heat transfer to fluids without phase change

HEAT TRANSFER

CONVECTION

CONVECTION• Convection is the transfer

of heat by the motion of liquids and gases.– Convection in a gas occurs

because gas expands when heated.

– Convection occurs because currents flow when hot gas rises and cool gas sink.

– Convection in liquids also occurs because of differences in density.

q = h A (T2 -T1)

Area contacting fluids (m2)

Heat transfer coefficient(watts/m2oC)

Heat flow (watts)

Temperaturedifference (oC)

CONVECTION EQUATION

What happens to the particles in a liquid or a gas when you heat them?

The particles spread out and become less dense.

CONVECTION

when air or water is heated, it expands,

CONVECTION

rises and is replaced by cold air from above.

Warm air is taken in by the air-conditioner & cooled.

An air-conditioner gives out cold air which falls.

This pushes the warm air up.

Why air-conditioners are usually set up high on the wall?

Where is the freezer

compartment put in a fridge?

Freezer compartment

It is put at the top, because cool air sinks, so it cools the food on the way down.

It is warmer at the

bottom, so this warmer air rises and a convection current is set

up.

CONVECTION

In a fire, why should you crawl close to the floor in a smoke-filled room?

Smoke is warmer than the surrounding air.

It rises & its places would be replaced by the surrounding cooler air.

Smoke is toxic.

We should crawl close to the floor to prevent smoke inhalation.

Air-conditioner is usually

installed high on the wall

immersion heater is fixed

near the bottom

• When the flow of gas or liquid comes from differences in density and temperature, it is called free convection.

• When the flow of gas or liquid is circulated by pumps or fans it is called forced convection.

CONVECTION

• Convection depends on speed.

• Motion increases heat transfer by convection in all fluids.

CONVECTION

• Convection depends on surface area.

• If the surface contacting the fluid is increased, the rate of heat transfer also increases.

• Almost all devices made for convection have fins for this purpose.

CONVECTION

• Both free and forced convection help to heat houses and cool car engines.

CONVECTION

• On a smaller scale near coastlines, convection is responsible for sea breezes.

• During the daytime, land is much hotter than the ocean.

• A sea breeze is created when hot air over the land rises due to convection and is replaced by cooler air from the ocean.

• At night the temperature reverses so a land breeze occurs.

CONVECTION

• Much of the Earth’s climate is regulated by giant convection currents in the ocean.

CONVECTION

FORCED CONVECTION: LAMINAR FLOW

The primary resistance to heat transfer by convection is normally controlled within a thin layer of the fluid, adjacent to the immersed body, in which viscous effects are important. The quantity of heat transferred is highly dependent upon the fluid motion within this boundary layer, being determined chiefly by the thickness of the layer.

Prandtl’s Boundary Layer EquationBy applying Newton’s second law and the continuity

equation to an infinitesimal, two dimensional control volume within the boundary layer, and assuming that

1. Fluid viscosity is constant2. Shear in the y-direction is negligible3. The flow is steady, and the fluid is

incompressible4. The vertical pressure gradient is negligible

FORCED CONVECTION: LAMINAR FLOW

THERMAL BOUNDARY LAYER: FLAT PLATE

When a fluid at one temperature flows along a surface which is at another temperature, a thermal boundary layer develops. The thermal boundary layer thickness is defined as the distance required for the temperature T to reach 99% of its free-stream value T∞. And assuming,1. steady, incompressible flow2. Constant fluid properties evaluated at film

temperature: Tw = Ts + T∞ 2

3. Negligible body forces, viscous heating, and conduction in the flow direction

PRANDTL NUMBER

Pr = ν = Cp μ α k is unity, which is approximately the case for most gases (0.6<Pr<1.0). The Prandtl number for liquids however, varies widely, ranging from large values for viscous oils to very small values for liquid metals which have high thermal conductivities.

The equation for heat transfer to flat plate;

Nu = 0.332 Re1/2 Pr 1/3

( 1 – ( xo /x) ¾) 1/3

Where: Nu = hx L / k = Nusselt number

Pr = Cp μ / k = Prandtl numberRe = u L ρ / μ = Reynolds number


When plate is heated over its entire length, xo = 0

Nu = 0.332 Re1/2 Pr 1/3

When average value of Nu over entire length, x1

Nu = 0.664 Re1/2 Pr 1/3

(for constant heat flux, the coefficient 0.332 becomes 0.453, effecting an increase of the coefficient over an entire length)


These equations are valid only for Prandtl numbers of 1.0 or greater, since the derivation assumes a thermal boundary layer no thicker than the hydrodynamic layer. However, they can be used for gases with Pr ≈ 0.70 with little error. The equations are also restricted to cases where the Nusselt number is fairly large, 10 or higher since axial conduction, which was neglected in the derivation, has a significant effect for thick boundary layers.


Castor oil at 38OC flows over a wide, 6 m long, heated plate at 0.06 m/s. For a surface temperature of 93OC, determine a) the hydrodynamic boundary layer thickness at the end of the plate b) the local heat transfer coefficient h, at the end of the plate c) the total heat rate from the surface per unit width. Assume the thermal diffusivity to be 7.22 x 10-8 m2/s and the thermal conductivity to be 0.213 w/m.K at the film temperature.

Laminar flow heat transfer to flat plate

a) δ = 5.0 L = ? √ReL

Re = u L ρ = u L = (0.06 m/s) (6.0 m) μ ν 6.0 x 10 -5 m2/s Re = 6000

δ = 5.0 ( 6.0 m) = 0.387 √ 6000

b) Pr = ν = 6.0 x 10 -5 m2/s = 8.31 x 102

α 7.22 x 10 -8 m2/s

h = 0.332 k u ½ Pr 1/3

ν Lh = 0.332 (0.213 w/m.K) 0.06 m/s ½ 8.31x102

6x10-5m2/s (6m)

h = 8.58 W/m2.K

c) ȟ = 2 hL ȟ = 2 (8.58 W/m2.K) = 17.16 W/m2.K

q = ȟ As (Ts – T∞) = 17.16 W/m2.K (6 m2/m) (93 - 38)OC q = 5665 W/m

Atmospheric air at 25OC flows over both the surface of a flat plate with a velocity of 5 m/s. the flat plate is maintained at a uniform temperature of 75OC. Determinea) the velocity boundary layer thickness and the

heat flux at the trailing edgeb) the total heat transfer from the plate to air

Laminar flow heat transfer to flat plate

Ťb = (75 + 25 )/2 = 50OC

ρ = 1.093 kg/m3 ν = 18.02 x 10-6 m2/sk = 0.028 W/m.K

Pr = Cpμ = 0.703 k Re = ρuL = uL = 5 m/s ( 1 m) = 2.775 x 105

μ ν 18.02 x 10-6 m2/s

a) δ = 5.0 L = 5.0 ( 1 m ) = 9.49 x10-3 m √ReL √2.775 x 105

Nu = 0.332 Re1/2 Pr 1/3

Nu = 0.332 (2.775 x 105)1/2 (0.703) 1/3

Nu = 155.5

Local heat transfer coefficient

hx = Nu k / L = 155.5 (0.028 W/m.K) / 1mh = 4.355 W/m2.K

q ’ = h (Ts –T) = 4.355 W/m2.K (75 – 25)OC

q’ = 217.74 W/m2

b) ȟ = 2 hL ȟ = 2 (4.355 W/m2.K) = 8.71 W/m2.K

q ’ = ȟ (Ts –T) = 8.71 W/m2.K (75 – 25)OC

q’ = 435.5 W/m2

A large heat transfer engineering problems and of importance involves the flow of fluids through pipes, particularly in heat exchangers.Conditions:1. The velocity of the fluid throughout the tube and

at all points in any cross section of the stream is constant, so that u = uO = V

2. Wall temperature is constant3. Properties of the fluid are independent of

temperature

LAMINAR FLOW HEAT TRANSFER INSIDE TUBES

Fo = αtT = 4ktT = 4kL rm2 CpρD2 CpρD2V

Gz = m Cp = Π Re Pr D ; Pe = Re Pr = DV k L 4 L L

Gz = m Cp = Π Re Pr D = Π D Pe = Π k L 4 L 4 L Fo

The Nusselt number for heat transfer to a fluid inside a pipe is

Nu = hi D kwhere the film coefficient hi is the average value over the length of the pipe and is calculate for the case of constant wall temperature:


hi = m Cp (Tb – Ta) ΠDL ΔŤL

Since ΔŤL = (Tw – Ta) – (Tw – Ťb)

ln (Tw – Ta / Tw – Ťb)

hi = m Cp ln Tw – Ta ΠDL Tw – Ťb

Then Nu = mCp ln Tw – Ta ΠkL Tw – Ťb

Or Nu = Gz ln Tw – Ta Π Tw – Ťb


Asymptotic Limitation: Ťb = Tw

Nu = 2 Gz and Gz ≈ 10 ΠFor laminar flow of fluids inside horizontal tubes, correction factor for heating and cooling

Nu = 2 m Cp 1/3 μ 0.14 = 2 Gz1/3 Фᵥ kL μw


For liquids, μw < μ and Фᵥ > 1 when the liquid is being heated and μw > μ and Фᵥ < 1 when the liquid is being cooled. For gases, the viscosity increases with temperature, so the inequalities are reversed. However, the change in viscosity is relatively small and that the term Фᵥ is usually omitted when dealing with gases.

Convective correlations for constant q in laminar flow with fully developed velocity and thermal profiles;

NuD = 4.364Convective correlations for constant surface temperature laminar flow with fully developed velocity and thermal profiles

NuD = 3.66

For heating water from 20OC to 60OC an electrically heated tube resulting in a constant heat flux of 10 kW/m2 is proposed. The mass flow rate is to be such that ReD = 2000, and consequently the flow must remain laminar. The tube inside diameter is 25 mm. the flow is fully developed. Determine the length of the tube required.

Ťb = (20 + 60) OC = 40OC 2

At 40OC; ρ = 994.6 kg/m3 ν = 0.658 x 10-6 m2/sCp = 4.1784 kJ/kg.K k = 0.628 W/m.K

q = q (A) = q’ ; q = m Cp (Tb – Ta) A

q’ (A) = q’ (ΠDL) = m Cp (Tb – Ta)

L = m Cp ( Tb – Ta) and m = ρu(ΠD2) q’ (ΠD) 4L = ρuDCp (Tb – Ta) but u = Re ν 4 q’ D

u = Reν = 2000(0.658 x 10-6 m2/s) = 0.0526 m/s D 0.025 m

L = ρuDCp (Tb – Ta) 4 q’

= 994.6kg (0.0526m)(0.025 m)(4178.4 J )(60 – 20)OC m3 s kg.K 4 (10000 J /s.m2)

L = 5.46 m

Air at 1.0 atmospheric pressure and 77OC enters a 5.0 mm ID tube with a bulk average velocity of 2.5 m/s. The velocity profile is developed and the thermal profile is “developing”. The tube length is 1.0 m, and a constant heat flux is imposed by the tube surface on the air over the entire length. An exit air bulk average temperature, Tb = 127OC, is required. Determinea) The exit h value, hL

b) The uniform heat flux c) The exit tube surface temperature

Ťb = 77 + 127 = 102OC 2

ρ = 0.9403 kg/m3 ν = 23.33 x 10-6 m2/sCp = 1.0115 kJ/kg.K k = 0.03184 W/m.Kμ = 2.1805 x 10-5 kg/m.s

ReD = Du = 0.005 m (2.5 m/s)

ν 23.33 x 10-6 m2/sReD = 536 (laminar)

Pr = Cpμ = 1.0115 kJ/kg.K(2.1805 x 10-5 kg/m.s) = 0.693 k 0.03184 W/m.K

Since L/D = 1.0/0.005 = 200, Consider the velocity and temperature profile to be fully developed over the tube’s entire length

Nu = 4.364

Nu = hD / k ; h = 4.364 k/D

h = 4.364 (0.0318 W/m.K) 0.005 m

h = 27.79 W/m2.K = hL = constant

q = q’ As = m Cp (Tb – Ta)

q’ = ρ u (Π/4D2) Cp (Tb –Ta) As

ṁ = 0.9403 kg/m3(2.5 m/s) Π/4(0.005m)2

ṁ = 4.615 x 10-5 kg/s

q’ = 4.615 x 10-5 kg/s (1011.5 J/kg.K)(127 – 77)OC Π (0.005 m)(1m)

q’ = 148.7 W/m2

q = h A (Ts – T∞) q’ = h (Ts – T∞)

Ts = 148.7 W/m2 + 127 OC 27.79 W/m2.K

Ts = 132.4OC

FORCED CONVECTION: TURBULENT FLOW

For turbulent flow of fluids inside a pipe, the relationship is given by the Sieder-Tate equation;

Nu = 0.023 Re0.80Prn Фᵥ

The effect of fluid properties on hi can be shown by condensing the above equation assuming that (μ/μw)0.14 = Фᵥ = 1

hi = 0.023 Gz0.80k2/3Cp1/3

D0.2μ0.47

ΔTi = ΔT Do 1 1 Di hi Uo

ΔTi = 1/hi ΔT 1/hi + Di/Do (1/ho)

ΔTi = inside resistance ΔT overall resistance

For heating : Tw = T + ΔTiFor cooling : Tw = T - ΔTi

FORCED CONVECTION: TRANSITION REGION BETWEEN LAMINAR AND

TURBULENT FLOW (6000 > Re > 2100)

Nu = 2 ΠD Re Pr 1/3 μ 0.14

4L μw

hi = Cp μ 2/3 μw 0.14

k μ

jH = 1.86 D 1/3 DG 2/3

L μ

Water at 30OC with a mass flow rate of 2 kg/s enters a 2.5 mm ID tube whose wall is maintained at a uniform temperature of 90OC calculate the length of the tube required to heat the water to 70OC.

Ťb = (30 + 70 ) OC /2 = 50OC

ρ = 990 kg/m3 μ = 0.547 x 10-3 kg/m.sCp = 4184 J/kg.K k = 0.644 W/m.K

u = 4m = 4 (2 kg/s) ρΠd2 990 kg/m3 Π (0.0025 m)2 u = 411.55 m/s

Re = duρ = (0.0025 m)(411.55m/s)(990 kg/m3) μ 0.547 x 10-3 kg/m.s

Re = 1862141.93 (turbulent)

Pr = Cp μ = (4184 J/kg.K)(0.547 x 10-3 kg/m.s) = 3.55 k 0.644 W/m.K

Nu = 0.023 Re0.80Pr0.4

Nu = 0.023 (1862141.93)0.80 (3.5538)0.4

Nu = 3962.91952

h = Nu k = 3962.91952(0.644 W/m.K) D (0.0025 m) h = 1020848.07 W/m2.K

q = hAΔTm = h (ΠDL) ΔTm = mCp (Tb –Ta)

ΔTm = ΔTi – Δto / ln (Δti/ Δto) ΔTi = 90 – 30 = 60OCΔTi = 90 – 70 = 20OC

ΔTm = LMTD = 60 – 20 = 36.41 OC ln (60/20)

L = mCp(Tb – Ta) hΠD ΔTm L = 2 kg/s (4184 J/kg.K) (70 – 30)OC 1020848.07 W/m2.K Π (0.0025 m) (36.41OC)

L = 1.15 m

Forced Convection on a Flat Plate

Assuming the transition from Laminar to turbulent flow takes place at a Reynolds's Number of 5 x105, determine the distance from the leading edge of a flat plate at which transition occurs for the flow of each of the following fluids with a velocity of 2 m/s at 40OC. Comment on the type of flow for the 5 fluids if the total length of the plate is 1 m.

• A) Air at atmospheric pressure• B) hydrogen at atmospheric pressure• C) Water• D) engine oil• E) Mercury

Given: Re = 5 x105

Ťb = 40OCu = 2 m/s L = 1 m

Forced Convection through TubesAir at 206.8 kPa and an average of 477.6 K is being heated as it flows through a tube of 25.4 mm inside diameter at a velocity of 7.62 m/s. The heating medium is 488.7 K steam condensing on the outside of the tube. Since the heat transfer coefficient of condensing steam is several thousand W/m2.K and the resistance of the metal wall is very small, it will be assumed that the air is 488.7 K. Calculate the heat transfer coefficient for an L/D > 60 and also the heat transfer flux.

Water is flowing in a horizontal 1-in schedule 40 steel pipe at an average temperature of 65.6OC and a velocity of 2.44 m/s. It is being heated by condensing steam at 107.8OC on the outside of the pipe wall. The steam side coefficient has been estimated as ho = 105000 W/m2.K.a) Calculate the convective coefficient hi for water

inside the pipeb) Calculate the overall coefficient based on inside

surface area.c) Calculate the heat transfer rate for 0.305 m pipe

1. Water at 30OC with a mass flow rate of 2 kg/s enters a 2.5 cm ID tube where wall is maintained at a uniform temperature of 90OC. Calculate the length of the tube required to heat the water to 70 OC.

2. Three kg/min of liquid sodium is heated from a bulk mean temperature of 400OC to 500OC, as it flows through a stainless steel tube of 5cm ID and is 2mm thick. The sodium is heated by a constant wall heat flux, which maintain tube wall temperature at 30OC above the bulk temperature of sodium all along the length of the tube. Calculate the length of the tube required.

heat transfer to fluids without phase change

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