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Sizing Indirect Fired Water Bath Heater with Example and Free Spreadsheet


In this post I want to share with you how to size indirect fired water bath heater.

Indirect Fired Water Bath Heaters (IFWBH) safely heat various mediums by transferring energy indirectly. In this system, the process coils are submerged in a heated bath solution—typically a water-glycol mixture—which absorbs heat from the firetube and transfers it to the process media.

How Indirect Fired Water Bath Heater Works

Fuel gas burns inside a firebox submerged in the lower section of the water bath, transferring heat through the firebox wall to maintain the bath at the desired temperature. The process fluid (such as well fluids, natural gas, or oil) flows through a coil immersed in the upper section of the bath, where it is heated indirectly through the tube walls. A temperature controller regulates the fuel gas supply to the firebox to keep the water bath at an optimum operating temperature of 190°F. Operating above this threshold is inefficient, as it leads to excessive water loss through evaporation.

Indirect Fired Water Bath Heater Components
Indirect Fired Water Bath Heater Components

Applications

Indirect fired water bath heaters are usually used for the following applications:

  • Heating natural gas before pressure reduction to prevent freezing of valve and instrumentation
  • Heating well stream fluids before phase separation
  • Heating fuel gas for power generation station
  • Fuel gas dew point control
  • Heating high pressure hydrocarbon gas streams

Standard Feature

Most IFWBH are designed in accordance with API 12K.

Sizing of Indirect Fired Water Bath Heater with Example

The target of sizing of IFWBH are to obtain the following parameters:

  • Heat load of firebox
  • Number of coils
  • Type of coil

The sizing of IFWBH in this post is for academic purposes only. For real application, you need to ask the specific vendors.

Below is the example of IFWBH sizing for well gas heating.

Data:

  • Natural gas flow rate = 4 MMscfd
  • Gas specific gravity = 0.7
  • Gas flowing pressure = 3350 psig
  • Shut-in pressure = 4500 psig
  • Gas flowing temperature = 85oF
  • Heater outlet pressure = 1000 psig

[1] Determination of Firebox Capacity/Heat Load

Determination of firebox heat load is necessary for identifying standard heater sizes. Before determining the firebox heat load, we need to perform the following two steps:

  • Whether hydrate is formed at inlet and outlet condition of heaters
  • Determine the temperature drop because of choking or pressure reduction

[1a] Hydrate formation checking at inlet condition

To check hydrate formation, we need the following parameters:

  • Flowing pressure = 3350 psig (3364.7 psia)

By using the following figure, we can estimate the hydrate formation temperature at the flowing pressure. At 3364.7 psia, hydrate formation temperature is 75oF. Because the flowing temperature is 85oF and the difference between flowing temperature and formation temperature is 10oF (85oF-75oF), therefore it is expected that hydrate will not form.

Hydrate formation temperature
Hydrate formation temperature

[1b] Hydrate formation checking at outlet condition

By using the same method as in step 1a, the hydrate formation temperature at outlet condition (pressure 1000 psig or 1014.7 psia) is 64oF. Giving the margin of 10oF, so the minimum required outlet temperature is 74oF (64oF + 10oF).

Hydrate formation temperature_2
Hydrate formation temperature_2

[1c] Temperature drop checking

Temperature drop can be estimated using the following chart. To estimate, we need the following data:

  • Initial pressure (before choking) = 3350 psig
  • Final pressure (after choking) = 1000 psig

Based on the chart, the temperature drop is 80oF. When initial temperature or flowing temperature is 85oF and the pressure is 3350 psig, then final temperature will be 5oF (85oF-80oF) when the pressure is reduced to 1000 psig.

Temperature drop as function of pressure drop
Temperature drop as function of pressure drop

 [1d] Firebox heat load calculation

To avoid low temperature after the choke, the heater coil will be split into a coil upstream of the choke valve and coil downstream the choke valve. For practical purpose, the temperature of the stream before choke will be raised to 130oF (with water bath temperature 190oF maximum, 130oF is within 60oF of the bath temperature).

The simplified flow will be like this:

Operating pressure and temperature
Operating pressure and temperature

Note:

  1. Based on the data
  2. Selected value
  3. 45oF is value in stream 2 minus temperature drop (see Step 1c) (130oF-85oF)
  4. 74oF is minimum required outlet temperature (see Step 1b)

The illustration of split pass coil is as follows:

Split-pass-coil with operating condition
Split-pass-coil with operating condition

To estimate heat load, we need to estimate enthalpy at specific pressure and temperature. Figures below can be used, depending on the gas specific gravity.

Entalphy at 0.7 SG
Entalphy at 0.7 SG
Entalphy at 0.8 SG
Entalphy at 0.8 SG

We will use gas specific gravity 0.7 since it is the case. For upstream coil, the following operating condition are used to estimate enthalpy:

Enthalpy of upstream coil
Enthalpy of upstream coil

Heat load at upstream coil is estimated = (Houtlet – Hinlet) x flowrate (in MMscfd)* 1,000,000 / 24 = (7-5.2) x 4 x 1,000,000 / 24 = 300,000 Btu/h

During choking process, it is assumed that movement of natural gas takes place so rapidly so that there is no heat lost or heat gained passing through the choke valve seat. Therefore, there is no changes in enthalpy.

Use the same method to estimate heat load at downstream coil. The following operating condition are used to estimate enthalpy:

Enthalpy of downstream coil
Enthalpy of downstream coil

Heat load at downstream coil is estimated = (Houtlet – Hinlet) x flowrate (in MMscfd)* 1,000,000 / 24 = (8.2-7) x 4 x 1,000,000 / 24 = 200,000 Btu/h

Heat loss is assumed to be 10% of total process heat load = 10% x (300,000 + 200,000) = 50,000 Btu/h

Therefore, total heat load will be heat load for upstream coil + heat load for downstream coil + heat loss = 300,000 + 200,000 + 50,000 = 550,000 Btu/h

Based on typical commercial IFWBH specification, we will select heat load 750,000 Btu/h.

Standard indirect heater and coil size
Standard indirect heater and coil size

[2] Coil Heat Transfer Area Determination

To estimate heat transfer area, we need to estimate “U” (overall heat transfer coefficient), mean temperature difference (MTD), and heat load. We already calculated heat load at Step 1.

Because shut-in pressure is 4500 psig, we will use 2” XXtra heavy coil.

Overall heat transfer coefficient can be estimated using the following chart. We need the data of flow rate, coil type, and operating pressure.

Oveall heat transfer coefficient chart for upstream coil
Oveall heat transfer coefficient chart for upstream coil

For upstream coil, the following data is used:

Data used to estimate upstream coil overall heat transfer coefficient
Data used to estimate upstream coil overall heat transfer coefficient

MTD (mean temperature difference) is estimated using the following chart.

Mean Temperature Difference (MTD) estimation
Mean Temperature Difference (MTD) estimation

For upstream coil, the following data is used to estimate mean temperature difference (MTD).

Since U, MTD, and heat load are already defined we can estimate the heat transfer area at upstream section coil.

A = Q/(U x MTD)

We get overall heat transfer area for upstream coil is 28.8 ft2.

We do the same method to estimate heat transfer area for downstream coil. The overall heat transfer coefficient for downstream coil is estimated as below.

Oveall heat transfer coefficient chart for downstream coil
Oveall heat transfer coefficient chart for downstream coil

For downstream coil, the following data is used:

MTD (mean temperature difference) is estimated using the following chart.

Mean Temperature Difference (MTD) estimation for downstream coil
Mean Temperature Difference (MTD) estimation for downstream coil

If the chart is not available, the MTD can be calculated using arithmetic MTD.

Arithmatic MTD
Arithmatic MTD

Where:

ΔT1 = Temperature of water bath – final gas temperature (190oF – 74oF)

ΔT2 = Temperature of water bath – initial gas temperature (190oF – 45oF)

We get mean temperature difference is 130oF

Using the same equation, we get heat transfer area for downstream coil is 13 ft2.

Total heat transfer area is 28.8 + 13 = 41.9 ft2.

[3] Selection of Standard Heater Specification

The standard heater specification is as follows.

Specification of heater
Specification of heater

Because we need heater with heat load 550,000 Btu/h (we select heater with heat load 750,000 Btu/h) with coil type 2”XXtra Heavy Coil, then number of coil will be 10, with selected heat transfer area 58 ft2 (calculated heat transfer area 41.9 ft2).

We also can estimate quantity of coil at upstream coil and downstream, respectively. Table below summarizes number of coil using heat transfer area approach.

Number of coil
Number of coil

[4] Flow Coil Pressure Drop Calculation

Coil pressure drop can be estimated by using the following chart.

Pressure drop chart
Pressure drop chart

Coil length can be checked by using standard specification of heater.

Standard indirect heater and coil specification
Standard indirect heater and coil specification

Pressure drop at upstream and downstream coil are estimated below.

Pressure drop at upstream and dowstream of coil
Pressure drop at upstream and dowstream of coil

[5] Conclusion

Based on case study above, the key specification required for IFWBH are:

  • Heat load = 750,000 Btu/h
  • Number of coil = 10
  • Coil type = 2” XXtra heavy coil
  • Type of pass = split pass

Free Spreadsheet

Due to high capacity, please email me if you need the spreadsheet!

I hope you find this post useful.

References:

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