Block 2 Engineering Principles & Heat Transfers

7/22/2019 Block 2 Engineering Principles & Heat Transfers

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The Steam and Condensate Loop 2.8.1

Block 2 Steam Engineering Principles and Heat Transfer Thermal Rating Module 2.8

Module 2.8

Thermal RatingSC-G

CM-12

CM

Issue2

C

opyright2007Spirax-SarcoLimited


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The Steam and Condensate Loop2.8.2


Thermal Rating

Some items of manufactured plant are supplied with information on thermal output. Thesedesign ratings can be both helpful and misleading. Ratings will usually involve raising a statedamount of air, water or other fluid through a given temperature rise, using steam at a specifiedpressure. They are generally published in good faith with a reasonable allowance for fouling of

the heat transfer surface.It must be clear that changing any factor at all will alter the predicted heat output and therebythe steam consumption. A secondary fluid which is colder than specified will increase the demand,while steam at less than the specified pressure will reduce the ability to transfer heat.

Temperature and pressure can often be measured easily so that corrections can be applied.However, flowrates of air, water and other fluids may be far more difficult to measure. Undetectedfanbelt slip or pump impeller wear can also lead to discrepancies, while lower than expectedresistances applied to pumps and fans can cause flowrates to be higher than the design values.

A more common source of error arises from the assumption that the manufacturers rating equatesto actual load. A heat exchanger may be capable of meeting or exceeding a given duty, but the

connected load may often only be a fraction of this. Clearly it is useful to have information on thethermal rating of plant, but care must be taken when relating this to an actual heat load.

If the load is quoted in kW, and the steam pressure is given, then steam flowrate may be determinedas shown in Equation 2.8.1:

Fig. 2.8.1 Typical heat exchanger manufacturers name-plate

Equation 2.8.1Load in kW x 3 600

Steam flowrate (kg h)h at operating pressure

=

fg


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The Steam and Condensate Loop 2.8.3


Questions

1. What is the result of using a heat exchanger rating to calculate its steam consumption?

a| The true connected heat load may be different from the rated figure

b| The rating does not take account of the temperature of the secondary medium

c| The rating is based on a steam pressure of 1.0 bar d| The rating does not allow for condensate forming in the heat exchanger

2. A heat exchanger has a design rating based on a working pressure of 7 bar g.What would be the effect of supplying the exchanger with steam at 3 bar g?

a| The heat output would be greater because the enthalpy of evaporationat 3 bar g is higher than at 7 bar g

b| The heat output would be greater because steam at 3 bar g has a greater volumethan steam at 7 bar g

c| Less weight of steam would be required because steam at 3 bar g has a higherenthalpy of evaporation than at 7 bar g

d| The output would be reduced because the difference in temperature between thesteam and product is reduced

1:a,2:dAnswers


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The Steam and Condensate Loop2.8.4



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Top

Sides

Base

10 50 90 130 170

25

20

15

10

5

0

Temperature difference C

W/m

C


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30

18 000

40 50 60 70 80 90 100

00

10 000 20 000 30 000

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1500

TemperaturetinC

Enthalpy h in kJ/ kg

Dry saturatedsteam line

Superheatedsteam region

100

400

300

200

00 500 1000 2 500 3 0002000

Saturatedwater line

Evaporation

lines

Criticalpoint

Lines ofconstant

pressure

Dryness fraction lines


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275 K

274 K

273 K

4.228 / 273.5

4.228 / 274.5

Temperature

Change in enthalpy

Mean temperature

H

T(mean)=


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= 0.70

= 0.75

= 0.80

= 0.85

= 0.90

= 0.95

250C

400 bar 200 bar 100 bar 50 bar 20 bar 10 bar 5 bar 2 bar

1800

2000

2200

2400

2600

2800

3000

3200

3400

3600

3800

300C

350C400C

450C

500C

550C

600C

650C

50C100C

150C

6.0 6.5 7.0 7.5 8.0 8.5 9.06.0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

200CSaturationline

Specificenthalpy(kJ/kg)

Specific entropy (kJ/ kg K)

Temperature(T)

373 K

473 K

273 KEntropy (S)

1 2

3

4


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= 0.70

= 0.75

= 0.80

= 0.85

= 0.90

= 0.95

250C


1800

2000

2200

2400

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2800

3000

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3800

300C

350C400C

450C

500C

550C

600C

650C

50C100C

150C

6.0 6.5 7.0 7.5 8.0 8.5 9.06.0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

200CSaturationline



= 0.70

= 0.75

= 0.80

= 0.85

= 0.90

= 0.95

250C


1800

2000

2200

2400

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3000

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300C

350C400C

450C

500C

550C

600C

650C

50C100C

150C

6.0 6.5 7.0 7.5 8.0 8.5 9.06.0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

200CSaturationline

Spe

cificenthalpy(kJ/kg)



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= 0.70

= 0.75

= 0.80

= 0.85

= 0.90

= 0.95

250C


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2200

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2600

28003000

3200

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300C

350C400C

450C

500C

550C

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650C

50C100C

150C

6.0 6.5 7.0 7.5 8.0 8.5 9.06.0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

200CSaturationline

Specificenthalpy

(kJ/kg)



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= 0.70

= 0.75

= 0.80

= 0.85

= 0.90

= 0.95

250C


1800

2000

2200

2400

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3000

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3800

300C

350C400C

450C500

C

550C

600C

650C

50C100C

150C

6.0 6.5 7.0 7.5 8.0 8.5 9.06.0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

200CSaturationline




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10 bar a180C

T(C)

6 bar a159C

2.1389 2.1389 + 4.0024 = 6.1413

1.9316 + 4.2097 = 6.1413

1.9316

2.1389 + 4.4471 = 6.586

1.9316 + 4.8285 = 6.76

0.8718 dry

0.9 dry

6.1413 S (kJ/kg C)


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s - u

Velocity

1.0 0.8 0.58 P/P1

P1 P

u

s


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10 bar a180C

T(C)

6 bar a159C

2.1389 2.1389 + 4.0024 = 6.1413

1.9316 + 4.2097 = 6.1413

1.9316

2.1389 + 4.4471 = 6.586

1.9316 + 4.8285 = 6.76

6.1413 S (kJ/kg C)

0.8718 dry

0.9 dry


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Block 2 Engineering Principles & Heat Transfers

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