of 40 /40
UNIT – IV CONVECTION PART – A 1. What is dimensional analysis? Dimensional analysis is a mathematical method which makes use of the study of the dimensions for solving several engineering problems. This method can be applied to all types of fluid resistances, heat flow problems in fluid mechanics and thermodynamics. 2. State Buckingham p theorem. Buckingham p theorem states as Follows: “If there are n variables in a dimensionally homogeneous equation and if these contain m fundamental dimensions, then the variables are arranged into (n – m) dimensionless terms. These dimensionless terms are called p terms. 3. What are all the advantages of dimensional analysis? 1. It expresses the functional relationship between the variables in dimensional terms. 2. It enables getting up a theoretical solution in a simplified dimensionless form. 3. The results of one series of tests can be applied to a large number of other similar problems with the help of dimensional analysis. 4. What are all the limitations of dimensional analysis? 1. The complete information is not provided by dimensional analysis. It only indi- cates that there is some relationship between the parameters. 2. No information is given about the internal mechanism of physical phenomenon. 3. Dimensional analysis does not give any clue regarding the selection of variables. 5. Define Reynolds number (Re). It is defined as the ratio of inertia force to viscous force. 6. Define prandtl number (Pr).

# APTH UNIT4

Embed Size (px)

DESCRIPTION

ewe

Citation preview

• UNIT IV

CONVECTION

PART A

1. What is dimensional analysis?

Dimensional analysis is a mathematical method which makes use of the study ofthe dimensions for solving several engineering problems. This method can be applied toall types of fluid resistances, heat flow problems in fluid mechanics andthermodynamics.

2. State Buckingham p theorem.

Buckingham p theorem states as Follows: If there are n variables in adimensionally homogeneous equation and if these contain m fundamental dimensions,then the variables are arranged into (n m) dimensionless terms. These dimensionlessterms are called p terms.

3. What are all the advantages of dimensional analysis?

1. It expresses the functional relationship between the variables in dimensionalterms.

2. It enables getting up a theoretical solution in a simplified dimensionless form. 3. The results of one series of tests can be applied to a large number of other similar

problems with the help of dimensional analysis.

4. What are all the limitations of dimensional analysis?

1. The complete information is not provided by dimensional analysis. It only indi-cates that there is some relationship between the parameters.

2. No information is given about the internal mechanism of physical phenomenon.3. Dimensional analysis does not give any clue regarding the selection of variables.

5. Define Reynolds number (Re).

It is defined as the ratio of inertia force to viscous force.

6. Define prandtl number (Pr).

• It is the ratio of the momentum diffusivity of the thermal diffusivity.

7. Define Nusselt number (Nu).

It is defined as the ratio of the heat flow by convection process under an unittemperature gradient to the heat flow rate by conduction under an unit temperaturegradient through a stationary thickness (L) of metre.

Nusselt number (Nu) =

8. Define Grash of number (Gr).

It is defined as the ratio of product of inertia force and buoyancy force to thesquare of viscous force.

9. Define Stanton number (St).

It is the ratio of nusselt number to the product of Reynolds number and prandtlnumber.

10. What is meant by Newtonion and non Newtonion fluids?

The fluids which obey the Newtons Law of viscosity are called Newtonion fluidsand those which do not obey are called non newtonion fluids.

11. What is meant by laminar flow and turbulent flow?

Laminar flow: Laminar flow is sometimes called stream line flow. In this type of flow,the fluid moves in layers and each fluid particle follows a smooth continuous path. Thefluid particles in each layer remain in an orderly sequence without mixing with eachother.

• Turbulent flow: In addition to the laminar type of flow, a distinct irregular flow isfrequency observed in nature. This type of flow is called turbulent flow. The path of anyindividual particle is zig zag and irregular. Fig. shows the instantaneous velocity inlaminar and turbulent flow.

12. What is hydrodynamic boundary layer?

In hydrodynamic boundary layer, velocity of the fluid is less than 99% of freestream velocity.

13. What is thermal boundary layer?

In thermal boundary layer, temperature of the fluid is less than 99% of freestream velocity.

14. Define convection.

Convection is a process of heat transfer that will occur between a solid surfaceand a fluid medium when they are at different temperatures.

15. State Newtons law of convection.

Heat transfer from the moving fluid to solid surface is given by the equation Q = h A (Tw T)

This equation is referred to as Newtons law of cooling. Where

h Local heat transfer coefficient in W/m2K.A Surface area in m2

Tw Surface (or) Wall temperature in K

• T - Temperature of fluid in K.

16. What is meant by free or natural convection?

If the fluid motion is produced due to change in density resulting fromtemperature gradients, the mode of heat transfer is said to be free or naturalconvection.

17. What is forced convection?

If the fluid motion is artificially created by means of an external force like ablower or fan, that type of heat transfer is known as forced convection.18. According to Newtons law of cooling the amount of heat transfer from a solidsurface of area A at temperature Tw to a fluid at a temperature T is given by_____________.

Ans : Q = h A (Tw T)

19. What is the form of equation used to calculate heat transfer for flow throughcylindrical pipes?

Nu = 0.023 (Re)0.8 (Pr)n

n = 0.4 for heating of fluidsn = 0.3 for cooling of fluids

20. What are the dimensionless parameters used in forced convection?

1. Reynolds number (Re)2. Nusdselt number (Nu)3. Prandtl number (Pr)

21. Define boundary layer thickness.

The thickness of the boundary layer has been defined as the distance from thesurface at which the local velocity or temperature reaches 99% of the external velocity ortemperature.

PART B

• 1. Air at 20C, at a pressure of 1 bar is flowing over a flat plate at a velocity of 3 m/s. ifthe plate maintained at 60C, calculate the heat transfer per unit width of the plate.Assuming the length of the plate along the flow of air is 2m.

Given : Fluid temperature T = 20C, Pressure p = 1 bar, Velocity U = 3 m/s,

Plate surface temperature Tw = 60C, Width W = 1 m, Length L = 2m.

Solution : We know, Film temperature

Properties of air at 40C:

Density r = 1.129 Kg/m3

Thermal conductivity K = Kinematic viscosity v = Prandtl number Pr = 0.699

We know, Reynolds number Re =

Reynolds number value is less than 5 105, so this is laminar flow.

For flat plate, Laminar flow, Local Nusselt Number Nux = 0.332 (Re)0.5 (Pr)0.333

Local heat transfer coefficient hx = 2.327 W/m2KWe know,

Average heat transfer coefficient h = 2 hx

• h = 4.65 W/m2K

Heat transfer Q = h A (Tw - T)

2. Air at 20C at atmospheric pressure flows over a flat plate at a velocity of 3 m/s. ifthe plate is 1 m wide and 80C, calculate the following at x = 300 mm.

1. Hydrodynamic boundary layer thickness, 2. Thermal boundary layer thickness, 3. Local friction coefficient,4. Average friction coefficient, 5. Local heat transfer coefficient6. Average heat transfer coefficient,7. Heat transfer.

Given: Fluid temperature T = 20C Velocity U = 3 m/s Wide W = 1 m

Surface temperature Tw = 80CDistance x = 300 mm = 0.3 m

Solution: We knowFilm temperature

We know, Reynolds number Re =

Since Re < 5 105, flow is laminar For Flat plate, laminar flow,

1. Hydrodynamic boundary layer thickness:

2. Thermal boundary layer thickness:

• 3. Local Friction coefficient:

4. Average friction coefficient:

5. Local heat transfer coefficient (hx):

Local Nusselt NumberNux = 0.332 (Re)0.5 (Pr)0.333

We know

Local Nusselt Number

6. Average heat transfer coefficient (h):

7. Heat transfer:

We know that,

3. Air at 30C flows over a flat plate at a velocity of 2 m/s. The plate is 2 m long and1.5 m wide. Calculate the following:

1. Boundary layer thickness at the trailing edge of the plate,2. Total drag force, 3. Total mass flow rate through the boundary layer between x = 40 cm and x = 85

cm.

Given: Fluid temperature T = 30C Velocity U = 2 m/s Length L = 2 m Wide W W = 1.5 m

To find:1. Boundary layer thickness

• 2. Total drag force.3. Total mass flow rate through the boundary layer between x = 40 cm and x = 85

cm.

Solution: Properties of air at 30C

We know, Reynolds number

For flat plate, laminar flow, [from HMT data book, Page No.99]

Hydrodynamic boundary layer thickness

Thermal boundary layer thickness,

We know,Average friction coefficient,

We know

Total mass flow rate between x = 40 cm and x= 85 cm.

Hydrodynamic boundary layer thickness

4. Air at 30C, Flows over a flat plate at a velocity of 4 m/s. The plate measures 50 30cm and is maintained at a uniform temperature of 90C. Compare the heat loss fromthe plate when the air flows (a) Parallel to 50 cm,

• (b) Parallel to 30 cmAlso calculate the percentage of heat loss.

Given: Fluid temperature T = 30C Velocity U = 4 m/s Plate dimensions = 50 cm 30 cm

Surface temperature Tw = 90C

Solution: Film temperature

Properties of air at 60C,

Case (i) : When the flow is parallel to 50 cm.Reynolds number

Local nusselt number NUx= 0.332(Re)0.5(Pr)0.333

NUx =0.332

Local nusselt number NUx =95.35We know

We know

Average heat transfer coefficient

Case (ii) : When the flow is parallel to 30 cm side. Reynolds number Re =

For flat plate, laminar flow, Local Nusselt Number

We know that, NUx =

• Average heat transfer coefficient h = 2hx

Case (iii):

5. Air at 40C is flows over a flat plate of 0.9 m at a velocity of 3 m/s. Calculate thefollowing:

1. Overall drag coefficient2. Average shear stress, 3. Compare the average shear stress with local shear stress (shear stress at the

trailing edge)Given : Fluid temperature T = 40C

Length L = 0.9 m Velocity U = 3 m/s.

Solution:

Properties of air at 40C:

We know, Reynolds number

For plate, laminar flow,

Drag coefficient (or) Average skin friction coefficient

We know Average friction coefficient

We know,

• Local skin friction coefficient

we know Local skin friction coefficient

6. Air at 290C flows over a flat plate at a velocity of 6 m/s. The plate is 1m long and0.5 m wide. The pressure of the air is 6 kN/2. If the plate is maintained at atemperature of 70C, estimate the rate of heat removed from the plate.

Given : Fluid temperature T = 290C Velocity U = 6 m/s. Length L = 1 m

Wide W = 0.5 m Pressure of air P = 6 kN/m2

Plate surface temperature Tw = 70C

To find: Heat removed from the plate

Solution:

We know Film temperature

Properties of air at 180C (At atmospheric pressure)

Note: Pressure other than atmospheric pressure is given, so kinematic viscosity willvary with pressure. Pr, K, Cp are same for all pressures. Kinematic viscosity

We know, Reynolds number

• For plate, laminar flow, Local nusselt number

We knowNUx =

We know

Average heat transfer coefficient h = 2hx

Heat transfer from both side of the plate = 2 254.1

= 508.2 W.

7. Air at 40C flows over a flat plate, 0.8 m long at a velocity of 50 m/s. The platesurface is maintained at 300C. Determine the heat transferred from the entire platelength to air taking into consideration both laminar and turbulent portion of theboundary layer. Also calculate the percentage error if the boundary layer is assumedto be turbulent nature from the very leading edge of the plate.

Given : Fluid temperature T = 40C Length L = 0.8 m Velocity U = 50 m/s

Plate surface temperature Tw = 300C

To find :

1. Heat transferred for: i. Entire plate is considered as combination of both laminar and turbulentflow.ii. Entire plate is considered as turbulent flow.

2. Percentage error.

• Solution: We know

Film temperature T

We know

Case (i): Laminar turbulent combined. [It means, flow is laminar upto Reynoldsnumber value is 5 105, after that flow is turbulent]

Average nusselt number = Nu = (Pr)0.333 (Re)0.8 871 Nu = (0.6815)0.333 [0.037 (1.26 106)0.8 871 Average nusselt number Nu = 1705.3

Case (ii) : Entire plate is turbulent flow:

Local nusselt number} Nux = 0.0296 (Re)0.8 (Pr)0.333 NUx = 0.0296 (1.26 106)0.8 (0.6815)0.333

NUx = 1977.57We know

Local heat transfer coefficient hx = 91.46 W/m2K

Average heat transfer coefficient (for turbulent flow)

h = 1.24 hx = 1.24 91.46

Average heat transfer coefficient} h = 113.41 W/m2K

We know Heat transfer Q2 = h A (Tw + T)

= h L W (Tw + T) = 113.41 0.8 1 (300 40)Q2 = 23589.2 W

• 8. Air at 20C flows over a flat plate at 60C with a free stream velocity of 6 m/s.Determine the value of the average convective heat transfer coefficient upto a lengthof 1 m in the flow direction.

Given : Fluid temperature T = 20C Plate temperature Tw = 60C Velocity U = 6 m/s Length L = 1 m

To find : Average heat transfer coefficient

Solution : We know

We know Reynolds number Re =

For flat plate, laminar flow

Local nusselt number} Nux = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (3.53 105)0.5 (0.699)0.333

NUx = 175.27

We know, Local nusselt number}

9. Air at 25C at the atmospheric pressure is flowing over a flat plate at 3 m/s. If theplate is 1 m wide and the temperature Tw = 75C. Calculate the following at a locationof 1m from leading edge. 1.i. Hydrodynamic boundary layer thickness, 1.ii. Local friction coefficient,1.iii. Thermal boundary layer thickness,1.iv. Local heat transfer coefficient

Given : Fluid temperature T = 25C Velocity U = 3 m/s Wide W = 1 m

• Plate surface temperature Tw = 75CDistance = 1 m

To find:

1.1.1. Hydrodynamic boundary layer thickness.1.1.2. Local friction coefficient 1.1.3. Thermal boundary layer thickness1.1.4. Local heat transfer coefficient

Solution: We know

We know,Reynolds number Re=

For flat plate, laminar flow, 1. Hydrodynamic boundary layer thickness,

2. Local friction coefficient

3. Thermal boundary layer thickness,

4. Local heat transfer coefficient (hx):

We know

Local nusselt number} NUx = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (1.67 105)0.5 (0.698)0.333

• NUx = 120.415

10. Atmospheric air at 300 K with a velocity of 2.5 m/s flows over a flat plate of lengthL = 2m and width W = 1m maintained at uniform temperature of 400 K. Calculate thelocal heat transfer coefficient at 1 m length and the average heat transfer coefficientfrom L = 0 to L = 2m. Also find the heat transfer,

Given : Fluid temperature T = 300 KVelocity U = 2.5 m/sTotal Length L = 2 mWidth W = 1 m

Surface temperature Tw = 400 K

To find:

1. Local heat transfer coefficient at L = 1 m2. Average heat transfer coefficient at L = 2 m3. Heat transfer Q

Solution:

Case (i): Local heat transfer coefficient at L = 1m

For flat plate, laminar flow,

Local Nusselt number} NUx = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (118539.5)0.5 (0.692)0.333

NUx = 101.18We know, Local nusselt number} 101.18 = hx = 3.0832 W/m2K

• Local heat transfer coefficient} hx = 3.08 W/m2K

Case (ii): Average heat transfer coefficient at L = 2mReynolds number Re =

For flat plate, laminar flow,

NUx = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (237079.18)0.5 (0.692)0.333

NUx = 143

Local heat transfer coefficient} hx = 2.17 W/m2KWe know that,

Average heat transfer coefficient} h = 2 hxh = 2 2.17h = 4.35 W/m2K

Average heat transfer coefficient} h = 4.35 W/m2K

Case (iii) : Heat transfer Q = h A (Tw - T)= 4.35 2 1 (400 300)

Q = 870 W.

11. For a particular engine, the underside of the crank case can be idealized as a flatplat measuring 80 cm 20 cm. The engine runs at 80 km/hr and the crank case iscooled by air flowing past it at the same speed. Calculate the loss of heat from thecrank case surface of temperature 75C to the ambient air temperature 25C. Assumethe boundary layer becomes turbulent from the loading edge itself.

Given : Area A = 80 cm 20 cm= 1600 cm2 = 0.16m2

Velocity U = 80 Km/hr

Flow is turbulent from the leading edge, i.e,. flow is fully turbulent.

• To find:1. Heat loss

We know

For flat plate, turbulent flow, [Fully turbulent from leading edge given]

Local Nusselt number} NUx = 0.0296 (Re)0.8 (Pt)0.333

= 0.0296 [9 105]0.8 (0.698)0.33

NUx = 1524.6We know that,

Local heat transfer coefficient} hx = 53.85 W/m2KFor turbulent flow, flat plate

Average heat transfer coefficient} h = 1.24 hxh = 1.24 53.85h = 66.78 W/m2K

We know, Heat loss Q = h A (Tw - T)= 66.78 0.16 (75 25)Q = 534.2 W

Formula used for Flow over cylinders and spheres

1. Film temperature Where T - Fluid temperature C

Tw Plate surface temperature C

2. Reynolds number Where U Velocity, m/s

D - Diameter, m

• n - Kinematic viscosity, m2/s

3. Nusselt number NU = C (Re)m (Pr)0.333

4. Nusselt number NU =

5. Heat transfer Q = h A (Tw - T)Where A =

For sphere:

Nusselt number NU = 0.37 (Re)0.6

Heat transfer Q = h A (Tw - T)Where A 4pr2

12. Air at 15C, 30 km/h flows over a cylinder of 400 mm diameter and 1500 mmheight with surface temperature of 45C. Calculate the heat loss.

Given : Fluid temperature T = 15C Velocity U = 30 Km/h

Diameter D = 400 mm = 0.4 mLength L = 1500 mm = 1.5 mPlate surface temperature Tw = 45C

To find: Heat loss.Solution: We know

Film temperature

Properties of air at 30C : [From HMT data book, Page No.22]Density r = 1.165 Kg/m3

Kinematic viscosity v = 16 10-6 m2/sPrandtl Number Pr = 0.701

Thermal conductivity K = 26.75 10-3 W/mK

We know

• Reynolds Number Re =

We know

Nusselt Number Nu = C (Re)m (Pr)0.333

[From HMT data book, Page No.105]

ReD value is 2.08 105, so C value is 0.0266 and m value is 0.805.

[From HMT data book, Page No.105] NU = 0.0266 (2.08 105)0.805 (0.701)0.333

We know that, Nusselt Number NU =

13. Air at 30C, 0.2 m/s flows across a 120W electric bulb at 130C. Find heat transferand power lost due to convection if bulb diameter is 70 mm.

Given : Fluid temperature T = 30C Velocity U = 0.2 m/s Heat energy Q1 = 120 W

Surface temperature Tw = 130CDiameter D = 70 mm = 0.070 m

To find:1. Heat Transfer2. Power lost due to convection Solution:1. Film temperature

We know

Nusselt Number Nu = 0.37 (Re)0.6

= 0.37 (663.82)0.6

• Nu = 18.25We know Nusselt number

Heat transfer coefficient h = 7.94 W/m2KWe know

Heat transfer Q2 = h A (Tw - T)

2. % of heat lost =

14. Air at 40C flows over a tube with a velocity of 30 m/s. The tube surfacetemperature is 120C. Calculate the heat transfer for the following cases.

1. Tube could be square with a side of 6 cm.2. Tube is circular cylinder of diameter 6 cm

Given : Fluid temperature T = 40C Velocity U = 30 m/s

Tube surface temperature Tw = 120C

To find: Heat transfer coefficient (h)

Solution: We know

Case (i): Tube is considered as square of side 6 cmi.e., L = 6cm = 0.06mReynolds number Re =

Case (ii)

Tube diameter D = 6cm = 0.06 m

• so corresponding C and m values are 0.0266 and 0.805 respectively.

Formulae Used for Flow Over Bank of Tubes

1. Maximum velocity Umax = Where Sn Transverse pitch, m.

2. Reynolds Number Re =

3. Nusselt Number, NU = 1.13 (Pr)0.33 [C Ren][From HMT data book, Page No.114]

15. In a surface condenser, water flows through staggered tubes while the air ispassed in cross flow over the tubes. The temperature and velocity of air are 30C and8 m/s respectively. The longitudinal and transverse pitches are 22 mm and 20 mmrespectively. The tube outside diameter is 18 mm and tube surface temperature is90C. Calculate the heat transfer coefficient.

Given: Fluid temperature T = 30C Velocity U = 8 m/s

Longitudinal pitch, Sp = 22mm = 0.022 mTransverse pitch, Sn = 20mm = 0.020 mDiameter D = 18mm = 0.018 mTube surface temperature Tw = 90C

Solution:

We know Maximum velocity Umax =

We know

• corresponding C, n values are 0.518 and 0.556 respectively.[From HMT data book, Page No.114]

C = 0.518n = 0.556

We know,

Nusselt Number Nu = 1.13 (Pr) 0.333[C (Re)n][From HMT data book, Page No.114]

We know Nusselt Number

Heat transfer coefficient h = 428.6 W/m2K.

Formulae used for flow through Cylinders (Internal flow)

1. Bulk mean temperature

Tmi = Inlet temperature C,Where

Tmo = Outlet temperature C.2. Reynolds Number If Reynolds number value is less than 2300, flow is laminar. If Reynolds number valuesis greater than 2300, flow is turbulent.

3. Laminar Flow: Nusselt Number NU 3.66

[From HMT data book, Page No.116]

4. Turbulent Flow (General Equation)Nusselt Number Nu = 0.023 (Re)0.8 (Pr)n

n = 0.4 Heating processn = 0.3 Cooling process

[From HMT data book, Page No.119]

This equation is valid for

• 0.6 < Pr < 160,Re < 10000

For turbulent flow,

This equation is valid for

5. Equivalent diameter for rectangular section,

Where A Area, m2,P Perimeter, mL Length, m,W Width, m.

6. Equivalent diameter for hollow cylinder

7. Heat transfer

Q = h A (Tw Tm) where A = p D L (or) Q = m Cp (Tmo Tmi)

Where Tw Tube wall temperature C, Tm Mean temperature C. Tmi Inlet temperature C Tmo Outlet temperature C.

8. Mass flow ratem - r A U Kg/s

Where r - Density, Kg/m3

A Area, U Velocity, m/s

16. When 0.6 Kg of water per minute is passed through a tube of 2 cm diameter, it isfound to be heated from 20C to 60C. The heating is achieved by condensing steamon the surface of the tube and subsequently the surface temperature of the tube ismaintained at 90C. Determine the length of the tube required for fully developed

• flow.Given : Mass m = 0.6 Kg/min = = 0.01 Kg/s

To find: length of the tube (L)

Solution:Bulk mean temperature

Let us first determine the type of flow

For laminar flow,

Nusselt number NU = 3.66We know

17. Water at 50C enters 50 mm diameter and 4 m long tube with a velocity of 0.8 m/s.The tube wall is maintained at a constant temperature of 90C. Determine the heattransfer coefficient and the total amount of heat transferred if exist water temperatureis 70C.

Given:

Inner temperature of water Tmi = 50CDiameter D = 50mm = 0.05 mLength L = 4 mVelocity U = 0.8 m/s

Total wall temperature Tw = 90C

• Exit temperature of water Tmo = 70C

To find:

1. Heat transfer coefficient (h)2. Heat transfer (Q)

Solution:Bulk mean temperature

Let us first determine the type of flow:

ratio is greater than 60. Re value is greater than 10,000 and Pr value is in between 0.6and 160 so,

Nusselt number NU = 0.023 (Re)0.8 (Pr)n

[Inlet temperature 50C, Exit temperature 70C Heating Process, So n = 0.4]

Heat transfer coefficient h = 4039.3 W/m2KHeat transfer Q = h A (Tw Tm)

18. What flows through 0.8 cm diameter, 3m long tube at an average temperature of40C. The flow velocity is 0.65 m/s and tube wall temperature is 140C. Calculate theaverage heat transfer coefficient.

Given : Diameter of tube D = 0.8 cm = 0.008 m Length L = 3 m Average temperature Tm = 40C Velocity U = 0.65 m/s Tube wall temperature Tw = 140C

• To find: Heat transfer coefficient (h)

ratio is in between 10 and 400, Re < 10000, so Nusselt Number Nu = 0.036 (Re)0.8 (Pr)0.33

We know Nusselt number NU=

19. Air at 15C, 35 m/s, flows through a hollow cylinder of 4 cm inner diameter and 6cm outer diameter and leaves at 45C. Tube wall is maintained at 60C. Calculate theheat transfer coefficient between the air and the inner tube.

Given: Inner temperature of air Tmi = 15C Velocity U = 35 m/s Inner diameter Di = 4 cm = 0.04m Outer diameter Do = 6 cm = 0.06m Exit temperature of air Tmo = 45C Tube wall temperature Tw = 60C

To find: Heat transfer coefficient (h)

Solution: We know

Hydraulic of Equivalent diameter

Since Re > 2300, flow is turbulent

For turbulent flow, general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

This is heating process so, n = 0.4

• 20. Air at 30C, 6 m/s flows over a rectangular section of size 300 800 mm. Calculatethe heat leakage per meter length per unit temperature difference.

Given : Air temperature Tm = 30C

Velocity U = 6 m/s Area A = 300 800 mm2

A = 0.24 m2

To find: 1. Heat leakage per metre length per unit temperature difference.

Solution:

Equivalent diameter for 300 800 mm2 cross section is given by

We know

Since Re > 2300, flow is turbulent.

For turbulent flow general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

Assuming the pipe wall temperature to be higher than a temperature. So heatingprocess n = 0.4

We know

Heat transfer coefficient h = 18.09 W/m2KHeat leakage per unit per length per unit temperature difference Q = h P =Q = 39.79 W

• 21. Air at 333K, 1.5 bar pressure, flow through 12 cm diameter tube. The surfacetemperature of the tube is maintained at 400K and mass flow rate is 75 kg/hr.calculate the heat transfer rate for 1.5 m length of the tube.

Given : Air temperature Tm = 333 K = 60C Diameter D = 12 cm = 0.12 m

Surface temperature Tw = 400 K = 127CMass flow rate m = 75 kg/hr = m = 0.020 Kg/sLength L = 1.5 m

To find:1. Heat transfer rate (Q)

Solution:

Since the pressure is not much above atmospheric, physical properties of air may betaken at atmospheric condition

We know

Since Re > 2300, so flow is turbulent For turbulent flow, general equation is (Re>10000)

Nu = 32.9

22. 250 Kg/hr of air are cooled from 100C to 30C by flowing through a 3.5 cm innerdiameter pipe coil bent in to a helix of 0.6 m diameter. Calculate the value of air sideheat transfer coefficient if the properties of air at 65C are

K = 0.0298 W/mKm = 0.003 Kg/hr m

• Pr = 0.7r = 1.044 Kg/m3

Given : Mass flow rate in = 205 kg/hr

Inlet temperature of air Tmi = 100COutlet temperature of air Tmo = 30CDiameter D = 3.5 cm = 0.035 mMean temperature

To find: Heat transfer coefficient (h)

Solution:Reynolds Number Re = Kinematic viscosity

Since Re > 2300, flow is turbulent For turbulent flow, general equation is (Re > 10000)

We know that,

Heat transfer coefficient h = 2266.2 W/m2K

23. In a long annulus (3.125 cm ID and 5 cm OD) the air is heated by maintaining thetemperature of the outer surface of inner tube at 50C. The air enters at 16C andleaves at 32C. Its flow rate is 30 m/s. Estimate the heat transfer coefficient betweenair and the inner tube.

Given : Inner diameter Di = 3.125 cm = 0.03125 m

• Outer diameter Do = 5 cm = 0.05 mTube wall temperature Tw = 50CInner temperature of air Tmi = 16COuter temperature of air tmo = 32CFlow rate U = 30 m/sTo find: Heat transfer coefficient (h)

Solution:Mean temperature Tm =

We know,

Hydraulic or equivalent diameter

= 0.05 0.03125Dh = 0.01875 m

Re = 35.3 10-6

Since Re > 2300, flow is turbulent

For turbulent flow, general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

This is heating process. So n = 0.4

24. Engine oil flows through a 50 mm diameter tube at an average temperature of147C. The flow velocity is 80 cm/s. Calculate the average heat transfer coefficient ifthe tube wall is maintained at a temperature of 200C and it is 2 m long.

Given : Diameter D = 50 mm = 0.050 mAverage temperature Tm = 147C

Velocity U = 80 cm/s = 0.80 m/sTube wall temperature Tw = 200C

Length L = 2m

To find: Average heat transfer coefficient (h)

• Solution : Properties of engine oil at 147C

We know

Since Re < 2300 flow is turbulent

For turbulent flow, (Re < 10000)

25. A system for heating water from an inlet temperature of 20C to an outlettemperature of 40C involves passing the water through a 2.5cm diameter steel pipe.The pipe surface temperature is maintained at 110C by condensing steam on itssurface. For a water mass flow rate of 0.5 kg/min, find the length of the tube desired.

Given : Inlet temperature Tmi = 20C Outlet temperature Tmo = 40C Diameter D = 2.5 cm = 0.025 m

Piper surface temperature Tw = 110CMass flow rate m = 0.5 Kg/min = 8.33 10-3 Kg/s

To find: Length of the tube (L)

Solution: We know

We know

For laminar flow,

Nusselt number Nu = 3.66

Heat transfer coefficient h = 89.3 W/m2K

• 1. Film temperature where Tw Surface temperature in C T - Fluid temperature in C

2. Coefficient of thermal expansion

3. Nusselt Number Nu =

Where h Heat transfer coefficient W/m2K L Length, m

K Thermal conductivity, W/mK4. Grashof number for vertical plate

5. If GrPr value is less than 109, flow is laminar. If GrPr value is greater than 109, flow isturbulent.

i.e., GrPr > 109, Laminar flow GrPr > 109, Turbulent flow

6. For laminar flow (Vertical plate):

Nusselt number Nu = 0.59 (GrPr)0.25

This expression is valid for,104 < Gr Pr < 109

7. For turbulent flow (Vertical plate):Nusselt Number Nu = 0.10 [Gr Pr]0.333

8. Heat transfer (vertical plate):Q = h A (Tw - T)

9. Grashof number for horizontal plate:

Where Lc Characteristic length = W Width of the plate.

10. For horizontal plate, upper surface heated, Nusselt number Nu = 0.54 [Gr Pr]0.25

• This expression is valid for

11. For horizontal plate, lower surface heated

Nusselt Number Nu = 0.27 [Gr Pr]0.25

This expression is valid for 105 < Gr Pr < 1011

12. Heat transfer (Horizontal plate)

Q = (hu + hj) A (Tw - T)Where hu Upper surface heated, heat transfer coefficient W/m2 KHi Lower surface heated, heat transfer coefficient, W/m2K

13. For horizontal cylinder

Nusselt number Nu = C [Gr Pr]m

14. For horizontal cylinder,

Heat transfer Q = h A (Tw - T)Where A - pDL

15.For sphere,

Nusselt number Nu = 2 + 0.43 [Gr Pr]0.25

Heat transfer Q = h A (Tw - T)Where A - 4pr2

16. Boundary layer thickness

26. A vertical plate of 0.75 m height is at 170 C and is exposed to air at a temperatureof 105C and one atmosphere calculate:1. Mean heat transfer coefficient,2. Rate of heat transfer per unit width of the plate

Given : Length L = 0.75 m

• Wall temperature Tw = 170CFluid temperature T = 105C

To find: 1. Heat transfer coefficient (h)2. Heat transfer (Q) per unit width

Solution: Velocity (U) is not given. So this is natural convection type problem.

We know that

We know Grahsof number Gr =

Since Gr Pr < 109, flow is laminar

Gr Pr value is in between 104 and 109 i.e., 104 < Gr Pr < 109

So, Nusselt Number

Nu = 0.59 (Gr Pr)0.25

We know Nusselt number Nu =

Heat transfer coefficient h = 4.24 W/m2KWe know

27. A large vertical plate 4 m height is maintained at 606C and exposed toatmospheric air at 106C. Calculate the heat transfer is the plate is 10 m wide.

Given : Vertical plate length (or) Height L = 4 mWall temperature Tw = 606CAir temperature T = 106C

• Wide W = 10 m

To find: Heat transfer (Q)

Solution:

Gr = 1.61 1011

Gr Pr = 1.61 1011 0.676Gr Pr = 1.08 1011 Since Gr Pr > 109, flow is turbulent For turbulent flow, Nusselt number Nu = 0.10 [Gr Pr]0.333

We know that, Nusselt number

Heat transfer coefficient h = 5.78 W/m2K

Heat transfer Q = h A DT

28. A thin 100 cm long and 10 cm wide horizontal plate is maintained at a uniformtemperature of 150C in a large tank full of water at 75C. Estimate the rate of heat tobe supplied to the plate to maintain constant plate temperature as heat is dissipatedfrom either side of plate.

Given :

Length of horizontal plate L = 100 cm = 1mWide W = 10 cm = 0.10 mPlate temperature Tw = 150CFluid temperature T = 75C

• To find: Heat loss (Q) from either side of plate

Solution:

Lc = 0.05 m

Gr Pr = 5.29 109

Gr Pr value is in between 8 106 and 1011

i.e., 8 106 < Gr Pr < 1011

For horizontal plate, upper surface heated:

Nusselt number Nu = 0.15 (Gr Pr)0.333

We know that,

Upper surface heated, heat transfer coefficient hu = 3543.6 W/m2K

For horizontal plate, lower surface heated:

Nusselt number Nu = 0.27 [Gr Pr]0.25

Lower surface heated, heat transfer coefficient h1 = 994.6 W/m2K

Total heat transfer Q = (hu + h1) A DT

= (hu + h1) W L (Tw - T)= (3543.6 + 994.6) 0.10 (150 75)Q = 34036.5 W

29. A hot plate 20 cm in height and 60 cm wide is exposed to the ambient air at 30C.Assuming the temperature of the plate is maintained at 110C. Find the beat loss

• from both surface of the plate. Assume horizontal plate.

Given :

Height (or) Length of the Plate L = 20 cm = 0.20 mWide W = 60 cm = 0.60 mFluid temperature T = 30CPlate surface temperature Tw = 110C

To find:

Heat loss from both the surface of the plate (Q)

Solution:

We know Coefficient of thermal expansion}

We know

Gr Pr value is in between 8 106 and 1011

i.e., 8 106 < Gr Pr < 1011

For horizontal plate, Upper surface heated,

Nusselt number Nu = 0.15 (Gr Pr)0.333

Upper surface heated, heat transfer coefficient hu = 6.99 W/m2K

For horizontal plate, lower surface heated:

Nusselt number Nu = 0.27 (Gr Pr)0.25

= 0.277 [1.06 108]0.25

• Nu = 28.15

We know that,

Lower surface heated, heat transfer coefficient h = 2.78 W/m2K

30. A vertical pipe 80 mm diameter and 2 m height is maintained at a consenttemperature of 120C. The pipe is surrounded by still atmospheric air at 30C. Findheat loss by natural convection.

Given : Vertical pipe diameter D = 80 mm = 0.080 mHeight (or) Length L = 2 mSurface temperature Tw = 120 CAir temperature T = 30C

To find: Heat loss (Q)

Solution: We know

We know

For turbulent flow,

Nu = 0.10 [Gr Pr]0.333

= 0.10 [3.32 1010]0.333

Nu = 318.8

We know that,