Class VI - Forced Convection - Internal Flow - Formulae & Problems

Chapter 3

Forced convection – Internal flow

Flow through cylinders

(Formulae & Problems)

Heat Transfer from circular surfaces – Flow through a cylinder (Internal flow)

Forced convection

Retardation of fluid flow near the walls causes Boundary layer development (Shown by dotted line in the fig)

Thickness of the BL is limited to radius of the pipe since the flow is within a confined passage

BL from the pipe walls meet at the centre of the pipe and the entire flow region acquires the same pattern of the flow Once the BL thickness becomes equal to the radius of the pipe there will be no further changes in the velocity distribution. This invariant velocity profile distribution is called fully developed velocity profile i.e Poiseulle flow

FORCED CONVECTIONHeat Transfer from circular surfaces – Flow through cylinder (Internal flow)

i) Mean film temperature,

– Fluid inlet temperature – Fluid outlet temperature

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

ii) Reynolds number,

Criteria for flow type

Re < 2300 - Laminar flowRe > 2300 - Turbulent flow

2mimo

f

TTT

moT miT

UD

Re

Laminar Flow: (HMT Data book, Pg no: 123)

Nu = 3.66

Turbulent Flow: (HMT Data book, Pg no: 125)

i) For 0.6 < Pr < 160, Re > 10,000 and ( L / D) > 60Nu = 0.023 (Re)0.8 (Pr)n

n = 0.4 – Heating processn = 0.3 – Cooling process

ii) For 10 < ( L / D) < 400 and Re < 10,000Nu = 0.036 (Re)0.8 (Pr)0.33 (D/L)0.055

Equivalent diameter for rectangular section

Dh (or) De =(4A / P) = [4LW/2(L+W)]

Where A – Area in m2, P – Perimeter in m, L – Length in m, W – Width in m.

FORCED CONVECTION

Equivalent diameter for hollow cylinder (annular spaces)

Dh (or) De = (4A / P) =

Where Do – Outer diameter in m, Di – Inner diameter in m

Heat transferQ = h A (Tw - Tm) where A = π D L (m2)

Tm – Mean temperature oC, = Tmi – Inlet temperature oC,Tmo – Outlet temperature oC. Tw – Tube wall temperature oC

Q = m Cp(Tmo - Tmi)

Mass flow ratem = ρ x A x U (kg/s)

where ρ – Density in kg/ m3 , A – Area, (π/4)D2 in m2

, U – Velocity in m/s.

FORCED CONVECTION

io

io

DD

DD

22

44

2mimo TT

1) Water flows inside a tube of 20 mm diameter and 3 m long flows at a velocity of 0.03 m/s. The water gets heated from 40o C to 120o C while passing through the tube. The tube wall is maintained at constant temperature of 160o C. Find heat transfer rate

Flow through Cylinders – Internal Flow

Given: Diameter, D = 0.020 m Length, L = 3 m Velocity, U = 0.03 m/sInner temperature of water, Tmi = 40o C Outer temperature of water, Tmo = 120o CWall temperature, Tw = 160o C

To find: Heat transfer rate ( Q )

Solution:Bulk mean Temp, Tm = (Tmi+T mo) / 2 = 80o C

Properties of water at 80o C(HMT Data book, Pg: 21)ρ = 974 kg/m3

γ = 0.364 x 10-6 m2/sPr = 2.22k = 0.6687 W/ mK

Re = (UD/γ) = 1648.35Since Re < 2300, flow is laminar.

For laminar flow, Nu = 3.66 (HMT Data book, Pg: 123)

Average heat transfer coefficient, h = Nu k / D h =122.39 W/m2K

Heat transfer rate, Q = h A (Tw - Tm) = h πDL (Tw-Tm)

Q = 1845.29 W

2) When 0.6 kg of water per minute is passed through a tube of 2 cm diameter, it is found to be heated from 20o C to 60o C. The heating is achieved by condensing steam on the surface of the tube and subsequently the surface temperature of the tube is maintained at 90o C. Determine the length of the tube required for fully developed flow.


Given: Mass, m = 0.01 kg/s Diameter, D = 0.02 mInlet temperature, Tmi = 20o C Outlet temperature, Tmo = 60o CTube surface temperature, Tw = 90o C

To find: Length of the tube, ( L )



γ = 0.657 x 10-6 m2/sPr = 4.340k = 0.628 W/mKCp = 4178 J/kg K

Mass flow rate, m = ρAU Velocity, U = m/ ρA = 0.031 m/s

Q = hA ∆T = h πDL (Tw-Tm)

Re = (UD/γ) = 943.6Since Re < 2300, flow is laminar.

Average heat transfer coefficient, h = Nu k / D h = 114.9 W/m2K

For laminar flow, Nu = 3.66 (HMT Data book, Pg: 123)

Heat transfer, Q = m Cp ∆T m Cp (Tmo - Tmi) = 1671.2 W

=> L = Q/ h πD (Tw-Tm)

Length of the tube, L = 4.62 m

3) Water at 50o C 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 90o C. Determine the heat transfer coefficient and the total amount of heat transferred if exit water temperature is 70o C.


Given: Diameter, D = 0.05 m Length, L = 4 m Velocity, U = 0.8 m/sInner temperature, Tmi = 50o C Exit temperature, Tmo = 70o CTube wall temperature, Tw = 90o C

To find: i) Heat transfer coefficient ( h ), ii) Heat transfer rate ( Q )



γ = 0.478 x 10-6 m2/sPr = 3.020k = 0.6513 W/mK

L/D ratio is greater than 60. Re value is greater than 10000 and Pr value is in between 0.6 and 160.

So, Nu =0.023(Re)0.8(Pr)n (HMT Data book, Pg: 125)The process involved is heating, hence n = 0.4

Since Re > 2300, flow is turbulent.L/D = 80 > 60, Re = 8.36 x 104 > 10000, Pr = 3.020 => 0.6 < Pr <160

Re = (UD/γ) = 8.36 x 104

Nu = 310

Heat transfer coefficient, h = Nu k / D = 4039.3 W/m2K

Heat transfer, Q = h A (Tw-Tm)= h πDL (Tw-Tm) = 76,139 W

4) Water flows through 0.8 cm diameter, 3 m long tube at an average temperature of 40o C. The flow velocity is 0.65 m/s and tube wall temperature is 140o C. Calculate the average heat transfer coefficient.


To find: Heat transfer coefficient, ( h )


γ = 0.657 x 10-6 m2/sPr = 4.340k = 0.628 W/mK

L/D = 375, 10 < L/D < 400

L/D ratio is in between 10 and 400, Re < 10000, so,

Nu = 0.036 (Re)0.8(Pr)0.33(D/L)0.055 (HMT Data book, Pg: 125)

Given: Diameter, D = 0.008 m Length, L = 3 m Average temperature, Tm = 40o CVelocity, U = 0.65 m/s Tube wall temperature, Tw = 140o C

Re = (UD/γ) = 7914.76Since Re > 2300, flow is turbulent.

Nu = 55.44

Heat transfer coefficient, h = Nu k / D h = 4352.3 W/m2K

5) Water at 30o C, 20 m/s flows through a straight tube of 60 mm diameter. The tube surface is maintained at 70o C and outlet temperature of water is 50o C. Find the heat transfer coefficient from the tube surface to the water, heat transferred and the tube length.


Given:

Inlet temperature, Tmi = 30o C Velocity, U = 20 m/s Diameter, D = 0.060 mOutlet temperature, Tmo = 50o C Tube surface temperature, Tw = 70o C

To find: i) Heat transfer coefficient ( h ), ii) Heat transferred ( Q ), 3. Tube Length ( L )



γ = 0.657 x 10-6 m2/sPr = 4.340k = 0.628 W/mKCp = 4178 J/kg K

Heat transfer coefficient, h = Nu k / D = 43726.59 W/m2K

Re = (UD/γ) = 1.8 x 106

Flow is turbulent

For turbulent flow, general equation is (Re >10000) Nu =0.023(Re)0.8(Pr)n (HMT Data book, Pg: 125) This is heating process, so n=0.4

Nu = 4177.7

Mass flow rate, m = ρAU = ρ((π/4) x D2)U = 56.2 kg/s

Heat transfer, Q = m Cp (Tmo - Tmi) = 4.69 x 106 W

Heat transfer, Q = h A (Tw – Tm) = h (πDL) (Tw – Tm)

So, L = Q/ h πD (Tw-Tm) = 18.96 m

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


Given: Inlet temperature, Tmi = 15o C Inner Diameter, Di = 0.04 m Velocity, U = 35 m / s Exit temperature, Tmo = 45o C Outer Diameter, Do = 0.06 m, Wall temp, Tw = 60o C

To find: Heat transfer coefficient ( h )

Solution:Mean Temp, Tm = (Tmi+T mo) / 2 = 30o CProperties of air at 30o C(HMT Data book, Pg: 33)

ρ = 1.165 kg/m3

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

Heat transfer coefficient, h = Nu k / De

h = 137.7 W/m2KHydraulic or Equivalent diameterDe = 4A/P = (4(π/4)[Do

2 – Di2])/(π )[Do + Di])

= 0.02 m

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

ReDe = (U De /γ) = 43750

Nu =0.023(Re)0.8(Pr)n (HMT Data book, Pg: 125)

This is heating process, so n=0.4

Nu = 102.9

7) Air at 30o C, 6 m/s flows in a rectangular section of size 300 x 800 mm. Calculate the heat leakage per meter length per unit temperature difference.


Given:

Air temperature, Tm = 30o C Velocity, U = 6 m/s Area = 0.3 x 0.8 m2 = 0.24 m2

To find: Heat leakage per meter length per unit temperature difference.

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

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

Heat transfer coefficient, h = Nu k / De = 18.09 W/m2K

Equivalent diameter (De)De = 4A/P =4(0.3 x 0.8) / 2( 0.3 + 0.8) where P – Perimeter = 2 (H + W)De = 0.436 m


Re = (U De /γ) = 1.63 x 104

Nu =0.023(Re)0.8(Pr)n (HMT Data book, Pg: 125)Assuming heating process, => n=0.4

Nu = 294.96

Q = h A (Tw – Tm) = h (PL) ΔT(Q / L ΔT) = h P = 39.79 W

8) Air at 333 K, 1.5 bar pressure, flow through 12 cm diameter tube. The surface temperature of the tube is maintained at 400 K 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= 60o C, Diameter, D=0.12 m, Wall temperature, Tw =400K=127o CMass flow rate, m = 75 kg/hr = 75 kg / 3600 s = 0.020 kg/s Length = 1.5 mTo find: Heat transfer rate ( Q )

Solution:Since the pressure is not much above atmospheric, physical properties of air may be taken at atmospheric condition.

Properties of air at 60o C(HMT Data book, Pg: 33)ρ = 1.060 kg/m3

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

Heat transfer coefficient, h = Nu k / D h = 7.94 W/m2K


Nu =0.023(Re)0.8(Pr)n (HMT Data book, Pg: 125)This is heating process, so n=0.4

Re = (U D /γ) = 10551.3

m = ρAU = ρ ((π/4) D2) UVelocity, U = m/ ρ ((π/4) D2) = 1.668 m/s

Nu = 32.9

Heat transfer rate, Q = h A (Tw - Tm) = h πDL (Tw-Tm)

Q = 300.82 W

Air at 2 bar pressure and 60o C is heated as it flows through a tube of diameter 25 mm at a velocity of 15 m/s. If the wall temperature is maintained at 100o C, find the heat transfer per unit length of the tube. How much would be the bulk temperature increase over one meter length of the tube.

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Class VI - Forced Convection - Internal Flow - Formulae & Problems