Heat 4e SM Chap11

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Solutions Manual

for

Heat and Mass Transfer: Fundamentals & Applications

Fourth Edition

Yunus A. Cengel & Afshin J. Ghajar

McGraw-Hill, 2011

Chapter 11 HEAT EXCHANGERS

PROPRIETARY AND CONFIDENTIAL

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Types of Heat Exchangers

11-1C Heat exchangers are classified according to the flow type as parallel flow, counter flow, and cross-flow arrangement.

In parallel flow, both the hot and cold fluids enter the heat exchanger at the same end and move in the same direction. In counter-flow, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite direction. In cross-flow, the hot and cold fluid streams move perpendicular to each other.

11-2C A heat exchanger is classified as being compact if β > 700 m2/m3 or (200 ft2/ft3) where β is the ratio of the heat

transfer surface area to its volume which is called the area density. The area density for double-pipe heat exchanger can not be in the order of 700. Therefore, it can not be classified as a compact heat exchanger.

11-3C Regenerative heat exchanger involves the alternate passage of the hot and cold fluid streams through the same flow

area. The static type regenerative heat exchanger is basically a porous mass which has a large heat storage capacity, such as a ceramic wire mash. Hot and cold fluids flow through this porous mass alternately. Heat is transferred from the hot fluid to the matrix of the regenerator during the flow of the hot fluid and from the matrix to the cold fluid. Thus the matrix serves as a temporary heat storage medium. The dynamic type regenerator involves a rotating drum and continuous flow of the hot and cold fluid through different portions of the drum so that any portion of the drum passes periodically through the hot stream, storing heat and then through the cold stream, rejecting this stored heat. Again the drum serves as the medium to transport the heat from the hot to the cold fluid stream.

11-4C In the shell and tube exchangers, baffles are commonly placed in the shell to force the shell side fluid to flow across

the shell to enhance heat transfer and to maintain uniform spacing between the tubes. Baffles disrupt the flow of fluid, and an increased pumping power will be needed to maintain flow. On the other hand, baffles eliminate dead spots and increase heat transfer rate.

11-5C Using six-tube passes in a shell and tube heat exchanger increases the heat transfer surface area, and the rate of heat

transfer increases. But it also increases the manufacturing costs.

11-6C Using so many tubes increases the heat transfer surface area which in turn increases the rate of heat transfer.

11-7C In counter-flow heat exchangers, the hot and the cold fluids move parallel to each other but both enter the heat

exchanger at opposite ends and flow in opposite direction. In cross-flow heat exchangers, the two fluids usually move perpendicular to each other. The cross-flow is said to be unmixed when the plate fins force the fluid to flow through a particular interfin spacing and prevent it from moving in the transverse direction. When the fluid is free to move in the transverse direction, the cross-flow is said to be mixed.

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The Overall Heat Transfer Coefficient

11-8C Heat is first transferred from the hot liquid to the wall by convection, through the wall by conduction and from the

wall to the cold liquid again by convection.

11-9C When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, which is usually

the case, the thermal resistance of the tube is negligible.

11-10C The heat transfer surface areas are Ai =πD1L and Ao =πD2L. When the thickness of inner tube is small, it is

reasonable to assume A_i ≅ A_o ≅ A_s.

11-11C The effect of fouling on a heat transfer is represented by a fouling factor Rf. Its effect on the heat transfer coefficient

is accounted for by introducing a thermal resistance Rf /As. The fouling increases with increasing temperature and decreasing

velocity.

11-12C None.

11-13C When one of the convection coefficients is much smaller than the other , and . Then we have

( ) and thus o i h h << A_i ≈A₀ ≈A_s o i h h >>1/ / 1 U_i =U₀ =U ≅h_i.

11-14C The most common type of fouling is the precipitation of solid deposits in a fluid on the heat transfer surfaces.

Another form of fouling is corrosion and other chemical fouling. Heat exchangers may also be fouled by the growth of algae in warm fluids. This type of fouling is called the biological fouling. Fouling represents additional resistance to heat transfer and causes the rate of heat transfer in a heat exchanger to decrease, and the pressure drop to increase.

11-15C When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, the thermal

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11-16 The heat transfer coefficients and the fouling factors on tube and shell side of a heat exchanger are given. The thermal

resistance and the overall heat transfer coefficients based on the inner and outer areas are to be determined.

Assumptions 1 The heat transfer coefficients and the fouling factors are constant and uniform. Analysis (a) The total thermal resistance of the heat exchanger per unit length is

C/W 0.1334° = ° + ° + ° + ° + ° = + + + + = m)] m)(1 016 . 0 ( [ C) . W/m 240 ( 1 m)] m)(1 016 . 0 ( [ C/W) . m 0002 . 0 ( m) C)(1 W/m. 380 ( 2 ) 2 . 1 / 6 . 1 ln( m)] m)(1 012 . 0 ( [ C/W) . m 0005 . 0 ( m)] m)(1 012 . 0 ( [ C) . W/m 800 ( 1 1 2 ) / ln( 1 2 2 2 2 π π π π π π R A h A R kL D D A R A h R o o o fo i o i fi i i Outer surface D0, A0, h0, U0 , Rf0 Inner surface Di, Ai, hi, Ui , Rfi

(b) The overall heat transfer coefficient based on the inner and the outer surface areas of the tube per length are

C . W/m 149 C . W/m 199 2 2 ° = ° = = ° = ° = = = = = m)] m)(1 016 . 0 ( [ C/W) 1334 . 0 ( 1 1 m)] m)(1 012 . 0 ( [ C/W) 1334 . 0 ( 1 1 1 1 1 π π o o i i o o i i RA U RA U A U A U UA R

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11-17 EES Prob. 11-16 is reconsidered. The effects of pipe conductivity and heat transfer coefficients on the thermal

resistance of the heat exchanger are to be investigated.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" k=380 [W/m-C] D_i=0.012 [m] D_o=0.016 [m] D_2=0.03 [m] h_i=800 [W/m^2-C] h_o=240 [W/m^2-C] R_f_i=0.0005 [m^2-C/W] R_f_o=0.0002 [m^2-C/W] "ANALYSIS" R=1/(h_i*A_i)+R_f_i/A_i+ln(D_o/D_i)/(2*pi*k*L)+R_f_o/A_o+1/(h_o*A_o) L=1 [m] “a unit length of the heat exchanger is considered"

A_i=pi*D_i*L A_o=pi*D_o*L U_i=1/(R*A_i) U_o=1/(R*A_o) k [W/m-C] R [C/W] 10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400 0.1379 0.1348 0.1342 0.1339 0.1338 0.1337 0.1336 0.1336 0.1336 0.1335 0.1335 0.1335 0.1335 0.1335 0.1334 0.1334 0.1334 0.1334 0.1334 0.1334 0 50 100 150 200 250 300 350 400 0.133 0.134 0.135 0.136 0.137 0.138 k [W/m-C] R [ C /W ]

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hi [W/m2_-C] R _[C/W] 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 0.1533 0.1485 0.1445 0.1411 0.1381 0.1356 0.1334 0.1315 0.1297 0.1282 0.1268 0.1255 0.1244 0.1233 0.1224 0.1215 0.1207 0.1199 0.1192 0.1185 0.1179 500 700 900 1100 1300 1500 0.115 0.12 0.125 0.13 0.135 0.14 0.145 0.15 0.155 h_i = W/m2-C R [ C /W ] ho [W/m2_-C] R _[C/W] 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850 1900 1950 2000 0.07041 0.06947 0.06861 0.06782 0.0671 0.06644 0.06582 0.06526 0.06473 0.06424 0.06378 0.06335 0.06295 0.06258 0.06222 0.06189 0.06157 0.06127 0.06099 0.06072 0.06047 1000 1200 1400 1600 1800 2000 0.06 0.062 0.064 0.066 0.068 0.07 0.072 R [ C /W ] h_o = W/m2-C

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11-18E Water is cooled by air in a cross-flow heat exchanger. The overall heat transfer coefficient is to be determined.

Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its

thickness is negligible. 2 Both the water and air flow are fully developed. 3 Properties of the water and air are constant.

Properties The properties of water at 180°F are (Table A-9E)

15 . 2 Pr s / ft 10 825 . 3 F Btu/h.ft. 388 . 0 2 6 = × = ° = − ν k _Water 180°F 4 ft/s Air 80°F 12 ft/s The properties of air at 80°F are (Table A-15E)

7290 . 0 Pr s / ft 10 697 . 1 F Btu/h.ft. 01481 . 0 2 4 = × = ° = − ν k

Analysis The overall heat transfer coefficient can be determined from

o i h h U 1 1 1 ₌ ₊

The Reynolds number of water is

360 , 65 s / ft 10 825 . 3 ft] /12 ft/s)[0.75 4 ( Re ₆ ₂ = × = = ₋ ν h avgD V

which is greater than 10,000. Therefore the flow of water is turbulent. Assuming the flow to be fully developed, the Nusselt number is determined from

222 ) 15 . 2 ( ) 360 , 65 ( 023 . 0 Pr Re 023 . 0 0.8 0.4 = 0.8 0.4 = = = k hD Nu h and (222)=1378Btu/h.ft .F ft 12 / 75 . 0 F Btu/h.ft. 388 . 0 ° 2_° = = Nu D k h h i

The Reynolds number of air is

4420 s / ft 10 697 . 1 ft] 12) ft/s)[3/(4 12 ( Re ₄ ₂ = × × = = ₋ ν VD

The flow of air is across the cylinder. The proper relation for Nusselt number in this case is

(

)

[

]

(

)

[

]

34.86 000 , 282 4420 1 7290 . 0 / 4 . 0 1 ) 7290 . 0 ( ) 4420 ( 62 . 0 3 . 0 000 , 282 Re 1 Pr / 4 . 0 1 Pr Re 62 . 0 3 . 0 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . 0 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . 0 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + = = k hD Nu

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11-19 Water flows through the tubes in a boiler. The overall heat transfer coefficient of this boiler based on the inner surface

area is to be determined.

Assumptions 1 Water flow is fully developed. 2 Properties of the water are constant. Properties The properties of water at 110°C are (Table A-9)

Outer surface D0, A0, h0, U0 , Rf0 Inner surface Di, Ai, hi, Ui , Rfi 58 . 1 Pr K . W/m 682 . 0 /s m 10 268 . 0 / 2 2 6 = = × = = − k ρ µ ν

Analysis The Reynolds number is

600 , 130 s / m 10 268 . 0 m) m/s)(0.01 5 . 3 ( Re avg ₆ ₂ = × = = ₋ ν h D V

which is greater than 10,000. Therefore, the flow is turbulent. Assuming fully developed flow, 9 . 341 ) 58 . 1 ( ) 600 , 130 ( 023 . 0 Pr Re 023 . 0 0.8 0.4 ₌ 0.8 0.4 ₌ = = k hD Nu h and C . W/m 23,320 = (341.9) m 01 . 0 C W/m. 682 . 0 ° 2 _° = = Nu D k h h

The total resistance of this heat exchanger is then determined from

C/W 001185 . 0 = ] m) m)(7 (0.014 C)[ . W/m 7200 ( 1 m)] C)(7 W/m. 2 . 14 ( 2 [ ) 1 / 4 . 1 ln( ] m) m)(5 (0.01 C)[ . W/m 320 , 23 ( 1 1 2 ) / ln( 1 2 2 ° ° + ° + ° = + + = + + = = π π π πo i _o _o i i o wall i total _kL _h _A D D A h R R R R R and C . W/m 3838 2 ° = ° = = ⎯→ ⎯ = ] m) m)(7 (0.01 C/W)[ 001185 . 0 ( 1 1 1 π i i i i RA U A U R

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11-20 Water is flowing through the tubes in a boiler. The overall heat transfer coefficient of this boiler based on the inner

surface area is to be determined.

Assumptions 1 Water flow is fully developed. 2 Properties of water are constant. 3 The heat transfer coefficient and the

fouling factor are constant and uniform.

Properties The properties of water at 110°C are (Table A-9)

Outer surface D0, A0, h0, U0 , Rf0 Inner surface Di, Ai, hi, Ui , Rfi 58 . 1 Pr K . W/m 682 . 0 /s m 10 268 . 0 / 2 2 6 = = × = = − k ρ µ ν

Analysis The Reynolds number is

600 , 130 s / m 10 268 . 0 m) m/s)(0.01 5 . 3 ( Re avg ₆ ₂ = × = = ₋ ν h D V

which is greater than 10,000. Therefore, the flow is turbulent. Assuming fully developed flow, 9 . 341 ) 58 . 1 ( ) 600 , 130 ( 023 . 0 Pr Re 023 . 0 0.8 0.4 = 0.8 0.4 = = = k hD Nu h and C . W/m 23,320 = (341.9) m 01 . 0 C W/m. 682 . 0 ° 2 _° = = Nu D k h h

The thermal resistance of heat exchanger with a fouling factor of 0.0005m2. C/W is determined from

,i = ° f R C/W 003459 . 0 m)] m)(7 014 . 0 ( [ C) . W/m 7200 ( 1 m) C)(7 W/m. 2 . 14 ( 2 ) 1 / 4 . 1 ln( m)] m)(7 01 . 0 ( [ C/W . m 0005 . 0 m)] m)(5 01 . 0 ( [ C) . W/m 320 , 23 ( 1 1 2 ) / ln( 1 2 2 2 , ° = ° + ° + ° + ° = + + + = π π π π π R A h kL D D A R A h R o o i o i i f i i Then, C . W/m 1315 2 ° = ° = = ⎯→ ⎯ = ] m) m)(7 (0.01 C/W)[ 003459 . 0 ( 1 1 1 π i i i i RA U A U R

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11-21 EES Prob. 11-20 is reconsidered. The overall heat transfer coefficient based on the inner surface as a function of

fouling factor is to be plotted.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" T_w=110 [C] Vel=3.5 [m/s] L=7 [m] k_pipe=14.2 [W/m-C] D_i=0.010 [m] D_o=0.014 [m] h_o=7200 [W/m^2-C] R_f_i=0.0005 [m^2-C/W] "PROPERTIES" k=conductivity(Water, T=T_w, P=300) Pr=Prandtl(Water, T=T_w, P=300) rho=density(Water, T=T_w, P=300) mu=viscosity(Water, T=T_w, P=300) nu=mu/rho "ANALYSIS"

Re=(Vel*D_i)/nu "Re is calculated to be greater than 10,000. Therefore, the flow is turbulent."

Nusselt=0.023*Re^0.8*Pr^0.4 h_i=k/D_i*Nusselt A_i=pi*D_i*L A_o=pi*D_o*L R=1/(h_i*A_i)+R_f_i/A_i+ln(D_o/D_i)/(2*pi*k_pipe*L)+1/(h_o*A_o) U_i=1/(R*A_i) Rf,i [m2_-C/W] U_[W/mi 2_-C] 0.0001 0.00015 0.0002 0.00025 0.0003 0.00035 0.0004 0.00045 0.0005 0.00055 0.0006 0.00065 0.0007 0.00075 0.0008 2769 2433 2169 1957 1782 1636 1513 1406 1314 1233 1161 1098 1040 989 942.4 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 750 1200 1650 2100 2550 3000 R_fi [m2-C/W] Ui [W /m 2 -C ]

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11-22E The overall heat transfer coefficient of a heat exchanger and the percentage change in the overall heat transfer

coefficient due to scale built-up are to be determined.

Assumptions 1 Steady operating condition exists. 2 The heat transfer coefficients and the fouling factors are constant and

uniform.

Analysis When operating at design and clean conditions, the overall heat transfer coefficient is given as

F ft Btu/hr 50 2 scale w/o = ⋅ ⋅° U

(a) After a period of use, the overall heat transfer coefficient due to the scale built-up is

F/Btu ft hr 022 . 0 F/Btu ft hr 002 . 0 F ft Btu/hr 50 1 1 1 2 2 2 scale w/o scale w/ ° ⋅ ⋅ = ° ⋅ ⋅ + ° ⋅ ⋅ = + = R_f U U or F ft Btu/hr 45.5 ⋅ 2⋅° = scale w/ U

(b) The percentage change in the overall heat transfer coefficient due to the scale built-up is

9% = × − = × − 100 50 5 . 45 50 100 scale w/o scale w/ scale w/o U U U

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11-23E The overall heat transfer coefficients based on the outer and inner surfaces for a heat exchanger are to be determined.

Assumptions 1 Steady operating condition exists. 2 Thermal properties are constant. Properties The conductivity of the tube material is given to be 0.5 Btu/hr·ft·°F. Analysis The overall heat transfer coefficient based on the outer surface is

o o i o i i o o h A D D kL A h A U 1 ) / ln( 2 1 1 1 ₌ ₊ ₊ π o i o o i o i o o o i o o i i o o D h D k D D D h A h A D D kL A A h A U 1 ln 2 1 ln 2 1 ₊ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = π Thus F ft Btu/hr 4.32 ⋅ 2⋅° = ° ⋅ ⋅ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = − − F ft Btu/hr 10 1 2 3 ln ) 5 . 0 ( 2 12 / 3 2 3 50 1 1 ln 1 2 1 1 o i o o i o i o _h D_D D_k D_D _h U

The overall heat transfer coefficient based on the inner surface is

o o i o i i i i h A D D kL A h A U 1 ) / ln( 2 1 1 1 ₌ ₊ ₊ π o i o i o i i o o i i o i i i i i D D h D D k D h A h A D D kL A A h A U 1 ln 2 1 ln 2 1 ₊ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = π Thus F ft Btu/hr 6.48 ⋅ 2⋅° = ° ⋅ ⋅ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = − − F ft Btu/hr 3 2 10 1 2 3 ln ) 5 . 0 ( 2 12 / 2 50 1 1 ln 2 1 2 1 1 o i o i o i i i _h D_k D_D _h _DD U

Discussion The two overall heat transfer coefficients differ significantly with Ui larger than Uo by a factor of 1.5. The overall

heat transfer coefficient ratio can be expressed as

i i o oA U A U 1 1 ₌ _→ ₌ ₌ ₌₁_.₅ i o i o o i D D A A U U

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11-24 Refrigerant-134a is cooled by water in a double-pipe heat exchanger. The overall heat transfer coefficient is to be

determined.

Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its

thickness is negligible. 2 Both the water and refrigerant-134a flow are fully developed. 3 Properties of the water and refrigerant-134a are constant.

Properties The properties of water at 20°C are (Table A-9) Cold water

Di D0 01 . 7 Pr C . W/m 598 . 0 /s m 10 004 . 1 / kg/m 998 2 6 3 = ° = × = = = − k ρ µ ν ρ

Analysis The hydraulic diameter for annular space is

Hot R-134a m 015 . 0 01 . 0 025 . 0 − = = − = _o _i h D D D

The average velocity of water in the tube and the Reynolds number are

m/s 729 . 0 4 m) 01 . 0 ( m) 025 . 0 ( ) kg/m 998 ( kg/s 3 . 0 4 2 2 3 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ = = π π ρ ρ i o c avg D D m A m V & & 890 , 10 s / m 10 004 . 1 m) m/s)(0.015 729 . 0 ( Re 2 6 = × = = ₋ ν h avgD V

which is greater than 4000. Therefore flow is turbulent. Assuming fully developed flow,

0 . 85 ) 01 . 7 ( ) 890 , 10 ( 023 . 0 Pr Re 023 . 0 0.8 0.4 = 0.8 0.4 = = = k hD Nu h and C . W/m 3390 = (85.0) m 015 . 0 C W/m. 598 . 0 ° 2_° = = Nu D k h h o

Then the overall heat transfer coefficient becomes

C . W/m 1856 2 ° = ° + ° = + = C . W/m 3390 1 C . W/m 4100 1 1 1 1 1 2 2 o i h h U

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11-25 Refrigerant-134a is cooled by water in a double-pipe heat exchanger. The overall heat transfer coefficient is to be

determined.

Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its

thickness is negligible. 2 Both the water and refrigerant-134a flows are fully developed. 3 Properties of the water and refrigerant-134a are constant. 4 The limestone layer can be treated as a plain layer since its thickness is very small relative to its diameter.

Properties The properties of water at 20°C are (Table A-9) Cold water

D0 Hot R-134a Limestone Di 01 . 7 Pr C . W/m 598 . 0 /s m 10 004 . 1 / kg/m 998 2 6 3 = ° = × = = = − k ρ µ ν ρ

Analysis The hydraulic diameter for annular space is

m 015 . 0 01 . 0 025 . 0 − = = − = _o _i h D D D

The average velocity of water in the tube and the Reynolds number are

m/s 729 . 0 4 m) 01 . 0 ( m) 025 . 0 ( ) kg/m 998 ( kg/s 3 . 0 4 2 2 3 2 2 avg = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ = = π π ρ ρ i o c D D m A m V & & 890 , 10 s / m 10 004 . 1 m) m/s)(0.015 729 . 0 ( Re 2 6 avg ₌ × = = ₋ ν h D V

which is greater than 10,000. Therefore flow is turbulent. Assuming fully developed flow,

0 . 85 ) 01 . 7 ( ) 890 , 10 ( 023 . 0 Pr Re 023 . 0 0.8 0.4 ₌ 0.8 0.4 ₌ = = k hD Nu h and C . W/m 3390 = (85.0) m 015 . 0 C W/m. 598 . 0 ° 2_° = = Nu D k h h o

Disregarding the curvature effects, the overall heat transfer coefficient is determined to be

C . W/m 481 2 ° = ° + ° + ° = + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = C . W/m 3390 1 C . W/m 3 . 1 m 002 . 0 C . W/m 4100 1 1 1 1 1 2 2 limeston o i k h L h U

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11-26 EES Prob. 11-25 is reconsidered. The overall heat transfer coefficient as a function of the limestone thickness is to be

plotted.

"GIVEN" D_i=0.010 [m] D_o=0.025 [m] T_w=20 [C] h_i=4100 [W/m^2-C] m_dot=0.3 [kg/s] L_limestone=2 [mm] k_limestone=1.3 [W/m-C] "PROPERTIES" k=conductivity(Water, T=T_w, P=100) Pr=Prandtl(Water, T=T_w, P=100) rho=density(Water, T=T_w, P=100) mu=viscosity(Water, T=T_w, P=100) nu=mu/rho "ANALYSIS" D_h=D_o-D_i Vel=m_dot/(rho*A_c) A_c=pi*(D_o^2-D_i^2)/4 Re=(Vel*D_h)/nu

"Re is calculated to be greater than 10,000. Therefore, the flow is turbulent."

Nusselt=0.023*Re^0.8*Pr^0.4 h_o=k/D_h*Nusselt U=1/(1/h_i+(L_limestone*Convert(mm, m))/k_limestone+1/h_o) Llimestone [mm] U [W/m2_-C] 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 762.4 720.2 682.4 648.3 617.5 589.5 564 540.5 518.9 499 480.6 463.4 447.5 400 450 500 550 600 650 700 750 800

U [

W

/m

2

-C]

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11-27 A water stream is heated by a jacketted-agitated vessel, fitted with a turbine agitator. The mass flow rate of water

is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings

is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant.

Properties The properties of water at 54°C are (Table A-9)

Water 10 54ºC Steam 100ºC 54ºC ºC 31 . 3 Pr s kg/m 10 513 . 0 kg/m 8 . 985 C W/m. 648 . 0 3 -3 = ⋅ × = = ° = µ ρ k

The specific heat of water at the average temperature of (10+54)/2=32°C is 4178 J/kg.°C (Table A-9)

Analysis We first determine the heat transfer coefficient on the inner wall of the vessel

76,865 s kg/m 10 513 . 0 ) kg/m 8 . 985 ( m) )(0.2 s (60/60 Re ₃ 3 2 -1 2 = ⋅ × = = ₋ µ ρ a D n& Nu=0.76Re2/3Pr1/3=0.76(76,865)2/3(3.31)1/3 =2048 (2048) 2211 W/m .C m 6 . 0 C W/m. 648 . 0 ° ₌ 2_° = = Nu D k h t j

The heat transfer coefficient on the outer side is determined as follows h_o =13,100(T_g −T_w)−0.25 =13,100(100−T_w)−0.25 C 2 . 89 ) 54 ( 2211 ) 100 ( 100 , 13 ) 54 ( 2211 ) 100 ( ) 100 ( 100 , 13 ) 54 ( ) ( 75 . 0 25 . 0 ° = → − = − − = − − − = − − w w w w w w w j w g o T T T T T T T h T T h C W/m 7226 ) 2 . 89 100 ( 100 , 13 ) 100 ( 100 , 13 ₋ 0.25 ₌ ₋ 0.25 ₌ 2_⋅ = _w − − o T h

Neglecting the wall resistance and the thickness of the wall, the overall heat transfer coefficient can be written as

C W/m 1694 7226 1 2211 1 1 1 1 2 1 ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − − o j h h U

From an energy balance

kg/h 1725 = = − × × = − ∆ = − kg/s 479 . 0 ) 54 100 )( 6 . 0 6 . 0 )( 1694 ( ) 10 54 )( 4178 ( )] ( [ water w w in out m m T UA T T c m & & & π

(17)

f & , a

Analysis of Heat Exchangers

11-28C The heat exchangers usually operate for long periods of time with no change in their operating conditions, and then

they can be modeled as steady-flow devices. As such , the mass flow rate of each fluid remains constant and the fluid properties such as temperature and velocity at any inlet and outlet remain constant. The kinetic and potential energy changes are negligible. The specific heat of a fluid can be treated as constant in a specified temperature range. Axial heat conduction along the tube is negligible. Finally, the outer surface of the heat exchanger is assumed to be perfectly insulated so that there is no heat loss to the surrounding medium and any heat transfer thus occurs is between the two fluids only.

11-29C When the heat capacity rates of the cold and hot fluids are identical, the temperature rise of the cold fluid will be

equal to the temperature drop of the hot fluid.

11-30C The product of the mass flow rate and the specific heat of a fluid is called the heat capacity rate and is expressed as

. When the heat capacity rates of the cold and hot fluids are equal, the temperature change is the same for the two fluids in a heat exchanger. That is, the temperature rise of the cold fluid is equal to the temperature drop of the hot fluid. A heat capacity of infinity for a fluid in a heat exchanger is experienced during a phase-change process in a condenser or boiler.

p

c m C= &

11-31C The mass flow rate of the cooling water can be determined from . The rate of condensation o

the steam is determined from Q& nd the total thermal resistance of the condenser is determined from . water cooling ) ( = mc T Q& & p∆ steam ) ( = mhfg T Q R= /& ∆

11-32C That relation is valid under steady operating conditions, constant specific heats, and negligible heat loss from the

(18)

The Log Mean Temperature Difference Method

11-33C ∆Tlm is called the log mean temperature difference, and is expressed as

) / ln( ₁ ₂ 2 1 T T T T T_lm ∆ ∆ ∆ − ∆ = ∆ where

for parallel-flow heat exchangers and

out c out h in c in h T T T T T T₁= _, - _, ∆ ₂= _, - _, ∆

for counter-flow heat exchangers

in c out h out c in h T T T T T T = _, - _, ∆ ₂= _, - _, ∆

11-34C The temperature difference between the two fluids decreases from ∆T1 at the inlet to ∆T2 at the outlet, and arithmetic

mean temperature difference is defined as

2 +

= 1 2

am T T

T ∆ ∆

∆ . The logarithmic mean temperature difference ∆Tlm is obtained

by tracing the actual temperature profile of the fluids along the heat exchanger, and is an exact representation of the average temperature difference between the hot and cold fluids. It truly reflects the exponential decay of the local temperature difference. The logarithmic mean temperature difference is always less than the arithmetic mean temperature.

11-35C ∆Tlm cannot be greater than both ∆T1 and ∆T2 because ∆Tln is always less than or equal to ∆Tm (arithmetic mean)

which can not be greater than both ∆T1 and ∆T2.

11-36C In the parallel-flow heat exchangers the hot and cold fluids enter the heat exchanger at the same end, and the

temperature of the hot fluid decreases and the temperature of the cold fluid increases along the heat exchanger. But the temperature of the cold fluid can never exceed that of the hot fluid. In case of the counter-flow heat exchangers the hot and cold fluids enter the heat exchanger from the opposite ends and the outlet temperature of the cold fluid may exceed the outlet temperature of the hot fluid.

11-37C First heat transfer rate is determined from Q&=m&c_p[T_in-T_out], ∆Tln from

) / ln( 1 2 2 1 T T T T T_lm ∆ ∆ ∆ − ∆ = ∆ , correction factor

from the figures, and finally the surface area of the heat exchanger from Q&=UAFDTlm,CF

11-38C The factor F is called as correction factor which depends on the geometry of the heat exchanger and the inlet and the

outlet temperatures of the hot and cold fluid streams. It represents how closely a heat exchanger approximates a counter-flow heat exchanger in terms of its logarithmic mean temperature difference. F cannot be greater than unity.

(19)

11-40C The ∆Tlm will be greatest for double-pipe counter-flow heat exchangers.

11-41 A counter-flow heat exchanger has a specified overall heat transfer coefficient operating at design and clean

conditions. After a period of use built-up scale gives a fouling factor, (a) the rate of heat transfer in the heat exchanger and (b) the mass flow rates of both hot and cold fluids are to be determined.

Assumptions 1 Steady operating condition exists. 2 The heat transfer coefficients and the fouling factors are constant and

uniform. 3 Fluid properties are constant.

Properties The specific heat of both hot and cold fluids is

given as 4.2 kJ/kg·K.

Analysis When operating at design and clean conditions, the

overall heat transfer coefficient is given as K W/m 284 2 scale w/o = ⋅ U

(a) After a period of use, the overall heat transfer coefficient due to the scale built-up is

K/W m 00392 . 0 K/W m 0004 . 0 K W/m 284 1 1 1 2 2 2 scale w/o scale w/ ⋅ = ⋅ + ⋅ = + = R_f U U or K W/m 255 2 scale w/ = ⋅ U

The log mean temperature difference is

C 3 . 49 C ] ) 27 71 ( )/ 38 93 ln[( ) 27 71 ( ) 38 93 ( ) / ln( ₁ ₂ 2 1 lm ₋ ₋ ° = ° − − − = ∆ ∆ ∆ − ∆ = ∆ T T T T T

Then, the rate of heat transfer in the heat exchanger is

W 10 1.17× 6 = ⋅ = ∆ =UA T_lm (255W/m2 K)(93m2)(49.3K) Q& _s

(b) The mass flow rate of the hot fluid is

→ ) ( h ,in h ,out ph hc T T m Q&= & − ) ( _h_,_in _h_,_out ph h _c _T Q _T m − = & & kg/s 12.7 = − ⋅ × = K ) 71 93 )( K J/kg 4200 ( J/s 10 17 1_. 6 m&_h

(20)

11-42E A single-pass cross-flow heat exchanger is used to cool jacket water using air. The log mean temperature difference

for the heat exchanger is to be determined.

Assumptions 1 Steady operating condition exists. 2 The heat exchanger is well-insulated so that heat loss to the surroundings

is negligible. 3 Fluid properties are constant. 4 Changes in the kinetic and potential energies of fluid streams are negligible.

Properties The specific heats of both water and air are given to be cph = 1.0 Btu/lbm·°F and cpc = 0.245 Btu/lbm·°F,

respectively.

Analysis The rate of heat transfer in the heat exchanger is

Btu/hr 10 6 . 4 F ) 140 190 )( F Btu/lbm 0 . 1 )( lbm/hr 000 , 92 ( ) ( 6 out , in , × = ° − ° ⋅ = − =m_hc_ph T_h T_h Q& &

Since heat transfer from the hot fluid is equal to the heat transfer to the cold fluid, we have → ) ( _c_,_out _c_,_in pc cc T T m

Q&= & − _,_out _c_,_in

pc c c _mQ_c T T = + & & F 9 . 136 F 90 ) F Btu/lbm 245 . 0 )( lbm/hr 000 , 400 ( Btu/hr 10 6 . 4 6 out , _⋅_° + ° = ° × = c T

Thus, the log mean temperature difference for the counter-flow arrangement is

F 5 51 F ] ) 90 140 ( / ) 9 . 136 190 ( ln[ ) 90 140 ( ) 9 . 136 190 ( ) / ln( ₁ ₂ 2 1 CF lm, ₋ ₋ ° = ° − − − = ∆ ∆ ∆ − ∆ = ∆ . T T T T T

Using Fig. 11-18c, the correction factor can be determined to be

0.92 94 . 0 190 140 9 . 136 90 50 . 0 190 90 190 140 1 2 2 1 1 1 1 2 ≈ ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 11-18c)

F 47.4° = ° = ∆ = ∆T_lm, F T_lm,_CF 0.92(51.5 F)

Discussion The correction factor (F) represents how closely the cross-flow heat exchanger approximates a counter-flow heat

(21)

11-43 Ethylene glycol is heated in a tube while steam condenses on the outside tube surface. The tube length is to be

determined.

Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the tubes are smooth. 3 Heat transfer to the

surroundings is negligible.

Properties The properties of ethylene glycol are given to be ρ = 1109 kg/m3, cp = 2428 J/kg⋅K, k = 0.253 W/m⋅K, µ

= 0.01545 kg/m⋅s, Pr = 148.5. The thermal conductivity of copper is given to be 386 W/m⋅K.

Analysis The rate of heat transfer is

W 420 , 36 C ) 25 40 )( C J/kg. 2428 )( kg/s 1 ( ) ( − = ° − ° = =mc_p T_e T_i Q& & Tg = 110ºC

The fluid velocity is

Ethylene glycol 1.5 kg/s 25ºC L

[

]

4.305m/s 4 / m) (0.02 ) kg/m 1109 ( kg/s 5 . 1 2 3 = = = π ρA_c m V &

The Reynolds number is

6181 s kg/m 01545 . 0 m) m/s)(0.02 )(4.305 kg/m (1109 Re 3 = ⋅ = = µ ρVD

which is greater than 2300 and smaller than 10,000. Therefore, we have transitional flow. We assume fully developed flow and evaluate the Nusselt number from turbulent flow relation:

3 . 183 ) 5 . 148 ( ) 6181 ( 023 . 0 Pr Re 023 . 0 0.8 0.4 = 0.8 0.4 = = = k hD Nu

Heat transfer coefficient on the inner surface is

C . W/m 2319 ) 3 . 183 ( m 02 . 0 C W/m. 253 . 0 ₂ ° = ° = = Nu D k h_i

Assuming a wall temperature of 100°C, the heat transfer coefficient on the outer surface is determined to be C . W/m 5174 ) 100 110 ( 9200 ) ( 9200 − 0.25 = − 0.25 = 2 ° = − − w g o T T h

Let us check if the assumption for the wall temperature holds:

C 55 . 89 ) 110 ( 025 . 0 5174 ) 5 . 32 ( 02 . 0 2319 ) ( ) ( ) ( ) ( avg , avg , ° = ⎯→ ⎯ − × = − × − = − − = − w w w w g o o b w i i w g o o b w i i T T T T T L D h T T L D h T T A h T T A h π π

Now we assume a wall temperature of 85°C:

C . W/m 4114 ) 85 110 ( 9200 ) ( 9200 − 0.25= − 0.25= 2° = − − w g o T T h Again checking, 2319×0.02(Tw−30)=4114×0.025(110−Tw)⎯⎯→Tw=85.9°C

which is sufficiently close to the assumed value of 90°C. Now that both heat transfer coefficients are available, we use thermal resistance concept to find overall heat transfer coefficient based on the outer surface area as follows:

C W/m 1267 4114 1 ) 386 ( 2 ) 2 / 5 . 2 ln( ) 025 . 0 ( ) 02 . 0 )( 2319 ( 025 . 0 1 1 2 ) / ln( 1 2 copper 1 2 ⋅ = + + = + + = o o i i o o h k D D D D h D U

(22)

11-44 During an experiment, the inlet and exit temperatures of water and oil and the mass flow rate of water are measured.

The overall heat transfer coefficient based on the inner surface area is to be determined.

is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.

Properties The specific heats of water and oil are given to be 4180 and 2150 J/kg.°C, respectively. Analysis The rate of heat transfer from the oil to the water is

kW 438.9 = C) 20 C C)(55 kJ/kg. kg/s)(4.18 3 ( )] ( [ − _water = ° ° − ° = mc_p T_out T_in Q& &

The heat transfer area on the tube side is

55°C 20°C Water 3 kg/s Oil 120°C 1 24 tubes 45°C 2 m 1.8 = m) m)(2 012 . 0 ( 24π π = =n D L A_i _i

The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are

C 25 = C 20 C 45 C 65 = C 55 C 120 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T C 9 . 41 ) 25 / 65 ln( 25 65 ) / ln( ₁ ₂ 2 1 , = ° − = ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm_CF 70 . 0 14 . 2 20 55 45 120 35 . 0 20 120 20 55 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P

C . kW/m 8.31 2 ° = ° = ∆ = ⎯→ ⎯ ∆ = C) 9 . 41 )( 70 . 0 )( m 8 . 1 ( kW 9 . 438 2 , , CF lm i i CF lm i i _A_F _T Q U T F A U Q& &

(23)

11-45 A stream of hydrocarbon is cooled by water in a double-pipe counterflow heat exchanger. The overall heat transfer

coefficient is to be determined.

Properties The specific heats of hydrocarbon and water are given to be 2.2 and 4.18 kJ/kg.°C, respectively. Analysis The rate of heat transfer is

kW 48.4 = C) 40 C C)(150 kJ/kg. kg/s)(2.2 3600 / 720 ( )] ( [ − _HC= ° ° − ° = mc_p T_out T_in Q& &

The outlet temperature of water is

40°C HC 150°C Water 10°C C 87.2 = C) 10 C)( kJ/kg. kg/s)(4.18 3600 / (540 kW 4 . 48 )] ( [ out w, out w, w ° ° − ° = − = T T T T c m

Q& & _p _out _in

The logarithmic mean temperature difference is

C 30 = C 10 C 40 C 62.8 = C 2 . 87 C 150 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T and C 4 . 44 ) 30 / 8 . 62 ln( 30 8 . 62 ) / ln( ₁ ₂ 2 1 ₌ − ₌ _° ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm

The overall heat transfer coefficient is determined from

K kW/m 2.31 2⋅ = ° × × = ∆ = U U T UA Q _lm C) )(44.4 0 . 6 0.025 ( kW 4 . 48 π &

(24)

11-46 Oil is heated by water in a 1-shell pass and 6-tube passes heat exchanger. The rate of heat transfer and the heat transfer

surface area are to be determined.

Properties The specific heat of oil is given to be 2.0 kJ/kg.°C. Analysis The rate of heat transfer in this heat exchanger is

kW 728 = C) 20 C C)(46 kJ/kg. kg/s)(2.0 14 ( )] ( [ − _oil = ° ° − ° = mc_p T_out T_in Q& & Oil 20°C 14 kg/s Water 80°C

60°C 6 tube passes 1 shell pass The logarithmic mean temperature difference for counter-flow

arrangement and the correction factor F are _46°C

C 40 = C 20 C 60 C 34 = C 46 C 80 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T C 92 . 36 ) 40 / 34 ln( 40 34 ) / ln( ₁ ₂ 2 1 , = ° − = ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm_CF 94 . 0 77 . 0 20 46 60 80 43 . 0 20 80 20 46 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P

Then the heat transfer surface area on the tube side becomes

2 m 21.0 = ° ° = ∆ = ⎯→ ⎯ ∆ = C) 92 . 36 ( C)(0.94) . kW/m 0 . 1 ( kW 728 2 , , CF lm s CF lm s _UF _T Q A T F UA Q& &

(25)

11-47 Water is heated in a double-pipe parallel-flow heat exchanger by geothermal water. The required length of tube is to

be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat

exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant.

Brine 140°C

Water 25°C

60°C

Properties The specific heats of water and geothermal fluid are

given to be 4.18 and 4.31 kJ/kg.°C, respectively.

Analysis The rate of heat transfer in the heat exchanger is

kW 29.26 = C) 25 C C)(60 kJ/kg. kg/s)(4.18 2 . 0 ( )] ( [ − _water = ° ° − ° = mc_p T_out T_in Q& &

Then the outlet temperature of the geothermal water is determined from

117.4 C C) kJ/kg. kg/s)(4.31 3 . 0 ( kW 26 . 29 C 140 )] ( [ _geot.water = ° ° − ° = − = ⎯→ ⎯ − = p in out out in p _m_c Q T T T T c m Q & & & &

C 57.4 = C 60 C 4 . 117 C 115 = C 25 C 140 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ out c out h in c in h T T T T T T and 82.9 C ) 4 . 57 / 115 ln( 4 . 57 115 ) / ln( ₁ ₂ 2 1 ₌ − ₌ _° ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm

The surface area of the heat exchanger is determined from

₂ 0.642m2 C) 9 . 82 )( kW/m 55 . 0 ( kW 26 . 29 ₌ ° = ∆ = ⎯→ ⎯ ∆ = lm s lm s _U _T Q A T UA Q& &

Then the length of the tube required becomes

m 25.5 = = = ⎯→ ⎯ = m) 008 . 0 ( m 642 . 0 2 π π π D A L DL A_s s

(26)

11-48 EES Prob. 11-47 is reconsidered. The effects of temperature and mass flow rate of geothermal water on the length of

the tube are to be investigated.

"GIVEN" T_w_in=25 [C] T_w_out=60 [C] m_dot_w=0.2 [kg/s] c_p_w=4.18 [kJ/kg-C] T_geo_in=140 [C] m_dot_geo=0.3 [kg/s] c_p_geo=4.31 [kJ/kg-C] D=0.008 [m] U=0.55 [kW/m^2-C] "ANALYSIS" Q_dot=m_dot_w*c_p_w*(T_w_out-T_w_in) Q_dot=m_dot_geo*c_p_geo*(T_geo_in-T_geo_out) DELTAT_1=T_geo_in-T_w_in DELTAT_2=T_geo_out-T_w_out DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=U*A*DELTAT_lm A=pi*D*L Tgeo,in [C] L [m] 100 53.73 105 46.81 110 41.62 115 37.56 120 34.27 125 31.54 130 29.24 135 27.26 140 25.54 145 24.04 150 22.7 155 21.51 160 20.45 165 19.48 170 18.61 175 17.81 180 17.08 185 16.4 190 15.78 195 15.21 200 14.67 100 120 140 160 180 200 10 15 20 25 30 35 40 45 50 55 T_geo,in [C] L [m ]

(27)

geo m& [kg/s] L [m] 0.1 46.31 0.125 35.52 0.15 31.57 0.175 29.44 0.2 28.1 0.225 27.16 0.25 26.48 0.275 25.96 0.3 25.54 0.325 25.21 0.35 24.93 0.375 24.69 0.4 24.49 0.425 24.32 0.45 24.17 0.475 24.04 0.5 23.92 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 20 25 30 35 40 45 50 m_geo [kg/s] L [m ]

(28)

11-49E Glycerin is heated by hot water in a 1-shell pass and 8-tube passes heat exchanger. The rate of heat transfer for the

cases of fouling and no fouling are to be determined.

is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Heat transfer coefficients and fouling factors are constant and uniform. 5 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive.

Properties The specific heats of glycerin and water are given to be 0.60 and 1.0 Btu/lbm.°F, respectively.

Analysis (a) The tubes are thin walled and thus we assume the inner surface area of the tube to be equal to the outer surface

area. Then the heat transfer surface area of this heat exchanger becomes

2 ft 9 . 418 ft) ft)(400 12 / 5 . 0 ( 8 = = =nπDL π A_s 120°F 175°F Hot Water Glycerin 80°F 140°F The temperature differences at the two ends of the heat exchanger are

F 40 = F 80 F 120 F 35 = F 140 F 175 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T and 37.44 F ) 40 / 35 ln( 40 35 ) / ln( 1 2 2 1 , = ° − = ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm_CF The correction factor is

50 . 0 09 . 1 175 120 140 80 58 . 0 175 80 175 120 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P

In case of no fouling, the overall heat transfer coefficient is determined from F . Btu/h.ft 704 . 3 F . Btu/h.ft 4 1 F . Btu/h.ft 50 1 1 1 1 1 2 2 2 ° = ° + ° = + = o i h h U

Then the rate of heat transfer becomes

Btu/h 29,040 = ° ° = ∆ =UA_sF T_lm_,CF (3.704Btu/h.ft2. F)(418.9ft2)(0.50)(37.44 F) Q&

(b) The thermal resistance of the heat exchanger with a fouling factor is

F/Btu h. 0006493 . 0 ) ft 9 . 418 ( F) . Btu/h.ft 4 ( 1 ft 9 . 418 F/Btu . h.ft 002 . 0 ) ft 9 . 418 ( F) . Btu/h.ft 50 ( 1 1 1 2 2 2 2 2 2 ° = ° + ° + ° = + + = o o i fi i i A h A R A h R

The overall heat transfer coefficient in this case is

F . Btu/h.ft 676 . 3 ) ft .9 F/Btu)(418 h. 0006493 . 0 ( 1 1 1 2 2 = ° ° = = ⎯→ ⎯ = s s RA U UA R

(29)

11-50 Water is heated in a double-pipe, parallel-flow uninsulated heat exchanger by geothermal water. The rate of heat

transfer to the cold water and the log mean temperature difference for this heat exchanger are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic and potential energies of fluid streams are

negligible. 4 There is no fouling. 5 Fluid properties are constant.

Properties The specific heat of hot water is given to be 4.25 kJ/kg.°C. Analysis The rate of heat given up by the hot water is

Hot water 85°C Cold water 50°C kW 208.3 = C) 50 C C)(85 kJ/kg. kg/s)(4.25 4 . 1 ( )] ( [ _{hot water} ° − ° ° = − = _p _in _out h mc T T Q& &

The rate of heat picked up by the cold water is

kW 202.0 = − = − =(1 0.03) _h (1 0.03)(208.3kW) c Q Q& &

C 43.9° = ° ⋅ = == ∆ ⎯→ ⎯ ∆ = ) m 4 )( C kW/m 15 . 1 ( kW 0 . 202 2 2 UA Q T T UA Q& _lm _lm &

(30)

11-51 Oil is cooled by water in a thin-walled double-pipe counter-flow heat exchanger. The overall heat transfer coefficient

of the heat exchanger is to be determined.

is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive.

Properties The specific heats of water and oil are given to be 4.18 and

2.20 kJ/kg.°C, respectively.

Analysis The rate of heat transfer from the water to the oil is

kW 550 = C) 50 C C)(150 kJ/kg. kg/s)(2.2 5 . 2 ( )] ( [ _oil ° − ° ° = − = mc_p T_in T_out Q& &

The outlet temperature of the water is determined from

C 7 . 109 C) kJ/kg. kg/s)(4.18 5 . 1 ( kW 550 + C 22 )] ( [ _water ° = ° ° = + = ⎯→ ⎯ − = p in out in out p T T T T _mQ_c c m Q & & & & 50°C Hot oil 150°C 2.5 kg/s Cold water 22°C 1.5 kg/s

C 28 = C 22 C 50 C 40.3 = C 7 . 109 C 150 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T C 8 . 33 ) 28 / 8 . 40 ln( 28 3 . 40 ) / ln( 1 2 2 1 ₌ − ₌ _° ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm

C . kW/m 34.6 2 ° = ° = ∆ = C) 8 . 33 m)( 6 )( m 025 . 0 ( kW 550 π lm s T A Q U &

(31)

11-52 EES Prob. 11-51 is reconsidered. The effects of oil exit temperature and water inlet temperature on the overall heat

transfer coefficient of the heat exchanger are to be investigated.

"GIVEN" T_oil_in=150 [C] T_oil_out=50 [C] m_dot_oil=2.5 [kg/s] c_p_oil=2.20 [kJ/kg-C] T_w_in=22 [C] m_dot_w=1.5 [kg/s] C_p_w=4.18 [kJ/kg-C] D=0.025 [m] L=6 [m] "ANALYSIS" Q_dot=m_dot_oil*c_p_oil*(T_oil_in-T_oil_out) Q_dot=m_dot_w*c_p_w*(T_w_out-T_w_in) DELTAT_1=T_oil_in-T_w_out DELTAT_2=T_oil_out-T_w_in DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=U*A*DELTAT_lm A=pi*D*L Toil,out [C] U [kW/m2_-C] 30 32.5 35 37.5 40 42.5 45 47.5 50 52.5 55 57.5 60 62.5 65 67.5 70 99.27 82.18 69.89 60.57 53.21 47.25 42.3 38.13 34.56 31.47 28.77 26.38 24.26 22.37 20.65 19.11 17.7 30 35 40 45 50 55 60 65 70 10 20 30 40 50 60 70 80 90 100

T

_oil,out

[C]

U [

k

W

/m

2

-C

]

(32)

Tw,in [C] U [kW/m2_C] 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 22.93 23.39 23.88 24.38 24.9 25.45 26.02 26.61 27.24 27.89 28.58 29.31 30.07 30.87 31.72 32.61 33.56 34.56 35.63 36.77 37.98 5 9 13 17 21 25 22 24 26 28 30 32 34 36 38

T

_w,in

[C]

U

[k

W

/m

2

-C

]

(33)

11-53 Water is heated by ethylene glycol in a 2-shell passes and 12-tube passes heat exchanger. The rate of heat transfer and

the heat transfer surface area on the tube side are to be determined.

Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.68 kJ/kg.°C, respectively. Analysis The rate of heat transfer in this heat exchanger is :

kW 160.5 = C) 22 C C)(70 kJ/kg. kg/s)(4.18 8 . 0 ( )] ( [ − _water = ° ° − ° = mc_p T_out T_in Q& &

The logarithmic mean temperature difference for counter-flow

arrangement and the correction factor F are _Ethylene

110°C (12 tube passes) C 38 = C 22 C 60 C 40 = C 70 C 110 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T 70°C 39 C ) 38 / 40 ln( 38 40 ) / ln( ₁ ₂ 2 1 , = ° − = ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm_CF Water 22°C 0.8 kg/s 0.92 04 . 1 22 70 60 110 55 . 0 22 110 22 70 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P 60°C

Then the heat transfer surface area on the tube side becomes

2 m 16.0 = ° ° = ∆ = ⎯→ ⎯ ∆ = C) 39 ( C)(0.92) . kW/m 28 . 0 ( kW 5 . 160 2 , , CF lm i i CF lm i i _U _F _T Q A T F A U Q& &

(34)

11-54 EES Prob. 11-53 is reconsidered. The effect of the mass flow rate of water on the rate of heat transfer and the

tube-side surface area is to be investigated.

"GIVEN" T_w_in=22 [C] T_w_out=70 [C] m_dot_w=0.8 [kg/s] c_p_w=4.18 [kJ/kg-C] T_glycol_in=110 [C] T_glycol_out=60 [C] c_p_glycol=2.68 [kJ/kg-C] U=0.28 [kW/m^2-C] "ANALYSIS" Q_dot=m_dot_w*c_p_w*(T_w_out-T_w_in) Q_dot=m_dot_glycol*c_p_glycol*(T_glycol_in-T_glycol_out) DELTAT_1=T_glycol_in-T_w_out DELTAT_2=T_glycol_out-T_w_in DELTAT_lm_CF=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) P=(T_w_out-T_w_in)/(T_glycol_in-T_w_in) R=(T_glycol_in-T_glycol_out)/(T_w_out-T_w_in)

F=0.92 "from Fig. 11-18b of the text at the calculated P and R"

Q_dot=U*A*F*DELTAT_lm_CF w m& [kg/s] Q& [kW] A [m2_] 0.4 80.26 7.99 0.5 100.3 9.988 0.6 120.4 11.99 0.7 140.4 13.98 0.8 160.5 15.98 0.9 180.6 17.98 1 200.6 19.98 1.1 220.7 21.97 1.2 240.8 23.97 1.3 260.8 25.97 1.4 280.9 27.97 1.5 301 29.96 1.6 321 31.96 1.7 341.1 33.96 1.8 361.2 35.96 1.9 381.2 37.95 2 401.3 39.95 2.1 421.3 41.95 2.2 441.4 43.95 0.25 0.65 1.05 1.45 1.85 2.25 50 100 150 200 250 300 350 400 450 5 10 15 20 25 30 35 40 45 50

m

_w

[kg/s]

Q

[

k

W

]

A

[m

2

]

heat area

(35)

11-55 A single-pass cross-flow heat exchanger with both fluids unmixed, the value of the overall heat transfer coefficient is

to be determined.

Assumptions 1 Steady operating condition exists. 2 The heat exchanger is well-insulated so that heat loss to the surroundings

is negligible. 3 Fluid properties are constant. 4 Changes in the kinetic and potential energies of fluid streams are negligible.

Properties The properties of oil are given to be cph = 1.93 kJ/kg·K and ρ = 870 kg/m3.

Analysis The mass flow rate of oil (hot fluid) is

kg/s 755 . 2 ) min/s 60 / 1 )( /min m 19 . 0 )( kg/m 870 ( 3 3 ₌ = = V& &h ρ m

Using energy balance on the hot fluid, we have

W 10 785 . 4 K ) 29 38 )( K J/kg 1930 )( kg/s 755 . 2 ( ) ( 4 out , in , × = − ⋅ = − =m_hc_ph T_h T_h Q& &

Using Fig. 11-18c, the correction factor can be determined to be

0.85 53 . 0 16 33 29 38 77 . 0 16 38 16 33 1 2 2 1 1 1 1 2 ≈ ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 11-18c)

The log mean temperature difference for the counter-flow arrangement is

C 372 . 8 C ] ) 16 29 ( / ) 33 38 ( ln[ ) 16 29 ( ) 33 38 ( ) / ln( ₁ ₂ 2 1 CF lm, ₋ ₋ ° = ° − − − = ∆ ∆ ∆ − ∆ = ∆ T T T T T

Thus, the overall heat transfer coefficient can be determined using

→ CF lm, T F UA Q&= _s ∆ CF lm, T F A Q U s ∆ = & K W/m 336 2⋅ = × = ) K 372 . 8 )( 85 . 0 )( m 20 ( W 10 785 . 4 2 4 U

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11-56E A 1-shell and 2-tube heat exchanger has specified overall heat transfer coefficient, inlet and outlet temperatures, and

mass flow rates, (a) the log mean temperature difference and (b) the surface area of the heat exchanger are to be determined.

Assumptions 1 Steady operating condition exists. 2 The

heat exchanger is well-insulated so that heat loss to the surroundings is negligible. 3 Fluid properties are constant.

4 Changes in the kinetic and potential energies of fluid

streams are negligible.

Properties The specific heat of water is given to be

cpc = 1.0 Btu/lbm·°F.

Analysis (a) Using Fig. 11-18a, the correction factor can be

determined to be 0.94 0 . 3 80 100 120 180 2 . 0 80 180 80 100 1 2 2 1 1 1 1 2 ≈ ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 11-18a)

The log mean temperature difference for the counter-flow arrangement is

F 7 . 57 C ] ) 80 120 ( / ) 100 180 ( ln[ ) 80 120 ( ) 100 180 ( ) / ln( ₁ ₂ 2 1 CF lm, ₋ ₋ ° = ° − − − = ∆ ∆ ∆ − ∆ = ∆ T T T T T

Hence, the log mean temperature difference is

F 54.2° = ° = ∆ = ∆T_lm, F T_lm,_CF 0.94(57.7 F)

(b) The surface area of the heat exchanger can be determined using

→ CF lm, T F UA Q&= _s ∆ CF lm, in , out , CF lm, ) ( T UF T T c m T UF Q A_s c pc c c ∆ − = ∆ = & & 2 ft 184 = ° ° ⋅ ⋅ ° − ° ⋅ = ) F 7 . 57 )( 94 . 0 )( F ft Btu/hr 40 ( F ) 80 100 )( F Btu/lbm 0 . 1 )( lbm/hr 000 , 20 ( 2 s A

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11-57 Engine oil is heated by condensing steam in a condenser. The rate of heat transfer and the length of the tube required

are to be determined.

is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive.

Properties The specific heat of engine oil is given to be 2.1 kJ/kg.°C. The heat of condensation of steam at 130°C is given to

be 2174 kJ/kg.

Analysis The rate of heat transfer in this heat exchanger is

kW 25.2 = C) 20 C C)(60 kJ/kg. kg/s)(2.1 3 . 0 ( )] ( [ − _oil = ° ° − ° = mc_p T_out T_in Q& &

The temperature differences at the two ends of the heat exchanger are

60°C Steam 130°C Oil 20°C 0.3 kg/s C 110 = C 20 C 130 C 70 = C 60 C 130 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T and C 5 . 88 ) 110 / 70 ln( 110 70 ) / ln( ₁ ₂ 2 1 ₌ − ₌ _° ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm The surface area is

2 2_._C)₍₈₈_.₅ _C) 0.44m kW/m 65 . 0 ( kW 2 . 25 ₌ ° ° = ∆ = lm s _U _T Q A &

Then the length of the tube required becomes

m 7.0 = = = ⎯→ ⎯ = m) 02 . 0 ( m 44 . 0 2 π π π D A L DL A s s

(38)

11-58E Water is heated by geothermal water in a double-pipe counter-flow heat exchanger. The mass flow rate of each fluid

and the total thermal resistance of the heat exchanger are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is

well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant.

Properties The specific heats of water and geothermal fluid are

given to be 1.0 and 1.03 Btu/lbm.°F, respectively.

Analysis The mass flow rate of each fluid is determined from

lbm/s 0.667 = F) 140 F F)(200 Btu/lbm. (1.0 Btu/s 40 ) ( )] ( [ water water ° − ° ° = − = − = in out p in out p T T c Q m T T c m Q & & & & 200°F 180°F Cold Water 140°F Hot brine 270°F lbm/s 0.431 = F) 180 F F)(270 Btu/lbm. (1.03 Btu/s 40 ) ( )] ( [ water geo. water geo. ° − ° ° = − = − = in out p in out p T T c Q m T T c m Q & & & &

The temperature differences at the two ends of the heat exchanger are

F 40 = F 140 F 180 F 70 = F 200 F 270 , , 2 , , 1 ° ° − ° = − = ∆ ° ° − ° = − = ∆ in c out h out c in h T T T T T T and 53.61 F ) 40 / 70 ln( 40 70 ) / ln( ₁ ₂ 2 1 ₌ − ₌ _° ∆ ∆ ∆ − ∆ = ∆ T T T T T_lm Then F/Btu s 1.34 ⋅° = ° = = ⎯→ ⎯ = = ° = ∆ = ⎯→ ⎯ ∆ = F Btu/s. 7462 . 0 1 1 1 F Btu/s. 7462 . 0 F 53.61 Btu/s 40 o s s lm s lm s UA R RA U T Q UA T UA Q& &