fundamentals of heat and mass transfer solutions manual phần 6 docx

SCHEMATIC: ASSUMPTIONS: 1 Steady-state conditions, 2 All heat dissipated by transformer transferred to glycerin, 3 Fully developed flow part a, 4 Negligible kinetic and potential energy

PROBLEM 8.28KNOWN: Inlet temperature, flow rate and properties of hot fluid Initial temperature, volume and

properties of pharmaceutical Heat transfer coefficient at outer surface and dimensions of coil

FIND: (a) Expressions for Tc(t) and Th,o(t), (b) Plots of Tc(t) and Th,o(t) for prescribed conditions.Effect of flow rate on time for pharmaceutical to reach a prescribed temperature

SCHEMATIC:

ASSUMPTIONS: (1) Constant properties, (2) Negligible heat loss from vessel to surroundings, (3)

Pharmaceutical is isothermal, (4) Negligible work due to stirring, (5) Negligible thermal energygeneration (or absorption) due to chemical reactions associated with the batch process, (6) Negligiblekinetic energy, potential energy and flow work changes for the hot fluid, (7) Negligible tube wallconduction resistance

ANALYSIS: (a) Performing an energy balance for a control surface about the stirred liquid, it

or, Th,o( ) t = Tc+ ( Th,i− Tc) exp ( − UA / m cs h p,h) (4) <

Substituting Eqs (2) and (4) into Eq (1),

Continued …

Eq (5) may be used to determine Tc(t) and the result used with (4) to determine Th,o(t).

(b) To evaluate the temperature histories, the overall heat transfer coefficient, ( )1

the pharmaceuticals increases with time due to heat transfer from the hot fluid, approaching the inlettemperature of the hot fluid (and its maximum possible temperature of 200°C) at t = 3600s

PROBLEM 8.28 (Cont.)

With increasing Tc, the rate of heat transfer from the hot fluid decreases (from 4.49 × 105 W at t = 0

to 6760 W at 3600s), in which case Th,o increases (from 125.2°C at t = 0 to 198.9°C at 3600s) Thetime required for the pharmaceuticals to reach a temperature of Tc = 160°C is

COMMENTS: Although design changes involving the length and diameter of the coil can be used to

alter the heating rate, process control parameters are limited to Th,i and m h

KNOWN: Tubing with glycerin welded to transformer lateral surface to remove dissipated power.

Maximum allowable temperature rise of coolant is 6°C

FIND: (a) Required coolant rate m , tube length L and lateral spacing S between turns, and (b) Effect of

flowrate on outlet temperature and maximum power

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) All heat dissipated by transformer transferred to

glycerin, (3) Fully developed flow (part a), (4) Negligible kinetic and potential energy changes, (5)Negligible tube wall thermal resistance

PROPERTIES: Table A.5, Glycerin ( mT ≈ 300 K): ρ = 1259.9 kg/m3, cp = 2427 J/kg⋅K, µ = 79.9 × 102

N⋅s/m2, k = 286 × 10-3 W/m⋅K, Pr = 6780

ANALYSIS: (a) From an overall energy balance assuming the maximum temperature rise of the

glycerin coolant is 6°C, find the flow rate as

PROBLEM 8.29 (Cont.)

(b) Parametric calculations were performed using the IHT Correlations Toolpad based on Eq 8.56 (a

thermal entry length condition), and the following results were obtained

With Ts maintained at 47°C, the maximum allowable transformer power (heat rate) and glycerin outlettemperature increase and decrease, respectively, with increasing m The increase in q is due to an

increase in NuD (and hence h) with increasing ReD The value of NuD increased from 5.3 to 9.4 withincreasing m from 0.05 to 0.25 kg/s

DD

Gz− = L D Re Pr = (15.3 m/0.02 m)/(5.47 × 6780) = 0.0206 < 0.05,entrance length effects are significant, and Eq 8.56 should be used to determine NuD

KNOWN: Diameter and length of copper tubing Temperature of collector plate to which tubing is

soldered Water inlet temperature and flow rate

FIND: (a) Water outlet temperature and heat rate, (b) Variation of outlet temperature and heat rate with

flow rate Variation of water temperature along tube for the smallest and largest flowrates

SCHEMATIC:

ASSUMPTIONS: (1) Straight tube with smooth surface, (2) Negligible kinetic/potential energy and

flow work changes, (3) Negligible thermal resistance between plate and tube inner surface, (4) ReD,c =2300

PROPERTIES: Table A.6, water (assume mT = (Tm,i + Ts)/2 = 47.5°C = 320.5 K): ρ = 986 kg/m3, cp =

4180 J/kg⋅K, µ = 577 × 10-6 N⋅s/m2, k = 0.640 W/m⋅K, Pr = 3.77 Table A.6, water (Ts = 343 K): µs =

400 × 10-6 N⋅s/m2

ANALYSIS: (a) For m = 0.01 kg/s, ReD = 4 m  π µ D = 4(0.01 kg/s)/π(0.01 m)577 × 10-6 N⋅s/m2 =

2200, in which case the flow may be assumed to be laminar With fd,tx D ≈ 0.05ReDPr =

0.05(2200)(3.77) = 415 and L/D = 800, the flow is fully developed over approximately 50% of the tubelength With ( ) 1/ 3( )0.14

DD

Hence, q = mc  p( Tm,o− Tm,i) = 0.01kg s 4186 J kg K 36.3 K ( ⋅ )( ) = 1519 W <

(b) The IHT Correlations, Rate Equations and Properties Tool Pads were used to determine the

parametric variations The effect of m  was considered in two steps, the first corresponding to m  <0.011 kg/s (ReD < 2300) and the second for m  > 0.011 kg/s (ReD > 2300) In the first case, Eq 8.57 wasused to determine h, while in the second Eq 8.60 was used The effects of m  are as follows

Continued

Laminar flow (ReD < 2300)

Turbulent flow (ReD>2300)

Turbulent flow (ReD>2300)

The outlet temperature decreases with increasing m  , although the effect is more pronounced for laminarflow If q were independent of m , (Tm,o - Tm,i) would decrease inversely with increasing m In turbulent

flow, however, the convection coefficient, and hence the heat rate, increases approximately as m 0.8,thereby attenuating the foregoing effect In laminar flow, q ~ m 0.5 and this attenuation is not as

0 2 4 6 8

Axial location, x(m) 20

30 40 50 60 70

mdot = 0.05 kg/s

The more pronounced increase for turbulent flow is due to the much larger value of h (4300 W/m2⋅K for

m = 0.05 kg/s relative to 217 W/m2⋅K for m  = 0.05 kg/s)

PROBLEM 8.31KNOWN: Diameter and surface temperature of ten tubes in an ice bath Inlet temperature and flowrate

per tube Volume (∀) of container and initial volume fraction, fv,i, of ice

FIND: (a) Tube length required to achieve a prescribed air outlet temperature Tm,o and time to

completely melt the ice, (b) Effect of mass flowrate on Tm,o and suitable design and operating conditions

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3)

Constant properties, (4) Fully developed flow throughout each tube, (5) Negligible tube wall thermalresistance

PROPERTIES: Table A.4, air (assume mT = 292 K): cp = 1007 J/kg⋅K, µ = 180.6 × 10-7 N⋅s/m2, k =0.0257 W/m⋅K, Pr = 0.709; Ice: ρ = 920 kg/m3, hsf = 3.34 × 105 J/kg

ANALYSIS: (a) With ReD = 4 m  /πDµ = 4(0.01 kg/s)/π(0.05 m)180.6 × 10-7 N⋅s/m2 = 14,100 for m  =0.01 kg/s, the flow is turbulent, and from Eq 8.60,

0 0.01 0.02 0.03 0.04 0.05

Mass flowrate per tube, mdot(kg/s) 12

13 14 15 16 17

Although heat extraction from the air passing through each tube increases with increasing flowrate, theincrease is not in proportion to the change in m  and the temperature difference (Tm,i - Tm,o) decreases If0.05 kg/s of air is routed through a single tube, the outlet temperature of Tm,o = 16.2°C slightly exceedsthe desired value of 16°C The prescribed value could be achieved by slightly increasing the tube length.However, in the interest of reducing pressure drop requirements, it would be better to operate at a lowerflowrate per tube If, for example, air is routed through four of the tubes at 0.01 kg/s per tube and thedischarge is mixed with 0.01 kg/s of the available air at 24°C, the desired result would be achieved

COMMENTS: Since the flow is turbulent and L/D = 31, the assumption of fully developed flow

throughout a tube is marginal and the foregoing analysis overestimates the discharge temperature

PROBLEM 8.32KNOWN: Thermal conductivity and inner and outer diameters of plastic pipe Volumetric flow rate and inlet and outlet temperatures of air flow through pipe Convection coefficient and temperature of water.

FIND: Pipe length and fan power requirement.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from air in vertical legs of pipe, (3)

Negligible flow work and potential and kinetic energy changes for air flow through pipe, (4) Smoothinterior surface, (5) Constant properties

PROPERTIES: Table A-4, Air (Tm,i = 29°C): ρi= 1.155 kg / m 3 Air (T m = ° 25 C :) cp = 1007J/kg⋅K, µ = 183.6 × 10-7 N⋅s/m2, ka = 0.0261 W/m⋅K, Pr = 0.707

ANALYSIS: From Eq (8.46a)

where f = 0.316 ReD−1/ 4 = 0.0294 from Eq (8.20a)

COMMENTS: (1) With L/Di = 91, the assumption of fully developed flow throughout the pipe is

justified (2) The fan power requirement is small, and the process is economical (3) The resistance

to heat transfer associated with convection at the outer surface is negligible

KNOWN: Flow rate, inlet temperature and desired outlet temperature of water passing through a tube of

prescribed diameter and surface temperature

FIND: (a) Required tube length, L, for prescribed conditions, (b) Required length using tube diameters

over the range 30 ≤ D ≤ 50 mm with flow rates m  = 1, 2 and 3 kg/s; represent this design informationgraphically, and (c) Pressure gradient as a function of tube diameter for the three flow rates assuming thetube wall is smooth

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible potential energy, kinetic energy and flow

work changes, (3) Constant properties

PROPERTIES: Table A.6, Water ( mT = 323 K): cp = 4181 J/kg⋅K, µ = 547 × 10-6 N⋅s/m2, k = 0.643W/m⋅K, Pr = 3.56

ANALYSIS: (a) From Eq 8.6, the Reynolds number is

(b) Using the IHT Correlations Tool, Internal

Flow, for fully developed Turbulent Flow, along

with appropriate energy balance and rate

equations, the required length L as a function of

flow rate is computed and plotted on the right

Tube diameter, D (mm) 5

10 15

PROBLEM 8.33 (Cont.)

(c) From Eq 8.22a the pressure drop is

2m

Using IHT with these equations and Eq (1), the

pressure gradient as a function of diameter for

the selected flow rates is computed and plotted

on the right

Tube diameter, D (mm)

0 1000 2000 3000 4000 5000

(2) The IHT Workspace used to generate the graphical results are shown below

// Rate Equation Tool - Tube Flow with Constant Surface Temperature:

/* For flow through a tube with a uniform wall temperature, Fig 8.7b, the

overall energy balance and heat rate equations are */

q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37

(Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * hDbar / (mdot * cp)) // Eq 8.42b

// where the fluid and constant tube wall temperatures are

Ts = 100 + 273 // Tube wall temperature, K

Tmi = 25 + 273 // Inlet mean fluid temperature, K

Tmo = 75 + 273 // Outlet mean fluid temperature, K

// The tube parameters are

mdot = 1 // Mass flow rate, kg/s

// Correlation Tool - Internal Flow, Fully Developed Turbulent Flow (Assumed):

NuDbar = NuD_bar_IF_T_FD(ReD,Pr,n) // Eq 8.60

n = 0.4 // n = 0.4 or 0.3 for Ts>Tm or Ts<Tm

NuDbar = hDbar * D / k

ReD = um * D / nu

/* Evaluate properties at the fluid average mean temperature, Tmbar */

Tmbar = Tfluid_avg (Tmi,Tmo)

// Properties Tool - Water:

// Water property functions :T dependence, From Table A.6

// Units: T(K), p(bars);

x = 0 // Quality (0=sat liquid or 1=sat vapor)

rho = rho_Tx("Water",Tmbar,x) // Density, kg/m^3

cp = cp_Tx("Water",Tmbar,x) // Specific heat, J/kg·K

nu = nu_Tx("Water",Tmbar,x) // Kinematic viscosity, m^2/s

k = k_Tx("Water",Tmbar,x) // Thermal conductivity, W/m·K

Pr = Pr_Tx("Water",Tmbar,x) // Prandtl number

// Pressure Gradient, Equations 8.21, 8.22a:

dPdx = f * rho * um^2 / ( 2 * D )

f = ( 0.790 * ln (ReD) - 1.64 ) ^ -2

KNOWN: Flow rate and inlet temperature of water passing through a tube of prescribed length,

diameter and surface temperature

FIND: (a) Outlet water temperature and rate of heat transfer to water for prescribed conditions, and (b)

Compute and plot the required tube length L to achieve Tm,o found in part (a) as a function of the surfacetemperature for the range 85 ≤ Ts≤ 95°C

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy,

potential energy and flow work effects, (4) Fully developed flow conditions

PROPERTIES: Table A.6, Water ( mT ≈ 325 K): cp = 4182 J/kg⋅K, µ = 528 × 10-6 N⋅s/m2, k = 0.645W/m⋅K, Pr = 3.42

ANALYSIS: (a) From Eq 8.6, the Reynolds number is

(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the

energy balance and rate equations used above, the required length, L, to achieve Tm,o = 44.9°C (seecomment 1 below) as a function of tube surface temperature is computed and plotted below

Continued

PROBLEM 8.34 (Cont.)

Tube temperature, Ts (C) 40

42 44 46 48 50

From the plot, the outlet temperature increases nearly linearly with the surface temperature Theconvection coefficient and heat rate show similar behavior for this range of conditions

COMMENTS: (1) The mean temperature Tm = 325 K was overestimated in part (a) Another iteration

is recommended and the results with mT = 309 K are: h = 6091 W/m2⋅K, Tm,o = 44.9°C and q = 167kW

(2) The IHT Workspace used to generate the graphical results are shown below.

// Rate Equation Tool - Tube Flow with Constant Surface Temperature:

/* For flow through a tube with a uniform wall temperature, Fig 8.7b, the overall energy balance and heat

rate equations are */

q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37

(Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * hDbar / (mdot * cp)) // Eq 8.42b

// where the fluid and constant tube wall temperatures are

Ts = 90 + 273 // Tube wall temperature, K

Ts_C = Ts - 273

Tmi = 25 + 273 // Inlet mean fluid temperature, K

//Tmo = // Outlet mean fluid temperature, K

L = 4 // Tube length, m; unknown

// The tube mass flow rate and fluid thermophysical properties are

mdot = rho * um * Ac

mdot = 2 // Mass flow rate, kg/s

// Correlation Tool - Internal Flow, Fully Developed Turbulent Flow (Assumed):

NuDbar = NuD_bar_IF_T_FD(ReD,Pr,n) // Eq 8.60

n = 0.4 // n = 0.4 or 0.3 for Ts>Tm or Ts<Tm

NuDbar = hDbar * D / k

ReD = um * D / nu

/* Evaluate properties at the fluid average mean temperature, Tmbar */

Tmbar = Tfluid_avg (Tmi,Tmo)

// Properties Tool - Water:

// Water property functions :T dependence, From Table A.6

// Units: T(K), p(bars);

x = 0 // Quality (0=sat liquid or 1=sat vapor)

rho = rho_Tx("Water",Tmbar,x) // Density, kg/m^3

cp = cp_Tx("Water",Tmbar,x) // Specific heat, J/kg·K

mu = mu_Tx("Water",Tmbar,x) // Viscosity, N·s/m^2

nu = nu_Tx("Water",Tmbar,x) // Kinematic viscosity, m^2/s

k = k_Tx("Water",Tmbar,x) // Thermal conductivity, W/m·K

Pr = Pr_Tx("Water",Tmbar,x) // Prandtl number

KNOWN: Diameters and thermal conductivity of steel pipe Temperature and velocity of water flow

in pipe Temperature and velocity of air in cross flow over pipe Cost of producing hot water

FIND: Daily cost of heat loss per unit length of pipe.

SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Negligible radiation from outer

surface, (4) Fully-developed flow in pipe

PROPERTIES: Table A-4, air (p = 1 atm, Tf≈ 300K): ka = 0.0263 W/m⋅K, νa = 15.89 × 10-6 m2/s,

Pra = 0.707 Table A-6, water (Tm = 323 K): ρw = 988 kg/m3, µw = 548 × 10-6 N⋅s/m2, kw = 0.643W/m⋅K, Prw = 3.56

ANALYSIS: The heat loss per unit length of pipe is

The daily energy loss is then Q ′ = 0.346 kW / m 24 h / d × = 8.22 kW h / d m ⋅ ⋅

and the associated cost is C ′ = ( 8.22 kW h / d m $0.05 / kW h ⋅ ⋅ )( ⋅ = ) $0.411/ m d ⋅ <COMMENTS: Because Rcnv,a′ >>R ′cnv,w, the convection resistance for the water side of the pipecould have been neglected, with negligible error The implication is that the temperature of the pipe’sinner surface closely approximates that of the water If R ′cnv,w is neglected, the heat loss is

q ′ = 346 W / m.

PROBLEM 8.36 KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing

uniform heat generation Flow rate and inlet temperature of water flowing through the pipe.

FIND: (a) Pipe length required to achieve desired outlet temperature, (b) Location and value of

maximum pipe temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic

energy, potential energy and flow work changes, (4) One-dimensional radial conduction in pipe wall, (5) Outer surface in adiabatic.

PROPERTIES: Table A-1, Stainless steel 316 (T ≈ 400K): k = 15 W/m ⋅ K; Table A-6, Water

The temperature distribution and the maximum wall temperature (r = ro) are

si

COMMENTS: The physical situation corresponds to a uniform surface heat flux, and Tm

increases linearly with x In the fully developed region, Ts also increases linearly with x.

PROBLEM 8.37KNOWN: Dimensions and thermal conductivity of concrete duct Convection conditions of ambient

air Flow rate and inlet temperature of water flow through duct

FIND: (a) Outlet temperature, (b) Pressure drop and pump power requirement, (c) Effect of flow rate

and pipe diameter on outlet temperature

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Fully developed flow throughout duct, (3) Negligible pipe

wall conduction resistance, (4) Negligible potential energy, kinetic energy and flow work changes forwater, (5) Constant properties

PROPERTIES: Table A-6, water (T m ≈ 360 K :) 3

(b) With f = 0.0206 from Fig 8.3 and um = m /  ρπD / 42 = 0.117 m / s,

3 2

m 967 kg / m 0.117 m / su

in the residence time of the water in the pipe (um increases with increasing m  for fixed D) Withincreasing D for fixed m  and w, Tm,o decreases due to an increase in the residence time, as well as areduction in the conduction resistance, Rcnd.

COMMENTS: (1) Use of Tm=360 K to evaluate properties of the water for Parts (a) and (b) isreasonable, and iteration is not necessary (2) The pressure drop and pump power requirement aresmall

PROBLEM 8.38KNOWN: Water flow through a thick-walled tube immersed in a well stirred, hot reaction tank

maintained at 85°C; conduction thermal resistance of the tube wall based upon the inner surface area

is Rcd′′ =0.002 m2⋅K / W

FIND: (a) The outlet temperature of the process fluid, Tm,o; assume, and then justify, fully

developed flow and thermal conditions within the tube; and (b) Do you expect Tm,o to increase ordecrease if the combined thermal entry condition exists within the tube? Estimate the outlet

temperature of the process fluid for this condition

SCHEMATIC:

ASSUMPTIONS: (1) Flow is fully developed, part (a), (2) Constant properties, (3) Negligible

kinetic and potential energy changes and flow work, and (4) Constant wall temperature heating

PROPERTIES: Table A-6, Water (Tm = (Tm,o + Tm,i)/2 = 337 K): cp = 4187 J/kg⋅K, µ = 4.415 ×

That is, the length is only twice that required to reach fully developed conditions

(b) Considering combined entry length conditions, estimate the convection coefficient using theSieder-Tate correlation, Eq 8.56,

COMMENTS: (1) The thermophysical properties for the fully developed correlation are evaluated at

the mean fluid temperature Tm = (Tm,o + Tm,i)/2 The values are shown above in the propertiessection

(2) For the Sieder-Tate correlation, the properties are also evaluated at Tm, except for µs, which isevaluated at Ts

(3) For this case where the tube length is about twice xfd,t, the average heat transfer coefficient islarger as we would expect, but amounts to only a 10% increase

PROBLEM 8.39KNOWN: Flow rate and temperature of atmospheric air entering a duct of prescribed diameter, length

and surface temperature

FIND: (a) Air outlet temperature and duct heat loss for the prescribed conditions and (b) Calculate and

plot q and ∆p for the range of diameters, 0.1 ≤ D ≤ 0.2 m, maintaining the total surface area, As = πDL, atthe same value as part (a) Explain the trade off between the heat transfer rate and pressure drop

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy and

potential energy changes, (4) Uniform surface temperature, (5) Fully developed flow conditions

PROPERTIES: Table A.4, Air ( mT ≈ 310 K, 1 atm): ρ = 1.128 kg/m3, cp = 1007 J/kg⋅K, µ = 189 × 107

and for the smooth surface conditions, Eq 8.21 can be used to evaluate the friction factor,

10 20 30 40 50

From above, as D increases, L decreases so that

As remains unchanged The decrease in heat

rate with increasing diameter is nearly linear,

while the pressure drop decreases markedly

This is the trade off: increased heat rate requires

a more significant increase in pressure drop, and

hence fan blower power requirements

Tube diameter, D (mm) 0

0.5 1 1.5 2

PROBLEM 8.40KNOWN: Inlet temperature, pressure and flow rate of air Tube diameter and length Pressure of

saturated steam

FIND: Outlet temperature and pressure of air Mass rate of steam condensation.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Outer surface of annulus is adiabatic, (3) Negligible potential

energy, kinetic energy and flow work changes for air, (4) Fully-developed flow throughout the tube,(5) Smooth tube surface, (6) Constant properties

PROPERTIES: Table A-4, air (T m ≈ 325 K, p = 5 atm :) ( ) 3

KNOWN: Duct diameter and length Thermal conductivity of insulation Gas inlet temperature and

velocity and minimum allowable outlet temperature Temperature and velocity of air in cross flow

FIND: Minimum allowable insulation thickness.

SCHEMATIC:

ASSUMPTIONS: (1) Negligible potential and kinetic energy and flow work changes for gas flow

through duct, (2) Fully developed flow throughout duct, (3) Negligible duct wall conduction

resistance, (4) Negligible effect of insulation thickness on outer convection coefficient and thermalresistance, (5) Properties of gas may be approximated as those of air

PROPERTIES: Table A-4, air (p = 1 atm) Tm,i = 1600K: (ρi = 0.218 kg/m3) Tm = (Tm,i +Tm,o)/2

KNOWN: Flow rate, inlet temperature and desired outlet temperature of liquid mercury flowing

through a tube of prescribed diameter and surface temperature

FIND: Required tube length and error associated with use of a correlation for moderate to large Pr

fluids

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy,

potential energy and flow work effects, (4) Fully developed flow

PROPERTIES: Table A-5, Mercury c Tm= 350 K h : cp = 137.7 J/kg⋅K, µ = 0.1309 × 10-2 N⋅s/m2,

COMMENTS: Such good agreement between results does not occur in general For example, if

ReD = 2 × 104, h = 1463 from Eq 8.66 and 2417 from Eq 8.60 Large errors are usually associatedwith using conventional (moderate to large Pr) correlations with liquid metals

PROBLEM 8.43 KNOWN: Surface temperature and diameter of a tube Velocity and temperature of air in cross

flow Velocity and temperature of air in fully developed internal flow.

FIND: Convection heat flux associated with the external and internal flows.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform cylinder surface temperature, (3) Fully

developed internal flow.

PROPERTIES: Table A-4, Air (298K): ν = 15.71 × 10-6 m2/s, k = 0.0261 W/m ⋅ K, Pr = 0.71.

ANALYSIS: For the external and internal flows,

4m

KNOWN: Diameter, length and surface temperature of condenser tubes Water velocity and inlet

ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy,

potential energy and flow work changes

PROPERTIES: Table A.6, Water ( mT = 300 K): ρ = 997 kg/m3, cp = 4179 J/kg⋅K, µ = 855 × 10-6kg/s⋅m, k = 0.613 W/m⋅K, Pr = 5.83

ANALYSIS: (a) From Equation 8.42b

(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the

energy balance and rate equations above, the calculation of part (a) is repeated with mT = (Tm,i + Tm,o)/2giving these results:

m

T = 307.3 K Tm,o = 51.7°C = 324.7 K <

(c) Using the IHT model developed for the part (b) analysis, the coolant mean velocity, um, as a function

of tube length L with Tm,o = 51.7°C is calculated and the results plotted below

Continued

PROBLEM 8.44 (Cont.)

Tube length, L (m) 0

1 2 3 4 5 6

COMMENTS: (1) Using mT = 300 K vs mT = (Tm,i + Tm,o)/2 = 307 K for this application resulted in

a difference of Tm,o = 50°C vs.Tm,o = 51.7°C While the difference is only 1.7°C, it is good practice touse the proper value for mT

(2) Note that um must be increased markedly with increasing length in order that Tm,o remain fixed

KNOWN: Gas turbine vane approximated as a tube of prescribed diameter and length maintained at a

known surface temperature Air inlet temperature and flowrate

FIND: (a) Outlet temperature of the air coolant for the prescribed conditions and (b) Compute and plot

the air outlet temperature Tm,o as a function of flow rate, 0.1 ≤ m  ≤ 0.6 kg/h Compare this result withthose for vanes having passage diameters of 2 and 4 mm

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes.

PROPERTIES: Table A.4, Air (assume mT = 780 K, 1 atm): cp = 1094 J/kg⋅K, k = 0.0563 W/m⋅K, µ

DD

(b) Using the IHT Correlations Tool, Internal Flow, for Laminar Flow with combined entry length, along

with the energy balance and rate equations above, the outlet temperature Tm,o was calculated as a function

of flow rate for diameters of D = 2, 3 and 4 mm The plot below shows that Tm,o decreases nearly

linearly with increasing flow rate, but is independent of passage diameter

Continued

COMMENTS: (1) Based upon the calculation for Tm,o = 578°C, mT = 775 K which is in good

agreement with our assumption to evaluate the thermophysical properties

(2) Why is Tm,o independent of D? From Eq (3), note that h is inversely proportional to D, h ~ D-1.From Eq (1), note that on the right-hand side the product P⋅h will be independent of D Hence, Tm,o willdepend only on m This is, of course, a consequence of the laminar flow condition and will not be the

same for turbulent flow

KNOWN: Gas-cooled nuclear reactor tube of 20 mm diameter and 780 mm length with helium heated

from 600 K to 1000 K at 8 × 10-3 kg/s

FIND: (a) Uniform tube wall temperature required to heat the helium, (b) Outlet temperature and

required flow rate to achieve same removal rate and wall temperature if the coolant gas is air

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic energy and potential energy

changes, (3) Fully developed conditions

PROPERTIES: Table A-4, Helium (Tm =800K, 1 atm :) ρ = 0.06272 kg/m3, cp = 5193 J/kg⋅K, k

Results of the iterative solution are

COMMENTS: To achieve the same cooling rate with air, the required mass rate is 6.5 times that

obtained with helium

KNOWN: Air at prescribed inlet temperature and mean velocity heated by condensing steam on its

outer surface.

FIND: (a) Air outlet temperature, pressure drop and heat transfer rate and (b) Effect on parameters

of part (a) if pressure were doubled.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes,

(3) Thermal resistance of tube wall and condensate film are negligible.

PROPERTIES: Table A-4, Air (assume Tm = 450K, 1 atm = 101.3 kPa): ρ = 0.7740 kg/m3,

cp = 1021 J/kg ⋅ K, µ = 250.7 × 10-7 N ⋅ s/m2, k = 0.0373 W/m ⋅ K, Pr = µ cp/k = 0.686 Note that only ρ is pressure dependent; i.e., ρ α P; Table A-6, Saturated water (20 bar): Tsat = Ts = 485K.

ANALYSIS: (a) For constant wall temperature heating, from Eq 8.46 but with U ≈ h sincei

Part (a) Part (b)

KNOWN: Diameter, length and surface temperature of tubes used to heat ambient air Flowrate and

inlet temperature of air

FIND: (a) Air outlet temperature and heat rate per tube, (b) Effect of flowrate on outlet temperature.

Design and operating conditions suitable for providing 1 kg/s of air at 75°C

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3)

Negligible tube wall thermal resistance

PROPERTIES: Table A.4, air (assume mT = 330 K): cp = 1008 J/kg⋅K, µ = 198.8 × 10-7 N⋅s/m2, k =0.0285 W/m⋅K, Pr = 0.703

ANALYSIS: (a) For m  = 0.01 kg/s, ReD = 4m  π µ D = 0.04 kg/s/π(0.05 m)198.8 × 10-7 N⋅s/m2 =12,810 Hence, the flow is turbulent If fully developed flow is assumed throughout the tube, the Dittus-Boelter correlation may be used to obtain the average Nusselt number

Hence, q = mc  p( Tm,o− Tm,i) = 0.01kg s 1008 J kg K 65.6 K ( ⋅ ) = 661W <

(b) The effect of flowrate on the outlet temperature was determined by using the IHT Correlations and Properties Toolpads.

0 0.01 0.02 0.03 0.04 0.05

Mass flowrate, mdot(kg/s) 75

80 85 90

Continued

PROBLEM 8.48 (Cont.)

Although h and hence the heat rate increase with increasing m  , the increase in q is not linearly

proportional to the increase in m and Tm,o decreases with increasing m.

A flowrate of m  = 0.05 kg/s is not large enough to provide the desired outlet temperature of

75°C, and to achieve this value, a flowrate of 0.0678 kg/s would be needed At such a flowrate,

N = 1 kg/s/0.0678 kg/s = 14.75 ≈ 15 tubes would be needed to satisfy the process air

requirement Alternatively, a lower flowrate could be supplied to a larger number of tubes and the discharge mixed with ambient air to satisfy the desired conditions Requirements of this option are that

amb

Nm  + m  = 1kg / s

( Nm  + m amb) cp( Tm,o− Tm,i) = 1kg s 1008 J kg K 75 20 K × ⋅ ( − ) = 55, 400 W

where m  is the flowrate per tube Using a larger number of tubes with a smaller flowrate per

tube would reduce flow pressure losses and hence provide for reduced operating costs.

COMMENTS: With L/D = 5 m/0.05 m = 100, the assumption of fully developed conditions

throughout the tube is reasonable.