Heat Transfer

Fitra Dani, Dwi Laura Pramita, Indah Zuliarti & Yohana Siregar

Kelompok 8 Kelas C Labtek II

4.5-5. Cooling and Overall U. Oil flowing at the rate of 7258 kg/h with a cpm

2.01 kJ/kg K is cooled from 394.3 K to 338.9 K in a counterflow heat exchanger

by water entering at 294.3 K and leaving at 305.4 K. Calculate the flow rate of the

water and the overall Ui if the Ai is 5.11 m2.

Solution

Assume cpm water is 4.187 kJ/kg K.

Heat balance

Qoil = Qwater

(m cpm ΔT)oil = (m cpm ΔT)water

7258 . 2.01 . (394.3-338.9) = m . 4.187 . (305.4-294.3)

m = 17389 kg/h

LMTD

Hot fluid (K) Cold fluid (K) Difference

394.3 Higher temperature 305.4 88.9

338.9 Lower temperature 294.3 44.6

ΔTm =

Q = Ui Ai ΔTm

808207.332 . = Ui 5.11 . 64.22

Ui = 684 W/m2 K

4.5-6. Laminar Flow and Heating of Oil. A hydrocarbon oil having the same

physical properties as the oil in Example 4.5-5 enters at 1750F inside a pipe

having an inside diameter of 0.0303 ft and a length of 15 ft. The inside pipe

surface temperature is constant at 3250F. The oil is to be heated to 2500F in the

pipe. How many lbm/h oil can be heated?

Solution

From Example 4.5-5, properties of the oil are cpm 0.5 btu/lbm 0F and km 0.083 btu/h

ft 0F. The viscosity of oil varies with temperature as follows: 1500F, 6.50 cp;

2000F, 5.05 cp; 2500F, 3.80 cp; 3000F, 2.82 cp; 3500F, 1.95 cp.

The bulk mean temperature of the oil is (175+250)/2 or 212.50F. The viscosity at

212.50F is 4.7375 cp (with interpolation) and at 3250F is 2.385 cp (with

interpolation). Assume flow rate 84.2 lbm/h.

b = 4.7375 . 2.4191 = 11.46 lbm/ft h

w = 2.385 . 2.4191 = 5.77 lbm/ft h

The cross section area of the pipe A is

A ft2

G = = = 116620 lbm/ft2 h

NRe = = =

NPr = =

Since Reynold number below 2100, so Eq. (4.5-4) will be used.

=

=

ha = 19.6 btu/h ft2 0F

Heat balance

Q = m cpm ΔT = 84.2 . 0.5 . (250-175) = 3157.5 btu/h

ΔTa = = = 112.5

Q = ha A ΔTa = 19.6 . . 0.0303 . 15 . 112.5 = 3148.42 btu/h

Both Q are same, so assume is correct, m = 84.2 lbm/h.

4.3-3. Heat Loss Through Thermopane Double Window. A double window

called thermopane is one in which two layers of glass are used separated by a

layer of dry stagnant air. In a given window, each of the glass layers is 6.35 mm

thick separated by a 6.35 mm space of stagnant air. The thermal conductivity of

the glass is 0.869 W/m K and that of air is 0.026 over the temperature range used.

For a temperature drop of 27.8 K over the system, calculate the heat loss for a

window 0.914 m x 1.83 m.

Solution

ΔT = 27.8 K

W

4.3-10. Effect of Convective Coefficients on Heat Loss in Double Window.

Repeat Problem 4.3-3 for heat loss in the double window. However, include a

convective coefficient of h 11.35 W/m2 K on the one outside surface of one side

of the window and an h 11.35 on the other outside surface. Also calculate the

overall U.

Solution

From Problem 4.3-3

ΣR = 0.1547

Then, R will added by R4 and R5

New of ΣR = 0.1547+0.05267+0.05267 = 0.26

W

W/m2 K

4.4-3. Heat Loss from a Buried Pipe. A water pipe whose wall temperature is

300 K has diameter of 150 mm and a length of 10 m. It is buried horizontally in

the ground at a depth of 0.4 m measured to the center line of pipe. The ground

surface temperature is 280 K and k 0.85 W/m K. Calculate the loss of heat from

the pipe.

Solution

From Table 4.4-1

q = kS (T-T0)

q = 0.85 . 26.54 . (300-280) = 451.24 W

4.5-1. Heating Air by Condensing Steam. Air is flowing through a tube having

an inside diameter of 38.1 mm at a velocity of 6.71 m/s, average temperature of

449.9 K and pressure of 138 kPa. The inside wall temperature is held constant at

477.6 K by steam condensing outside the tube wall. Calculate the heat transfer

coefficient for a long tube and the heat transfer flux.

Solution

kg/m3

From Appendix A.3

b = 0.000025 kg/m s

w = 0.000026 kg/m s

k = 0.03721 W/m K

NPr = 0.687

NRe = 10921 (Turbulen)

h = 39.35 W/m2 K

4.6-2. Chilling Frozen Meat. Cold water at -28.90C and 1 atm is recirculated at a

velocity of 0.61 m/s over the exposed top flat surface of a piece of frozen meat.

The sides and bottom of this rectangular slab of meat are insulated and the top

surface is 254 mm by 254 mm square. If the surface of the meat is at -6.7 0C,

predict the average heat transfer coefficient to the surface. As an approximation

assume that either Eq. (4.6-2) can be used.

Solution

T average = -17.80C

From Appendix A.3

k = 0.0225 W/m K

NPr = 0.72

b = 0.0000162 kg/m s

= 1.379 kg/m3

NRe = 13189 (Turbulen)

h = 6.05 W/m2 K

4.6-3. Heat Transfer to an Apple. It is desired to predict the heat transfer

coefficient for air being blown by an apple lying on a screen with large openings.

The air velocity is 0.61 m/s at 101.32 kPa pressure and 316.5 K. The surface of

the apple is at 277.6 K and its average diameter is 114 mm. Assume that it is

sphere.

Solution

T average = 297.05 K

From Appendix A.3

k = 0.02512 W/m K

NPr = 0.709

b = 0.000018 kg/m s

= 1.201 kg/m3

NRe = 5860 (Turbulen)

h = 9.988 W/m2 K

4.2-2 (heat removal of a cooling coil)

Diket :

T 1 = 40°F

T 1 = 80°F

ID = 0,25 inchi

OD = 0,40 inchi

K = 7,75+7,78. 10−3 inchi

L=1 ft

Jawab:

r1 = 0,0125 inchi=0,010416666667 ft

r1 = 0,2 inchi=0,0166666666 ft

A1 = 2πLr1 =2(3,14)(1)(0,010416) ft2

A2 = 2πLr2 =2(3,14)(1)(0,0166) ft2

Aℑ=A2−A1

ln ¿¿

K = 7,75+7,78. 10−3 T 1+T 2

2=8,2168btu /hft ° F

q=k Aℑ

T1−T 2❑

r2/r1¿¿

q=8,2168btu

h ft ° F.0,083 ft2 .

(−40)° F(0,0166−0,0104 ) ft

q=439,583501btu /h=1,2198btu /s

q=1286,167037W

4.2-5. Temperature Distribution in Hollow Sphere.

qA

=−kdTdR

A❑ = 4π r2❑

qA∫r1

r 2drr 2 =−∫

T 1

T 2

dt

q=4 π k2(T 1−T 2)

¿¿

4.3-7. Convection, Conduction and Overall U.

Pipa 2ln schedule 40

ID = 2,067 inchi

OD = 2,357 inchi

Tebal isolasi= 51mm = 2,007874 inchi

T i = 450 K=350,33°F

T o = 300 K=80,33°F

ro=2,007874016+¿

hi=30,7W

m2 k=5,406547734btu /hft ° F

h0=10,8W

m2 k=1,901977705btu /hft ° F

k a=0,0623Wm❑k

=0,0359963btu/hft ° F

k b=45Wm❑ k

=26,00058935btuh ft

° F(¿App A 3−16 steel)

Jawab:

ri =2,067

2=1,0335 ln=0,086125 ft

r1 =2,375

2=1,1875 ln=0,09895833 ft

r0=2,007874016+1,1875=3,19537401inc h i=0,266281168 ft

Ai = 2πLr i =2(3,14)(3,280839895) (0,086125 )=1,17449147 ft2

A1 = 2πLr1 =2(3,14)(3,280839895) (0,0989583333 )= ,03890293 ft2

A0 = 2πLr 0=2(3,14)(3,280839895) (0,266281168 )=5,486370522 ft 2

AAlm=A1−A i

ln ¿¿

ABlm=A0−A i

ln ¿¿

K i=1❑

h1 A1

= 15,406547734 ×1,77449147

=0,1042331931

RA=r1−riK A A Alm

=(1,18775−1,0335) /12

0,03599637147.1,903639794=0,187282071

RB=r0−riKB ABlm

=(−3,195374016−1,1875)/12

26,00058935.3,482802133=1,847750037×10−3

R0=1❑

h0 A0

= 11,901977705×5,486370522

=0,09583175549

RA=T 1−T 0

R i+R A+RB+R0

=

350,33−80,3312

0,1042331931+0,187282071+1,847750037×10−3+0,09583175549=693,7400527

btuh

U 0=U 0 . A0 (T i−T 0 )=(T¿¿ i−T 0)

∑R

❑¿

U 0=1

∑R

Ao

= 15,486370522,0,389847696

=0,4683255701

q=U 0 . A0 (T i−T0 )=0,4683255701.5,468325570 (350,33−80,33 )=691,4583063btuh

4.5-8 Heat Transfer with a Liquid Metal

Diket:

m=2kg /s

D=35 mm=0,035m

T 1 = 425°C=698,15 K

T 2 = 430°C=703,15 K

T❑ = 25°C=298,15 K

k❑=15,6Wm❑k

cb=149J

kg❑k

ce=1,34×10−3Pa . s

A=π D2

4=9,6125×10−4m2

G= 29,6125

×10−4=2079,812817kg

m2R

N ℜ=DGμ

=0,035×2079,812817

1,34×10−3=0,0127987195

hl= kd=(0,625 ) ℜ0,4= 15,6

0,035× (0,625 ) (54323,469 10,4 )=21823,86014

W

m2k

q=m .Cp .∆T=2 (149 )(703,15−798,15)=1490 W

qA

= 14906566783,901

=hL (T w−T )=21823,56014.298,15=650678,901W

m2

A=1490

6566783,901=2,289917758×10−4

m

A=πDL=3,14×0,035×L

L=2,08363732×10−3m

4.1-1 Insulation in a cold Room. Calculate the heat loss per m2 of surface area

for a temporary insulating wall of a food cold storage room where the

outside temperature is 299,90 K and the inside temperature 276,50 K. The

wall is composed of 25,4 mm of corkboard having k of 0,0433 W/mK

Penyelesaian :

Dik : T2 = 299,9 0K

T1 = 276,5 0K

X2-X1 = 0,0254 m

K = 0,0433 w/m K

Dit : q/A = ?

qA

= KX 2−X 1

(T 1−T 2)

= 0,04330,0254

(276,5−299,9)

= -39,9 W

4.1-2 Determination of Thermal Conductivity. In determining the thermal

conductivity of an insulating material, the temperatures were measured

on both sides of a flat slab of 15 mm of the material and were 318,4 and

303,2 0K. The heat flux was measured as 35,1 W/m2. Calculate the thermal

conductivity in Btu/h ft 0F and in W/mK.

Penyelesaian :

Dik : X2-X1 = 0,025 m

T1 = 318,4 0K

T2 = 303,20 K

jawab :

qA

= KX 2−X 1

(T 1−T 2 )

K =qA

(X 2−X 1 )

(T 1−T 2)

= 35,1(0,025)

(318,4−303,2)

= 0,0577 W/m K

4.3-1 Insulation needed for Food Cold Storage Room. A food cold storage

room is to be constructed of an inner layer of 19,1 mm of pine wood, a

middle layer of cork board, and an outer layer of 50,8 mm of concrete.

The inside wall surface temperature is-17,8 0C and the outside surface

temperature is 29,4 0C. At the outer concrete surface. The mean

conductivities are for pine, 0,151 ; cork, 0,0433; and concrete, 0,762 W/m

K. The total inside surface area of the roomto use in the calculation is

approximately 39 m2. What thickness of cork board is needed to keep the

heat loss to 586 W???

Penyelesaian :

Dik : ∆ XA=19,1mm

∆ Xc=50,8mm

T1 = -17,8 0C

T4 = 29,4 0C

kA = 0,151

kB = 0,0433

kC = 0,762

q = 586 W

A = 39 m2

Jawab :

RA = ∆ XAkA A

= 0,0191

0,151 (39 ) = 3,243335031 X 10−3

RC = ∆ XBkB A

= 0,05080,762 (39 ) = 1,7049401709 X 10−3

Q = T 1−T 4

RA+RB+RC

-RA + RB + RC = (−17,8−29,4)

−586

= 0,08054607509

RB = 0,07559333835

XB = 0,07559333835 (0,0433 X 39)

= 0,1276544705 m

4.3-2 Insulation of a Furnace. A wall of furnace 0,244 m thick is constructed of

material having a thermal conductivity of 1,30 W/m K. The wall will be

insulated on the outside wiht the material having an average k of 0,346

W/m K, so the heat loss from the furnace will be equal to or less than 1830

W/m2. The inner surface temperature is 1588 K and the outer 299 K.

Calculated the thickness of insulation required.

Penyelesaian :

Dik : ∆ XA=0,244m

kA = 1,3

kB = 0,346

q = 1830

T1 = 1588

T2 = 299

Dit : x = ??

Jawab :

RA = ∆ XAkA x A

= 0,2441,3 (1 ) = 0,1876923077

Q = T 1−T 2RA+RB

RA + RB = 0,7043715847

RB = 0,516679277

XB = 0,1787710298 m

4.5-8 Heat transfer with a Liquid metal. The liquid metal bismuth at a flow

rate of 2.00 kg/s enters a tube having an inside diameter of 35 mm at 425

0C and is heated to 430 0C in the tube. The tube wall is maintained at a

temperature of 25 0C above the liquid bulk temperature. Calculated the

tube length require. The physical properties are as follows (H1) ; k = 15,6

W/m K. K, cp = 149 J/kg K, μ=1,34 x10−3Pa s

Peneyelesaian :

Dik : m = 2 kg/s

ID = 35 mm

K = 15,6 W/m K

Cp = 149 J/kg K

φ=1,34 x10−3 pa s

T1 = 4250C

T2 = 430 0C

Tw = 25 0C

Dit : L = ?

Jawab :

A = π D 24

=3,14 (0,035 ) 2

4 = 9,61625 x 10-4 m2

G = VA

= 2

9,61625 X10−4=2079,812817

kgm 2S

Nre = DGφ

=0,035(2079,812817)

1,34 X10−3 = 54323,4691

Npr = Cpφk

=149 (1,34 x10−3)

15,6 = 0,012798795

hL = (k ( 0,625) nre0,4 )/D

¿15,6 (0,625 ) (54323,4691 x 0,01279871795 )0,4

0,035

= 3817,667996 W/m2 K

Q = m cp ∆T = 2 (149) (5) = 1490 W

qA

=1490A

=h L (Tw−T )

= 3817,667996 (25)

= 95441,6999

A = 1490

95441,6999=0,0156116247

A = 0,0156116247 = π DL=π (0,035 )L

L = 0,142053 m

4.6-1 Heat transfer from a Flat palte. Air at a pressure of 101,3 kpa and a

temperature of 188,8 k is flowing over a thin, smooth falt plate at 3,05

m/s. The plate length in teh direction of flow is 0,305 m and is at 333,2 K.

Calculated the heat transfer coeffisient assuming laminar flow.

Penyelesaian :

Dik : Tw = 288,8 K

Tb = 333,2 K

L = 0,305 m

V = 3,05 m/s

Tf = Tw+Tb

2=288,8+333,2

2

= 311 K = 37,85 0C

K = 0,027 W/m K

ρ=1,137kg /m 3

φ=1,9 x10−5Pa S

Npr = 0,705

Cp = 1,0048 kj/kg K

Dit : h = ?

Jawab

Nre x L = LV ρφ

= 0,305 (3,05 )(1,137)

1,9 x10−5 = 55668,11842

Nμu=h Lk

=0,664 Nre0,5 LNpr

13

= h (0,305 )(0,027)

= 0,664 (55668,11842¿¿0,5¿

h = 12,34331032 W/m2

4.7-2 Losses by Natural Convection from a Cylinder. A vertikal cylinder 76,2

mm in diameter and 121,9 mm high is maintained at 397,1 0K. At its

surface. It losses heat by natural convection to air at 294,3 K. Heat is

lossses from the cylindrical side and the flat circular and at the top.

Calculate the heat loss neglecting radiation losses. Use the simplified

equation of table 4.7-2 and those equation for the lowest range of NGr

Npr, the equivalent L to use for the top flat surface it 0,9 times the

diameter.

Penyelesaian :

Dik : L = 76,2 mm

D = 121,9 mm

Tw = 397,1 0K

Tb = 294,3 0K

T = 0,9 times the diameter

Tf = Tw+Tb

2=397,1+294,3

2 = 345,7 0K

K = 0,0297232852 W/mK

ρ=1,0233213 kg/m 3

Npr = 0,7000072202

φ=2,059891696 x10−5Pa s

β=2,897689531 x10−3

∆T=397,1−294,3=102,8 K

Jawab :

NGR = L3 ρ g β∆T

φ =

(0,9 )3 (1,0233212 )2 (9,806 ) (2,897689531 x10−3 )(102,8)(2,059891697 x10−5)

= 5255325707

NGR. NpR = (5255325707) (0,7000072202) = 3678765939 =

3,6 x 10 9

L3 ∆T=¿ (0,9)3 (102,8) = 74,9412

H = 1,37 (∆rL )

14 = 1,37( 102,8

0,9 )1/4

Maka :

A = π ( D2 )2

+ π DL= (3,14 )( 0,12192 )

2

+ 3,14 (0,1219)(0,0762)

= 0,04083156305 m2

Q = hA (Tw-Tb)

= 18,799545 W

4.3-13. Temperature Rise in Heating Wire. A current of 250 A is passing

through a stainless steel wire having a diameter of 5.08 mm. The wire is 2.44 m

long and has resistance of 0.0843 ohm. The outer surface is held constant at 427.6

K. The thermal conductivity is k 22.5 W/m K. Calculate the center-line

temperature at steady state.

Solution

I2R = q r2L

2502 . 0.0843 = q 0.002542 . 2.44

q = 3.35 . 108 W/m3

T = 3.35 . 108 . 0.002542 / 90 + 427.6 = 451.6 K

4.7-5. Natural Convection on Plate Spaces.

From Example 4.7-3

T average = 380.4 K

NGr = 3.423 . 104

h = 2.01 W/m2 K

Area = 0.6 . 0.4 = 0.24 m2

q = h A (T–T0)

q = 2.01 . 0.24 . (394.3-366.5) = 12.53 W

4.3-8. Heat Transfer in Steam Heater.

Ri = 2.067 / (2 . 12) = 0.086125 ft

Thickness = 0.154 in = 0.01283 ft

R1 = 0.086125+0.01283 = 0.099 ft

Ai = 2 . 3.14 . 1 . 0.086125 = 0.54 ft2

A1 = 2 . 3.14 . 1 . 0.099 = 0.62 ft2

AAlm =

Ri = 1/hi Ai = 1/(500 . 0.54) = 0.0037

R1 = 0.01283/(26 . 0.58) = 0.00085

Ro = 1/ho A1 = 1/(1500 . 0.62) = 0.001075

= 0.005625

q = (220-70)/0.005625 = 26665 btu/h

Ui = 26665/(0.54 . (220-70)) = 329.2 btu/h ft2 0F

Uo = 26665/(0.62 . (220-70)) = 286.4 btu/h ft2 0F