (kg/m³), v = bulk velocity (m/s), d = diameter (m), m = dynamic viscosity (kg/m s)

Re = ( 1.009 x 8 x 0.040 ) / 2.075 x 10^-5 = 15,560

The Prandtl number can be found from the steam tables page 16, at 350 K = 0.697

Therefore; Nu _d = 0.023 Re_d ^0.8 Pr ^0.4

Nu _d = 0.023 x 15,560 ^0.8 x 0.697 ^0.4

Nu _d = 0.023 x 2257.5 x 0.866

Nu _d = 44.96

Also Nusselt number Nu = h_c . L / k

Where; hc = Surface heat transfer coefficient for convection (kW/m²K)

L = Characteristic dimension

k = Thermal conductivity of fluid (kW/mK) from Steam tables page 16, k_f = 3.003 x 10^-5 kW/m K @ 350 K.

Therefore; h_c = Nu . k / L

h_c = 44.96 x 3.003 x 10^-5 / 0.040

h_c = 0.03375 kW/m² K

Example 4 – Forced Convective Heat Transfer

Find the heat transfer to the air by forced convection in example 3, if the pipe is 5 metres long.

Answer:

The equation for heat transfer by convection is: Q = hc . A . dT

Where;

Q = Heat transfer by convection (Watts) or (kW)

hc = Heat transfer coefficient or film coefficient (W/m²degC) or (kW/m²degC)

A = Surface area (m²)

dT = Temperature difference, fluid to surface or surface to fluid (degC) or (K)

In this case it is the log mean temperature difference (lmtd)

lmtd ( q _m ) = ( q ₁ - q ₂ ) / ( log_e q ₁ / q ₂ )

lmtd ( q _m ) = ( q ₁ - q ₂ ) / ( log_e q ₁ / q ₂ )

lmtd ( q _m ) = ( 100 - 20 ) - ( 100 - 88 ) / ( log_e 80 / 12 )

lmtd ( q _m ) = 68 / 1.8971

lmtd ( q _m ) = 35.84 K

Pipe surface area = p x d x l

Pipe surface area = p x d x l

Pipe surface area = p x 0.040 x 5

Pipe surface area = 0.6283 m².

Therefore Q = hc . A . q _m

Q = 0.03375 x 0.6283 x 35.84

Q = 0.760 kW or 760 Watts