Liquid Level Measurements

If the liquid level in the vessel rises to 7 feet, the transmitter should (ideally) output a final signal corresponding to 7 feet of level. However, since the transmitter in question is a hydrostatic pressure type, and the liquid density has increased as well, it will not register accurately anymore.

The old specific gravity was 1.12, and the new specific gravity is 1.35. The error in span will be the ratio of the new specific gravity to the old, or 1.35/1.12 = 1.2054. In other words, the new acid/water mixture is 1.2054 times denser than it is supposed to be, according to how the transmitter was calibrated. Thus, the transmitter output will correspond to a liquid level of 7 feet times the error factor of 1.2054, or 8.4375 feet.

A good way to solve this problem is to apply the problem-solving technique of {\it simplification}. Imagine the two specific gravities being much simpler numbers: 1 and 2 instead of 1.12 and 1.35, respectively. It should be obvious now that a doubling of specific gravity will result in a doubling of level indication (i.e. the transmitter will register twice as much liquid level as there actually is inside the vessel). It should also be obvious that the error factor is 2:1, which is precisely the ratio of new:old specific gravity.

Alternatively, liquid level could be measured with a float, a capacitive sensor, ultrasonic or radar gauge, or some other instrument functioning on the detection of the liquid/vapor interface.

The transmitter output will stay the same.

If you thought that the transmitter output would decrease due to the water becoming less dense, I recommend you explore the concept of density a little deeper with the following ``thought experiment:’‘

{\bullet} Imagine the vessel filled half-way with liquid.
{\bullet} Imagine that liquid heating up and expanding until it is only {\it half} as dense as it was at the beginning of the experiment.
{\bullet} Calculate the new hydrostatic pressure with the expanded, less-dense liquid.

In a way, this is a ``trick’’ question. As any given mass of water heats up, its volume will increase. This is what makes it less dense than before: increased volume with the same mass (weight). Knowing how to calculate hydrostatic pressure from height and specific gravity, you might have approached this problem by assuming the water column height would have stayed the same while its density decreased, resulting in a lesser hydrostatic pressure and decreased transmitter output. However, this answer is incorrect.

If the water volume expands as a result of a temperature increase, the level inside the vessel {\it must rise}, because it will require a higher vessel level to contain a greater volume of water. The level will rise by the same percentage that the density decreases, resulting in a cancellation of level increase with density decrease. As a result, the hydrostatic pressure at any point in the vessel remains constant.

If you have difficulty picturing this effect, consider an exaggerated example to make things simpler. Suppose that a vessel has been filled to a height of 10 feet with cold liquid:

Now, imagine that same vessel being heated until the liquid expands to exactly {\it twice} its former volume:

The liquid level must double to accommodate twice the volume in the same vessel, assuming a vessel of constant diameter, and its density will be cut in half (twice the volume with the same mass). Consequently, the two factors of liquid level change and liquid density change cancel each other out to give the exact same hydrostatic pressure as before. In other words, the transmitter will register the same liquid level as it did when the liquid was cold, and will output the exact same signal, even though the actual vessel level is quite a bit more than it was when cold. Pretty tricky, huh?

Any vapor pressure will be sensed by the transmitter and interpreted as increased liquid level!

Two solutions to this problem:

(1) Use two transmitters (one at top of vessel, one at bottom) and electronically subtract their output signals.

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(2) Connect the ``Low’’ side of the one $\Delta$P transmitter to the top of the vessel to naturally compensate for vapor pressure.

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Since the transmitter infers level from the amount of pressure sensed at the bottom of the vessel, any vapor pressure buildup inside the vessel will be falsely interpreted as additional liquid level, since the transmitter senses the {\it sum total} of hydrostatic pressure plus any vapor pressure trapped inside the vessel.

For example, if the process liquid level were at 10 feet (50

In many applications, the gas pressure inside a vessel far exceeds the hydrostatic pressure generated by the column of liquid inside of it, so this problem can be a very serious one in level measurement. We need to fully understand the nature of the problem and how to solve it in order to successfully measure liquid level in many industrial applications.

The solution to this dilemma is to measure the {\it difference} in pressure between the top and bottom of the vessel. We could do this by using two pressure transmitters, one at the top and one at the bottom, and detect hydrostatic pressure by subtracting the top transmitter’s measurement from the bottom transmitter’s measurement. A computer may be used to perform the mathematical subtraction of signals:

Another, more elegant method, incorporates a {\it differential} pressure transmitter with pipe connections to both ends of the vessel. This solution performs the subtraction mechanically, by directly exposing the transmitter’s sensing element to the {\it difference} of two applied pressures:

Because the differential pressure transmitter solution requires only one sensing instrument rather than two, and results in better accuracy because we are only dealing with the inaccuracies of a single instrument rather than the compounded inaccuracies of two instruments (three, if you include the subtraction unit), the differential, or ``d/p’’ solution is the one more widely used.

In this particular level measurement application, a differential pressure instrument would have the exact same calibration points (lower and upper range-values) as a ``normal’’ pressure transmitter connected to an open vessel.

One way to simulate a full condition is to use a {\it hand scale}:

$$\begin{array} {|l|l|} \hline Process & Percent of & $\Delta$ pressure & Output signal & Output signal & Output signal \\ \hline level (ft) & span ( \\ \hline \8 & 0 & 156 & 3 & 2.88 & 3.12 \\ \hline 8.4 & 10 & 160.8 & 4.2 & 4.08 & 4.32 \\ \hline 9 & 25 & 168 & 6 & 5.88 & 6.12 \\ \hline 10 & 50 & 180 & 9 & 8.88 & 9.12 \\ \hline 11 & 75 & 192 & 12 & 11.88 & 12.12 \\ \hline 11.6 & 90 & 199.2 & 13.8 & 13.68 & 13.92 \\ \hline 12 & 100 & 204 & 15 & 14.88 & 15.12 \\ \hline \end{array}$$

The elevation for this transmitter (i.e. the total differential pressure applied by the height of fill fluid on both sides) is equal to the total height difference between the remote seal diaphragms multiplied by the specific gravity of the fill fluid:

$$P_{elevation} = (44 \hbox{ in})(0.934) = 41.1 \hbox{ "WC}$$

In the LRV condition, this is the only pressure seen by the transmitter. Therefore, 41.1 “WC is the appropriate LRV setting for this transmitter. If we assume that the ``H’’ port of this DP transmitter connects to the lower nozzle, the LRV will be -44.1 “WC. If we assume the ``H’’ port connects to the upper nozzle, the LRV will be +41.4 “WC.

In the URV condition, we have the exact same amount of elevation (the fill fluid inside the capillary tubes) but on the lower nozzle we have the hydrostatic pressure of 44 vertical inches of water (i.e. the water inside the seal drum). Thus, in the URV condition the transmitter sees a differential pressure of:

$$P_{differential} = 44 \hbox{" WC} - 41.1 \hbox{ "WC} = 2.9 \hbox{ "WC}$$

If we assume the ``H’’ port of this DP transmitter connects to the lower nozzle, the URV will be +2.9 “WC. If we assume the ``H’’ port of this DP transmitter connects to the upper nozzle, the URV will be -2.9 “WC.

Hydrostatic pressure when completely full of water = 4.769 PSI

Hydrostatic pressure when completely full of gasoline = 3.208 PSI

Hydrostatic pressure when water-gasoline interface is at the 50

{\bullet} {\bf (6 points)} Transmitter $\Delta$P at 0
{\bullet} {\bf (6 points)} Transmitter $\Delta$P at 50
{\bullet} {\bf (6 points)} Transmitter $\Delta$P at 100

The liquid has a density of 66 pounds per cubic foot.

$$\begin{array} {|l|l|} \hline Process & Percent of & Buoyant & Output signal \\ \hline level (in) & span ( \\ \hline \ 0 & 0 & 0 & 4 \\ \hline 10 & 25 & 1.135 & 8 \\ \hline 20 & 50 & 2.270 & 12 \\ \hline 30 & 75 & 3.405 & 16 \\ \hline 40 & 100 & 4.540 & 20 \\ \hline \end{array}$$

$$\begin{array} {|l|l|} \hline Process & Percent of & Buoyant & Output signal \\ \hline level (in) & span ( \\ \hline \ 0 & 0 & 0 & 4 \\ \hline 5 & 25 & 0.567 & 8 \\ \hline 10 & 50 & 1.135 & 12 \\ \hline 15 & 75 & 1.702 & 16 \\ \hline 20 & 100 & 2.270 & 20 \\ \hline \end{array}$$