Soils 456 Laboratory 6


Mariotte Siphon

by Brian Wahlin and John Replogle, U.S. Water Conservation Laboratory

Subject Areas:  

Physics, Hydraulics, Mathematics, Environment


To develop a device that will apply fertilizers to a field in a constant and uniform manner


2-liter bottle with cap, exacto knife, small drill, Shoe Goo, small round tube about 2 inches long (e.g., a straw), plastic tube from Windex bottle, stop watch, graduated cylinder, water

Time Needed:  

About half an hour to build the siphon (not including drying time for Shoe Goo). Experiments can be run in under a half an hour.


Mariotte siphons or Mariotte bottles (figure 1) are devices that provide a constant pressure that will deliver a constant rate of flow from closed bottles or tanks. The flow rate will depend upon the head as defined in figure 1 and not on the height of the free water surface. These devices find many uses where a nonchanging pressure is needed. One application in agriculture is for applying liquid fertilizer to field crops. Because the fertilizer is applied at a constant rate to the field with this device, it is applied more uniformly and there is less of a chance of overfertilizing which can result in contaminating the groundwater.

Figure 1. Schematic diagram of a Mariotte Siphon


Figure 1 is a schematic diagram of a simple Mariotte siphon that can be built using a 2-liter bottle. First, peel the label off of the 2-liter bottle. Then, use the exacto knife to cut a small hole in the side of the bottle about 0.5 to 1 inch above the black base of the bottle. The hole should be just big enough so the small straw can fit inside it. Seal the edges around the straw with Shoe Goo. Next, drill a hole in the cap of the bottle. This hole should be just big enough for the plastic Windex tube. This hole should be very snug around the Windex tube. Ideally, the Windex tube should be able to move if you pull up and down on it, but it should stay in place when you let go of it. Fill the bottle with water, and put the cap with the Windex tube on the bottle. Note:  the outlet tube (or straw) will have to be plugged while the bottle is being filled. This can be done using your finger or by placing a small piece of rubber tubing that has been clamped at one end over the straw.

1. Close the outlet to the siphon, and fill the siphon to any given water level above the end of the vent tube. Have the students predict what will happen after the outlet is opened.

2. Construct a Mariotte siphon with two or three vent tubes of different lengths. Insert one of the vent tubes into the siphon, and measure the flow rate using a graduated cylinder and a stop watch. Repeat the experiment using the other vent tubes, and then compare the flow rates. This can also be done by raising and lowering the vent tube in the siphon. Measure the head

3. The constant flow of water out of a Mariotte siphon can also be visually demonstrated. This can be done by watching the shape of the jet of water exiting the Mariotte siphon. As long as the water surface is above the end of the vent tube, the water jet should have the same profile regardless of how full the tank is. However, if the vent tube is removed, the flow rate at the exit will depend upon the water surface, and the shape of the outlet jet will change as the water surface drops.

4. Attach a long, flexible hose to the end of the outlet. Raise the end of this tube above the end of the vent tube but below the water surface in the siphon. The flow of water out of the hose will stop even though it is below the water surface in the siphon. If this were done on a regular bucket of water, the water would come out of the hose until it was raised above the water surface.

Discussion Questions

1. In hydraulics, what is the definition of head?

2. Why does the flow rate change in the siphon without the vent but does not change in the siphon with the vent?

3. What are some other applications where a constant flow rate would be beneficial?

4. In procedure #4, why doesn't water come out of hose when it is above the vent even though it is below the water surface?

Part Two Added by Ted Sammis

The flow rate from the bottle is a function of the velocity of the water going though the outlet and the area of the outlet.



Q= quantity of flow
A= cross-sectional area of water
V= the mean velocity of water perpendicular to the cross- sectional area.

The theoretical discharge through an orifice ( which would be a simple hole in the bottom of the bottle) with out any head loss due to friction may be determined by substituting the value of V = (2gh)0.5

This is the theoretical velocity of water where:

g= acceleration due to gravity 981 cm/s
h=head in cm

Combining the two equations results in the orifice equation. The standard submerged orifice equation is:

Q=C A (h)0.5


Q is discharge in l/s
A= area of orifice in cm2 = pie r^2
h= head in cm
C= coefficient of discharge and accounts for the friction loss of head as the water enters the outlet and includes the constant (2g) 0.5

The coefficient of discharge C is normally 0.027 but can range from 0.027 to 0.035. The power of h should be 0.5 but may vary slightly.

1.Measure the inside diameter of the straw outlet and using the flow and head data in part 1 section 2 calculate the submerged orifice equation constants assuming the straw acts as a submerged orifice .Use the measured data and regression analysis (Reges.exe) or do a log transformation and uses linear regression in Excel  to calculate the power h is taken to and the coefficient of discharge for the Mariotte bottle. Compare the answers to the standard equation and explain why they are different