| |
|
How to Heat Liquids in Vessels |
There are many advantages to using eductors
for heating liquids in open vessels. These give the vessel heating
eductor a place as a viable option for heating in many types
of vessels.
The eductor heater provides direct contact of the steam into
the liquid. This assures complete transfer of the energy in the
steam into the liquid being heated. Other types of heating lose
efficiency as the interior of the heat exchanger builds up a
scale. With eductors, the velocity of the steam being injected
into the vessel also causes the liquid contents of the vessel
to be agitated while heating occurs, without the need for
other types of mixers in the vessel. This provides for more even
heating of the vessel contents. They also permit the steam to
be dispersed over more of the liquid volume, resulting in a more
homogenous heating than with other methods of injecting steam.
These designs of eductors allow steam to be used from 10 to
140 PSIG for heating. Because of the nature of direct steam injection,
heating vessels at atmospheric pressure beyond 140°F should
not be attempted. Exceeding this temperature could result in
uncondensed steam evolving from the liquid. |
| |
| Models for Heating Vessels |
The standard Jerguson/Jacoby-Tarbox models for
heating in vessels are the TLA and ULJ. The model TLA is well
suited to providing strong tank agitation while heating.
Compared with other heaters, the cost per application is small.
The model ULJ is designed to provide vigorous circulation of
the liquid with low pressure steam inputs.
TLAs should be located with the outlet pointed toward the most
remote portion of the tank to provide the best agitation possible. |
| |
| Sg = Specific gravity of tank liquid Sh = Specific
beat of tank liquid |
| |
| How to Size Eductors for Heating Vessels |
| Information needed to size includes the following: |
| What is the tank liquid? (If it is not water (Sg
=1.0, Sh = 1.0), contact your representative.) |
| What temperature rise (∆T) is needed? |
| What is the final tank temperature? |
| What is the vessel capacity? |
| Time available to heat the vessel (t)? |
| Steam pressure available? |
| |
| Step 1 To determine the amount of steam required
to heat the liquid in the vessel, multiply the gallons in the vessel
to be heated x8.33 x Sg x Sh x temperature rise AT desired, divided
by 1100 (BTUs per Lb steam). |
| Lb steam required (Wm) = Gal x 8.33 x Sg x Sh x AT/1100 |
| |
| Step 2 To calculate the flow of steam required
per minute, divide the steam flow from Step 1 by the time you need
to complete the heating process. |
| Lb steam per minute (Qm)= Wm/minutes (t) |
| |
| Step 3 If multiple units are going to be used, divide
the number from Step 2 by the number of units to be used. |
| |
| Step 4 Go to the TLA Steam Flow
table. Find the amount of steam flow Qm (Lb/Min) at the steam pressure
available. This is the steam flow for a 1-1/2" unit. Take
the steam flow obtained in Step 3 divided by the steam flow from
the Steam Flow table. This will give the Sizing Factor
(S.F.) needed to heat the vessel in the time required. |
| |
| S.F. = Desired Steam Flowrate/ ated Steam Flowrate |
| Step 5 Choose the eductor size that has at
meets or exceeds the number ined in Step 4 |
| |
- The liquid in the vessel is Water (Sg =1.0)
- The temperature
rise desired is ∆T = 50°F
- The final tank temperature is
120°F
- The vessel holds 550 Gal
- The time to heat it is 20 Min
- Steam is
available at 40 PSIG
- Use two eductors
|
| |
| Step 1 Wm = 550 x 8.33 x 1.0 x 50/1100 = 208 Lb of
steam required |
| Step 2 Qm = Lb steam per minute = 208/20 = 10.4 Lb
steam per minute |
| Step 3 Are multiple units going to be used? If so
how many? In this case, we will use two eductors. 10.4 Lb steam
per minute/2 = 5.2 Lb/min per eductor |
| Step 4 S.F. = 5.2/13.4 = 0.39 desired S.F |
| Step 5 Choose the model TLA 3/4" with
a S.F. of 0.50 as this is the smallest unit that meets or exceeds
the desired S.F |
| |
| Calculating Actual Performance |
| Qm per unit =13.4 x .50 = 6.7 Lb/Min |
| Qm for installation = 6.7 x 2 =13.4 Lb/Min |
| Time to heat tank = 208 Lb (Step 1)/13.4 = minutes |
| |
| If two TLAs 3/4" are installed and operated
at 40 PSIG of steam pressure, they will heat the liquid in 15.5
minutes. |
| |
| |
| Previous Page |
| |