WORKED PROBLEMS HEAT EXCHANGERSWORKED PROBLEMS HEAT EXCHANGERS

Supplement to Heat Exchanger Seminar Held in Edmonton 21/11/2004

WORKED EXAMPLE 1

PARALLEL flow - Counter flow Heat Transfer Area Requirement

Problem:

A heat exchanger heats 25,000 kg/hr of water entering at 80oC while cooling 20,000 kg/hr of water from 100oC to 80oC. Determine the heat transfer area necessary for (i) Parallel flow arrangement (ii) Counter flow arrangement. Given Overall heat transfer coefficient, U= 1,500 W/m2.

Solution:

Heat of heat transferred by the cooling water = Q = mCpT = 20000 x 4.184 x (100 - 80) = 1673600 kJ/hr

This will be the amount of heat transferred to the water that is getting heated up. Therefore, temperature change is = 1673600 / (25000 x 4.184) = 16oC

i.e., outlet temperature = 80 + 16 = 96oC

Parallel flow - Counter flow Heat Transfer Area Requirement, modified data for the previous problem

A heat exchanger heats 25,000 kg/hr of water entering at 30oC while cooling 20,000 kg/hr of water from 100oC to 80oC. Determine the area necessary for (i) Parallel flow arrangement(ii) Counter flow arrangement. Overall heat transfer coefficient may be assumed as 1,600 W/m2K.

Calculations:

Heat transferred by the cooling water = Q = mCpT = 20000 x 4.184 x (100 - 80) = 1673600 kJ/hr = 464.89 kW

This will be the amount of heat transferred to the water that is getting heated up. Therefore, temperature change is = 1673600 / (25000 x 4.184) = 16oC

i.e., outlet temperature = 30 + 16 = 46oC

With these data, the following temperature-length diagram is drawn.

(i) Parallel flow arrangement:

To = 100 - 30 = 70oC

TL = 80 - 46 = 34oC

LMTD = Tln = (To - TL) / ln (To/TL)

= (70 - 34) / ln (70/34) = 49.85oC

A = Q / (UTln) = 464.89 x 103 / (1600 x 49.85) = 5.8286 m2

Area required for parallel flow = 5.8286 m2

(ii) Counter flow arrangement:

To = 80 - 30 = 50oC

TL = 100 - 46 = 54oC

LMTD = Tln = (To - TL) / ln (To/TL)

= (50 - 54) / ln (50/54) = 51.97oC

A = Q / (UTln) = 464.89 x 103 / (1600 x 51.97) = 5.5908 m2

Area required for counter flow = 5.5908 m2

THERMAL DESIGN OF HEAT EXCHANGERS

Worked example 2:

Design a shell-and-tube heat exchanger to heat 2.5 kg/s of water from 15 to 85C. The heating is to be accomplished by passing hot oil, which is available at 160C, through the shell side of the exchanger. The oil is known to provide an average convection coefficient of h = 400 W/m2K on the outside of the tubes. Ten (10) tubes pass the water through the shell. Each tube is thin walled of diameter D = 25 mm, and makes eight (8) passes through the shell. If the oil leaves the exchanger at 100C, what is its flow rate? How long must the tubes be to accomplish the desired heating?

Solution

Known: fluid inlet and outlet temperatures for a shell-and-tube heat exchanger with 10 tubes making 8 passes.

N = 10 tubes, (OD = ID = 25 mm),

M = 8 passes

Required:

1. Oil flow rate required to achieve specified outlet temperature

2. Tube length required to achieve specified water heating.

Assumptions:

Negligible heat loss to the surroundings and kinetic and potential energy changes.

Constant properties.

Negligible tube wall thermal resistance and fouling effects.

Fully developed water flow in tubes.

Properties:

oil (average Th = 130C): cp = 2350 J/kg.K. water (average Tc = 50C): cp = 4181 J/kg.K, = 54810-6 N.s/m2, k = 0.643 W/m.K,

Pr = 3.56.

Analysis: 1. Oil flow rate can be obtained from the heat balance:

Q' = mccp,c(TCo- TCi) = 2.5 kg/s 4181 J/kg.K (85 - 15)C Q' = 7.317 105 W

2. The required tube length may be obtained from:

Q' = U.A.F.TLM

where hi may be obtained by first calculating ReD. With m1 mc/N = 0.25 kg/s defined as the water flow rate per tube,

Hence the water flow is turbulent,

NuD = 0.023 ReD4/5 Pr 0.4 = 0.023 (23234)4/5(3.56)0.4 = 119 hi = (k/D) NuD = 119 (0.643 W/m.K 0.025m) =3061 W/m2K

Hence

U = 1/[(1/400) + (1/3061)] = 354 W/m2K

The correction factor F may be obtained from F Chart, corresponding to:

R = (160-100)/(85-15) = 0.86 P = (85-15)/(160-15) = 0.48

Hence F 0.87.

= 79.9 deg C

Since A = N DL, where N = 10 is the number of tubes,

L = Q'/UN DFTLM = (7.317 105)/(354 10 0.025 0.87 79.9) L = 37.9 m

Comments:

With (L/D) = 37.9 m/0.025 m = 1516, the assumption of fully developed conditions throughout the tube is justified.

With eight passes, the shell length is approximately L/M = 4.7 m.

WORKED EXAMPLE 3:

Design of a Parallel Tube Heat Exchanger

(Double pipe counterflow exchanger)

The general heat transfer equation is:

Problem:

Find the Required Length of a Heat Exchanger with Specified Flows:

Turbulent Flow in Both Streams

The design constraints are given in the schematic above. We show this as a countercurrent configuration.

The benzene flow is specified as a mass flow rate (in pound mass units), and the water flow is given as a linear velocity.

The inside pipe :

Schedule 40 1 1/4 pipe

do = 1.66 in. = 0.138 ft. di = 1.38 in = 0.115 ft.

The outside pipe :

Schedule 40 2 pipe

Di = 2.07 in = 0.115 ft.

Sc = D2/4 = 0.0104 ft2

To calculate the heat exchanger area, we must find Ao = DL.

The general heat transfer equation:

The overall heat transfer coefficient, Uo, is given by

In another form,

To evaluate the parameters of the problem, we need the physical and thermal properties and conditions for flow in the system

Internal Film Resistance

The Nusselt number on the inside of the inner pipe is given by the Dittus-Boelter equation:

Thus

The heat transfer area per unit length is

so inner film resistance is

The other tube dimensions are:

Calculation of the Water Flow Rate

The hydraulic diameter is:

Given the water velocity of 5 ft/s, we can solve for the water flow rate

The overall heat balance:

Solving for the outlet water temperature:

External Film Resistance

The physical properties of the water must be estimated in order to determine the film heat transfer coefficient in the annular shell. The average water temperature Tb is calculated as 84.7 F

From the Dittus-Boelter equation, the Nusselt number is given as:

thus

The external area/length is

so that the external film resistance is

Conduction Resistance

The last term in the equation for the overall heat transfer coefficient is:

Overall Heat Transfer Coefficient

The overall resistance is

Log-Mean T

Heat Load

Heating Rate/unit Length

Given the heat load, we can calculate the length of tubing so that

Co-current Flow Heat Exchanger

The heat loads are identical, the Overall Resistances to heat transfer (UA)-1 are no different since the film coefficients do not change, but the Tlm are different.

There are two observations to be made:

1. The tube length required for co-current flow is more than twice as long.

2. The approach temperature for co-current flow becomes diminishingly small.

Question 1. If we use more tubes, how do we estimate the number of tubes and the required area for a single pass heat exchanger.

If we use N identical tubes,

From the Dittus-Boelter equation we have

The required length is

so that

WORKED EXAMPLE 4:

Sensible heat Transfer Recycle Oil Cooler

Problem:

385,000 lb/hr of 31oAPI (s = 0.87 at 60oF) recycle hydrocarbon oil at 250oF to be cooled to 150oF by 572,000 lb/hr of water having a range of 85 to 120oF.

Pressure drops of 15 psi for oil and 5 psi for water are allowed.

Fouling factors are 0.004 hr.ft2 oF/Btu for oil and 0.001 hr.ft2 oF/Btu for water.

Tubes to be -in. O.D. x 14 BWG min wall C. steel, 20 ft long on 1 square pitch.

Viscosity of 31oAPI distillate:

At 100oF = 1.15 cPs = 0.858 = 1.34 cSt

150 oF = 0.85 cPs = 0.839 = 1.013

300 oF = 0.45 cPs = 0.785 = 0.573

The average temperatures will be used as caloric mean temperatures for determination of physical properties to be used in calculations.

Calculations:

With oil on shellside and water on tubeside

TSM = (250 + 150)/2 = 200oF .

Thus, c = 0.52 Btu/lb.oF; = 0.805cP cSt; s = 0.82; = 0.66 x 2.42 = 1.597 lb/ft.hr

kPr1/3 = 0.17 Btu/hr.ft.oF

TTM = (120 + 85)/2 = 102.5oF

Thus, c = 1.0 Btu/lb.oF; = 0.75 cP; s = 0.82; = 0.75 x 2.42 = 1.815 lb/ft.hr

kPr1/3 = 0.591 Btu/hr.ft.oF

A. Heat Balance at Mean Temperatures

Q = W c t

Oil: Q = 385,000 x 0.52(250 150) = 20 x 106 Btu/hr

Water: Q = 572,000 x 1.0 (120 85) = 20 x 106 Btu/hr

B. LMTD

OilWater

250120130

1508565

1003565

LMTD = (GTTD LTTD)/ln(GTTD/LTTD) = 65 / ln(130/65) = 93.8oF

R = (T1 T2)/(t2 t1) = 100/35 = 2.86

S = (t2 t1)/(T1 t1) = 35/165 = 0.212

Fn = 0.93 (Chart for 1-2n STHE)

Thus

t = F x LMTD = 0.93 x 93.8 = 87.2oF

C. Exchanger Sizing Trial

Assume Ud = 90 Btu/hr.ft2.oF

A = Q/Ud.T = 20 x 106 /( 90 x 87.2) = 2548 ft2

Tube data

O.D. = 0.75 in = 0.0625 ft

Ao = 0.1963 ft2/ft

I.D. = 0.584 in = 0.048667 ft

Ac = 0.268 in2

Rmio = 0.000298

Ao/Ai = 0.75/0.584 = 1.284

Assume two (2) tube passes

No. of tubes = 2548/(20 x 0.1963) = 649 tubes

From tube count table (for 2 tube passes, -in. tubes on 1 square pitch), nearest count:

664 tubes in 32 I.D. shell

Also from empirical formuls, Ds = (Nt x P2/0.71)0.5 = (664 x 12/0.71)

= 30.58 select 32

Thus,

Nt = 664

A = 664 x 0.1963 = 2607 ft2

Corrected Ud = 20 x 106/(2607 x 87.2) = 88 Btu/hr.ft2.oF

D. Tubeside Rating

Act = 664 x 0.268/(2 x 144) = 0.618 ft2

Gt = Wt/Act = 572,000/0.618 = 925,732 lb/ft2.hr

Vt = Gt/225000 s = 4.11 ft/s

NRet = I.D. Gt/ = 0.048667 x 925,732/1.815 = 24822

Using the tubeside heat transfer curve (Sieder and Tate):

jHt = 81

hio = jHt x kPr1/3/(I.D. x Ao/Ai) = 81 x 0.591 /(0.048667 x 1.284)

= 766 Btu/hr.ft2.oF

Rio = 1/hio = 0.001316

Rfio = 0.001 x 1.284 = 0.001284

Rmio = 0.000298

Rtio = 0.002898

E. Shellside Rating

Acs = Ds x C x Bp / 144 x P

Assume 20 baffle spaces at 20% cut

Bp = 20 x 12/20 = 12 in. = 0.375 (which is > min of 0.2 Ds, thus OK)

C = p OD = 1.0 0.75 = 0.25

(see shellside transfer curve)

Acs = 32 x 0.25 x 12/(144 x 1) = 0.667 ft2

Gs = Ws/Acs = 385,000/0.667 = 577,500 lb/ft2.hr

Des = 4 x free area /wetted perimeter

For square pitch, p

Des = 4 x (p2 OD2/4)/(12 OD)

For triangular pitch, p

Des = 4{p/2 x .866 p (1/2 x OD2)/4}/(12 x OD/2)

NOTE

In case of finned tubing, OD in the Des formulas is calculated as the diameter for a hypothetical plain tube having the volumetric displacement of the fin tube metal, abbreviation is usually De. Some Des for plain and 19 FPI Lofin are shown on hellside heat transfer curve.

For this example (square pitch)

Des = 0.95/12 = 0.079 ft

NRes = Des x Gs / = 0.079 x 577,500 / 1.597 = 28,567

From heat transfer chart

jHs = 110

ho = (jHs x kPr1/3 /Des)

where = (/w)0.14

ho/ = 110 x 0.17/0.079 = 236.7

Ro = 1/ (ho/ ) = 0.004225

Rfo = 0.0040

Rto = 0.008225

F. Trial Tube Wall Temperature

tw = Tsm {Rto/(Rto +Rtio)} (Tsm Ttm)

tw = 200 {0.008225/(00.8225 + 0.002898)} (200 102.5) = 127.9oF

For tw = 127.9

= 1.13 cSt

= s x = 0.833 x 1.13 = 0.94129 cP = 2.42 x 0.9412 = 2.278 lb/fr.hr

Thus,

= (1.597/2.278)0.14 = 0.951

ho = (ho/ ) = 236.7 x 0.951 = 225.22 Btu/hr.ft2.oF

Ro = 0.00444

Rfo = 0.004

0.00844

Rio = 0.002898

R = 0.011338

Thus

Ud = 88.2 > 88 assumed.

G. Shellside Pressure Drop

For NRes = 28,567 and BC = 20%

From shellside pressure drop chart)

fs = 0.00265 ft2/in2

Number of bundle crosses (Nb + 1), wher Nb is th number of baffles

Nb + 1 = 12 x 20/12 = 20

Using the Fanning equation,

Ps = (fs Gs2 Ds (Nb + 1)/(5.22 x 1010 x Des x s x )0

Ps = (0.00265 x 5.772 x 1010 x 32/12 x 20)/(5.22 x 1010 x 0.079 x 0.82 x 0.951)

Ps = 14.65 psi < 15 psi allowed.

H. Tubeside Pressure Drop

For NRet = 24,822

From tubeside friction factor Chart

ft = 0.00022 ft2/in2

Return loss (per pass) = 4 velocity heads

= ft Leq/ID

Leq = ID/ft = 2 x 4 x 0.584/(0.00022 x 123) = 12.9 ft

Pt = ft x Gt2 x L /(5.22 x 1010 x ID x s x )

= 1

S = 1

Pt = 0.00022 x 9.2572 x 1010 (2 x 20 + 12.9)/(5.22 x 1010 x 0.048667 x 1 x 1) = 3.92

Pt = 4.0 psi < 5.0 psi allowed

WORKED EXAMPLE 5:

Effectiveness of exchangers

Problem:

A process liquor at 299.5 K is to be heated using water at 366 K available from another part of the plant. A heat exchanger having an installed area of 8.1 m is to be used for this operation, with the liquid flowing in true counter-current flow. The flow rates of the process liquor and water will be 3.1 and 1.1 kg/s respectively. Previous experience indicates that an overall heat transfer coefficient of 454 W/mK will apply. The specific heat of the process liquor is 2.1 kJ/kg K.

Calculate the outlet temperatures of water and process liquor using the number of transfer units and the effectiveness of the exchanger.

Solution:

Extracting the relevant data from the question: A = 8.1 m2 U = 454 W/m2K m'liq = 3.1 kg/s Cpliq = 2100 J/kg.K m'wat = 1.1 kg/s Cpwat = unknown

A precise value for the specific heat capacity of water cannot be fixed as the average water temperature is not yet known, it is best to use a guess of 4200 J/kg.K and then reach an answer by iteration. CH = 1.1 4200 = 4620 W/K CC = 3.1 2100 = 6510 W/K Q'max = Cmin(Ti - ti) = 4620(366 - 299.5) = 307 230 W CR = 4620 6510 = 0.7096774

We now have to find the appropriate formula for effectiveness of the exchanger. Although we are not told if the unit is a concentric tube or shell and tube design, CONCENTRIC TUBE COUNTERFLOW is the best description. (All the shell and tube formulae are for multiple tube side passes). = 0.472423 Q' = Q'max = 0.472423 307 230 = 145 142.52 Watts Q' = m'watCpwat(Ti - To)

Now we can calculate an average water temperature and use a more accurate value for Cpwat by using physical property tables.

At 350 K, the specific heat of water is given as 4195 J/kg.K (Incropera and de Witt): CH = 1.1 4195 = 4614.5 W/K Q'max = Cmin(Ti - ti) = 4614.5(366 - 299.5) = 306 864.25 W CR = 4614.5 6510 = 0.7088325 = 0.4728427 Q' = Q'max = 0.4728427 306 864.25 = 145 098.52 Watts

i.e. answer converges on water outlet temperature of 334.6 K. As for the process liquor: Q' = m'liqCpliq(to - ti)

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