**UFRGS**

# Kreith SI - Ch-06

(Parte **4** de 5)

These estimates vary by about 3% around an average value of 1880 W/(m2 K). But the Sieder-Tate correlation is more applicable in this case because it takes the large variation of the viscosity with temperature into account.

Note that the above correlations require that all properties (except μs) be evaluated at the bulk temperature.

PROBLEM 6.21

In a refrigeration system, brine (10 per cent NaCl) by weight having a viscosity of 0.0016 (Ns)/m2 and a thermal conductivity of 0.85 W/(m K) is flowing through a long 2.5-cm-ID pipe at 6.1 m/s. Under these conditions, the heat transfer coefficient was found to be 16,500 W/(m2 K). For a brine temperature of –1°C and a pipe temperature of 18.3°C, determine the temperature rise of the brine per meter length of pipe if the velocity of the brine is doubled. Assume that the specific heat of the brine is 3768 J/(kg K) and that its density is equal to that of water.

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 546

• Brine flowing through a pipe

• Brine properties Viscosity (μ) = 0.0016 Ns/m2 Thermal conductivity (k) = 0.85 W/(m K)

10% NaCl by weight

• Pipe inside diameter (D) = 2.5 cm = 0.025 m

• Brine velocity (V) = 6.1 m/s

• Heat transfer coefficient (ch) = 16,500 W/(m2 K)

• Brine temperature (Tb) = –1°C • Pipe temperature (Ts) = 18.3°C

• Temperature rise of the brine per meter length (ΔTb/m) if the velocity is doubled (V = 12.2 m/s) ASSUMPTIONS

• Steady state • Fully developed flow

• Constant and uniform pipe wall temperature

• Density of the brine is the same as water density

Brine

Ts = 18.3°C

PROPERTIES AND CONSTANTS From Appendix 2, Table 13, density (ρ) of water ≈ 1000 kg/m3

SOLUTION The Reynolds number at the original velocity is

The thermal conductivity of the fluid can be calculated from the given heat transfer coefficient using the Dittus-Boelter correlation of Equation (6.63)

DNu = chD k

= 0.023 ReD 0.8 Prn where n = 0.4 for heating cRe k k =

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 547 k = ( )

= 0.852 W/(m K)

The Prandtl number is

The Reynolds number for the new velocity is twice the original Reynolds number: ReD = 190,626. From Equation (6.63): For fully developed flow

The temperature after one meter is given by Equation (6.36) outinTTΔ Δ

= ,out

,in bc o bc oTTTT −

,out

,inbsbsTTTT − −

= exp ( )

= 0.9048 (per m length)

ΔTb = Tb,out – Tb,in = (),out ,in

,in bs bs s bs

– Tb,in

ΔTb = [0.9043 (–1°C – 18.3°C) + 18.3°C] + 1°C = 1.84°C per meter length

PROBLEM 6.2

Derive an equation of the form hc = f(T, D, V) for turbulent flow of water through a long tube in the temperature range between 20° and 100°C.

• Turbulent water flow through a long tube • Water temperature range (Tb) = 20°C to 100°C

• An expression of the form ch = f(T, D. V)

• Steady state • Variation of properties with temperature can be approximated with a power law

• Fully developed flow

• Water is being heated

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 548

D Water

PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water

Temperature (°C) 20 100 Temperature (K) 293 373 Density, ρ (kg/m3) 998.2 958.4

Thermal conductivity, k ()W/(mK) 0.597 0.682

Absolute viscosity, ()2(Ns)m/ 993 × 10–6 277.5 × 10–6 Prandtl number, Pr 7.0 1.75

SOLUTION Applying the Dittus-Boelter expression of Equation (6.63) for the Nusselt number

DNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating ch = DNukD = 0.023

To put this in the required form, the fluid properties must be expressed as a function of temperature. Assuming the power law variation

Property = ATR where A and n are constant evaluated from the property values.

For density ρ(293) = 998.2 kg/m3 = A(293)n ρ(373) = 958.2 kg/m3 = A(373)n

Solving these simultaneously A = 2613 n = – 0.1694

Therefore, ρ(T) = 2613 T –0.1694 Applying a similar analysis for the remaining properties yields the following relationships

Substituting these into the expression for the heat transfer coefficient

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 549

Note that in equations of the type derived, the coefficient has definite dimensions. Hence, the use of such equations is limited to the conditions specified and are not recommended.

PROBLEM 6.23

The intake manifold of an automobile engine can be approximated as a 4 cm ID tube, 30 cm in length. Air at a bulk temperature of 20°C enters the manifold at a flow rate of 0.01 kg/s. The manifold is a heavy aluminum casting and is at a uniform temperature of 40°C. Determine the temperature of the air at the end of the manifold.

• Air flow through a tube • Tube inside diameter (D) = 4 cm = 0.04 m

• Tube length (L) = 30 cm = 0.30 m

• Inlet bulk temperature (Tb,in) = 20°C • Air flow rate (m ) = 0.01 kg/s

• Tube surface temperature (Ts) = 40°C

• Outlet bulk temperature (Tb,out) ASSUMPTIONS

• Steady state • Constant and uniform tube surface temperature

Air = 20°C

= 0.01 kg/sTm b,in

Aluminum Manifold = 40°CTs

D =4c m

L =3 0c m Tb,out =?

From Appendix 2, Table 27, for dry air at the inlet bulk temperature of 20°C

Thermal conductivity (k) = 0.0251 W/(m K) Absolute viscosity (μ) = 18,240 × 10–6 (Ns)/m2

Prandtl number (Pr) = 0.71 Specific heat (c) = 1012 J/(kg K)

SOLUTION The Reynolds number is

ReD = VDρμ = 4m Dπ μ

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 550

The Nusselt number for fully developed flow can be estimated from the Dittus-Boelter correlation of Equation (6.36)

Nufd = 0.023 ReD 0.8 Prn where n = 0.4 for heating

Nufd = 0.023 (17,451)0.8 (0.71)0.4 = 49.62

Since L/D = 30cm/4 cm = 7.5 < 60, the flow is not fully developed and the fully developed heat transfer coefficient must be corrected using Equation (6.68) fdNuNu = ,,cL cf dh h = 1 + a bLD

= 47.60 W/(m2 K)

Applying Equation (6.36) to determine the outlet air temperature outinTTΔ Δ

= ,out

,inbsbsTTTT −

Tb,out = Ts – (Ts – Tb,in) expchDLmc π −

Tb,out = 40°C – (40°C – 20°C) exp ( )

The rise in air temperature is not large enough to require another iteration using new air properties at the average bulk air temperature.

PROBLEM 6.24

High-pressure water at a bulk inlet temperature of 93°C is flowing with a velocity of 1.5 m/s through a 0.015-m-diameter tube, 0.3 m long. If the tube wall temperature is 204°C, determine the average heat transfer coefficient and estimate the bulk temperature rise of the water.

• Water flowing through a tube

• Bulk inlet water temperature (Tb,in) = 93°C • Water velocity (V) = 1.5 m/s

• Tube diameter (D) = 0.015 m

• Tube length (L) = 0.3 m

• Tube surface temperature (Ts) = 204°C

(a) The average heat transfer coefficient (,cLh) (b) The bulk temperature rise of the water (ΔTb)

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 551

• Steady state • Constant and uniform tube temperature

• Pressure is high enough to supress vapor generation.

SKETCH Ts = 204°C

Water = 93°C

= 0.015 m/sTV b,in

L=0.3m PROPERTIES AND CONSTANTS

From Appendix 2, Table 13, for water at the inlet bulk temperature of 93°C

Density (ρ) = 963.0 kg/m3 Thermal conductivity (k) = 0.679 W/(m K)

Kinematic viscosity (ν) = 0.314 × 10–6 m2/s Prandtl number (Pr) = 1.8 Specific heat (c) = 4205 J/(kg K)

SOLUTION The Reynolds number is

(a) The Nusselt number for fully developed flow can be estimated from the Dittus-Boelter correlation of Equation (6.36) fdNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating

Since L/D = 0.3/0.015 = 20 < 60, the heat transfer coefficient must be corrected by Equations (6.68) and (6.69). Since L/D is at the upper end of the range for (6.68) and the lower end of the range for (6.69), the average of the two equations will be used.

From (6.68)

Nu = ,cLch h = 1+ a bLD b = 2.08 × 10–6 ReD – 0.815 = 2.08 × 10–6 (71,656) – 0.815 = – 0.6

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 552

From (6.69)

= ,,cL cf dh

The average of the two values is ,,/ccfdhLh = 1.27

∴ ,cLh = 1.27()210,265W/(mK) = 13,037 W/(m2 K) (b) The bulk temperature can be calculated from Equations (6.36) outinTTΔ Δ

= ,out

,insbsbTTTT− − = expcp PLhmc

Tb,out = Ts – (Ts – Tb,in) exp = 4chL VDcρ −

Tb,out = 204 °C – (204°C – 93°C)

The bulk temperature rise is ΔTb = Tb,out – Tb,in = 111°C – 93ºC = 18°C PROBLEM 6.25

Suppose an engineer suggests that air is to be used instead of water in the tube of Problem 6.24 and the velocity of the air is to be increased until the heat transfer coefficient with the air equals that obtained with water at 1.5 m/s. Determine the velocity required and comment on the feasibility of the engineer’s suggestion. Note that the speed of sound in air at 100°C is 387 m/s.

From Problem 6.24: Water at a bulk inlet temperature of 93°C is flowing with a velocity of 1.5 m/s through a 0.015-m-diameter tube, 0.3 m long. If the tube wall temperature is 204°C, determine the average heat transfer coefficient and estimate the bulk temperature rise of the water.

• Air flow through a tube

• Bulk inlet air temperature (Tb,in) = 93°C • Tube diameter (D) = 0.015 m

• Tube length (L) = 0.3 m

• Tube surface temperature (Ts) = 204°C

• From Problem 6.23: ,cLh = 13,037 W/(m2 K)

• The velocity (V) required to obtain ,cLh = 13,037 W/(m2 K)

• Steady state • Constant and uniform tube temperature

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 553

Air

L=0.3m PROPERTIES AND CONSTANTS

From Appendix 2, Table 27, for dry air at the inlet bulk temperature of 93°C Thermal conductivity (k) = 0.0302 W/(m K)

Kinematic viscosity (ν) = 2.9 × 10–6 m2/s Prandtl number (Pr) = 0.71

The flow must be turbulent, therefore, the heat transfer coefficient of the fully developed case must be 13,037 W/(m2 K) as shown in Problem 6.24. Therefore, the Nusselt number is fdNu = ,cfdhD k

Applying the Dittus-Boelter correlation of Equation (6.63) fdNu = 0.023 ReD 0.8 Prn = 5099 where n = 0.4 for heating

Solving for the Reynolds number

ReD = VDν =

Solving for the velocity

This velocity is obviously unrealistic because it corresponds to a Mach number of 30. Under such conditions when the speed of sound is reached, a shock wave will form and choke the flow.

PROBLEM 6.26

Atmospheric air at 10°C enters a 2 m long smooth rectangular duct with a

7.5 cm × 15 cm cross-section. The mass flow rate of the air is 0.1 kg/s. If the sides are at 150°C, estimate (a) the heat transfer coefficient, (b) the air outlet temperature, (c) the rate of heat transfer, and (d) the pressure drop.

• Atmospheric air flow through a rectangular duct

• Inlet bulk temperature (Tb,in) = 10°C • Duct length (L) = 2 m

• Cross-section = 7.5 cm × 15 cm = 0.075 m × 0.15 m

• Mass flow rate (m) = 0.1 kg/s

• Duct surface temperature (Ts) = 150°C

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 554

(a) The heat transfer coefficient ()ch

(b) The air outlet temperature (Tb,out) (c) The rate of heat transfer (q)

(d) The pressure drop (Δp)

• Steady state • The duct is smooth

Air = 10°CTb,in

Ts = 150°C

L =2m

0.15 m Tb,out

SOLUTION The hydraulic diameter of the duct is

DH = 4cA P

For the first iteration, let Tb,out = 50°C. For dry air at the average bulk temperature of 30°C Density (ρ) = 1.128 kg/m3

Thermal conductivity (k) = 0.0258 W/(m K)

Absolute viscosity (μ) = 18.68 × 10–6 (Ns)/m2 Prandtl number (Pr) = 0.71

Specific heat (cp) = 1013 J/(kg K)

ReD = HVD

= HmDA μ

= 47,585 > 10,0 (Turbulent)

= 2m 0.1m = 20

(a) Therefore, entrance effects may be significant — the correction Equations (6.68) and (6.69) will be applied to the Dittus Boelter correlation, Equation (6.63). From Equation (6.63) fdNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 5

Since L/D = 20 is on the low end of the range of Equation (6.69) and the high end of the range for Equation (6.68), the average of these two corrections will be applied

From Equation (6.68) h = 1 + abLD where a = 24 Re–0.23 = 24(47,585)–0.23 = 2.02 b = 2.08 × 10–6 Re – 0.815 = 2.08 × 10–6 (47,585) – 0.815 = – 0.716

From Equation (6.69)

= 1.3

The average of the two values is ,cLh/,cfdh = 1.27

∴ ,cLh = 1.27()28.56W/(mK) = 36.2 W/(m2 K) (b) The outlet temperature is found by rearranging Equation (6.63)

Tb,out = Ts – (Tb,in – Ts) expcPLh

Tb,out = 150°C – (10°C – 150°C) exp () ( ) ()

No further iteration is needed since the result is close to the initial guess. (c) The rate of heat transfer is given by

(d) The friction can be estimated from the lowest line for Figure 6.18: Re = 47,585 → f ≈ 0.021. The pressure drop is given by Equation (6.13) oΔp = f HLD22

= f HL D

HmD

= f HLD8 ρ 2

HmD π

So

Δp = 0.0212 m

PROBLEM 6.27

Air at 16°C and atmospheric pressure enters a 1.25-cm-ID tube at 30 m/s. For an average wall temperature of 100°C, determine the discharge temperature of the air and the pressure drop if the pipe is (a) 10 cm long and (b) 102 cm long.

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 556

• Atmospheric air flowing through a tube

• Entering air temperature (Tb,in) = 16°C • Tube inside diameter (D) = 1.25 cm = 0.0125 m

• Air velocity (V) = 30 m/s

• Average wall surface temperature (Ts) = 100°C

The discharge temperature (Tb,out) and the pressure drop (Δp) if the pipe length (L) is

(a) 10 cm (0.1 m) (b) 102 cm (1.02 m)

• Steady state • The tube is smooth

L = (a) 10 cm (b) 102 cm

=3 0m /sTV b,in

Ts = 100°C Tb,out =?

From Appendix 2, Table 27, for dry air at the entering bulk temperature of 16°C

Density (ρ) = 1.182 kg/m3 Thermal conductivity (k) = 0.0248 W/(m K)

Kinematic viscosity (ν) = 15.3 × 10–6 m2/s Prandtl number (Pr) = 0.71 Specific heat (c) = 1012 J/(kg K)

The discharge temperature will first be calculated using air properties evaluated at the entering temperature and will then be recalculated using the average bulk air temperature of the first iteration to evaluate the air properties.

The Reynolds number is

ReD = VD

(a) L/D = 0.1 m/0.0125 m = 8 < 20. Therefore, the flow is not fully developed and the heat transfer coefficient will have to be corrected with Equation (6.68). The Dittus-Boelter correlation of Equation (6.63) will be used to calculate the fully developed Nusselt number fdNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating

Applying Equation (6.68)

Nu = ,cLch h = 1 + abLD

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 557 b = 2.08 × 10–6 ReD – 0.815 = 2.08 × 10–6 (24,510) – 0.815 = – 0.7640

(Parte **4** de 5)