Suggested Response - Q1

*** A boiler is operating at 900 kPa. It produces saturated steam. The steam lines are long, so the steam loses 5% of its energy by the time it reaches the pressure reducing station. The pressure reducing station reduces the pressure to 200 kPa. From the pressure reducing station, the steam then supplies a heat exchanger at 200 kPa. The condensate leaving the heat exchanger is taken to an atmospheric condensate system.

a)    Using a Mollier diagram, indicate the following on the diagram:

i.    A – The steam leaving the boiler

      ii.    B – The steam entering the pressure reducing station (before the removal of condensate)

     iii.    C – The steam entering the pressure reducing valve

     iv.    D – The steam just after the pressure reducing valve

      v.    E – The condensate leaving the heat exchanger

     vi.    F – The condensate at the atmospheric condensate tank just before it flashes

    vii.    G – The condensate after flashing has occurred

Figure: Mollier diagram for water

Figure: Mollier diagram for water

b)    What is the state of the steam at each of the points in a)?

c)    Calculate how much condensate (as a percentage of total steam) will be drained before the pressure reducing station. Assume that you have 100% separation of vapour and liquid. Also assume that there are no pressure losses in the steam line (although we know that is not possible).

d)    To what temperature will the steam be superheated after the pressure is reduced if the specific heat of vapour (sg) is 7.7014 kJ kg-1 K-1?

e)    How much energy is available to perform heating in the heat exchanger?

f)     Calculate the amount of condensate that will flash off in the condensate tank.

g)    If the heat exchanger requires 1,000 kg of steam per hour, and the boiler feed water temperature is 20°C, what is the required gas flow rate for the boiler if it has an 80% boiler efficiency and the calorific value of the gas is 44 MJ/kg? Remember to take into account the condensate drained from the before the pressure reducing station calculated in (c) above. For the purposes of this calculation, assume there is no flash steam from this condensate draining point. Also take into account the condensate flash losses calculated in (f) above.

Assume condensate is returned to the boiler at 100°C.

Steam table information for SAQ 1 is as follows:

Table: Saturation temperature table for steam in SI units

T

Psat

hf

hg

hfg

°C

kPa

kJkg-1

kJkg-1

kJkg-1

0

0.6119

0.9007

2,500.02

2,499.12

10

1.2282

42.2477

2,520.42

2,477.27

20

2.3376

84.2043

2,538.85

2,454.65

30

4.242

125.811

2,556.81

2,431

40

7.3743

167.452

2,574.65

2,407.2

50

12.333

209.15

2,592.34

2,383.19

60

19.916

250.918

2,609.8

2,358.89

70

31.156

292.765

2,626.94

2,334.18

80

47.353

334.696

2,643.69

2,303.9

90

70.104

376.72

2,660.01

2,283.29

100

101.325

418.849

2,675.84

2,256.99

110

143.273

461.099

2,691.14

2,230.04

 

Table: Saturation pressure table for steam in SI units

P

Tsat

hf

hg

hfg

kPa

°C

kJkg-1

kJkg-1

kJkg-1

100

99.63

417.31

2,675.27

2,257.96

200

120.23

504.47

2,706.17

2,291.7

300

133.54

561.16

2,724.63

2,163.47

400

143.62

604.37

2,737.68

2,133.31

500

151.84

639.79

2,747.64

2,107.86

600

158.83

670.07

2,755.6

2,085.52

700

164.95

696.7

2,762.13

2,065.43

800

170.4

720.57

2,767.61

2,047.05

900

175.35

742.27

2,772.27

2,030

1000

179.87

762.23

2,776.28

2,014.05

1250

189.8

806.32

2,784.16

1,977.84

1500

198.28

844.32

2,789.84

1,945.52

 

Please note:

hf is the specific enthalpy of the water

hfg is the specific enthalpy of evaporation

hg is the specific enthalpy of steam

Response:

a)

Figure: Mollier diagram for water

Figure: Mollier diagram for water

 

b)   A –  Saturated vapour at 1 MPa a (900 kPa g) and 179.9°C

B –  Wet steam at 1MPa a (900 kPa g) and 179.9°C

Both condensate and vapour are present

93.1% dryness

C –  Saturated vapour at 1MPa a (900 kPa g) and 179.9°C

D –  Superheated steam at 300 kPa a (200 kPa g) and T > 133°C

E  –  Saturated liquid at 300 kPa a (200 kPa g) and 133°C

F  –  Wet steam at 101.325 kPa a (0 kPa g) and 100°C

G –  Saturated liquid at 101.325 kPa a (0 kPa g) and 100°C

 

c)    The enthalpy of saturated steam at 1 MPa a is:

H subscript A space equals space H subscript g space equals space 2 comma 776.28 space k J space k g to the power of negative 1 end exponent

If 5% losses occur, the enthalpy at point B will be:

H subscript B space equals space H subscript W left parenthesis w e t space s t e a m right parenthesis end subscript space equals space 0.95 space cross times space H subscript A space equals space 0.95 space cross times space 2 comma 776.28 space equals space 2 comma 637.47 space k J space k g to the power of negative 1 end exponent

Thus, this is now wet steam at 1 MPa a.

Reading from the Mollier diagram for 1 MPa a and 2,637.47 kJ kg-1, we have a dryness fraction (x) of 0.93.

Alternatively, if x is the fraction of the steam vapour in the water/steam mixture, we can calculate the following:

H subscript B space equals space x H subscript g space plus space left parenthesis 1 space minus space x right parenthesis space H subscript l

2 comma 637.47 space equals space x 2 comma 776.28 space plus space left parenthesis 1 space minus space x right parenthesis space space 762.23

x space equals space left parenthesis 2 comma 637.47 space minus space 762.23 right parenthesis space divided by space left parenthesis 2 comma 776.28 space minus space 762.23 right parenthesis space equals space 0.931

Therefore, under these conditions 6.9% (100-93.1%) of the steam by weight will be drained as condensate before the pressure reducing station.

 

d)    The vapour at 300 kPa a will still have an enthalpy of 2,776.28 kJ kg-1.

From the steam tables, using interpolation at 300 kPa a, the saturated vapour enthalpy of steam is 2,724.63 kJ kg-1, and the saturation temperature is 133.54°C.

The enthalpy change in superheating the steam is thus:

H subscript s u p e r h e a t space end subscript equals space 2 comma 776.28 space minus space 2 comma 724.63 space equals space 51.65 space k J space k g to the power of negative 1 end exponent

The specific heat of the vapour (sg) = 7.7014 kJ kg-1 K-1.

Therefore, the temperature increase from this change in enthalpy is:

T subscript s u p e r h e a t end subscript space equals space fraction numerator 51.65 over denominator 7.7014 end fraction space equals space 6.71 degree C

And so the temperature of the superheated steam is calculated as:

T space equals space 133.54 space space plus space space 6.71 space equals space 140.25 degree C

 

e)    The energy available for heating consists of the sensible heat in the steam, resulting from the superheat and the latent heat.

The condensate leaving the heat exchanger will be saturated liquid at 300 kPa a.

By looking up the value from the steam table, under these conditions Hl  561.16 kJ kg-1.

Therefore, the energy available

H subscript D E end subscript space equals space 2 comma 776.28 space minus space 561.16 space equals space 2 comma 215.12 space k J space k g to the power of negative 1 end exponent

 

f)    The condensate leaving the heat exchanger is saturated liquid at 300 kPa a. To calculate how much flashing occurs, we need to consider the values for 101.325 kPa (atmospheric pressure) and 100°C.

H subscript l space equals space 418.85 space k J space k g to the power of negative 1 end exponent

H subscript g space equals space 2675.84 space k J space k g to the power of negative 1 end exponent

If x is the fraction of flash steam that is generated:

H subscript B F end subscript space equals space x H subscript g space plus space left parenthesis 1 space minus space x right parenthesis H subscript l

561.16 space equals space x 2 comma 675.84 space plus space left parenthesis 1 space minus space x right parenthesis space 418.85

x space equals space left parenthesis 561.16 space minus space 418.85 right parenthesis space space divided by space left parenthesis 2 comma 675.84 space minus space 418.85 right parenthesis space equals space 0.063

Therefore, 6.3% of the condensate will flash off.

g)    If the heat exchanger requires 1,000 kg hr-1, then the boiler must produce more steam to cater for the condensate drainage prior to the pressure reducing station. As 6.9% is drained, the boiler must therefore produce:

m subscript s t e a m end subscript space equals space 1 comma 000 space cross times space fraction numerator 100 over denominator 100 space minus space 6.9 end fraction space equals space 1 comma 074 space k g space h r to the power of negative 1 end exponent

To calculate the amount of condensate returned:

We have 74 kg from prior to the pressure reducing station and 937 kg from the heat exchanger. The value 937 kg is calculated by subtracting 6.3% (the amount of condensate that flashes off) from 1,000 kg (1,000 – 63 = 937 kg). Therefore, the total condensate returned is

m subscript c o n d e n s a t e end subscript space equals space 74 space plus space 937 space equals space 1 comma 011 space k g space h r to the power of negative 1 end exponent

The makeup water requirement is:

m subscript m a k e space u p end subscript space equals space 1 comma 074 space minus space 1 comma 011 space equals space 63 space k g space h r to the power of negative 1 end exponent

First, let’s calculate the energy required to heat the makeup water to 100°C:

q space equals space m C p capital delta T

q space equals space 63 space cross times space 4.18 space cross times space left parenthesis 100 space – space 20 right parenthesis space equals space 21 comma 067 space k J space h r to the power of negative 1 end exponent space equals space 21 space M J space h r to the power of negative 1 end exponent

Now, let’s calculate the energy required to generate saturated steam from saturated water at 1 MPa a.

q space equals space m capital delta H

q space equals space 1 comma 074 space cross times space left parenthesis 2 comma 776.28 space minus space 418.73 right parenthesis space equals space 2 comma 532 comma 009 space k J space h r to the power of negative 1 end exponent space equals space 2 comma 532 space M J space h r to the power of negative 1 end exponent

Therefore, the total rate of energy required (q) to heat up the make up water is:

q space equals space 21 space plus space 2 comma 532 space equals space 2 comma 553 space M J space h r to the power of negative 1 end exponent

If the boiler has an 80% efficiency then the energy rate required from the fuel will be:

q subscript f u e l end subscript space equals space fraction numerator 2 comma 553 over denominator 0.8 end fraction space equals space 3 comma 191 space M J space h r to the power of negative 1 end exponent

And so

m subscript f u e l end subscript space equals space fraction numerator 3 comma 191 over denominator 44 end fraction space equals space 72.53 space k g space h r to the power of negative 1 end exponent

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