Drum Level Measurement

Drum Level Measurement

Dr. Teerawat Thepmanee

Automation Engineering, KMITL

What is boiler ?

Drum level measurement 2

What is boiler ?

What is a boiler ?When water is heated in a boiler, it begins to absorb energy. Depending on the pressure in the boiler, the water will evaporate at a certain temperature to form steam. The steam contains a large quantity of stored energy which will eventuallybe transferred to the process or the space to be heated

Heat exchangerSteam turbine3Drum level measurement

Types of boilers


Package boiler(Fire tube)

Utility boiler(water wall tube)


Boiler drum

Application overview

• Steam Drum Level is both a critical and difficult measurement to make. Control of the water level in the drum must be precise.

• A water level that is too high can result in water carryover into the steam piping. A level that is too low can expose the generating tubes (down comers), preventing the water in the drum from cooling the furnace tubes, possibly damaging them.

• Several factors make this measurement difficult to obtain. The steam drum itself may not be perfectly level, and even at steadystate conditions, considerable turbulence in the drum can cause the level to fluctuate.

• In addition, a changing rate of water inflow and steam outflow adds to the potential for measurement error.


Application overview

• Measurement of boiler steam drum level using a differential pressure transmitter must take into account certain physical properties of the fluid.– The Steam drum contains a two-phase mixture of water and steam at

saturation conditions.

– The densities of water and steam vary with saturation temperature or pressure.

– The density of saturated steam above water must be considered, as well as the density of saturated water in the drum.

• This technical note offers a method for calibrating transmittersthat takes into account these factors.


Water Property


1 lbm

320F 0.01602 ft3

ρ=

= 62.42 lbm/ft3

1 lbm

0.01602 ft3

ρ =62.42 lbm/ft3

Pressure (constant)

HEAT

Water Property


1 lbm

1200F 0.01620 ft3

ρ=

= 61.73 lbm/ft3

1 lbm

0.01620 ft3

ρ =61.73 lbm/ft3

Pressure (constant)

HEAT

Water Property


1 lbm

320F 0.01602 ft3

ρ= 1 lbm0.01672 ft3

1200F 0.01620 ft3ρ =61.73 lbm/ft3

ρ =62.42 lbm/ft3

2120F 0.01672 ft3ρ =59.81 lbm/ft3

Pressure (constant)

= 59.81 lbm/ft3

HEAT

Saturated Water


1 lbm

2120F 0.01672 ft3Boiling point

HEAT

Boiling

No bubble

Saturated Water


1 lbm

2120F 0.01672 ft3

swell

Pressure is increased

Water andSteam bubble

Boiling

2120F

BoilingWith

Bubble

Saturated Water


1 lbm

2120F 0.01672 ft3

more swell

Pressure is increased


Saturated Water


1 lbm

2120F 0.01672 ft3

shrink



Drum level measurement

ρWD

ρW

ρS

Pd

0

+ 300

- 300h

H

Saturated water

PH PL

Saturated Steam

Condensing pot

V



Define process variable

• ρS = steam density

• ρWD = drum water density

• ρW = water density

• g = gravity

• PH = Static pressure at the high side of the transmitter

• PL = Static pressure at the low side of the transmitter

Define process variable

• Pd = Static pressure in the steam drum at the top tap

• H = Distance between the high and low drum taps

• h = Drum water level (measured from the bottom tap)

• V =Vertical distance from the bottom tap to the transmitter



Solution

PH = ρw.g.V + ρWD.g.h + ρS.g.(H-h) + Pd

PL = ρw.g.V + ρW.g.H + Pd

∆P = ρw.g.V + ρWD.g.h + ρS.g.(H-h) + Pd - ρw.g.V - ρW.g.H - Pd

∆P = ρWD.g.h + ρS.g.H - ρS.g.h - ρW.g.H

∆P = ρWD.g.h - ρS.g.h - ρW.g.H + ρS.g.H

∆P = g.h(ρWD- ρS) - g.H(ρW - ρS)


Ex. Drum level measurement

Ex. operating condition 100 °°°°C at pressure 14.696 Psia

Determine hmin and hmax (±300 mm)

From steam table ρWD= 958.098 , ρS= 0.598, ρW(43 °C ) = 990.628

Level max.(100 %) h = H = 0.6 m

∆P(100%)= PH – PL

= g.h(ρWD – ρS) – g.H(ρW – ρS)

= 9.807×0.6×(958.098–0.598)–9.807×0.6×(990.628–0.598)

= 5633.93 – 5825.33

= – 191.40 Pa


Ex. Drum level measurement

Ex operating condition 100 °°°°C at pressure 14.696 Psia

Level min.(0 %) h = 0 m, H = 0.6 m

∆P(0%) = PH – PL


= 0 – 9.807 ×0.6 × (990.628 – 0.598)

= 0 – 5825.33

= – 5825.33 Pa


Calibration range

• Finally, check the zero elevation and span against the transmitter specifications to ensure the selected transmitter can be calibrated to the required values.

% Level (m) D/P (Pa) Signal (mA)

100 0.6 -191.40 20

0 0 -5825.33 4


Calibration range

Level %

DBR(m)

Range (m)

PH (Pa)

PL(Pa)

DP(Pa)

O/P (mA)

Controller(m)

100 0.60 0.30 5633.93 5825.33 -191.40 20.000 0.600

75 0.45 0.15 4225.44 5825.33 -1599.88 16.000 0.450

50 0.30 0.00 2816.96 5825.33 -3008.36 12.000 0.300

25 0.15 -0.15 1408.48 5825.33 -4416.85 8.000 0.150

0 0.00 -0.30 0.00 5825.33 -5825.33 4.000 0.000


Normalization Unit

Percent (0 - 100%) to level(0 – 0.6 m)

At 50 %, what is Level ?

Solution

sensitivity = span output / span input

= 0.6/100

= 0.006 m / %

At 50 %

level = 0.006 × 50

= 0.3 m


Normalization Unit

Level 0 – 0.6 m to DP (- 5825.33 – (-191.40))

At 0.3 m, What is differential pressure ?

Solution


= (-191.40 – (- 5825.33)) / (0.6 – 0)

= 9389.88 Pa/m

At 0.3 m

∆P = 9389.88 ×(0.3-0) + (- 5825.33)

= - 3008.36 Pa


Normalization Unit

DP (-191.40 – (- 5825.33)) to Signal (4-20 mA)

At (- 3008.36 Pa), What is Signal ?

Solution


= (20-4) / (-191.40 – (- 5825.33))

= 2.84 ×10-3 mA/Pa

At (- 3008.36 Pa)

signal = 2.84 ×10-3 × (- 3008.36 - (- 5825.33)) + (4)

= 12 mA


Level Transmitter

Transmitter Signal 12 mA ,What is water in drum level ?

Solution


= (20-4) / (0.6 – 0)

= 26.67 mA/m

At 12 mA

controller = [(12-4)/26.67] + (0)

= 0.3 m


Level & DP

Level

∆P

0% 100 %50%25% 75%

0 0.15 0.30 0.45 0.6 m

-191.40

-1599.88

-3008.36

-4416.85

-5825.33


DP Transmitter signal

-191.40

-1599.88-3008.36

-4416.85

-5825.33

mA %

∆P (Pa) 0% 100 %50%25% 75%

20

16

12

8

4

100%

0%

50%

25%

75%

0.60

0.45

0.30

0.15

0.0

Level (m)



At Operating condition 353 °°°°C at pressure 2498.1 Psia

Level min.(50 %) h = 0.3 m, H = 0.6 m

Solution

∆P(50%)= PH – PL


= 9.807×0.3×(560.478 – 122.4)– 9.807×0.3 ×(990.628 – 122.4)

= 1288.82 – 5108.65

= – 3819.83 Pa


PH PL

0%

50%

100%

PH PL

0%

50%

100%

Operating conditionTemp = 100 °CPressure = 14.696 Psia �P = - 3008.36 Pa Signal = 12 mA

Operating conditionTemp = 353 °CPressure = 2498.1 Psia �P = - 3819.83 Pa Signal = ???? mA

30Drum level measurement


Normalization Unit

DP (-191.40 – (- 5825.33)) to Signal (4-20 mA)

At (- 3819.83 Pa), What is Signal ?

Solution


= (20-4) / (-191.40 – (- 5825.33))

= 2.84 ×10-3 mA/Pa

At (- 3819.83 Pa)

Signal = 2.84 ×10-3 × (- 3819.83 - (- 5825.33)) + (4)

= 9.69 mA


Level Transmitter

At Transmitter Signal 9.69 mA ,What is water in drum level ?

Solution

Sensitivity = span output / span input

= (20-4) / (0.6 – 0)

= 26.67 mA/m

At 12 mA

Controller = (9.69-4)/26.67 + (0)

= 0.214 m



• Error = 0.214 - 0.3 = - 0.086 m

• Water Level at 35% from 50 % or reduce 15%

Condition % Level

(m)

DP

(Pa)

Signal

(mA)

Controller

(m)

100 °C

14.696 Psia

50 0.3 - 3008.36 12 0.3

353 °C

2498.1 Psia

50 0.3 - 3819.83 9.69 0.214


DP Transmitter signal

-191.40-3008.36-5825.33

mA %

∆P (Pa) 0% 100 %50%25% 75%

20

16

12

8

4

100%

0%

50%

25%

75%

9.69

-3819.83

0.214

0.60

0.45

0.30

0.15

0.0

Level (m)



Ex Operating condition 353 °°°°C at pressure 2498.1 Psia

Determine hmin and hmax (±300 mm)

From steam table ρWD= 560.478 , ρS= 122.400, ρW(43 °C ) = 990.628

Level max.(100 %) h = H = 0.6 m

Solution

∆P = PH – PL


= 9.807×0.6×(560.478 – 122.4)–9.807×0.6×(990.628 – 122.4)

= 2577.65 – 5108.65

= – 2531 Pa



Ex Operating condition 353 °°°°C at pressure 2498.1 Psia

Level min.(0 %) h = 0 m, H = 0.6 m

Solution

∆P = PH – PL


= 0 – 9.807 ×0.6 × (990.628 – 122.4)

= 0 – 5108.65

= – 5108.65 Pa


Calibration range

Level %

DBR(m)

range (m)

PH (Pa)

PL(Pa)

DP(Pa)

Output (mA)

Controller(m)

100 0.60 0.30 2577.65 5108.65 -2531.00 13.356 0.351

75 0.45 0.15 1933.23 5108.65 -3175.41 11.526 0.282

50 0.30 0.00 1288.82 5108.65 -3819.82 9.696 0.214

25 0.15 -0.15 644.41 5108.65 -4464.23 7.865 0.145

0 0.00 -0.30 0.00 5108.65 -5108.65 6.035 0.076


Density compensate

From differential pressure formula

∆P = g.h(ρWD- ρS) - g.H(ρW - ρS)

Liquid level in boiler drum

∆P + gH(ρw-ρs)

g(ρWD-ρs)h =


Controller or DCS Card

Density compensate

4-20mA to steam table (ρWD,ρS)

4-20mA to ∆P

∆P + gH(ρw-ρs)

g(ρWD-ρs)h =

ρWD

ρS

signal

signal

Exercise


ρWD

ρW

ρS

Pd

0h

H

Saturated water

PH PL

Saturated Steam

Condensing pot

V

What is calibration range of D/P transmitter at operating condition 200 °C and wet leg temperature 43.333 °C WhereH = 1000 mmh = 700mmV = 150 mm

Thank you

Question ?

Drum Level Measurement

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