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6.4 Empirical correlations for natural convection 193

(d)Hence, radiation plays a significant role in the heat transfer from the plate and about 35% error occurs in dTdt if radiation is not considered.

(e)Biot number (Bi) = hLk

(i) Without consideration of radiation

Bi =

4.81 × 0.5

= 0.011 < 0.1

205

 

 

Yes, the assumption of spatial isothermality is valid. (ii) With consideration of radiation

qradiation = σ (Tw4 T4 ) = hr (Tw T)

with hr being the radiative heat transfer coefficient. Introduction of this quantity is convenient but has to be used with caution as radiation is highly nonlinear.

htotal = hc + hr

htotal = hc + σ (Tw2 T2 )(Tw T)

htotal = 4.18 + 5.67 × 108 × 0.25 × (4002 + 3102 )(710)

h = 7.387W/m2 K

total

 

 

Bi =

7.387 × 0.5

= 0.018 < 0.1

205

 

 

Yes, the assumption of spatial isothermality is valid.

Example 6.2: Consider a cylindrical electric immersion heater with a rating of 1000W. The heater is 16 mm in diameter and 350 mm long. The heater is placed horizontally in a pool of water in a bucket at 37 °C.

(a) Estimate the temperature of the heater under steady state. The properties of the water at temperatures 37 °C and 67 °C are tabulated below in Table 6.1 (A more comprehensive Table of properties for water was given in chapter 5). Please note that only β has to be evaluated at Tand all other properties need to be evaluated at

Table 6.1  Thermophysical properties of water

S. No.

T(°C)

ρ(kg/m3)

Cp (J/

k

α(m2/s)

ν, (m2/s)

Pr

kg.K)

(W/m.K)

1

37

993.3

4179

0.6260

1.508 × 10–7

6.98 × 10–7

4.63

2

67

979.5

4189

0.6605

1.610 × 10–7

4.31 × 10–7

2.68

 

 

 

 

 

 

 

 

194 CHAPTER 6  Natural convection

Tf= (TW+ T)/2, where TW is the heater temperature. Since TW is unknown, assume TW initially to be 67 °C.

(b) What will be the heater temperature be if we forget to add water in the bucket and the heater is left to operate in air?

Solution

(a) Tw is assumed to be 67 ºC as given in the problem. Film temperature, Tf = (TW + T)/2 = (340 + 310)/2 = 650/2 = 325K.

Properties of water at Tf = 325K:

ν = 5.65 × 10–7 m2/s, α = 1.56 × 10–7 m2/s, k = 0.64325 W/mK, Pr = 3.655. β = 3.61 × 10–4 K–1 (at T)

 

RaD =

gβ(T T )L3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

w

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ν α

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RaD =

9.81× (340 310) × 0.0163 × 3.61×104

 

 

 

 

 

5.65 ×1.56 ×1014

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RaD = 4.94 ×106

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

From Eq. (6.103)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/6 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RaD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD = 0.60

+

0.387

 

 

 

 

 

 

 

 

 

 

 

 

9/16

 

16/9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.559

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/6 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4.94 106

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD = 0.60

+

0.387

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

16/9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.559

 

 

9/16

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 27.286

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h =

 

×

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h = 27.286 ×

0.64325

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.016

 

 

 

 

 

 

 

 

 

 

 

 

h = 1097W / m2 K.

 

 

 

 

 

 

 

 

 

The estimated wall temperature

T

= T

 

+

 

1000

 

= 361.8 K

 

 

 

h × π DL

 

 

 

 

 

 

 

w

 

 

 

 

 

 

 

 

 

 

 

 

6.4 Empirical correlations for natural convection 195

Iteration 2

Update wall temperature from 340K to 361.8K

Film temperature, Tf = (TW + T)/2 = (361.8 + 310)/2 = 335.9K. Properties of water at Tf = 325K:

ν = 4.67 ×10–7 m2/s, α = 1.6 × 10–7 m2/s, k = 0:66 W/mK, Pr = 2.95

 

RaD =

9.81× (361.8 310) × 0.0163 × 3.61×104

 

 

 

 

 

 

 

4.67 ×1.6 ×1014

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RaD = 1×107

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/6 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 107

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD =

0.60 + 0.387

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

16/9

 

 

 

 

 

 

 

 

 

 

1+

 

0.559

9/16

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.95

 

 

 

 

 

 

 

 

 

 

= 33.135

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h =

 

×

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h = 33.135 ×

0.66

 

 

 

 

 

 

 

 

 

 

 

 

0.016

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h = 1358.15 W/m2K

 

 

 

 

 

Estimated wall temperature

Tw = T+

 

 

1000

 

= 351.9 K.

 

 

 

 

h

× π DL

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

After seven iterations (see Table 6.2),

assumed TW = Estimated TW = 354.6 K = 81.5 °C.

(b) Now with air as the operating fluid and assuming TW = 360 K Film temperature, Tf = (TW + T)/2 = (360 + 310)/2 = 335 K. Properties of air at Tf = 335 K:

ν = 1.9 × 10–5 m2/s, α = 2.72 × 10–5 m2/s, k = 0.0285 W/mK, Pr = 0.7075

β = 1 = 3.2 × 103 K1

T

196 CHAPTER 6  Natural convection

Table 6.2  Iterative method to solve Example 6.2 (All temperatures in Kelvin)

S. No.

T

T

(T

- T )2

 

w,i

w,i+1

w;i+1

w;i

1

340.0

361.8

475.24

2

361.8

351.8

100.00

3

351.8

355.7

15.29

 

4

355.7

354.1

2.59

 

5

354.1

354.5

0.16

 

6

354.5

354.6

0.01

 

7

354.6

354.6

0.00

 

 

 

 

 

 

 

RaD =

9.81× (360 310) × 0.0163 × 3.2 ×103

 

 

 

 

1.9 × 2.72 ×1010

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RaL

= 12540.8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/6 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12540.8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD = 0.60

+ 0.387

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

16/9

 

 

 

 

 

 

 

 

 

 

 

0.559

 

9/16

 

 

 

 

 

 

 

 

 

 

 

 

1

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.7075

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 4.6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h =

 

×

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

NuD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h = 4.6 × 0.028545

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.016

 

 

 

 

 

 

 

 

 

 

 

h = 8.232 W/m2K.

 

 

 

 

 

 

Estimated wall temperature,

T

= T +

 

1000

 

 

= 7217.8K

 

 

h × π DL

 

 

 

 

 

 

w

 

 

 

 

 

 

 

As we can see, at the end of the first iteration itself, the temperature of the heater has reached an incredibly high value far beyond the melting point of the material of the heater (what it is made of is immaterial). Hence, even by inadvertence, we can ill-afford to operate an immersion heater designed for water, with air as the medium.

6.4 Empirical correlations for natural convection 197

Problems

6.1A 25 cm tall vertical plate infinitely deep in the direction perpendicular to the plane of the paper and negligible thickness is maintained at a temperature of 100 °C in quiescent air at 35 °C. Get an estimate of the boundary layer thickness at

(a)5 cm from the bottom.

(b)15 cm from the bottom.

(c)The top of the plate.

6.2Revisit Problem 6.1. Get an estimate of uref for the problem under question. Compare it with typical velocities encountered in forced convection, with air as the ambient medium.

6.3A heated sphere at TW = 100 °C made of copper has a diameter of 20 mm. It is kept in still air at 30 °C. The density and specific heat of copper are 8900 kg/m3 and 0.384 kJ/kg.K, respectively.

(a) Determine the heat transfer coefficient from the sphere.

(b)At time t = 0, when the heated sphere begins to cool, what is the value of dTdt assuming the sphere to be spatially isothermal.

6.4Consider Problem 2.4 of chapter 2 that concerned the cooling of a current carrying copper wire. In this problem, the free convection heat transfer coefficient was assumed to be 6 W/m2K. Is this justified?

6.5Consider the integral solution of natural convection over a flat plate for a fluid with Pr > 1. Considering a quadratic profile for the temperature and a cubic profile for velocity, determine the following.

(a)Expression for δ(x).

(b)Expression for Nux .

(c)Expression for NuL .

6.6Using the results from Problem 6.5, determine the location y (for any x) at which the vertical velocity is the maximum and hence determine its magnitude.

6.7Revisit Problem 6.1. Determine the local and average Nusselt numbers at the three locations mentioned in the problem.

198 CHAPTER 6  Natural convection

Table 6.3  Thermophysical properties of air at atmospheric pressure (101325 Pa) (Kadoya et al., 1985; Jacobsen et al., 1992)

T(K)

ρ(kg/m3)

cp (J/

µ(kg/m.s)

ν(m2/s)

k

α(m2/s)

Pr

kg.K)

(W/m.K)

100

3.605

1039

0.711 × 10–5

0.197 × 10–5

0.00941

0.251 ×

0.784

 

 

 

 

 

 

10–5

 

150

2.368

1012

1.035

0.437

0.01406

0.587

0.745

200

1.769

1007

1.333

0.754

0.01836

1.031

0.731

250

1.412

1006

1.606

1.137

0.02241

1.578

0.721

260

1.358

1006

1.649

1.214

0.02329

1.705

0.712

270

1.308

1006

1.699

1.299

0.02400

1.824

0.712

280

1.261

1006

1.747

1.385

0.02473

1.879

0.711

290

1.217

1006

1.795

1.475

0.02544

2.078

0.710

300

1.177

1007

1.857

1.578

0.02623

2.213

0.713

310

1.139

1007

1.889

1.659

0.02684

2.340

0.709

320

1.103

1008

1.935

1.754

0.02753

2.476

0.708

330

1.070

1008

1.981

1.851

0.02821

2.616

0.708

340

1.038

1009

2.025

1.951

0.02888

2.821

0.707

350

1.008

1009

2.090

2.073

0.02984

2.931

0.707

400

0.8821

1014

2.310

2.619

0.03328

3.721

0.704

450

0.7840

1021

2.517

3.210

0.03656

4.567

0.703

500

0.7056

1030

2.713

3.845

0.03971

5.464

0.704

550

0.6414

1040

2.902

4.524

0.04277

6.412

0.706

600

0.5880

1051

3.082

5.242

0.04573

7.400

0.708

650

0.5427

1063

3.257

6.001

0.04863

8.430

0.712

700

0.5040

1075

3.425

6.796

0.05146

9.498

0.715

750

0.4704

1087

3.588

7.623

0.05425

10.61

0.719

800

0.4410

1099

3.747

8.497

0.05699

11.76

0.723

 

 

 

 

 

 

 

 

References

Bejan, A., 2013. Convection Heat Transfer. John Wiley and Sons, Hoboken, NJ.

Churchill, S.W., 1983. Comprehensive, theoretically based, correlating equations for free convection from isothermal spheres. Chem. Eng. Commun. 24 (4–6), 339–352.

Churchill, Stuart W., Chu, Humbert H.S., 1975a. Correlating equations for laminar and turbulent free convection from a vertical plate. Int. J. Heat Mass Transf. 18 (11), 1323–1329.

Churchill, S.W., Chu, H.H.S., 1975b. Correlating equations for laminar and turbulent free convection from a horizontal cylinder. Int. J. Heat Mass Transf. 18 (9), 1049–1053.

Kadoya, K., Matsunaga, N., Nagashima, A., 1985. Viscosity and thermal conductivity of dry air in the gaseous phase. J. Phys. Chem. Ref. Data 14 (4), 947–970.

Jacobsen, R.T., Penoncello, S.G., Beyerlein, S.W., Clarke, W.P., Lemmon, E.W., 1992. A thermodynamic property formulation for air. Fluid Phase Equilibria 79, 113–124.

Lienhard IV, J.H., Lienhard V, J.H., 2020. A Heat Transfer Textbook. Courier Dover Publications, Mineola, NY.