1. in a counter flow double pipe heat exchanger, oil is cooled … year/heat and mass... ·...

1. In a counter flow double pipe heat exchanger, oil is cooled from 85˚C to 55˚C by water

entering at 25˚C. The mass flow rate of oil is 9,800 kg/h and specific heat f oil is 2000 J/kg K.

The mass flow rate of water is 8,000 kg/h and specific heat of water is 4180 J/kg K. Determine

the heat exchanger area and heat transfer rate for an overall heat transfer co-efficient of 280

W/m2 K.

Given:

Hot fluid – oil, Cold fluid - water --------------------------

(T1, T2) (t1, t2) Water

--------------------------

Entry temperature of oil, T1 = 85˚C Oil

--------------------------

Exit temperature of oil T2 = 55˚C Water

--------------------------

Mass flow rate of oil (Hot fluid), mh = 9,800 kg/h

skg /3600

800,9

Specific heat of oil, Cph = 2000 J/kg K

Mass flow rate of water (cold fluid), mc = 8,000 kg/h

skg /3600

000,8

Specific heat of water, Cpc = 4180 J/kg K

Overall heat transfer co-efficient, U = 280 W/ m2K

mh = 2.72 kg/s

mc = 2.22 kg/s

To find:

1. Heat exchanger area, (A)

2. Heat transfer rate, (Q)

Solution:

We know that,

Heat lost by oil (Hot fluid) = Heat gained by water (cold fluid)

Qh = Qc

Ct

t

tt

ttCmTTCm pccphh

5.42

109.2316.9279102.163

418022.25585200072.2

2

3

2

3

12

1221

Heat transfer,

WQ

Q

TTCmorttCmQ phhpcc

3

2112

10162

255.42418022.2

We know that,

Heat transfer, mTUAQ …….. (1)

Where

mT - Logarithmic Mean Temperature Difference. (LMTD)

For Counter flow,

Exit temperature of water, Ct 5.422

2555

5.4285

25555.4285

12

21

1221

In

T

tT

tTIn

tTtTT

m

m

Substitute mT U and Q values in Equn (1)

8.3528010162

)1(

3

A

TUAQ m

==>

Result:

1. Heat exchanger area, A = 16.16 m2

2. Heat transfer, Q = 162 103 W

2. Water flows at the rate of 65 kg/min through a double pipe, counter flow heat exchanger.

Water is heated from 50˚C to 75˚C by oil flowing through the tube. The specific heat of the oil

is 1.780 kJ/kg K. The oil enters at 115˚C and leaves at 70˚C. The overall heat transfer co-

efficient is 340 W/m2K. Calculate the following

1. Heat exchanger area

2. Rate of heat transfer

Given:

Hot fluid – oil, Cold fluid - water

(T1, T2) (t1, t2)

Mass flow rate of oil (cold fluid), mc = 65 kg/min

skg /60

65

mT = 35. 8˚C

A = 16.16 m2

mc = 1.08 kg/s

Entry temperature of water, t1 = 50˚C

Exit temperature of water, t2 = 75˚C

Specific heat of oil (Hot fluid) Cph = 1.780 kJ/kg K

= 1.780103 J/m

2K

Entry temperature of oil, T1 = 115˚C

Exit temperature of oil, T2 = 70˚C

Overall heat transfer co-efficient, U = 340 W/m2K

To find:

1. Heat exchanger area, (A)

2. Heat transfer rate, (Q)

Solution:

We know that,

Heat transfer, 2112 TTCmorttCmQ phhpcc

5075418608.1

12

Q

TTCmQ pch

We know that,

Heat transfer, mTAUQ …….(1)

Where


For Counter flow,

Specific heat of water, Cpc = 4186 J/kgK

5070

75115

507075115

12

21

1221

In

T

tT

tTIn

tTtTT

m

m

Substitute mT U and Q and U values in Equn (1)

8.2834010113

)1(

3

A

TUAQ m

==>

Result:

1. Heat exchanger area, A = 11.54 m2

2. Heat transfer, Q = 113 103 W

3. In a counter flow single pass heat exchanger is used to cool the engine oil from 150˚C to

55˚C with water, available at 23˚C as the cooling medium. The specific heat of oil is 2125 J/kg

K. The flow rate of cooling water through the inner tube of 0.4 m diameter is 2.2 kg/s. The

flow rate of oil through the outer tube of 0.75 m diameter is 2.4 kg/s. If the value of the overall

heat transfer co-efficient is 240 W/m2 K, how long must the heat exchanger be to meet its

cooling requirement?

Given:


(T1, T2) (t1, t2)



Entry temperature of water, T1 = 23˚C

Specific heat of oil (Hot fluid) Cph = 2125 kJ/kg K

mT = 28. 8˚C

A = 11.54 m2

Inner diameter, D1 = 0.4 m

Flow rate of water (cooling fluid), mc = 2.2 kg/s

Outer diameter, D2= 0.75 m

Flow rate of oil (Hot fluid), mh = 2.4 kg/s


To find:

Length of the heat exchanger, L

Solution:

We know that,

Heat lost by oil (Hot fluid) = Heat gained by water (cold fluid)

2341862.25515021254.2 2

1221

t

ttCmTTCm

QQ

pccphh

ch

3

2

3 102112.9209105.484 t

Heat transfer, 2112 TTCmorttCmQ phhpcc

236.7541862.2 Q

We know that,

Heat transfer mTUAQ …… (1)

Where


[ Specific heat of water, Cpc = 4186 J/kgK]

WQ 3104.484

For Counter flow,

2355

6.75150

23556.75150

12

21

1221

In

T

tT

tTIn

tTtTT

m

m

Substitute mT , U and Q values in Equn (1)

2.5024010104.484

)1(

33

A

TUAQ m

==>

We know that,

Area, A = LD 1

40.20 = 0.4 L

==>

Result:

Length of the heat exchanger, L = 31.9 m.

4. In an oil cooler for a lubrication system, oil is cooled from 70˚C to 40˚C by using a cooling

water flow at 25˚C. The mass flow rate of oil is 900 kg/h and the mass flow rate of water is 700

kg/h. Give your choice for a parallel flow or counter flow heat exchanger, with reasons. If the

overall heat transfers co-efficient is 20 W/m2K, find the area of the heat exchanger. Take

specific heat of oil is 2 kJ/ kg˚C.

Given:

Hot fluid – oil, (T1, T2) Cold fluid - water (t1, t2)



mT = 50.2˚C

A = 40.20 m2

L = 31.9 m


The mass flow rate of oil, hkgmh /900

skg

skg

/25.0

/3600

900

The mass flow rate of water, hkgmc /700

skg

skg

/197.0

/3600

700


Specific heat of oil, Cph = 2 kJ/kg˚C

= 2 103 J/kg˚C

To find:

1. Choice of heat exchanger (Whether parallel flow or counter flow)

2. Area of heat exchanger.

Solution:

We know that,

Heat lost by oil (Hot fluid) = Heat gained by water (Cold fluid)

254186194.0407010225.0 2

3

1221

t

ttCmTTCm

QQ

pccphh

ch

10.302,2008.812000,15 2 t

==> t2 = 43.47˚C

[ Specific heat of water, Cpc = 4186 J/kg K]

> T2

Since t2 > T2, counter flow arrangement should be used.

We know that,

Heat transfer mTUAQ …… (1)

For Counter flow,

569.0

53.11

2540

47.4370

254047.4370

12

21

1221

In

T

tT

tTIn

tTtTT

m

m

…….. (2)

We know that,

Heat transfer,

407010225.0 3

21

1221

TTCmQ

ttCormTTCmQ

phh

pccphh

…….. (3)

Substitute, Q, U, and mT values in equation (1)

26.2020000,15

A

TUAQ m

==>

mT = 20.26˚C

Q = 15,000 J/s

A = 37.02 m2

Exit temperature of water, t2 = 43.47˚C

Result:

1. Choice of heat exchanger – counter flow arrangement

2. Surface area, A = 37.02 m2.

5. In a refrigerating plant water is cooled from 20˚C to 7˚C by brine solution entering at - 2˚C

and leaving at 3˚C. The design heat load is 5500 W and the overall heat transfer co-efficient is

800 W/m2 K. What area required when using a shell and tube heat exchanger with the water

making one shell pass and the brine making two tube passes.

Given:

Hot fluid – water, Cold fluid – water

(T1, T2) (t1, t2)


Exit temperature of water, T2 = 7˚C

Entry temperature of water, T1 = - 2˚C

Exit temperature of brine solution, t2 = 3˚C

Heat load, Q = 5500 W


To find:

Area required (A)

Solution:

Shell and tube heat exchanger – One shell pass and two tube passes

For shell and tube heat exchanger (or) cross flow heat exchanger.

mTFUAQ [Counter flow] …….. (1)

Where

F – Correction factor

mT - Logarithmic mean temperature difference for counter flow.

For Counter flow,

27

320

27320

12

21

1221

In

tT

tTIn

tTtTT m

From graph,

22

5

220

23,

11

12

tT

ttPvalueX axis

Curve value, 5

13

23

720

12

21

tt

TTP

Xaxis value is 0.22, curve value is 2.6, and corresponding Y axis value is 0.94.

Substitute mT Q , U and F value in Equn (1)

57.1280094.05500

)1(

A

TFUAQ m

==>

Result:

Area of heat exchanger, A = 0.58 m2

mT = 12.57˚C

P = 0.22

A = 0.58 m2

R = 2.6

6. A parallel flow heat exchanger is used to cool 4.2 kg/min of hot liquid of specific heat 3.5

kJ/kg K at 130˚C. A cooling water of specific heat 4.18 kJ/kg K is used for cooling purpose at

a temperature of 15˚C. The mass flow rate of cooling water is 17 kg/min. calculate the flowing.

1. Outlet temperature of liquid

2. Outlet temperature of water

3. Effectiveness of heat exchanger

Take

Overall heat transfer co-efficient is 1100 W/m2K.

Heat exchanger area is 0.30 m2

Given:

Mass flow rate of hot liquid, mh = 4.2 kg/min

Specific heat of hot liquid, Cph = 3.5 kJ/kg K

Inlet temperature of hot liquid, T1 = 130˚C

Specific heat of water, Cpc = 4.18 kJ/kg K

Inlet temperature of cooling water, t1 = 15˚C

Mass flow rate of cooling water, mc = 17 kg/min


Area, A = 0.30 m2

To find:

1. Outlet temperature of liquid, (T2)

2. Outlet temperature of water, (t2)

mh = 0.07 kg/s

kgKJC ph /105.3 3

kgKJC pc /1018.4 3

mc = 0.28 kg/s

3. Effectiveness of heat exchanger,

Solution:

Capacity rate of hot liquid, phh CmC

3105.307.0

……….. (1)

Capacity rate of water, phc CmC

31018.428.0

………… (2)

From (1) and (2),

Cmin = 245 W/K

Cmax = 1170.4 W/K

209.04.1170

245

max

min C

C

……………. (3)

Number of transfer units, NTU = minC

UA

245

03.01100 NTU

…….. (4)

From graph,

Xaxis NYU = 1.34

C = 245 W/K

C = 1170.4 W/K

max

min

C

C= 0.209

NTU = 1.34

Curve 209.0max

min C

C

Corresponding Yaxis value is 64%

i.e.,

Maximum possible heat transfer

Qmax = Cmin (T1 – t1)

= 245 (130 – 15)

Actual heat transfer rate

175,2864.0

max

QQ

We know that,

Heat transfer, 12 ttCmQ pcc

64.0

Qmax = 28,175 W

Q = 18,032 W

Ct

t

tt

40.30

175564.11700323,18

1018.428.0032,18

2

2

52

3

We know that,

Heat transfer, 21 TTCmQ pch

Ct

T

T

4.56

245318500323,18

130105.307.0032,18

2

2

2

3

Result:

1. T2 = 56.4˚C

2. t2 = 30.40˚C

3. = 0.64

7. In a counter flow heat exchanger, water at 20˚C flowing at the rate of 1200 kg/h. it is heated

by oil of specific heat 2100 J/kg K flowing at the rate of 520 kg/h at inlet temperature of 95˚C.

Determine the following

1. Total heat transfer

2. Outlet temperature of water

3. Outlet temperature of oil

Take

Overall heat transfer co-efficient is 100 W/m2 K.

Heat exchangers are is 1 m2.

Given:

Capacity rate of hot oil, C = phh Cm

Outlet temperature of cold water, Ct 40.302

Outlet temperature of hot liquid, CT 4.562

2100144.0

………… (1)

Capacity rate of water, C = phh Cm

418633.0

………… (2)

Inlet temperature of hot liquid, T1 = 130˚C

Specific heat of water, Cpc = 4.18 kJ/kg K

From Equn (1) and (2),

Cmin = 302.4 W/K

Cmax = 1381.3 W/K

218.03.1381

4.302

max

min C

C

……………. (3)

Number of transfer units, NTU = minC

UA

4.302

11100 NTU

…….. (4)

KWC /3.1381

kgKJC pc /1018.4 3

max

min

C

C= 0.218

NTU = 3.3

C = 302.4 W/K

From graph,

Xaxis NYU = 3.3

Curve 218.0max

min C

C

Corresponding Yaxis value is 0.95

i.e.,

218.0max

min C

C

Maximum possible heat transfer

Qmax = Cmin (T1 – t1)

= 302.4 (95 – 20)

Actual heat transfer rate

680,2295.0

max

QQ

95.0

Qmax = 22,680 W

Q = 21,546 W

We know that,

Heat transfer, 12 ttCmQ pcc

20418633.0546,21 2 t

kgKJC pc /4186

6.627,2738.1381546,21 2 t

==>

We know that,

Heat transfer, 21 TTCmQ pch

2

2

4.302728,28546,21

952100144.0546,21

T

T

==>

Result:

1. T2 = 21,546W

2. T2 = 23.75˚C

3. t2 = 35.5˚C

8. In a counter flow heat exchanger, water is heated from 20˚C to 80˚C by an oil with a

specific heat of 2.5 kJ/kg – K and mass flow rate of 0.5 kg/s. The oil is cooled from 110˚C to

40˚C. If the overall heats transfer co-efficient is 1400 W/m2K, find the following by using

NTU method

1. Mass flow rate of water

2. Effectiveness of heat exchanger

3. Surface area

Outlet temperature of water, Ct 5.352

Outlet temperature of oil, CT 75.232

t2 = 35.5˚C

T2 = 23.75˚C

Given:


(T1, T2) (t1, t2)

Inlet temperature of water, t1 = 20˚C

Outlet temperature of water, t2 = 80˚C

Specific heat of oil, Cph = 2.5 kJ/kg - K

= 2.5 103 J/kg - K

The mass flow rate of oil, skgmh /5.0

Inlet temperature of oil, T1 = 110˚C

Outlet temperature of oil, T2 = 40˚C

Overall heat transfer co-efficient, U = 1400 W/m2 K

To find:

1. Mass flow rate of water, mc

2. Effectiveness of heat exchanger,

3. Surface area, A

Solution:

We know that,

Heat lost by oil = Heat gained by water

2080418640110105.25.0 3

1221

c

pccphh

ch

m

ttCmTTCm

QQ

==> mc = 0.348 kg/s

Mass flow rate of oil (Hot fluid), C = mhCph

[ Specific heat of water, Cpc = 4186 J/kg K]

3105.25.0

……… (1)

Capacity rate of water (Cold fluid), C = mcCpc

4186348.0

………… (2)

From Equn (1) and (2),

Cmin = 1250 W/K

Cmax = 1456.73 W/K

858.073.1456

1250

max

min C

C

………… (3)

We know that,

Effectiveness, 11

21

TT

TT

20110

40110

From graph,

Xaxis 77.0

Curve 858.0max

min C

C

C = 1250 W/K

C = 1456.73 W/K

max

min

C

C= 0.858

77.0

Corresponding Xaxis value is 3.4, i.e., NTU = 3.4

We know that,

NTU = minC

UA

minC

UANTU

==>

Result:

1. mc = 0.348 kg/s

2. = 0.77

3. A = 3.03 m2

A = 3.03 m2

1. in a counter flow double pipe heat exchanger, oil is cooled … year/heat and mass... ·...

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