class v - forced convection - external flow - cylinders & spheres - formulae & problems

Chapter 3

Forced convection – External flow

Flow over cylinders & spheres

(Formulae & Problems)

H1) Atmospheric Air at 275 K and a free stream velocity of 20 m/s flows over a flat plate 1.5 m long that is maintained at an uniform temperature of 325 K. Calculate the average heat transfer coefficient over the region where the boundary layer is laminar, the average heat transfer coefficient over the entire length of the plate and the total heat transfer rate from the plate to the air over the length 1.5 m and width 1 m. Assume the transition occurs at ReC=2 x 105.

(June 2006, Anna Univ)

Home work

Given: Fluid Temperature, Tά = 275 K Length, L = 1.5m Velocity, U = 20 m/sPlate surface temperature, Tw = 325 K Width W = 1m Critical Re = 3.5 x 105

To find: i) hl (laminar portion) ii) ht (entire plate) and iii) Q

Solution:Film Temp, Tf = (Tw+ Tά) /2Tf = 300K = 27o C ~ 25o C

Properties of air at 25o Cρ = 1.185 kg/m3

k = 0.02634 W/mKγ = 15.53 x 10-6 m2/sPr = 0.702

CASE I: For laminar flow portion of the plateTransition occurs at Rec=2x 105

Rex = (UL/γ)

Length corresponding to the transition (Lt)= (Rec γ / U)Lt=0.155 m

Forced convectionFor a flat plate laminar flow,Local Nux = 0.332 (Re)0.5 (Pr)0.333 where Re = 2 x 105

Nux = 131.97

Local hx = (Nux k / Lt) hx (laminar) = 22.42 W/m2K

Average h = 2 hx

h (laminar) = 44.84 W/m2K

CASE II: For entire length of the plateReL = (UL/γ)ReL = 1.93 x 106

Flow is combined laminar-turbulent flow

For a flat plate combined flow,Average Nu = (Pr)0.333 [0.037(Re)0.8 – 871] Nu = 2737.18

Average h = (Nu k / L)

h (entire plate) = 48.06 W/m2K

Rate of heat dissipation, Q = h A (Tw – Tά)Q = h (LW) (Tw – Tά)

Q = 3604.5 W

H2) Air at a pressure of 8 kN/m2 and a temperature of 250oC flows over a flat plate 0.3 m wide and 1 m long at a velocity of 8 m/s. If the plate is maintained at a temperature of 78oC, estimate the rate of heat to be removed continuously from the plate. (April 1997, Bharathiyar Univ)

Home work

Given: Fluid Temperature, Tά = 250oC Length, L = 1m Velocity, U = 8 m/sPlate surface temperature, Tw = 78oC Width W = 0.3m Pressure, p = 8 kN/m2

To find: i) Rate of heat transfer (Q)

Solution:Film Temp, Tf = (Tw+ Tά) /2Tf = 164o C ~ 160oC


k = 0.03640 W/mKγ = 30.09 x 10-6 m2/sPr = 0.682

Given pressure is above patm. So γ varies with pressure. (Value of Pr, k, Cp will remain same)

γ = γatm x (patm / pgiven)γ = 3.81 x 10-4 m2/s

ReL = (UL/γ)ReL = 2.1 x 105 – The flow is laminar

Forced convectionFor a flat plate laminar flow,Local Nux = 0.332 (Re)0.5 (Pr)0.333 Nux = 42.35

Local hx = (Nux k / L) hx (laminar) = 1.542 W/m2K

Average h = 2 hx

h (laminar) = 3.084 W/m2K

Rate of heat dissipation, Q = h A (Tw – Tά)Q = h (LW) (Tw – Tά)

Q = 159.1 W

Extended analysis:Rate of heat dissipation from both sides of plate, Qboth = 2 Q

Q = 318.2 W

Heat Transfer from circular surfaces – Flow over a cylinder (External flow)

Forced convection

Two regions Boundary layer region

near the surface Inviscid region

away from the surface

Pressure gradient along the cylinder is responsible for the development of a separated flow region on the backside of the cylinder

Separation of flow affects the drag force on a curved surface to great extent


i) Mean film temperature, All the thermo physical properties of the fluid (like density, viscosity, specific heat, thermal conductivity) should taken corresponding to mean film temperature – Plate surface temperature – Fluid temperature

ii) Reynolds number,

iii) Nusselt number, Nu = C (Re)m (Pr)0.333 (HMT Data book Page: 115)

Forced convection

2

TT

T wf wT T

UD

Re

ReD C m

0.4 – 4.0 0.989 0.330

4.1 – 40.0 0.911 0.385

40.1 – 4000 0.683 0.466

4001 – 40,000 0.193 0.618

40,001 – 400,000 0.0266 0.805


iv) Average heat transfer coefficient, h = (Nu k /D)

v) Heat transfer rate, Q = h A (Tw- Tά)

Where A = π DL

Forced convection

Heat Transfer from circular surfaces – Flow over a sphere (External flow)

i) Nusselt number, Nu = 0.37 (Re)0.6 (HMT Data book Page: 119)

17 < Re < 70,000

ii) Average heat transfer coefficient, h = (Nu k /D)

iii) Heat transfer rate, Q = h A (Tw- Tά)

Where A = 4π r2

1) Air at 15oC, 30 km/h flows over a cylinder of 400 mm diameter and 1500 mm height with surface temperature of 45oC. Calculate the heat loss.

Forced convection – Flow over Cylinders

Given: Fluid Temperature, Tά = 15o C Velocity, U=30 km/h=8.33 m/s Diameter, D = 0.4 mPlate surface temperature, Tw = 45o C Length, L = 1.5 m

To find: Heat loss (Q)

Solution:Film Temp, Tf = (Tw+ Tά) /2 Tf = 30o C

Properties of air at 30o C(HMT Data book, Pg: 33)ρ = 1.165 kg/m3

k = 0.02675 W/mKγ = 16 x 10-6 m2/sPr = 0.701

Re = (UD/γ) = 2.08 x 105

Average Nu = C (Re)m (Pr)0.333 Re value is 2.08 x 105 Corresponding ‘C’ value is 0.0266 and ‘m’ value is 0.805.

Nu = 451.3

Average h = (Nu k / D) h = 30.18 W/m2K

Heat loss, Q = h A (Tw – Tά)Q = h (π D L) (Tw – Tά)Q = 1706.6 W

2) Air at 30oC, 0.2 m/s flows across a 120 W spherical electric bulb at 130oC. Find the heat transfer rate and power lost due to convection if bulb diameter is 70 mm.


Given: Fluid Temperature, Tά = 30o C Velocity, U = 0.2 m/s Heat energy, Qbulb = 120 WSurface temperature, Tw = 130o C Diameter, D = 0.070 m

To find: i) Heat transfer rate (Q), 2. Power lost due to convection ((Q/ Qbulb)


Properties of air at 80o Cρ = 1 kg/m3

k = 0.03047 W/mKγ = 21.09 x 10-6 m2/sPr = 0.692

Re = (UD/γ) = 663.82

For sphere, Nu = 0.37 (Re)0.6 Nu = 18.25

Average h = (Nu k / D) h = 7.94 W/m2K

Heat transfer, Q = h A (Tw – Tά)Q = h (4 π r2 ) (Tw – Tά)

Q = 12.22 W

% heat lost = (Q/ Qbulb) x 100)

% heat lost = 10.18 %

3) Air at 40oC flows over a tube with a velocity of 30 m/s. The tube surface temperature is 120oC, calculate the heat transfer coefficient for the following cases. i) Tube could be square with a side of 6 cm ii) Tube is circular cylinder of diameter 6 cm.


Given: Fluid Temperature, Tά = 40o C Velocity, U = 30 m/sTube surface temperature, Tw = 120o C

To find: i) hsquare tube ii) hcircular tube


Properties of air at 80o Cρ = 1 kg/m3

k = 0.03047 W/mKγ = 21.09 x 10-6 m2/sPr = 0.692

Case (i) – Tube is considered as square of side 6 cmL = 0.06 m

Re = (UL/γ) Re = 0.853 x 105

For a flow over non-circular tube,Nu = C (Re)m (Pr)0.333

For square cross- section , n = 0.675, C = 0.092 (Pg: 118) Nu = 173.3

Forced convection – Flow over CylindersAverage h = (Nu k / L)

hsquare tube = 88 W/m2K

Case (ii) – Tube is circular cylinder of diameter 6 cmD = 0.06 m

For a flow over circular tube,Nu = C (Re)m (Pr)0.333

Re = (UD/γ) Re = 0.853 x 105

Re value is 0.853 x 105

Corresponding ‘C’ value is 0.0266 and ‘m’ value is 0.805.

Nu = 219.3Average h = (Nu k / D)

hcircular tube = 111.3 W/m2K

Flow over bank of tubesHeat transfer in flow over a bank or bundle of tubes has numerous industrial applications like

Steam generation in boiler Air conditioning cooling coil

One fluid moves over the tube while the second fluid (at different temperature) passes through the tubes

The tube rows of the bank may be either staggered or inline

Flow over bank of tubes

Configuration of bank of tubesCharacterized by

i) Tube diameter (D)ii) Transverse pitch (St)iii) Longitudinal pitch (Sl)iv) Diagonal pitch (SD) (used for staggered arrangement)

DS

SUU

DU

t

t

D

max

maxRe

Where,U – velocity of fluid (m/s)St – transverse pitch (m)D – Diameter of the tube (m)

Reynolds number is based on the largest velocity of the fluid flowing over the bank of tubes

Flow over bank of tubes

Formula used for flow over bank of tubes (HMT data book, Page no: 122)

i) Maximum velocity,

ii) Reynolds number,

iii) Nusselt number, Nu = C (Re)n (for air) Nu = 1.13 C (Re)n (Pr)0.333 (for other fluids)

DS

SUU

t

t

max

Value of C & n is based on St/D & Sl/D ratio

Where St – Transverse pitchSl – longitudinal pitchD – Tube diameter

DU

DmaxRe

1) In a surface condenser, hot water flows through staggered tubes while the liquid coolant is passed in cross flow over a tube bank. The temperature and velocity of the coolant are 30o C and 8 m/s resp. The longitudinal and transverse pitches are 22 mm and 20 mm resp. The tube outside diameter is 18 mm and tube surface temperature is 90o C. Calculate the heat transfer coefficient

Flow over Bank of Tubes

Given: Fluid Temperature, Tά = 30o C Velocity, U = 8 m/s Longitudinal pitch, Sl = 0.022 m Transverse pitch, St = 0.020 m Diameter, D = 0.018 mm Surface temperature, Tw = 90o C

To find: 1. Heat transfer coefficient ( h )



k = 0.02896 W/mKγ = 18.97 x 10-6 m2/sPr = 0.696

Max velocity, Umax = U x (St / (St – D))Umax = 80 m/s

Re = (Umax D/γ) Re = 7.5 x 104

For a flow over tube of banks,

Nu = 1,13 (Pr)0.333 C (Re)n (for other than air)


Average h = (Nu k / D)h = 428.6 W/m2K

Value of C & n is based on St/D & Sl/D ratio

St /D = 1.11 & Sl/D = 1.22

Corresponding C value is 0.518 & n value is 0.556. (HMT Data book, Pg: 122)

Nu = 1,13 (Pr)0.333 0.518 (Re)0.556

Nu = 266.3

Average heat transfer coefficient = 428.6 W/m2K

class v - forced convection - external flow - cylinders & spheres - formulae & problems

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