CHNG 3802 Heat Transfer Tutorial Answers

31
Q1 Q 6 T k ρ Exothermic Reaction in stirred tank water 20 0.597 1001 4156 1.01E-03 6.96E-04 Q 10000 W/m3 oil 20 0.145 888 1770 0.799 0.21 H 2m air 1 atm 20 0.0262 1.203 1004 1.84E-05 1.93E-05 D 2m V 6.28318 Q 62831.8 W Flow 2 m/s A 12.56636 m2 Diameter 0.2 m U 150 Tr 80 C 33.33333 C Re Pr Nu h Tj 46.66667 C water 397617 7.01 1512.1 4513.7 oil 445 9753.31 N/A Maximum diameter if Tj is limited to 30oC air 1 atm 26152 0.71 68.4 9.0 Q=UA(Tr-Tj) Q/A=U(Tr-Tj) Q2 A=piDH Q=10,000*pi/D^2/4*H Calculate the heat transfer coefficients in the previous question using the Sieder Tate Correlation Q/A=10,000*D/4 D=4*Q/A/10,000 D=4*U(Tr-Tj)/10,000 Tj 30 Tr 80 50 7500 D 3 In both cases assume the wall is at 40oC Nu h water 1631.12 4868.9 oil air 1 atm 81.73 10.7 Q3 If one side of an exchanger had air flowing at 2 m/s and the other water at 2 m/s which would be the controlling resistance? 1/U 0.093604 U 10.683 U is approximately equal to h for the Air. The air would be the controlling resistance Q4 Water flowing through smooth pipe, heated by condnesing steam, assume waters resistance is controlling 1 m pipe Flowrate 0.2 kg/s 0.1 kg/s diameter 0.02 m 0.01 m area 0.000314 m2 0.00007853975 m2 u 0.635984 m/s 1.27196865154 m/s Re 1.26E+04 1.26E+04 f 0.003734 0.00373443524 302.4 Pa/m 2419.2 Pa/m Pr 7.01 7.01 Nu 95.83517 95.8351703746 h 2860.7 W/m2K 5721.4 W/m2K hA 179.7 W/K 359.5 W/K Q5 Concentric Pipe Heat Exchanger Water Cold Stream T k ρ m Water Tin 20 oC 20 0.597 1001 4156 ### Tout 40 oC 40 0.628 995 4178 ### Flow 0.11 kg/s 100 0.68 961 4216 ### ID 0.015 m 120 0.685 945 4250 ### IA 0.000177 m2 r 998 u 0.62372 m/s m 0.000837 Pressure Drop Inside Pipe Cp 4167 k 0.6125 Re 1.12E+04 Re 11153 f 0.003853427 Pr 5.70 544 Nu 79.77307 h 3257.4 W/m2K Hot Stream Outer Pipe Water Tin 120 oC Tout 100 oC Flow 0.11 kg/s ID 0.02 m OD 0.03 m IA 0.000393 m2 dm 0.01 m r 953 u 0.293927 m/s m 0.000212 Pressure Drop Outside Pipe Cp 4233 k 0.6825 Re 1.32E+04 Re 13228 f 0.003692477 Pr 1.31 174 Nu 50.84908 h 3470.45 W/m2K Steel Wall k 60 W/m/K Ao/(hiAi) 0.000409 ln(ro/ri) 4.79E-05 1/ho 0.000288 Uo 1341.528 LMTD 80 K Q 9167.4 W A 0.085419 m2 L 1.36 m CP m 20oC m 40oC ( o C) (W m -1 K -1 ) (kg m -3 ) (J kg -1 K - 1 ) (kg m -1 s - 1 ) (kg m -1 s - 1 ) DT DT UDT DP CP ( o C) (W m -1 K -1 ) (kg m -3 ) (J kg - 1 K -1 ) (kg m -1 s -1 ) DP DP Nu= hL k =0.027 Re 0.8 Pr 1/ 3 ( μ μ w ) 0.14 Smooth pipes, 2 .5× 10 3 < Re< 10 5 φ=0.0396 Re 0.25 Smooth pipes, 2 .5× 10 3 < Re< 10 y φ=2.5ln(Reφ 0.5 ) +0.3 ΔP f =4φ l d ρu 2

description

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Transcript of CHNG 3802 Heat Transfer Tutorial Answers

Page 1: CHNG 3802 Heat Transfer Tutorial Answers

Q1 Q 6

T k ρ Exothermic Reaction in stirred tank

water 20 0.597 1001 4156 1.01E-03 6.96E-04 Q 10000 W/m3

oil 20 0.145 888 1770 0.799 0.21 H 2 m

air 1 atm 20 0.0262 1.203 1004 1.84E-05 1.93E-05 D 2 mV 6.28318Q 62831.8 W

Flow 2 m/s A 12.56636 m2

Diameter 0.2 m U 150Tr 80 C

33.33333 CRe Pr Nu h Tj 46.66667 C

water 397617 7.01 1512.1 4513.7oil 445 9753.31 N/A Maximum diameter if Tj is limited to 30oCair 1 atm 26152 0.71 68.4 9.0

Q=UA(Tr-Tj)Q/A=U(Tr-Tj)

Q2 A=piDHQ=10,000*pi/D^2/4*H

Calculate the heat transfer coefficients in the previous question using the Sieder Tate Correlation Q/A=10,000*D/4D=4*Q/A/10,000D=4*U(Tr-Tj)/10,000

Tj 30Tr 80

507500

D 3

In both cases assume the wall is at 40oCNu h

water 1631.12 4868.9oilair 1 atm 81.73 10.7

Q3If one side of an exchanger had air flowing at 2 m/s and the other water at 2 m/s which would be the controlling resistance?

1/U 0.093604U 10.683U is approximately equal to h for the Air. The air would be the controlling resistance

Q4Water flowing through smooth pipe, heated by condnesing steam, assume waters resistance is controlling1 m pipe

Flowrate 0.2 kg/s 0.1 kg/sdiameter 0.02 m 0.01 marea 0.000314 m2 0.00007853975 m2u 0.635984 m/s 1.27196865154 m/s

Re 1.26E+04 1.26E+04f 0.003734 0.00373443524

302.4 Pa/m 2419.2 Pa/m

Pr 7.01 7.01Nu 95.83517 95.8351703746h 2860.7 W/m2K 5721.4 W/m2K

hA 179.7 W/K 359.5 W/K

Q5 Concentric Pipe Heat Exchanger

Water

Cold Stream T k ρ m

Water

Tin 20 oC 20 0.597 1001 4156 1.01E-03Tout 40 oC 40 0.628 995 4178 6.96E-04Flow 0.11 kg/s 100 0.68 961 4216 2.82E-04ID 0.015 m 120 0.685 945 4250 1.59E-04IA 0.000177 m2

r 998u 0.62372 m/sm 0.000837 Pressure Drop Inside PipeCp 4167k 0.6125 Re 1.12E+04Re 11153 f 0.003853427Pr 5.70 544Nu 79.77307h 3257.4 W/m2K

Hot Stream Outer PipeWaterTin 120 oCTout 100 oCFlow 0.11 kg/s

ID 0.02 mOD 0.03 mIA 0.000393 m2dm 0.01 m

r 953u 0.293927 m/sm 0.000212 Pressure Drop Outside PipeCp 4233k 0.6825 Re 1.32E+04Re 13228 f 0.003692477Pr 1.31 174Nu 50.84908h 3470.45 W/m2K

Steel Wallk 60 W/m/K

Ao/(hiAi) 0.000409

ln(ro/ri)/kx 4.79E-05

1/ho 0.000288

Uo 1341.528

LMTD 80 K

Q 9167.4 W

A 0.085419 m2

L 1.36 m

CP m 20oC m 40oC(oC) (W m-1 K-1) (kg m-3) (J kg-1K-1) (kg m-1 s-1) (kg m-1 s-1)

DT

DTUDT

DP

CP

(oC) (W m-1 K-1) (kg m-3) (J kg-1K-

1)(kg m-1 s-

1)

DP

DP

Nu=h Lk

=0 . 027 Re0 .8 Pr1 /3 ( μμw )

0 .14

Smooth pipes, 2 .5× 103< Re< 105

φ=0 .0396 Re−0 .25

Smooth pipes, 2 .5× 103< Re< 10 y

φ=2 .5 ln (Reφ0 .5 )+0 .3

−ΔPf=4φldρu2

Page 2: CHNG 3802 Heat Transfer Tutorial Answers

Question 1

Estimate the heat transfer coefficient from a horizontal pipe 0.15m in diameter with a surface temperature of 400K to air at 294K

do 0.15Tpipe 400Tamb 294Tmean 347

Physical propertiesr 1.018533 kg/m3b 0.002882m 2.00E-05 Pa.sCp 1100 J/kg/Kk 0.02975 W/mKg 9.81 m/s2

Gr 2.62E+07Pr 7.39E-01GrPr 1.94E+07

C' 0.53n 0.25h 6.98 W/m2K

C" 1.32n 0.25h 6.81 W/m2K

q 348.5 W/m340.0 W/m

Gr= ρ2 L3 ΔT β gμ2 Pr=

cP μ

k

h=C left (ΔT right ) rSup { size 8{n} } l rSup { size 8{3n - 1} } } { ¿

Page 3: CHNG 3802 Heat Transfer Tutorial Answers

Question 2

Estimate the heat transfer coefficient and the rate of heating from a flat vertical convenction heater.

DimensionsWidth 2 mHeight 0.8 mTheater 60 CTroom 15 CTmean 310.65

Physical Propertiesr 1.1377143403 kg/m3b 0.0032190568m 1.93E-05 Pa.sCp 1100 J/kg/Kk 0.02702375 W/mKg 9.81 m/s2

Gr 2.53E+09Pr 7.86E-01GrPr 1.99E+09

C' 0.13n 0.33h 5.14 W/m2KQ 370.1 W

C" 1.24n 0.33h 4.36 W/m2KQ 314.3 W

Page 4: CHNG 3802 Heat Transfer Tutorial Answers

Question 3

Estimate the heat loss though a the gap between double glazed windows.

DimensionsGap (L) 0.015 mHeight (H) 2 mWidth 2 mThot 10 CTcold -10 CTmean 273.15

Physical Propertiesr 1.293907962 kg/m3b 0.003660992m 1.60E-05 Pa.sCp 1050 J/kg/Kk 0.02421125 W/mKg 9.81 m/s2

Gr 1.59E+04Pr 6.94E-01GrPr 1.10E+04Ra 1.10E+04500H/L 6.67E+04

Nu 1.11E+00h 1.80 W/m2/KR 0.555850012 Km2/WQ 143.9 W

Nu=hLk=1+

Ra720 ( LH )

Page 5: CHNG 3802 Heat Transfer Tutorial Answers

Question 4

If the thermal resistance of glass is 0.003R glass 0.003

DimensionsGap (L) 0.015 mHeight (H) 2 mWidth 2 mTroom 20 Delta T gla 0.234771Thot 19.77 C Delta T gla 0.234771Tcold -19.77 C diff 2.88E-14Toutside -20Tmean 273.15 Delta T air 40

Physical Propertiesr 1.293908 kg/m3b 0.003661m 1.60E-05 Pa.sCp 1050 J/kg/Kk 0.024211 W/mKg 9.81 m/s2

Gr 3.13E+04Pr 6.94E-01GrPr 2.17E+04Ra 2.17E+04500H/L 6.67E+04

Nu 1.23E+00h 1.98 W/m2/KR 0.505135 Km2/WRglass 0.006Rtotal 0.511135Q 313.0 W

Nu=hLk=1+

Ra720 ( LH )

Page 6: CHNG 3802 Heat Transfer Tutorial Answers

Question 1 Question 2

Exothermic Reaction in stirred tank h Jacket 2000 W/m2Kh Vessel 350 W/m2K

Q 10000 W/m3H 2 m Overall 298 W/m2KD 2 mV 6.28318 Effect of stirrer diameterQ 62831.8 W 1.5A 12.56636 m2 1.721719U 150 New h VesselTr 80 C 525 W/m2K

33.33333 C 603 W/m2KTj 46.66667 C

New OverallMaximum diameter if Tj is limited to 30oC 416 W/m2K

463 W/m2KQ=UA(Tr-Tj)Q/A=U(Tr-Tj) 40%A=piDH 55%Q=10,000*pi/D^2/4*HQ/A=10,000*D/4D=4*Q/A/10,000D=4*U(Tr-Tj)/10,000

Tj 30Tr 80

507500

D 3

DT

DTUDT

Page 7: CHNG 3802 Heat Transfer Tutorial Answers

Question 3 Question 4

Diameter 2 m Diameter 2Height 1 m Height 2.5

Volume 3.14159 m3 Impeller 0.5Area 6.28318 m2 150

2.5U 450 k 0.6Ti 65 cp 4Tf 43 4000Tj 37 mu 0.001

density 1000Density 900 kg/m3 Nbl 4Mass 2827.431 kg Theta 90

cp 4 kJ/kg/K4000 J/kg/K

ln term 1.540445

UA/Mcp 0.00025

t 6161.78 seconds1.711606 hours0.71160642.69634 Re 625000

Pr 6.666667

mu/muw

Nu 10180.94

h 3054.283

Assume constant Jacket Temperature

∫T i

TfdTT−T j

=∫0

tUAMc p

dt=UAMc p

∫0

t

dt

ln(T j−T i

T j−T f)=UA

Mcpt

Nu=0 .74 Re2

3 Pr1

3 ( μμw)0. 14

Page 8: CHNG 3802 Heat Transfer Tutorial Answers

mm

mrpmrev/sW/mKkJ/kg/KJ/kg/KPa.skg/m3

Re^2/3 7310.044

Pr^1/3 1.882072

1

W/m2K

Assume constant Jacket Temperature

∫T i

TfdTT−T j

=∫0

tUAMc p

dt=UAMc p

∫0

t

dt

ln(T j−T i

T j−T f)=UA

Mcpt

Nu=0 .74 Re2

3 Pr1

3 ( μμw)0. 14

Page 9: CHNG 3802 Heat Transfer Tutorial Answers

Question 1 Question 2 Question 3 Data for Q 4 Question 4 Example 12.1 C&R vol 6 Example 12.3 C&R vol 6Kern's Method

Oil Kerosene Oil Methanol Methanol Methanol Heat Transfer coefficient Shell Side Pressure DropFlow 2 kg/s Flow 10 kg/s Flow 10 kg/s Shell side Flowrate 100000 kg/hr Shell sidecp 2000 J/kg/K cp 2450 J/kg/K cp 2000 J/kg/K Flowrate 100000 kg/hr 27.78 kg/s Flowrate 100000 kg/hr As 0.063653 m2 Cross Flow ZoneInlet T 80 C Inlet T 100 C Inlet T 90 C Cp 2840 J/kg/K Cp 2840 J/kg/K Cp 2840 J/kg/K Gs 436.3952 kg/m2sOutlet T 40 C Outlet T 60 C Inlet Temp 95 C Inlet Temp 95 C Inlet Temp 95 C Re 25670 jF 5.60E-02 from fig 12.36

Water Outlet Temp 40 C Outlet Temp 40 C Outlet Temp 40 C Pr 5.08 us 0.58186 m/sQ 160000 W Q 980000 W Flow 20 kg/s density 750 kg/m3 Q 4338889 W density 750 kg/m3 jH 5.30E-03 from fig 12.31 1137.566 Pa

cp 4200 J/kg/K m 3.40E-04 Pa.s m 3.40E-04 Pa.s hoc 2210.17Water Water Inlet T 25 C k 0.19 W/m/K Water k 0.19 W/m/K Bypass Correctioncp 4200 J/kg/K cp 4200 J/kg/K fouling 5000 W/m2K Cp 4200 J/kg/K fouling 5000 W/m2K Tube Row Correction FactorInlet T 25 C Inlet T 20 C A 200 m2 Inlet Temp 25 C F'b 0.670063Outlet T 35 C Outlet T 90 C U 150 Water Outlet Temp 40 C Water Tube row vertical pitch

Tube side Flowrate 68.87125 kg/s Tube side p't 0.02175 Leakage CorrectionFlow 3.81 kg/s Flow 3.33 kg/s R 4.2 Cp 4200 J/kg/K Cp 4200 J/kg/K Baffle cut height

NTUc 0.357143 Inlet Temp 25 C 30.78621 Inlet Temp 25 C Hc 0.2235 AL/As 0.438321

27.30718 C or K 21.64043 C or K Outlet Temp 40 C Outlet Temp 40 C Height between baffle tips 0.52 From fig 12.38From Figure 3.22 m 8.00E-04 Pa.s R 3.666667 3.666667 m 8.00E-04 Pa.s 0.447 F'L 0.312718

R 4 R 0.571429 q 0.47 k 0.59 W/m/K P or S 0.214286 k 0.59 W/m/K Ncv 20.55172P 0.181818 P 0.875 P 0.17 density 995 kg/m3 density 995 kg/m3 Ncv 20 238.3665 PaF 0.85 From Figure 3.8 Hewitt F (1 exchanger) <0.3 From Figure 3.8 Hewitt F 0.9 fouling 3000 W/m2K F 0.85 From Figure 12.19 C&R vol6 fouling 3000 W/m2K Fn 1.03 from fig 12.32

F (2 exchanger) 0.3 From Figure 3.10 Hewitt 0.87 From Figure 3.8 Hewitt

23.2111 F (3 exchanger) 0.8 From Figure 3.10 Hewitt Tcout 36.05 U 600 W/m2K 26.16828 Number of tubes 918 Window ZoneThout 43.59 Tubes U 600 W/m2K Shell ID 0.894 m Window Correction factor

U 200 W/m2/K 17.31234 ID 0.016 m A 276.3453 m2 Bundle Diameter 0.826 m Ra 0.18

A 34.46627 m2 33.18836 OD 0.02 m Tube OD 0.02 m Hb 0.1895 Aw 0.061153 m2

U 350 W/m2/K 29.86952 L 4.88 m Allowing for tube sheets Pitch 1.25 D 0.025 m 0.1895 uw 0.605645 m/sA 161.7 m2 Cupronickel 50 W/m/K L 4.83 m Tube length 4.83 m Bundle cut 0.229419 uz 0.593634 m/s

A of each exchanger 53.9 m3 Q 928200 Triangular pitch A per tube 0.303478 m2 Baffle pitch 0.356 m R'a, Ra 0.18 from fig 12.41 Nwu 8.712644

Q 928200 Split Ring floating head Number or tubes 910.5955 Ns/Ncv 0.2 Nw 165.24 Nwu 8

Q 896085.6 Bundle cleara 0.068 m 911 tube to baffle clearance 0.0008 165 281.0157 PaBaffle spacing = Ds/5 Db 0.823348 m baffle to shell clearance 0.0048 Nc 588

25% baffle cut Ds 0.891348 m Rw 0.359477 End Zone

Fw 1.02 from fig 12.33Tube side coefficient (3 different methods can be used) Pressure Drop 1067.137 Pa

Bypass CorrectionMean water Temp 32.5 C jf 0.0043 From figure 12.24 Total Pressure DropX Area 0.000201 m2 7949.2 Pa Ab 0.024208 m2Tubes per pass 455.5 7.95 kPa Ab/As 0.380313 Number of BafflesFlow area 0.091584 m2 Ns/Ncv 0.2 Nb 12.56742mass velocity 752.0039 kg/sm2 Fb 0.873603 Nb 12density 955 kg/m3u 0.787439 m/s Leakage Correction 8128.494 Pa

8.128494 kPahi 3985.003 W/m2/K Eqn 12.17 of C&R vol 6 Atb 0.018925 m2

2.1 rads from fig 12.41 Assume fouledRe 15040.08 Asb 0.008975 m2Pr 5.694915 AL 0.0279 m2 Ds/lB 2.511236

AL/As 0.438321 1.4 inetrpoleted from table 12.7Nu 101.3265 0.3 from fig 12.35 11.37989 kPah 3736.413 W/m2/K Dittus Boelter FL 0.603491 Kern's predicted 69 kPa

jh 0.004 From figure 12.23 Shell side CoefficientNu 106.8122h 3938.7 W/m2/K hs 1224.186

Take lowest value 3736.413 W/m2/K

Shell Side Pressure Drop

Baffle Spacing 0.17827 m 1.165419Tube pitch 0.025 m jf 0.04 From figure 12.30

25 mm 277168.2 PaAs 0.03178 m2 277.2 kPaMass Velocity 874.0641 kg/m2/sde 0.014201 m C&R suggestes reducing the baffle pitch

Mean shell side temperature 67.5 doubling baffle pitch decreases pressure drop by(1/2)^22

Re 36507.6Pr 5.082105 69.3 kPajH 0.0033Nu 206.0 would reduce h by (1/2)^0.8h 2756.3

h 1583.078Overall Coefficient

1/Uo 0.001611/Uo 0.001341 Uo 620.987 W/m2/KUo 745.454 W/m2/K

DPi

DTLM

DTLM DTLM b'L

DPc

DTM DTM

DTM

DTLM

DTM

mcpDT

mcpDT

UADTM DPw

DPe

DP tube

DPs

qb

bL DPs

Gs/r

DP shell

DP shell

T22
John Kavanagh: > 0.8 not in cross over
Page 10: CHNG 3802 Heat Transfer Tutorial Answers

Q1 Concentric Pipe Heat Exchanger

Water

Cold Stream T k ρ m

Water

Tin 20 oC 20 0.597 1001 4156 1.01E-03Tout 40 oC 40 0.628 995 4178 6.96E-04Flow 0.25 kg/s 100 0.68 961 4216 2.82E-04ID 0.015 m 120 0.685 945 4250 1.59E-04IA 0.000177 m2

r 998u 1.417546 m/sm 0.000837Cp 4167k 0.6125Re 25348Pr 5.70 PaNu 154h 6282 W/m2K

Hot Stream Outer PipeWaterTin 90 oCTout 80 oCFlow 0.5 kg/s Pressure Drop Inside Pipe

ID 0.02 m Re 2.53E+04OD 0.03 m f 0.003138IA 0.000393 m2 3685 Padm 0.01 m

r 969.5u 1.313295 m/sm 0.000325 Pressure Drop Outside PipeCp 4233k 0.6825 Re 3.92E+04Re 39157 f 0.002815Pr 2.02 4085 PaNu 144h 9816 W/m2K

Steel Wallk 60 W/m/K

Ao/(hiAi) 0.000212

ln(ro/ri)/kxr 4.79E-05

1/ho 0.000102 0.635497

CP

(oC) (W m-1 K-1)

(kg m-3) (J kg-1K-

1)(kg m-1 s-

1)

DP

DP

0 20 40 60 80 100 120 140

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

f(x) = − 0.00762535606623447 x − 2.84190383966259R² = 0.987901336805838

Column J

Linear (Column J)

Page 11: CHNG 3802 Heat Transfer Tutorial Answers

Uo 2761.956 W/m2K

Delta T 1 50Delta T 2 60

LMTD 54.84815 K

Q 20835 W

A 0.137535

L 2.19 m

Page 12: CHNG 3802 Heat Transfer Tutorial Answers

Q2

log(mu) If the thermal resistance of glass is 0.003

R glass 0.003

-2.996971

-3.157391 Dimensions

-3.549751 Gap (L) 0.015 m

-3.798603 Height (H) 2 mWidth 2 mTroom 20 Delta T gla 0.09Thot 19.91 C Delta T gla 0.087163349Tcold -9.77 C diff 7.7202E-15Toutside -10Tmean 278.2238 Delta T air 30

Physical Propertiesr 1.270312 kg/m3b 0.003594m 1.60E-05 Pa.sCp 1050 J/kg/Kk 0.024592 W/mKg 9.81 m/s2

Gr 2.23E+04Pr 6.83E-01GrPr 1.52E+04Ra 1.52E+04500H/L 6.67E+04

Nu 1.16E+00h 1.90 W/m2/KR 0.526544 Km2/WRglass 0.006R curtain 0.5Rtotal 1.032544Q 116.2 W

0 20 40 60 80 100 120 140

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

f(x) = − 0.00762535606623447 x − 2.84190383966259R² = 0.987901336805838

Column J

Linear (Column J)

Nu=hLk=1+

Ra720 ( LH )

Page 13: CHNG 3802 Heat Transfer Tutorial Answers

Q3

Estimate the heat transfer coefficient from a horizontal pipe 0.15m in diameter

with water inside the pipe at 370K to air at 294K outside. The pipe is 100m long

Assume that natural convection is the limiting heat transfer resistance

do 0.15Tpipe 370Tamb 294Tmean 332

Physical properties airr 1.064551 kg/m3b 0.003012m 2.00E-05 Pa.sCp 1100 J/kg/Kk 0.028625 W/mKg 9.81 m/s2

Gr 2.15E+07Pr 7.69E-01GrPr 1.65E+07

C' 0.53n 0.25h 6.45 W/m2K

C" 1.32n 0.25h 6.26 W/m2K

q 230.9 W/m224.3 W/m

Water flow 1 m/sPhysical properties waterr 960.7m 2.91E-04Cp 4212k 0.6783

Re 4.95E+05Pr 1.81E+00Nu 988h 4466

Gr= ρ2 L3 ΔT β gμ2 Pr=

cP μ

k

h=C left (ΔT right ) rSup { size 8{n} } l rSup { size 8{3n - 1} } } { ¿

Page 14: CHNG 3802 Heat Transfer Tutorial Answers

U 6.437182

Area 47.12385

Assume temperature drop of water is smallQ 23087.48 W

23054.2 W

Temperature Decrease0.322871

Page 15: CHNG 3802 Heat Transfer Tutorial Answers

Estimate the heat transfer coefficient from a horizontal pipe 0.15m in diameter

with water inside the pipe at 370K to air at 294K outside. The pipe is 100m long

Assume that natural convection is the limiting heat transfer resistance

Page 16: CHNG 3802 Heat Transfer Tutorial Answers

Q1

T k ρ

water 20 0.597 1001 4156 1.01E-03 6.96E-04

air 1 atm 20 0.0262 1.203 1004 1.84E-05 1.93E-05

Flow 2 m/s

Diameter 0.2 m

Dittus BoelterRe Pr Nu h

water 397617 7.01 1512.139 4513.7

air 1 atm 26152 0.71 68.39655 9.0

Calculate the heat transfer coefficients in the previous question using the Sieder Tate Correlation

In both cases assume the wall is at 40oCNu h

water 1631.12 4868.9

air 1 atm 81.73 10.7

Question 2 Pressure drop

Water

phi 0.001577 estphi 0.00167

Delta P/l 133.7623 Pa/m

Air

phi 0.003114

CP m 20oC m 40oC

(oC) (W m-1 K-1)

(kg m-3) (J kg-1K-

1)(kg m-1 s-

1)(kg m-1 s-

1)

Smooth pipes, 2 .5× 103< Re< 105

φ=0 .0396 Re−0 .25

Smooth pipes, 2 .5× 103< Re< 107

φ−0.5=2 .5 ln (Reφ0 .5 )+0 .3

−ΔP f=4φldρu2

Page 17: CHNG 3802 Heat Transfer Tutorial Answers

Delta P/l 0.299691 Pa/m

Q3 Concentric Pipe Heat Exchanger

Water

Cold Stream T k ρ m

Water

Tin 20 oC 20 0.597 1001 4156 1.01E-03Tout 30 oC 40 0.628 995 4178 6.96E-04Flow 0.22 kg/s 100 0.68 961 4216 2.82E-04ID 0.015 m 120 0.685 945 4250 1.59E-04IA 0.000177 m2

r 999.5u 1.245568 m/sm 0.000918 Pressure Drop Inside PipeCp 4172.5k 0.62025 Re 2.03E+04Re 20338 f 0.003316Pr 6.18 1364 Pa/mNu 133.2549h 5510.089 W/m2K

Hot Stream Outer PipeWater 0.005835Tin 120 oCTout 100 oCFlow 0.11 kg/s

ID 0.02 mOD 0.03 mIA 0.000393 m2dm 0.01 m

r 953u 0.293927 m/sm 0.000212 Pressure Drop Outside PipeCp 4233k 0.6825 Re 1.32E+04Re 13228 f 0.003692Pr 1.31 121 Pa/mNu 50.84908h 3470.45 W/m2K

Steel Wallk 60 W/m/K

CP

(oC) (W m-1 K-1)

(kg m-3) (J kg-1K-

1)(kg m-1 s-

1)

DP

DP

Page 18: CHNG 3802 Heat Transfer Tutorial Answers

Ao/(hiAi) 0.000242

ln(ro/ri)/kxr 4.79E-05

1/ho 0.000288

Uo 1729.881

LMTD 84.90187 K

Q 9179.5 W

A 0.062501

L 0.99 m

Page 19: CHNG 3802 Heat Transfer Tutorial Answers

Question 4Estimate the heat transfer coefficient and the rate of heating

from a flat vertical convenction heater.

Dimensions

Width 2 m

Height 0.8 mTheater 60 CTroom 15 C

Tmean 310.65

Physical Propertiesr 1.137714 kg/m3b 0.003219m 1.93E-05 Pa.sCp 1100 J/kg/Kk 0.027024 W/mKg 9.81 m/s2

Gr 2.53E+09Pr 7.86E-01GrPr 1.99E+09

C' 0.13n 0.33h 5.14 W/m2KQ 370.1 W

Power (W) 370.1C" 1.24 Current (A) 1.541957n 0.33 cost 0.296056h 4.36 W/m2KQ 314.3 W

Question 5

h Jacket 2000 W/m2Kh Vessel 350 W/m2K

ReOverall 297.8723 W/m2K

NuEffect of 50% diameter increase

1.51.721719

New h Vessel525 W/m2K

602.6017 W/m2K

New Overall415.8416 W/m2K

Page 20: CHNG 3802 Heat Transfer Tutorial Answers

463.0764 W/m2K

Percentage increase40%55%

Question 6

Diameter 2 m

Height 1 m

Volume 3.14159 m3Area 6.28318 m2

U 400Ti 65Tf 43Tj 37

Density 900 kg/m3Mass 2827.431 kg

cp 4 kJ/kg/K4000 J/kg/K

ln term 1.540445

UA/Mcp 0.000222

t 6932.003 seconds115.5334 minutes1.925556 hours

2.777778

1.2 kg/m3 50 g1.2 g/L 2.777778 moles1.2 mg/mL60 mg of air 50 mL of air50 g of water 0.06 g

0.002069 moles50.06 g

180oC total moles10 bar 27 bar 2.779847

5.064 kg/m3 490

Page 21: CHNG 3802 Heat Transfer Tutorial Answers

5.064 g/L 490 all vapourised at 180oC and 100mLVolume9.885466 L 0.102163 L P 104695.8

104.6958500.6 kg/m3

Page 22: CHNG 3802 Heat Transfer Tutorial Answers

Estimate the heat transfer coefficient and the rate of heating

314.31.3093990.251405

2.25

1.5 1.721719

Page 23: CHNG 3802 Heat Transfer Tutorial Answers

all vapourised at 180oC and 100mL

PakPa

Page 24: CHNG 3802 Heat Transfer Tutorial Answers

Question 1 Question 2 Question 3 Condensation on a bank of horizontal tubes Question 4 Condensation of mixture Question 5 Desuperheating condensation and subcooling Question 6 Condenser Design

Calculate the average heat transfer coefficient for the condensation of steam at one bar on Condensation on horizontal tube (iterative) At what rate will one bar saturated steam condense a bank of 1000 tubes , Condenser at 2.01 bar, steam enters at 150oC, and is condensed against pipes at 120oC OHTCs Hysis Question

a) a vertical surface 1 m high 2.5cm in diameter, 5 m in length, in 25 vetrical rows at 60oC 50 kg/hr of steam and 0.001kg/hr of air enter the exchanger. U gas cooling 250 W/m2K

b) a horizontal rod Inside tube water Tsat 100 C Assuming that the steam and condensate are both at 120oC and are in equilibrium U condensation 2000 W/m2K

both surfaces are at 60oC hi 3400 W/m2K Tsurf 60 C And that the solubility of air in liquid water can be ignored U liq cooling 500 W/m2K

di 0.025 m 972 kg/m3 What is the air concentration in the outlet gas stream?

Tsat 100 C Ti 293 K 3.51E-04 Pa.s Organic Stream

Tsurf 60 C 0.67 W/m/K Data for Water/Steam Flowrate 3 kg/s

972 kg/m3 Outside 2.31E+06 J/kg Temp Psat Inlet Temp 150 C

3.51E-04 Pa.s 551 kg/m3 40 C (oC) (bar) Cp gas 1000 J/kg

0.67 W/m/K 1.29E-04 Pa.s L 1 m 120.23 2 Delta H evap 100 kJ/kg

2.31E+06 J/kg 8.20E-02 W/m/K W 1 m Cp liq 2000 J/kg40 C 2.84E+05 J/kg g 9.81 m/s2 Outlet Temp 70 C

L 1 m 1.41E+01 kg/m3 0.588 kg/m3 Total Pressure 2.01 Boiling Point 125 CW 1 m do 0.03 m do 0.025 mg 9.81 m/s2 g 9.81E+00 L1 5 m Mol wt Water

0.588 kg/m3 Tsat 400 K Nt 25 air 29 Inlet Temp 60 Cdo 0.05 m tubes 1000 water 18 Cp 4200 J/kgL1 1 m Flowrate 10 kg/s

Steam in steam out1a) Vertical 50 kg/hr 0.124137931034 kg/hr Q gas cooling 75000 W

2.77778 kmol/hr 0.006896551724 kmol/hr 75 kWair in air out

Iterations Solver 0.001 kg/hr 0.001 kg/hr Q condensation 300000 W3.44827586E-05 kmol/hr 3.44827586E-05 kmol/hr 300 kW

Twall 346.50 Twall 316.40183 condensate outhf 4953 W/m2K Inlet Air Mole fraction 49.875862069 kg/hr Q liq cooling 330000 W

hf 4362 W/m2K A 392.7 m2 0.0012% 2.77088 kmol/hr 330 kWhf 887 W/m2K hf 793.17329 Q 77803291 W

1b) Horizontal mcond 33.71 kg/s Mole fraction of water in condensate 1 Cooling Required1/U 0.001480583 1/U 0.0016137 Vapour pressure of water 2 bar 705000 WU 675 W/m2K U 620 Mole fraction of water in vapour 0.9950248756 705 kWq 72269 W/m2 q 66307.256 Mole fraction of air 0.0049751244 Oultet Water TTw new 318.51 Tw new 316.40256 Outlet Air Mole fraction 0.50% 76.8 C

Inlet Air Mole fraction 0.001%2nd Iterartion T diff ^2 5.33E-07 3 exchanger Model

hf 7122 W/m2Khf 798.2 W/m2K Condensation per length of tube

1c) 0.0220 kg/s/m

Q 174499 W 1/U 0.0016056879G 0.07560 kg/s/m U 623 W/m2KReL 861 q 66638 W/m2

Tw new 316.52 K1d)A 0.157 3rd IterationQ 44750 W 0.2564465G 0.1234 kg/s/m hf 793.5 W/m2KRe 1406 Gas cooling Condensation Liq Cooling

1/U 0.00161326 Q 75000 W Q 300000 W Q 330000 W

U 619.9 W/m2K Thi 150 Thi 120 Thi 120

q 66325.5 W/m2 Tho 125 Tho 120 Tho 70

Tw new 316.41 K Tci 75.0 Tci 67.9 Tci 60

Tco 76.8 Tco 75.0 Tco 67.9

Condensation per length of tube U 250 U 2000 U 500

0.0220 kg/s/m 60.9 K 48.5 K 25.5 K

A 4.928 A 3.094 A 25.863

A total 33.885

Part b Including CopperCopperk 401 W/mK Twall outside 316.75438R inside 0.0003529412R condensa0.0012607585R copper 6.820008E-06 hf 793.173291/U 0.00162051971/U 0.0016136997 1/U 0.0016205

U 617.08598U inc coppe 617.1 q 66028.2U without c 619.7 Tw new 316.75438

Makes a small difference in U. T diff ^2 0Condensation per length of tube

0.02193 kg/s/m

rL

mL

kL

rL DHevap

mL rL DTsat

kL mL

DHevap kL

DTsat DHevap

rG rG

rG

oC oC oCoC oC oCoC oC oCoC oC oC

W/m2K W/m2K W/m2K

DTLM DTLM DTLM

m2 m2 m2

m2

h f=0 . 943[ k L3 ρL( ρL− ρG ) gΔH LG

μL LΔT f]1

4

41

3 ˆ728.0

DD

fLo

LGGLLLf

Td

Hgkh

mrrr

41

3 ˆ728.0

DD

fLo

LGGLLLf

Td

Hgkh

mrrr

h f=0 . 728[ kL3 ρL (ρL− ρG )gΔ H LG

do μL ΔT f]1

4N

t

−16

1Uo

=Do

2kw

ln(Do

Di)+ 1

ho

+ 1hi (

Do

Di)

GasCooling Condensation

LiquidCooling

Page 25: CHNG 3802 Heat Transfer Tutorial Answers

Question 1 Pool Boiling Question 2 Critical Heat flux Question 3 Question 4 Reboiler Design

Calculate the critical heat flux for a saucepan on a stove water is being vapourised in the tubes of a forced convection boiler Hysis Questionestimate the local heat transfer coefficient at the point where 5% of the liquid has been vapourised

0.59 kg/m3 inlet liquid velocity is 2 m/s, operating pressure 1 bar, inside diameter 16mm, wall temperature is ~120oC

958 kg/m4s 0.0589 N/m Using Chen's method Diameter 0.016

2258000 J/kg Pressure 1g 9.81 m/s2 Tw 120

Bp 100 C Velocity 2 m/s

958.3 kg/m3 P/Pc 0.004519

2.78E-04 Pa.s

0.597 kg/m3 Vapourised 5% 20%

1.26E-05 Pa.s Liquid only 1.9 m/s 1.6

kL 0.597 W/mK ReL 1.05E+05 8.84E+04

cpL 4.22E+03 J/kg/K jH 2.80E-03 from figure 12.23 2.90E-03 from figure 12.23

q crit 1253305 W/m2 Pc 221.29 bar Pr 1.96 1.96

0.59 kg/m3 hfc 13698 W/m2/K 11947 W/m2/K

958 kg/m4 Diameter of bases 0.0589 N/m 0.2 m 1/Xtt 3.86 15.68

2258000 J/kg fc 6.7 From figure 12.56 16 From figure 12.56cpL 4220 J/kg/K Q 39373.7 W h'fc 91775 W/m2/K 191151 W/m2/K

2.78E-04 Pa.s 39.4 kWkL 0.68 W/m/K hn,b ^0.3 25.47418 25.47418Tw 120 C P=VI hn,b 48640.82 W/m2/K 48640.82 W/m2/KTs 100 C V 240 Vpw 200000 Pa I 164.06 amps Relfc^1.25 1.13E+06 2.83E+06ps 100000 Pa fs 0.03 0.025

Domestic power points are typically 10 Amp, For a stove a high amp outlet is generally used h'nb 1459.225 W/m2K 1216.02 W/m2Knumerator 1.112504 164 amps is more than the current supply to most homes.denominato 6.66E-01 hcb 93235 W/m2K 192368 W/m2K

1.67E+00 9.32E+04 W/m2K 1.92E+05 W/m2KT term 2.052P term 5623.413hnb 19277 W/m2/Kq 385534 W/m2

Experimental valuehnb 6.00E+04 p487 C&R vol 1q 1.00E+06Foster and Zuberhnb 32%q 39%

rv

rL

DHevap

rL

mL

rv

mv

rv

rL

DHevap

mL

Foster and Zuber ( very complex )

hn,b=0 . 00122[k L0 .79c pL

0 .45 ρL0.49

σ 0.5 μL0 .29Δ H

lg0.24 ρV

0 .24 ] (Tw−T s)0.24 ( pw−ps)0.75

Wherehn,b=nucleate pool boiling coefficientσ= surface tension (N/m )Tw=wall temperature (o C )T s=saturation temperature of boiling liquid ( o C )pw=saturation pressure corresponding to Tw(Pa )ps=saturation pressure corresponding to T s(Pa )

qcrit=QA=0 .149 ρv

12 Δ Hgl [σg (ρL−ρV )]

14

Mostinski

hn,b=0 . 104 (Pc )0 .69 (q )0.7 [1 . 8(PPc)0.17

+4 (PPc)1 .2

+10(PPc)10]

WhereP= operating pressure (bar )Pc= liquid critical pressure (bar )q= heat flux (W/m2 K )

(hn ,b )0 .3=0 .104 (Pc )0.69 (ΔT )0.7 [1 . 8(PPc)0.17

+4 (PPc)1 .2

+10(PPc)10]

Chen's method forced convection in vertical tubes .There are other methods, agreement between methods is poor .Assume that the foced convective boiling coefficient is made up of convective (h ' fc) and nucleate boiling (h 'nb)componentshcb=h' fc+h'nbThe convective coefficient (h ' fc ) can be estimated using single phase correlations (hfc ) and corrected ( f c ) to account for two phase flowh ' fc=hfc×f ch fc is calculated assuming that liquid is flowing the tube alone .The two phase correction factor is obtained from Figure 12 . 56 (C&R vol6 )

in which the term (1X tt

) is the Lockart-Martinelli two phase flow factor .

This parameter is given by

1X tt

=[ x1−x ]0 .9[ ρL

ρV ]0.5

[μL

μV ]0.1

Where x is the vapour quality, the mass fraction of vapour .The nucleate boiling coefficient can be calculated using correlations for pool boiling, modified by a factor f s to account for the fact that nucleate boiling is more difficult in a flowing liquid .h 'nb=hnb× f sThe suppresion factor is obtained from Figure 12 .57 and is a function of theliquid Reynolds (ReL) number and the forced conversion correction factor ( f s ).

ReL=(1−x )(M

¿

A)de

μL

Page 26: CHNG 3802 Heat Transfer Tutorial Answers

Question 1

Determine the temperature of an object which is emitting most strongly at 650nm

l 650 nm0.00000065 m

T 4458 K

λmaxT=2.898×10−3 mK

Page 27: CHNG 3802 Heat Transfer Tutorial Answers

Question 2

What is the temperature if the emissive power at 10um is 10^9W/m3

C1 3.742E-16C2 1.44E-02

l 0.000001 m

1.00E+09 W/m3

T 1121.4

What if there is a 3% error in E

1.03E+09T 1124.0

0.23%

El,b

El,b

E λ ,b=C1

λ5 [exp(C2

λT )−1][exp(C2

λT )−1]=C1

λ5 Eλ ,b

exp(C2

λT )=C1

λ5 Eλ , b

+1

C2

λT=ln(C1

λ5 Eλ ,b

+1)T=

C2

λ ln(C1

λ5 Eλ , b

+1)

Page 28: CHNG 3802 Heat Transfer Tutorial Answers

Question 3

Heating Elements

10mm diameter, 0.5 m long, T limited to 1700K, provide 300kW

D 0.01L 0.5T 1700s 5.67E-08Q 3.00E+05

Eb 4.74E+05A 0.015708q 7.44E+03N elements 40.32952

41

Page 29: CHNG 3802 Heat Transfer Tutorial Answers

Question 4 Real Suraces and View Factors

Thermos, Ignore the ends

s 5.67E-08

Inner Cylinder(surface 1)D 0.08L 0.2T 80e 0.1

Outer Cylinder(surface 2)D 0.1L 0.2T 20e 0.1

A1 0.050265A2 0.062832

F11 0F12 1 (convex)

F21 0.8F22 0.2

total resistancesurface 1 179.0495view factor 1surface 2 143.2396

323.289 m-2

Eb1 881.9011Eb2 418.7383

Q1->2 1.432659 W

Page 30: CHNG 3802 Heat Transfer Tutorial Answers

Question 5 Absorption and Emission by gases Question 6 Convection and Radiation

Walls of combusiton chamber Horizontal pipe 0.15 m in diameter0.5m diameter T =400K2 m long Air at 294K

s 5.67E-08 T 400 KD 0.5 m s 5.67E-08L 2 m Eb 1.45E+03 W/m2

h 13.7 W/m2KTwall 750 K q 684 W/mCO2 10%H2O 10% From tute 5P 150 kPaT gas 1250 K q 348.5 W/m

Pc 0.15 barPw 0.15 bar

Le 0.4 mpL c 0.06 bar mpL w 0.06 bar m(Pco2+Pw) 0.12 bar meg 0.15 from Fig 2.96 Hewitt

e'g 0.22 from Fig 2.96 Hewittag 0.291366q 1.55E+04 W/m2Q 5.49E+04 W