Distinguishing between salt poisoning and hypernatraemic dehydration in childrenBMJ 2003; 326 doi: https://doi.org/10.1136/bmj.326.7381.157 (Published 18 January 2003) Cite this as: BMJ 2003;326:157
Box A [published as supplied by the authors] Derivation of the fractional excretion of sodium, FENa, and water, FEH2O
Fractional excretions are the proportions of the salt and water filtered at the glomerulus that subsequently reach the urine.
Abbreviations: P = plasma concentration, U = urine concentration, Cr = creatinine, Na = sodium, GFR = glomerular filtration rate, V = urine flow rate.
Note: units of concentration must be the same in plasma and urine. Beware; plasma creatinine is usually reported in μ mol/l, while urine creatinine is often reported as mmol/l.
Worked example = case 2 on admission:
Fractional excretion of water, FEH2O
The FEH2O is the volume of water that appears as urine compared to the amount filtered.
Thus, FEH2O = V / GFR
Since GFR = UCr x V / PCr
FEH2O = V x PCr / UCr x V
Simplifying, FEH2O = PCr / UCr (then multiply by 100 to express as a percentage).
Example: Case 2 had a plasma creatinine of 57 μ mol/l and a urine creatinine of 5500 μ mol/l (5.50 mmol/l) has a FEH2O of (57 / 5500) x 100 = 1.04%. This means that his tubules are reabsorbing 98.96% of the water filtered by the glomerulus.
Fractional excretion of sodium, FENa
The FENa is the amount of sodium lost in the urine compared to the amount filtered.
Thus, FENa = urinary sodium excretion / filtered sodium
The urinary sodium excretion = UNa x V
And the filtered sodium = PNa x GFR
Since GFR = UCr x V / PCr
The filtered sodium = PNa x UCr x V / PCr
Therefore, FENa = UNa x V x PCr / PNa x UCr x V
Simplifying, FENa = UNa / PNa x PCr / UCr (then multiply by 100 to express as a percentage).
Example: Case 2 had a plasma sodium of 156 and creatinine of 57 μ mol/l, and a urine sodium of 152 mmol/l, and creatinine of 5500 μ mol/l (5.50 mmol/l) has a FENa of (152 / 156) x (57 / 5500) x 100 = 1.01%. This means that his tubules are reabsorbing 98.99% of the sodium filtered by the glomerulus.
Box B [published as supplied by the authors] Estimation of ‘spot’ losses and balances of water and sodium (published as supplied by the authors)
Worked example, case 2.
Abbreviations as in box A.
Rate of water loss
· At 65 cm in length, his admission creatinine of 57 μ mol/l is equivalent to a GFR of 55 ml/min/1.73m2, calculated by the Schwartz formula.15
· At 5 kg, his surface area was 0.3 m2.13 Thus, his absolute GFR was 9.5 ml/min, or 13,700 ml/day.
· At 5 kg, this equates to a GFR of 2,740 ml/kg/day.
· His FEH2O on admission was 1.04%, so his urine flow was 1.04% of his GFR, or 28 ml/kg/day.
Rate of sodium loss
· His filtered sodium on admission = GFR x PNa = 2.74 l/kg/day x 156 mmol/l = 427 mmol/kg/day.
· His FENa was 1.01%, so his excreted sodium was 1.01% x 427 = 4.3 mmol/kg/day.
His weights show he was 7.6% dehydrated on arrival, and became 6% overloaded by 17 hours, before returning to normal. Thus he gained 13.6% weight, or retained 136 ml/kg of water initially. He received 184 ml/kg of water during that time, so would be predicted to have voided urine at a rate of 68 ml/kg/day, ignoring insensible losses. The estimates of urine output rising from low levels to 76 ml/kg/day are close to that.
He received 3.2 mmol/kg of sodium during 17 hours of resuscitation. His estimated sodium excretion rates varied from 1.9 to 5.7 mmol/kg/day, so his net balance was approximately neutral, laying between the extremes of a gain of 1.9 mmol/kg, and a loss of 0.8 mmol/kg.
These balance estimates indicate that his hypernatraemia was corrected by retaining 136 ml/kg net of water, and little if any change in sodium. Assuming body water is 60% of body weight, retaining 136 ml/kg would dilute his plasma sodium from 162 to 133 mmol/l. The measured change was a little less than this, a difference that may be due to the sodium in his resuscitation fluid (mean 23 mmol/l). These figures confirm that he had been water depleted and not salt overloaded.
The exchangeable sodium space is about 50% of the body weight. Therefore, for the plasma sodium to fall after recovery from salt poisoning, each mmol/l fall in plasma concentration would result in a loss of 0.5 mmol/kg sodium into the urine. Thus, if case 2 had been salt poisoned, correction of his plasma sodium from 162 to 144 mmol/l would have been accompanied by the urinary loss of about 9 mmol/kg sodium in 17 hours, a rate of 12.7 mmol/kg/day, far higher than his estimated values.
- This Week In The BMJ Published: 18 January 2003; BMJ 326 doi:10.1136/bmj.326.7381.0/e
- Correction Published: 01 March 2003; BMJ 326 doi:10.1136/bmj.326.7387.497
- Maternal deaths from suicide must be tackled, say expertsBMJ December 07, 2016, 355 i6585; DOI: https://doi.org/10.1136/bmj.i6585
- Andrew Wakefield calls Trump “on our side” over vaccines after meetingBMJ December 05, 2016, 355 i6545; DOI: https://doi.org/10.1136/bmj.i6545
- The future of NHS dentistry in ScotlandBMJ December 05, 2016, 355 i6466; DOI: https://doi.org/10.1136/bmj.i6466
- South Africa begins first HIV vaccine trial in seven yearsBMJ December 01, 2016, 355 i6501; DOI: https://doi.org/10.1136/bmj.i6501
- Better evidence for smarter policy makingBMJ December 01, 2016, 355 i6399; DOI: https://doi.org/10.1136/bmj.i6399
- Non-accidental salt poisoning
- The epidemiology of hypernatraemia in hospitalised children in Lothian: a 10-year study showing differences between dehydration, osmoregulatory dysfunction and salt poisoning
- The kidney speaks: interpreting urinary sodium and osmolality
- An unusual case of extreme hypernatraemia
- Will changing maintenance intravenous fluid from 0.18% to 0.45% saline do more harm than good?
- Hypernatraemia: diagnosis and management
- Protecting children, supporting professionals