Tuesday, September 4, 2007

Polyuria, polydipsia

The set up

Patient is the ICU, you receive a consult for hypernatremia resistant to therapy for three days.

Na 174
Cl 128
BUN 38

K 3.2
HCO3 27
Cr 1.6

glucose 128

urine lytes:

Na 56
K 12

Body weight 45 kg
urine output 2,800

Step one

Calculate the fluid deficit:
We assume the increase in sodium is entirely due to loss of water. If the patient is hypotensive, tachycardic orthostatic or has other clues that he maybe volume depleted, the situation is more complex. We calculate the percentage that the sodium is elevated above an imagined "ideal sodium." Many equations use 140 as the idealized sodium, I tend to use 145. The percentage the sodium is above the ideal is identical to the percentage of the total body water that is needed to lower the sodium to the target level.

174-145 / 145 = 0.2 (the sodium is 20% above 145)

Multiple that by the estimated total body water (weight x % body water) = 5.4 liters

I used 60% for estimated total body water. The patient looks like a young boy. Rose suggests lowering the estimated % body water by 10% so 50% would be okay also. In the elderly and obese this number can go below 50%. 

Step two

Going from 174 to 154 is too much change in 24 hours. The speed limit is 12 mmol/L/day so we will target a change from 174 to 162.

I use a modified water deficit formula. I divide 12 (the change in sodium we are looking for) by the target sodium (current sodium minus 12):

12/(174-12) = 0.08 x total body water = 2.0 liters or just over 80 mL/hour

Step three

Calculate the electrolyte free water clearance. The reason the ICU team repeatedly failed to correct the sodium is that they corrected the deficit without accounting for the ongoing water losses. In this case it is renal losses. If you just add the urine volume to the water deficit you will over estimate the amount of water needed because the urine contains sodium and potassium. To account for the electrolyte content of the urine we calculate the free water clearance.

Urine Na + K divided by the serum Na will give the ratio of urinary electrolytes to serum solute. 

56 + 12 / 174 = 39%
The urine has 39% electrolyte content of plasma, another way of thinking about this is that 39% of the urine volume is isotonic and the remainder (61%) is pure water. The 61% is what we are interested in; multiply the urne output by 61% this is the volume of water we need to give the patient to account for his ongoing renal losses. 

Then multiply this by the urine output (2,800) to get the electrolyte free water clearance, 1705 ml. This is another 71 mL/hour

Wrap up

Final fluid orders: 154 mL/hour of water, this can be given as D5W in the IV or preferably oral flushes. This calculation does not insensible losses.

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