7. The pharmacist is asked to provide a cyclic rate for a home TPN. The dosage i

Question

7. The pharmacist is asked to provide a cyclic rate for a home TPN. The dosage is (a volume to be given in) ml of (certain percent) dextrose over (given numbers) hours with a titrate-up rate of (given volume) mL/h for (given # of) hours and a taper-down rate of (given volume) mL/h for 2 hours. Administration time is (time of day) PM to (time of day) AM. What is the infusion rate of the solution in milliliters per hour at (a given time) AM to the nearest milliliter? (Answer to the nearest milliliter, no decimal point)

8. Based on the infusion rate calculated above in question 7, how many calories to the nearest calorie would be administered to the patient by (a given time) AM? Assume 3.4Kcal/g of carbohydrate.(Answer to the nearest calorie, no decimal point)

Explanation / Answer

Let us assume that

7.The pharmacist is asked to provide a cyclic rate for a home TPN. The dosage is 2000mL of 50 % dextrose over 12 hours with a titrate up rate of 80 mL/hour for 1 hour and taper-down rate of 40 ml/h for 2 hour. Administration time is 10pm to 10am. What is the infusion rate of the solution in mL/hr at 1am rounder to the nearest whole millimeter ?

1.During the 1st hour, the titrate up rate is 80mL/hr, so 80mL is lost.

2.During the last 2 hour, the taper down rate is 40mL/hr, so 80mL is lost.

3. Thus, spent 3 out of the 12 hours "titrating" losing a total of 160 mL during those 3 "titration hours".

4.Now, only 9 hours remaining to the administered 1840 mL (2000 mL originally- 160 lost to titration).

5.1 Am falls within this 9 hours range.

6.So, 1840 mL / 9 hours gives 204 mL/hour. (infusion rate of the solution in milliliters per hour at (a given time) 1 AM to the nearest milliliter).

8.To calculate grams of dextrose and calories administered to the patient by 1 AM:

1.Multiply the total volume of dextrose solution (2000ml) supplied in a day by the dextrose concentration (50 %). This gives the grams of dextrose supplied in a day, which is 2000*0.50 = 1000 g of dextrose, whereas at the time of 1 AM the total volume is 1840 mL, thus it is 1840*0.50 = 920 g of dextrose.

2.Multiply the grams of dextrose by 3.4 (there are 3.4 kcal/g dextrose) to determine kcalories administered to the patient by 1 AM.

So, 920 g of dextrose multiplied by 3.4 kcal/g will give 3,128 kcal (This much calories would be administered to the patient by 1 AM)

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