Calculating Parenteral Feedings
HSCI 368, D. ChenMaynard, PhD, RD
Total Parenteral Nutrition (TPN) must be individualized for the patient. However, often there are standard concentrations of protein, carbohydrate, and fat available in standard volumes, which limits the degree of individualization using these standard solutions. However, mixing different formulas in calculated amounts can be used to develop feeding solutions for most patients. We can also calculate the nutrients to meet the needs of an individual patient using these standard solutions.
Dextrose solution and lipid emulsions are common macronutrients used to provide energy in parenteral solution. Nitrogen for protein synthesis is obtained from synthetic crystalline Lamino acid solutions.
When
designing a TPN solution, you need to consider these questions:
1. What is the patient’s energy need? (Kcal required/day) 2. How much protein/nitrogen does the patient need in a day? 3. How much fluid can the patient tolerate and need? 4. How much fat emulsion can be given/tolerated? 5. How much dextrose is needed? Dextrose concentration? 6. Which electrolytes are needed and how much?

Where do you start to determine the composition of parenteral feeding?
Nutrition assessment should tell you the status of the patient so you can tailor the needs accordingly. Also the TPN solution is going to fully supply the need or is it supplemental?
1. START BY DETERMINING ENERGY NEEDS:
You may use HarrisBenedict + activity + infection/fever factor + etc = Energy (Kcal)
or you may use Kcal/Kg based on the following scale:
Normal need: 2530 Kcal/kg/day
Elective surgery: 2830 kcal/kg/day
Severe injury: 3040 kcal/kg/day
Extensive trauma/burn: 4555 kcal/kg/day
Example: for a 70 Kg male with elective surgery X 30 Kcal/Kg = 2,100 Kcal/day
2. NEXT, DETERMINE THE PROTEIN NEEDS:
Protein needs are determined by the patients, usually start with 0.83 gm/Kg and add stress and other factors as needed. Some patients may need up to 2.5 g/Kg due to loss and other factors. For example a malnourished 70 Kg man may need 84 gm protein per day (70 Kg X 1.2 gm/Kg). 
Another way to estimate protein needs is to use the nonprotein energy (Kcal) to nitrogen (N) ratio. The nonprotein energy to N ratio is based on the premise that sufficient energy must be ingested before protein will be used for tissue maintenance and repair. The ratio of 100150 Kcal: 1 gm N in stressful condition promote anabolism and 250300 Kcal: 1 gm N is adequate for normal body maintenance. 
If your 50 Kg patient needs 1500 Kcal and you want an Energy:N ratio of 150:1, how many grams of N are you going to need? 1500 Kcal ¸ 150 Kcal/gm N = 10 gm N
Example of your patient, who needs 2,100 Kcal: 2100 ¸ 150 = 14 grams
Think back: with the first method of calculation of protein needs, we found that he needed 84 grams protein, while with the second method, you need 14 gm N X 6.25 gm prot./gm N = 87.5 gram protein. Either method is correct and both are estimation of need.
Note: Prolonged TPN solution should contain 2 amino acids: Glutamine (most abundant amino acid in the blood that protects the gut epithelia tissue lining. Choline helps protect the liver from hepatic fat deposits that hinder its function.
How much protein in the solution? Assume that we give this 70 kg person 85 gm of protein using 10% amino acid solution:
10% solution has 100 gm
protein/L so we can set up equation
85 gm / X ml = 100 gm/1000 ml 100X gm ml = 85000 gm ml solve for X: X = 850 ml, gm units cancel out 
There are 3 different methods to calculate fluid needs:
In general, patients need a minimum of 30 cc fluid/Kg body weight to maintain hydration: 3050 cc/Kg is a good amount to aim, however, needs may go up at high as 100 cc/Kg in severely catabolic state such as burns. If the patient is overhydrated, TPN can be used in concentrated form to "pull off" extracellular fluids (ECF).
b. Another method to calculate fluid needs is 100 cc free water/gram N intake + 1 cc/Kcal provided,
The third method to calculate fluid needs: 100 cc/Kg for first 10 Kg
+ 50 cc/Kg for the second 10 Kg
+ 20 cc/Kg for each additional Kg
Example: using method a: 70 Kg X 33 cc/Kg = 2100 cc/day 70 Kg X 50 cc/Kg = 3500 cc/day 
method
b: 14 g N X 100 = 1,400 cc + 2100 cc for each Kcal provided =
3600 cc/day 
method
c: 100 cc/Kg X 10 Kg = 1000 cc
50 cc/Kg X 10 Kg = 500 cc 20 cc/Kg X 50 Kg = 1000 cc total: 1000 + 500 + 1500 = 2,500 cc/day 
So far, our NUTRITION PRESCRIPTION READS:
2,100 Kcal /day (of nonprotein energy)
85 (range is 84.0 – 87.5) gram protein /day
2,100 – 3,500 cc fluid /day
Now, we need to calculate what is the contribution of total energy from fat and CHO?
At present, 10%, 20% and 30% fat emulsion are available.
10% fat emulsion has 1.1 Kcal/cc
20% fat emulsion has 2.0 Kcal/cc
30% fat emulsion has 3.0 Kcal/cc
Fat emulsions are available most commonly in 250cc and 500 cc bottle. However, in three in one admixture, add any volume of fat you need, the pharmacist can mix the appropriate amount in the bag.
Example: What is the energy content of a 500 cc bottle of 10% fat emulsion?
500cc X 1.1 Kcal/cc = 550 Kcal
Energy content of a 500 cc of 20% fat? 500 cc X 2.0 Kcal/cc = 1000 Kcal
Fat has 9 Kcal/gram so you can determine the number of fat grams if you need it.
e.g. How many grams of fat in 500 cc of 10% lipid fat? 500 cc X 1.1 Kcal/cc = 550 Kcal;
500 Kcal ¸ 9 Kcal/gram = 61 gram of fat , Actually, there are 55.6 gm of fat and 5.4 gram glycerol. Glycerol is not a fat. Purified egg lecithin phosphatides are used as emulsifier. Tg’s do not contribute to serum osmolarity, emulsions are made isotonic by adding glycerol. Lipid emulsions have osmolality of 270 to 340 mOsm/L. Soybean and safflowers are rich in linolenic acid.
The essential fatty acid (EFA) also has to be determined. It is estimated that 24% of Kcal must be from EFA and a per needs 1 to 2.5 gram of fat/Kg.
MAXIMAL tolerance level of lipid is considered to be 2.5 gm/Kg body weight and 60% of energy from fat is also considered to be upper limit. More than 60% of energy from fat may result in hyperlipidemia due to impaired lipid clearance by the body.
For our reference 70 Kg male, how much fat and EFA would you recommend?
2,100 Kcal X 0.04 = 84 Kcal
70 Kg X 2.5 = 175 gm fat
from our calculations, 500 cc of 10% lipid emulsion provides 61 grams fat and 550 kcal, so if you give 750 cc (1.5 bottle of 500 cc bottle) of Intralipid you will be adding: 61g/bottle X 1.5 bottle = 91.5 g of fat /day and 550 kcal/bottle X 1.5 bottle = 825 Kcal/day from fat
To find out fat content of two commonly used fat emulsions:
Safflower oil has 77% EFA fat: 550 Kcal (10% 500cc) bottle of LIPOSIN gives 423.5 Kcal of EFA fat: 550 Kcal X 0.77 = 423.5 Kcal
INTRALIPID(soy based emulsion) has 52% EFA fat: 550 Kcal bottle of INTRALIPID gives 286 Kcal of fat: 550 Kcal X 0.52 = 286 Kcal of fat
Your 70 Kg patient needs 84 Kcal of EFA energy, over the week, he will need:
84 Kcal X 7 days = 588 Kcal from EFA/week,
so how much INTRALIPID is he going to need in a week?
588 Kcal/week ¸ 286 Kcal from fat = 2.06 bottles of 10% fat emulsion.
Therefore he will need at least 1,000 cc or two 500 cc bottles per week for EFA. So he will need at least 1 L of Intralipid per week or 143 ml/day to prevent EFA deficiency.
So if we add 1.5 (500cc) bottle per day, we will have sufficient EFA and provides 91.5 g fat and 825 Kcal. This provides 825 kcal from fat / 2100 Kcal = 39.2 % of energy from fat and it is still way below the limit of 2.5 gm/Kg body weight (175g/day).
In TPN, CHO is given as dextrose monohydrate, which yields 3.4 Kcal/gram (recall, glucose provides 4.0 Kcal/gm). Dextrose is available in 50 to 70% solution (D50W, D 70 are the most common solutions used to prepare TPN solutions). 
The amount of fat used in solution will determine the amount of CHO needed.
Energy content of one liter (1000 cc) 50% Dextrose solution (D50W)?
1,000 cc X 50% = 500 gram dextrose X 3.4 Kcal/gram = 1,700 Kcal from CHO
Energy content of 1000 cc of D5W (5% dextrose solution)?
1,000 cc X 0.05 = 50 gm dextrose X 3.4 Kcal/gram = 170 Kcal
MAXIMUM rate of CHO administration should not exceed 5 mg CHO/Kg/min.
Excessive administration of CHO can lead to hyperglycemia, liver abnormalities, and increased ventilatory drive (increased CO2 production).
1,000 cc of D50W provides 500 gm CHO/L
500 gm CHO ¸ 70 Kg ¸ 24 hours ¸ 60 min/hr = 4.96 mg CHO/Kg/min, so maximum amount of CHO we can use is 500 gm/day
So how much CHO would your patient need?
2,100 Kcal/day with 85 gm protein, 750 cc 10% Intralipid, which provides 825 Kcal: 2,100 Kcal – 825 Kcal = 1,275 Kcal needed from CHO, to see how many gram of dextrose is this? 1275 Kcal/3.4 Kcal/gm = 375 g dextrose needed: we can calculate it two ways:
1. If we use D50W as starting solution, we will need to use X ml of this solution:
375 g / X ml = 500g/1000 ml, if we solve for X, we have 375000 / 500 = 750 ml
2. one Liter (1 L) of D50W gives 1,700 kcal, to obtain 1,275 Kcal from CHO, how much D50W should you add? 1275 Kcal ¸ 1700 Kcal = 750 cc of D50W
Either method gives you the same answer of 750 cc of D50W needed in your final solution.
Since the fluid needs are 2,100 – 3,600 cc/day, with protein needs of 85 gm/day: your formulation for your 70 Kg patient will be: 10% amino acids 850 cc
D50W dextrose 750 cc
10% Intralipid 750cc
total volume: 2,350 cc/day
You need to add 70150 cc fluid for electrolytes, vitamins, and additives to give a final volume of say 2500 ml
Vitamins/minerals:
electrolytes should be monitored and adjusted as needed
vitamin solutions include 12 vitamins except vitamin K which can be injected IM
trace elements: iron is not routinely added because it changes pH of solution, some minerals are incompatible with TPN solution due to their effects on pH.
Final concentration has: 2.65% amino acids
15 % dextrose
1000 cc of 10% Intralipid
total volume has about 2500 ml so to calculate rate/hr: 2500 ml/24 hr = 105 ml/hr so if we set the pump so that it will administer at 105 cc/hr for 24 hr, we are ok with the infusion. Note: This formulation will be administered by central line only since dextrose concentration is greater than 10%.
Osmolarity Calculation for a TPN solution:
e.g. 50 g dextrose X 5 = 250 mOsm/L by glucose e..g 30 g protein X 10 = 300 mOsm/L e.g. NaCl = 2mOsm per NaCl molecule 250 mOsm/L + 300 mOsm/l = 550 mOsm/L 

Milieqivalent vs mg of electrolytes
Conversion from mg to
mEq
mg/atomic wt X valance = mEq 
mineral element  chemical symbol  Atomic wt  valance 
chlorine  Cl  35.4  1 
potassium  K  39  1 
sodium  Na  23  1 
calcium  Ca  40  2 
magnesium  Mg  24.3  2 
sulfur  S  32  
sulfate  SO4  96  2 
Conversion of
specific wt of sodium to sodium chloride
multiply by 2.54 e.g. 1000 mg Na = 1000 mg X 2.54 = 2,540 mg NaCl (2.5 g) 
to convert specific
wt of sodium chloride to sodium
