Drum Coil Heater: Bath Heater Sizing

There are several applications where a coil immersed in hot water (such as in a boiler drum or a hot water tank) is used to heat up a thermic fluid or even cool steam for steam temperature control. Cold feed water from economizer inlet may also be preheated in a coil or exchanger tubes located inside the hot water bath to minimize acid dew point cor- rosion concerns. Drum heating coils using steam are also used to keep boiler warm in case

324 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

TABLE 6.5

Natural Convection Heat Transfer, hc

SI Units

Metric Units

British Units

Nu = 0.54[d 3 ρ 2 gbΔTC p /(µk)] 0.25 Nu = 0.54[d 3 ρ 2 gbΔTC p /(µk)] 0.25 Nu = 0.54[d 3 ρ 2 gbΔTC p /(µk)] 0.25 h c = 0.953[k 3 ρ 2 bΔT C p /(µd o )] 0.25 h c = 57.2[k 3 ρ 2 bΔTC p /(µd o )] 0.25 h c = 144[k 3 ρ 2 bΔTC p /(µd o )] 0.25 h c , W/m 2 K

Btu/ft 2 h °F d o is the tube OD, m

kcal/m 2 h °C

ft ρ is the density, kg/m 3 kg/m 3 lb/ft 3

g is the acceleration gravity 9.8 × 3600 2 m/h 2 32 × 3600 2 ft/h 2 b is the expansion coefficient, 1/K

Btu/ft h °F µ , kg/m s

kcal/m h °C

kg/m h

lb/ft h

C p , J/kg K kcal/kg °C

Btu/lb °F

of quick start applications. Superheated steam for steam temperature control purpose is also cooled in the boiler drum water using pipes immersed in the drum. In all of these applications, the heat transfer between the tubes and the liquid bath is through natural convection. The advantage of using this method for heating thermic fluids or viscous oils in a water bath rather than using a direct fired natural gas–fired heater is that the heat flux is much lower and is controlled unlike in a fired heater.

The outside natural convection heat transfer coefficient between the coil and the water or fluid in the drum may be obtained using the equation in Table 6.5.

() µ k  

where all the terms are in consistent units. Simplifying, the convection heat transfer coefficient in kcal/m 2 h °C is given by

C p is the specific heat, kcal/kg°C

d o is tube OD, m

g is the acceleration due to gravity, m/h 2 k is the thermal conductivity, kcal/m h °C ρ is the fluid density, kg/m 3

b is the volumetric expansion coefficient, 1/K Δ T is the temperature difference between tubes and fluid, K µ is the viscosity, kg/m h

All fluid properties are evaluated at the mean temperature between fluid and tubes, while expansion coefficient is at fluid temperature.

Miscellaneous Boiler Calculations 325

Example 6.10

A coil made of 33.5 mm OD pipe (ID = 26.6 mm) is immersed in a drum, which is maintained at 38°C using steam at 110°C inside tubes. The purpose of the coil is to keep the water at the desired temperature. Tube temperature is assumed to be 93°C. Estimate the outside convection heat transfer coefficient, and the overall heat transfer coefficient

assuming condensing coefficient is 11,000 kcal/m 2 h °C. Use a fouling factor of 0.0002 m 2 h

°C/kcal (0.001 ft 2 h °F/Btu) on shell side and tube side. First let us estimate the liquid properties at the film temperature of 66°C. We may have to correct the properties if the film temperature is different.

From Table F.13, at 66°C, k = 0.381 Btu/ft h°F = 0.5677 kcal/m h °C, b = 0.0002 1/°R =

0.00036 1/K, µ = 1.04 lb/ft h = 1.55 kg/m h, d o = 0.0335 m, ρ = 61.2 lb/ft 3 = 979 kg/m 3 , C p = 1 kcal/kg °C, ΔT = 55°C

h c = 57.2 × [k 3 ρ 2 b ΔT C p /(µ d o )] 0.25 = 57.2 × [0.5677 3 × 979 2 × 0.0.00036 × 55 × 1/ (1.55 × 0.0335)] 0.25 = 920 kcal/m 2 h °C (188 Btu/ft 2 h °F).

The overall heat transfer coefficient is given by 1/U o = 1/920 + 0.0002 + (0.0335/2/35) × ln(33.5/26.6) + 0.0002 × 33.5/26.6 + 33.5/ (11,000 × 26.6) = 0.001085 + 0.0002 + 0.00011 + 0.000252 + 0.000114 = 0.00176 or U o =

567 kcal/m 2 h °C (659 W/m 2 K)

Heat flux = 567 × ( 110 − 38 ) = , 40 824 kcal/m h 2 = . 47 4 kW/m 2

Tube wall temperature = 38 + 40,824 × (0.0002 + 0.001085) = 90°C. Properties may be corrected for the average temperature of 0.5 × (90 + 38) = 64°C. However, it is close to that assumed, and hence, we may proceed with the sizing of the coil.

If the duty and log-mean temperature difference are known, the required surface area may be determined. Say, the heat loss in the boiler or duty of the coil is 0.125 MM kcal/h.

LMTD = (110 − 38) = 72°C. Hence, surface area required = 0.125 × 10 6 /72/567 = 3.1 m 2

Bath heaters are often used to heat up viscous fluids or any fluid by immersing a coil or exchanger inside a bath of hot water or glycol maintained at the desired temperature by controlling the burner heat input. The heater looks like a fire tube boiler firing natural gas or oil and is of multipass design depending on the efficiency required. Since the bath is maintained at a low temperature, the fluid heated inside the tubes immersed in the bath is not subject to high film temperatures as in a fired heater. If a fired heater or a fire tube boiler is used for heating the liquid, then the high heat flux due to the high gas temperature profile in the boiler has to be considered, while with bath heaters, that concern is not pres- ent. Thermal expansion issues are also not a concern as both the hot and cold fluids are at reasonably low temperatures. Expansion of the hot water or bath fluid is handled by using an expansion tank built integral with the shell as shown in Figure 6.6. This type of heater is seen in oil fields.

As discussed in Chapter 3, superheated steam is sometimes cooled in a coil located inside the steam drum or mud drum. Its sizing may also be done in a similar manner.

Example 6.11

Saturated water temperature in a boiler steam drum is 190°C. For steam temperature control purposes as discussed in Chapter 3, 40,000 kg/h of superheated steam at

50 kg/cm 2 g and at 350°C is to be cooled to 300°C. Determine the coil size if drum length is about 8 m. Tubes of 50.8 × 43 mm are available for the exchanger.

326 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

Flue gas outlet

Expansion tank

Oil inlet

Oil outlet

Second pass First gas pass

Burner

Bath ﬂuid

FIGURE 6.6

Bath heater fired by gas or oil.

Solution

First determine the tube-side heat transfer coefficient for superheated steam at 325°C average temperature from Appendix B. Let us start with 2000 kg/h per tube. This gives

a steam velocity (see Table B.1), V = 0.0003461 wv/d i 2 . From steam tables, specific volume of steam is 0.04878 m 3 /kg. Steam velocity V = 0.0003461 × 2000 × 0.04878/0.043 2 = 18.3 m/s, which is a reasonable velocity.

Estimating h i

From Table B.4, at 5000 kPa and 325°C, C = 314. Hence, h i = 0.0278 × (2000/3600) 0.8 ×

314/0.043 1.8 = 1572 W/m 2 K = 1352 kcal/m 2 h °C.

Estimating h o

Assume that the tube OD is at 300°C. The average temperature of water is 245°C (473°F). From Table F.13, k = 0.3645 Btu/ft h°F, C p = 1.125 Btu/lb °F, µ = 0.275 lb/ft h, d o = 2 in.,

b = 0.00067/°R, ΔT = (300 − 190) = 110°C = 198°F. ρ = 50 lb/ft 3

h c = 144 × . 0 3645 3 × 50 2 .  0 25 × . 0 00067 × 198 × . 1 1125 / ( . 0 275 2 × )  = 34 3 3 Btu/ft h F 2 °

= 1680 kcal/m h C 2 °.

1/U o = (1/1680) + 0.0002 + (.0508/2 × 35)ln (50.8/43) + (0.0002 × 50.8/43) + 1/50.8/(43 × 1352)

= 0.000595 + 0.0002 + 0.000121 + 0.000236 + 0.000874 = 0.002026 or U o = 493 kcal/m 2 h °C

From steam table, the enthalpies of steam between 350°C and 300°C assuming a drop

of 0.5 kg/cm 2 pressure are 733.3 and 698.7 kcal/kg. Hence, duty of exchanger = 40,000 × (733.3 − 698.77) = 1.381 MM kcal/h.

Using a tube length of 3.5 m and say 10 tubes across and 2 starts (20 streams), surface

area required = 1.381 × 10 6 /[(325 − 190) × 493] = 20.7 m 2 = 3.14 × 0.0508 × 3.5 × 10 × N or

N = 3.71 rows. Use 4 rows of length 3.5 m with 20 streams in inlet and outlet headers. This provides some margin, and by controlling the steam flows, one may control the steam temperature. The 10 tubes in the header will need a width of about 700 mm, and the drum should be able to accommodate this coil. Figure 3.30 shows a similar coil with three starts, while this exchanger will have 10 tubes across with 2 starts, making the streams 20 as required with a total of four rows deep.

Miscellaneous Boiler Calculations 327

The steam pressure drop is (see Table B.6). Δ P = 0.6375 × 10 −12 fL e vw 2 /d i 5 = 0.6375 × 10 −12 × 0.02 × (3.5 × 4 + 3 × 32 × 0.0508) × 0.04877 × 2000 2 /0.043 5 = 0.32 kg/cm 2 . The average steam velocity = 18.3 m/s. The inlet

and exit losses will be about 1.5 times the velocity head = 1.5 × 18.3 × 18.3/(2 × 9.8 ×

0.04877) = 525 kg/m 2 = 0.0525 kg/cm 2 . Total pressure drop is about 0.38 kg/cm 2 .

Drum Coil Heater: Bath Heater Sizing