# THE HOLLIDAY-SEGAR 4-2-1 RULE.

This rule helps to easily calculate the maintenance fluid needs in hospitalized patients.

Maintenance fluid can be defined as the amount of fluid required to compensate for ongoing fluid losses (insensible and sensible), thus maintaining steady state in the body.

In a study published in 1957 in the Journal of Paediatrics, Malcolm Holiday and William Segar developed a simple rule that can be easily remembered for calculating maintenance fluid in sick children.

This rule is based on the assumption that there exists a DIRECT LINEAR RELATIONSHIP between physiologic water and energy metabolism (that is, under normal resting conditions, 1ml of body water is required to metabolize 1KCal).

These authors (Holliday and Segar), having also understood that there is a relationship (NON-LINEAR) between body weight and energy expenditure, therefore plotted a graph showing this relationship. By this, they arrived at a method that can help to calculate maintenance fluid using body weight, based on the the established DIRECT LINEAR relationship between body water and energy metabolism.

And here is the conclusion:

First 10kg (0-10kg) will require 100ml/kg/day
Second 10kg (11-20kg) will require 50ml/kg/day
Remaing kg (>20kg) will require 20ml/kg/day

This formula is then simplified to produce the 4-2-1 rule, by expressing the maintenance fluid in “per hour” instead of “per day”.

To achieve this, the amount of fluid is divided by 24 (since 1day= 24hours).

0-10 kg= 100ml/kg ÷ 24= 4.167 (approx. 4ml/kg/hr)
11-20kg = 50ml/kg ÷ 24= 2.083 (approx. 2ml/kg/hr)
>20kg= 20ml/kg ÷ 24= 0.83 (approx. 1ml/kg/hr)

And that’s the 4-2-1 rule for calculating HOURLY MAINTENANCE FLUID.

Example 1:
What is the daily maintenance fluid for a sick child weighing 12kg? Also, calculate the maintenace fluid to be administered to this child over a period of 8hrs.

Weight of the child is 12kg (10kg + 2kg)
First 10kg= 100ml/kg/day= 100ml*10= 1000ml/day
Second 2kg= 50ml/kg/day= 50ml*2= 100ml/day

The daily maintenance fluid for the child
= 1000ml + 100ml= 1100ml/day.

To calculate the 8hrs period maintenance fluid, convert daily maintenance to hourly maintenance, and multiply the result by 8.
1100ml ÷ 24= 45.83ml/hr
Maintenance fluid for a period of 8hrs
= 45.83ml*8
= 366.64 (approx. 367ml).

Example 2:
Using the 4-2-1 rule, calculate the hourly maintenance fluid for a sick child weighing 25kg.

Weight= 25kg (10kg + 10kg + 5kg)
First 10kg= 4ml*10= 40ml/hr
Second 10kg= 2ml*10= 20ml/hr
Last 5kg= 1ml*5= 5ml/hr

The hourly maintenance fluid
= 40+20+5
= 65ml/hr.

To calculate the daily maintenance fluid, simply multiply by 24hrs
Daily maintenance fluid for this child= 65ml*24= 1560ml per day.

OGUNLABI, David (RN).

# CONVERTING BETWEEN CELSIUS AND FAHRENHEIT SCALES.

Celsius and Fahrenheit scales are the two major scales for temperature measurement.

On the Celsius scale, the freezing point of water is 0°C while the boiling point is 100°C. Making a difference of 100°C.

On the Fahrenheit scale, 0°C (freezing point of water) is equivalent to 32°F, while 100°C (boiling point of water) is equivalent to 212°F. Making a difference of 180°F (that is, 212°F – 32°F).

It is therefore right to say:
100°C is equivalent to 180°F
One degree Celcius will therefore be equivalent to (1*180) ÷ 100= 1.8

This means; one degree Celsius is 1.8 times greater than degree Fahrenheit.

To convert from Celsius to Fahrenheit:
First MULTIPLY the Celsius temperature by 1.8, then ADD 32.
(°C*1.8) + 32

Example 1:
Convert 36°C to Fahrenheit

To convert from Celsius to Fahrenheit:
(°C*1.8) + 32
(36*1.8) + 32
64.8 + 32
96.8°F

To convert from Fahrenheit to Celsius:
First SUBTRACT 32 from Fahrenheit temperature, then DIVIDE by 1.8
(°F-32) ÷ 1.8

Example 2:
Convert 100°F to °C
(F-32) ÷ 1.8
(100-32) ÷ 1.8
68 ÷ 1.8
37.8°C

Try these.
1. Convert 108°F to °C
2. Convert 35.4°C to °F

Ogunlabi, David (RN).

# EPINEPHRINE DRIP RATE CALCULATION.

Epinephrine is usually ordered in microgram per minute (mcg/min), and is administered via infusion or syring pump due to its vasoactive property. The nurse is saddled with the responsibility of ensuring the right dose is set and administered to the patient.

The infusion rate equation for epinephrine is the same for other medications that are ordered in mcg/min.

Infusion rate equation for Epinephrine (in ml/hr)
= (ordered dose in mcg/min * 60) ÷ (dosage strength in solution in mg/ml * 1000)

NOTE: The weight of patient is insignificant in this equation.

Example:
Your patient has an order for Epinephrine 4mcg/min. Pharmacy has supplied a solution labelled ‘Epinephrine 1mg in D5W 500ml. You will set the infusion pump for how many ml/hr?

Ordered dose = 4mcg/min
Dosage strength = 1mg epinephrine in 500ml of 5% D/W = 1mg ÷ 500ml = 0.002mg/ml

Infusion rate equation for Epinephrine (in ml/hr)
= (ordered dose in mcg/min * 60) ÷ (dosage strength in solution in mg/ml * 1000)

= (4*60) ÷ (0.002*1000)
= 240 ÷ 2 = 120 ml/hr

Try this.
Your patient has an order for Epinephrine 2mcg/min. If 1mg of Epinephrine is diluted in 50ml of sterile water, and the patient weighs 50 lb. You will set the infusion pump for how many ml/min?

Ogunlabi, David (RN)

# DOPAMINE DRIP CALCULATION.

Dopamine is a positive inotropic agent (it increases the force of myocardial contractility). It is usually administered via syringe or infusion pump due to its vasoactive properties. Dopamine drip rate is therefore expressed in ml/hr.

STEPS INVOLVED IN CALCULATING DOPAMINE FLOW RATE ARE:
1. identify the prescribed or ordered dose of dopamine.
2. Identify the weight of the patient (weight must be in kilogram).
3. Identify the dosage strength or concentration of dopamine in solution (measured in mg/ml): This can be calculated by dividing dosage of dopamine added to diluent (in milligram) by the amount of diluent used (in milliliters). It is therefore expressed in mg/ml.
4. Set up the formula and insert the above parameters.

Dopamine drip (in ml/hr)
= (ordered dose in mcg/kg/min * weight in kg * 60) ÷ (dopamine strength in mg/ml * 1000)

NOTE:
The “60” in the above formula stands for 60min (1hr), which helps to convert minute to hour.

The “1000” in the above formula stands for 1000mcg (1mg), when helps to convert mcg to mg.

Example 1:

A patient is ordered to start an IV Dopamine drip at 5 mcg/kg/min. The patient weighs 180 lbs. You have a bag of Dopamine that reads 400 mg/250 mL. What will you set the IV pump drip rate (mL/hr) at?

Step 1: ordered dose= 5 mcg/kg/min
Step 2: weight= 180 lb (that is 180 pounds).
To convert pounds to kg; 1kg = 2.2 pounds
Therefore, 180 pounds = (180÷2.2)kg = 81.8kg
Step 3:
Dopamine strength= 400mg ÷ 250ml = 1.6 mg/ml
Step 4:
Dopamine drip (in ml/hr)
= (ordered dose in mcg/kg/min * weight in kg * 60) ÷ (dopamine strength in mg/ml * 1000)
= (5 * 81.8 * 60) ÷ (1.6 * 1000)
= 24540 ÷ 1600
= 15.3 ml/hr.

Try this.

A patient is ordered to start an IV Dopamine drip at 15 mcg/kg/min. The patient weighs 154 lbs. You have a bag of Dopamine that reads 400 mg/50 mL. What will you set the IV pump drip rate (mL/hr) at?

Ogunlabi, David (RN).

# RECONSTITUTION OF DEXTROSE SOLUTION INTO DESIRED STRENGTH.

Example 1:
Prepare 150ml of a 8.5% dextrose solution. You have D5W and D10W. How much of each solution is needed?

In the example above, there are three (3) different strengths/concentrations/percentages of dextrose solution:
5%, 8.5% and 10% dextrose solution.

The constituting strengths are 5% (solution A) and 10% (solution B).
The final strength is 8.5% (solution C).

METHOD 1.

The standard formula for calculating the strength and volume of constituted solution is:

(% of sol. A * Vol of sol. A) + (% of sol. B * vol of sol. B) = (% of sol. C * Vol of sol C)

% of Sol. A = 5%= 5÷100= 0.05
% of Sol B = 10%= 10÷100= 0.1
% of Sol C = 8.5%= 8.5÷100= 0.085
Vol of Sol A is unknown
Vol of Sol B is unknown
Vol of Sol C = 150ml

Vol A + Vol B = Vol C
A+B = 150
A= (150-B)

Insert the values into the formula.

(% of sol. A * Vol of sol. A) + (% of sol. B * vol of sol. B) = (% of sol. C * Vol of sol C)

(5%*A) + (10%*B) = (7.5%*C)
(5%)*(150-B) + (10%*B) = (8.5%*150)
0.05*(150-B) + (0.1*B) = (0.085*150)
(0.05*150 – 0.05*B) + 0.1B = 12.75
7.5 – 0.05B + 0.1B = 12.75
Collect like terms
0.1B – 0.05B = 12.75-7.5
0.05B = 5.25
B = 5.25 ÷ 0.05
B = 105ml

Remember that A = 150-B
A= 150-105
A= 45ml.

Therefore, 45ml of sol A (5% dextrose) will be added to 105ml of sol B (10% dextrose) to produce 150ml of sol C (8.5% dextrose).

ALTERNATIVE METHOD (METHOD 2).
Step 1: Find the difference between each constituting strength and the final strength.
Constituting strength= 5% and 10%
Final strength= 8.5%
Difference between 5% and 8.5%= 8.5%-5%= 3.5% (higher difference)
Difference between10% and 8.5%= 10%-8.5%= 1.5% (lower difference)

Step 2: Add the results of the difference in Step 1 above.
3.5+1.5= 5

Step 3: Divide each difference in step 1 by the sum of the difference in Step 2 and multiply by the volume of final solution.
(3.5÷5)*150ml
= 105ml
(1.5÷5)*150ml
= 45ml

Step 4: Assign the higher value to the constituting strength with lower difference (look at step 1); and the lower value to the constituting strength with higher difference (look at step 1)
There are two constituting strength: 5% and 10% dextrose.
So the higher value (105ml) is assigned to the constituting strength with lower difference (10% dextrose)
And the lower value (45ml) is assigned to the constituting strength with higher difference (5% dextrose)

In conclusion, 45ml of sol A (5% dextrose) will be added to 105ml of sol B (10% dextrose) to produce 150ml of sol C (8.5% dextrose).

Try this.
1. Prepare 20 mL of dextrose 7.5% weight in volume (w/v) using dextrose 5% and dextrose 50%. How many milliliters of each will be needed?

Ogunlabi, David (RN)

# PREPARATION OF INTRAVENOUS DEXTROSE SOLUTIONS OF VARYING STRENGTH- PART 1

Example 1:
A patient is to have 500ml of 5% dextrose in water, but the available strength is 100ml of 50% dextrose in water. As the nurse on duty, how would you prepare 500ml of 5% dextrose in water from 50% dextrose in water?

Desired strength = 5%
Available strength = 50%
Desired volume = 500ml

First calculate the Amount of the available strength (50% D/W) needed for the preparation.

Amount of the available strength needed = (desired strength ÷ available strength) * (desired volume)
A= (5% ÷ 50%) * 500ml
A= 0.1*500
A= 50ml
Therefore, 50ml of 50% dextrose in water will be needed..
It is then made up to 500ml by adding 450ml of sterile water (diluent)
The preparation = 50ml of 50% D/W + 450ml of sterile water = 5% dextrose in water.

Example 2:
A patient needs 7.5% dextrose 500ml. The pharmacy has 10% dextrose in stock.
A. What amount of concentrate is needed?
B. What amount of diluent is needed?

Desired strength= 7.5%
Strength in stock or available strength = 10%
Desired volume = 500ml

Amount of concentrate (10% dextrose) needed
A= (desired strength ÷ available strength) * (desired volume)
A= (7.5% ÷ 10%) * 500ml
A= 0.75 * 500ml
A= 375ml of 10% dextrose

Amount of diluent= 500ml – 375ml
= 125ml
The preparation= 375ml of 10% dextrose + 125ml of diluent.

Example 3:
How many ml of 50% dextrose should be added to 1L of 0.9 saline to make a 5% dextrose solution?

Assume the amount of 50% to be added is A ml
Desired strength= 5%
Available strength= 50%
Desired volume = (1000+A)ml
NOTE:
1L = 1000ml
The total volume desired will be the sum of 1000ml of 0.9% Saline and A ml which we are to calculate.

A= (Desired strength ÷ Available strength) * Desired volume
A= (5% ÷ 50%) * (1000 + A)
A= (0.1) * (1000 + A)
OPEN THE BRACKET BY MULTIPLYING 0.1 BY 1000, AND 0.1 BY A
A= 100 + 0.1A
COLLECT LIKE TERMS
1A – 0.1A = 100
0.9A= 100
DIVIDE BOTH SIDES BY 0.9

0.9A÷0.9 = 100÷0.9
A= 111.1ml
Therefore, 111.1ml of 50% dextrose will be added to 1L of 0.9% saline to make 5% dextrose solution.

Example 4:
You have a 70% dextrose (w/v) solution. How many grams of dextrose are in 400ml?

w/v means weight/volume
Weight is usually in gram (g)
Volume is usually in milliliters (ml)
70% dextrose (w/v) = 70g dextrose in 100ml of solution (v)

In 70% dextrose solution,
Each 100ml of solution contains 70g dextrose
So, 400ml of the solution will contain …..g dextrose
CROSS MULTIPLY
(400*70g) ÷ 100
= 28000g ÷ 100
= 280g
Therefore, 400ml of the solution contains 280g of dextrose.

Solve this exercise.
1. Prepare 500ml of 10% dextrose in water from 100ml of 50% dextrose solution.

2. How many ml of 50% dextrose should be added to 500ml of Ringer’s Lactate to make 4.3% dextrose solution?

3. A patient needs 45% dextrose 100ml The pharmacy has only 70% dextrose 500ml.
a) how much concentrate will you need?
b) how much diluent?

4. How many ml of 50% dextrose solution will supply 140g of dextrose?

1.
100ml of 50% dextrose + 400ml of diluent

2.
47ml of 50% dextrose will be added to 500ml of RL to make 4.3% dextrose solution.

3.
64.3ml of 70% dextrose + 35.7ml of diluent = 100ml of 45% dextrose solution.

4.
280ml of solution= 140g dextrose.

Ogunlabi, David (RN)

# HOW “NORMAL” IS NORMAL SALINE?

Normal saline is one of the mostly used intravenous solutions in medical practice. It is an isotonic solution made up of sodium (Na) and chloride (Cl) ions. Its osmolarity is almost the same as the plasma osmolarity. For this reason, it is the intravenous fluid/solution of choice in the management of hypovolemic shock. It remains in the intravascular space, maintains the intravascular volume, thus enhancing tissue and organ perfusion.

Normal saline is also referred to as 0.9% saline. It is not correct to say or write “0.9% normal saline” because the 0.9% is the concentration that makes the solution NORMAL. Therefore, the right thing to say or write is “0.9% Saline” or “Normal Saline”.

So let’s see how the osmolarity of normal saline is derived, and then compare it with plasma osmolarity.

OSMOLARITY OF 0.9% SALINE is the product of:
1.) Molar concentration of 0.9% Saline
2.) Number of moles of NaCl
3.) Osmotic coefficient of 0.9% Saline

= (molar concentration of 0.9% Saline)*(number of moles of NaCl)*(osmotic coefficient of 0.9% saline)

How to calculate the MOLAR CONCENTRATION OF 0.9% Saline

MOLAR MASS OF NaCl = Atomic mass of Na (23) + Atomic mass of Cl (35.5)
= 23+35.5 = 58.5g per mole

0.9% saline means:
Each 100ml of solution contains 0.9g NaCl
1000ml (1L) of solution will contain …….g NaCl
(CROSS MULTIPLY)

= (1000*0.9) ÷ 100
= 9g per liter (9g/L)

If 58.5g NaCl = 1mole
9g/L NaCl = …….mole
(CROSS MULTIPLY)

= (9*1) ÷ 58.5 = 9 ÷ 58.5
= 0.154 mole/L

To convert 0.154 mole to milli mole, multiply by 1000
= 0.154 * 1000 = 154 mmol/L

Molar concentration of 0.9% Saline = 154 mmol/L or 154 mEq/L

1 molar of NaCl dissociates into 2, that is, Na and Cl ions in solution.

Number of moles in solution= 2

Osmotic coefficient of normal saline= 0.93 (this is a constant value)

Osmolarity of normal saline = (molar concentration of 0.9% Saline)*(number of moles of NaCl)*(osmotic coefficient of 0.9% saline)

= 154*2*0.93
= 286 mOsm/L

The normal plasma osmolarity ranges between 275-295 mOsm/L

Placing the concentration of normal saline side by side with that of plasma, it is obvious that normal saline has almost the same concentration as the plasma concentration.

NB:
1.) SALINE is sodium chloride (NaCl).

2.)Milli equivalent (mEq) is the same as Milli mole (mmol).

3.) mOsm means milli osmole

4.)Osmolarity and osmolality both refer to osmotic concentration. The slight difference between the two is in their units of measurement:

Osmolarity is measured in mOsm per liter of solvent
Osmolality is measured in mOsm per kilogram of solvent.

Ogunlabi, David (RN).