Example

If a person is excreting urine with an osmolality of 50 mOsm/kg H2O at a rate of 10 mL/min, it is obvious that there is a greater than normal proportion of water to solute in the urine, and therefore this production of dilute urine is acting to concentrate the plasma. If the plasma osmol-ality is 300 mOsm/kg H2O, the osmolal clearance can be calculated to be 1.67 mL/min. In other words, a urine flow of 1.67 mL/min is what would be required to excrete this quantity of solutes with a urine osmolality of 300 mOsm/kg H2O. The actual urine flow of 10 mL/min is mathematically equivalent to this flow of 1.67 mL/min plus a flow of 8.33 mL/min of pure water. In this case, we consider the 8.33 mL/min to be the free water clearance, i.e., the volume of urine flow in excess of the amount required to excrete the solutes in the urine with the same osmolality as the plasma.

To compare the rate of solute excretion with the rate of water excretion, the concept of free water clearance has been developed. To understand the meaning of free water, consider the following example.

The free water clearance (Ch2o) is calculated as a difference between the rate of urine volume flow and the osmolal clearance:

Using Eq. [2], Eq. [3] can be rearranged to a form that is sometimes more convenient:

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