Lecture 7: IV Flow Rate and Infusion/Completion Times
Calculating Flow Rates
In order to set up a IV, we need to know the flow rate; the flow rate is the speed at which the fluid infuses into the patient, and it is expressed in volume over time. For a manual IV setup, the flow rate is expressed in gtt/min, because we measure the rate by counting the number of drops which fall in the drip chamber each minute. For an IV setup on a volumetric pump or other electronic infusion device, we use the flow rate measured in mL/h because we can type this amount into the device manually without having to count drops in a drip chamber. Notice that both of these rates measure volume over time.
We can convert between a flow rate measured in gtt/min and a flow rate measured in mL/h. We already know how to convert between hours and minutes because we know that 1 h = 60 min. We can also convert between drops and milliliters, because both of these units measure volume, but we have to be careful, because drops come in many different sizes. In fact, any medications which are measured in drops always come with their own calibrated dropper, and the dropper from one medication cannot be used to dose another medication because the drops indicated on the labels are probably not going to be the same size. So, how do we measure the size of drops in an IV setup?
IV Tubing Calibration
The size of a drop in an IV setup depends upon the width of the IV tubing: thinner tubing produces smaller drops and wider tubing produces larger ones. IV tubing is pretty standard, and so there are two major categories of IV tubing:

Macrodrip tubing is wider and so produces larger drops. It is the tubing most commonly used for routine IV administration, such as infusion of IV fluids that do not contain sensitive medication.
Macrodrip tubing comes in 3 sizes: 10 gtt/mL, 15 gtt/mL, and 20 gtt/mL. (Note: 10 gtt/mL, for example, means 10 gtt = 1mL.)

Microdrip tubing is narrower and so produces smaller drops. It is used for children and infants, or to infuse sensitive medications where precision in the flow rate is essential.
Microdrip tubing (sometimes called minidrip) comes in only one size: 60 gtt/mL.
Notice that the smaller the drops, the more accurate the measurement of the medication; if we are setting the rate on a manual IV setup by counting the number of drops per minute, there is a certain amount of error introduced into the calculation  for example, I will have to round off to the nearest whole drop, because I cannot count partial drops. So, the smaller the drops, the less error is introduced when I round off to the nearest whole drop.
Also note that the larger the drop, the fewer are needed to make a whole mL.
Calculating Flow Rate from Medication Orders
So, whenever we receive an IV order, we will need to calculate the required flow rate before we can set up the IV. For this section, we will always be using a manual IV setup, so our flow rates in this section will always be in gtt/min.
Let's look at some example problems:
(The mention of the "drop factor method" in any of the problems below refers to a shortcut mentioned in the book; you do not need to know this method, but you are free to use it if you choose.)
Example:
The order is for 2 L of D10W to infuse over 16 h. Set calibration is 10 gtt/mL.
What flow rate will we need to set on a manual IV in order to correctly give this dosage over the specified time?
We want our answer to be in
, because all the problems in this chapter involve a manual IV setup, so the speed must be in gtt/min, because the way we measure the flow rate on a manual IV is to count the number of drops which fall in the drip chamber per minute.
_____
=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, since we know that we must infuse 2 L over 16 h, we can put this total volume over total time to get a rate expressed in volume over time.
_____
= We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.
But we notice that the problem does not give us any other information in L; it does, however, give us a fraction we could use to cancel out mL:
So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction
will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:
_____
= ×Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 10 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____
= ××Now we notice that we want our answer to be in gtt/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation.
(Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other  we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.)
We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××× = ××× = ××× Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 10 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ×××
(Notice that if we want to use the drop factor method, all we have done here is to calculate the drop factor 6. So if we wanted to use the drop factor method we could have simply skipped this step and gone directly to:
_____ = ×÷ 6)
We can divide both 6 and 1000 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ××× We can divide both 16 and 500 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ××× We can divide both 4 and 2 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ××× Now we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____
= ××× = Dividing the top by the bottom and then rounding to the nearest whole drop yields:
= 20.833333333333 = 21 So the correct flow rate is 21
. Example:
The order is for 500 mL of 1/2NS to infuse over 2 h. Set calibration is 15 gtt/mL.
What flow rate will we need to set on a manual IV in order to correctly give this dosage over the specified time?
We want our answer to be in
, because all the problems in this chapter involve a manual IV setup, so the speed must be in gtt/min, because the way we measure the flow rate on a manual IV is to count the number of drops which fall in the drip chamber per minute.
_____
=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, since we know that we must infuse 500 mL over 2 h, we can put this total volume over total time to get a rate expressed in volume over time.
_____
= Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 15 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____
= ×Now we notice that we want our answer to be in gtt/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation.
(Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other  we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.)
We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×× = ×× = ×× Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 15 and 60 by 15. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ×× = ××
(Notice that if we want to use the drop factor method, all we have done here is to calculate the drop factor 4. So if we wanted to use the drop factor method we could have simply skipped this step and gone directly to:
_____ = ÷ 4 )
We can divide both 4 and 500 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ×× = ×× Now we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____
= Dividing the top by the bottom and then rounding to the nearest whole drop yields:
= 62.5 = 63 So the correct flow rate is 63
. Example:
The order is for 15 mL of an IV antibiotic to infuse over 7 min. You will use a microdrip set.
What flow rate will we need to set on a manual IV in order to correctly give this dosage over the specified time?
We want our answer to be in
, because all the problems in this chapter involve a manual IV setup, so the speed must be in gtt/min, because the way we measure the flow rate on a manual IV is to count the number of drops which fall in the drip chamber per minute.
_____
=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, since we know that we must infuse 15 mL over 7 min, we can put this total volume over total time to get a rate expressed in volume over time.
_____
= Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 60 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation.
(Where did we get that the tubing calibration is 60 gtt/mL? Notice that our problem told us that we would be using a microdrip set; remember that microdrip tubing is always 60 gtt/mL.)So our equation becomes:
_____
= ×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= × = × = × Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
Now we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____
= × = Dividing the top by the bottom and then rounding to the nearest whole drop yields:
= 128.57142857143 = 129 So the correct flow rate is 129
. Example:
The order is for 50 mL of D2.5LR to infuse over 15 min. Set calibration is 20 gtt/mL.
What flow rate will we need to set on a manual IV in order to correctly give this dosage over the specified time?
We want our answer to be in
, because all the problems in this chapter involve a manual IV setup, so the speed must be in gtt/min, because the way we measure the flow rate on a manual IV is to count the number of drops which fall in the drip chamber per minute.
_____
=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, since we know that we must infuse 50 mL over 15 min, we can put this total volume over total time to get a rate expressed in volume over time.
_____
= Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 20 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____
= ×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= × = × = × Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 20 and 15 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= × = × Now we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____
= × = Dividing the top by the bottom and then rounding to the nearest whole drop yields:
= 66.666666666667 = 67 So the correct flow rate is 67
. Example:
The order is for 1 L of whole blood to infuse over 10 h. Set calibration is 10 gtt/mL.
What flow rate will we need to set on a manual IV in order to correctly give this dosage over the specified time?
We want our answer to be in
, because all the problems in this chapter involve a manual IV setup, so the speed must be in gtt/min, because the way we measure the flow rate on a manual IV is to count the number of drops which fall in the drip chamber per minute.
_____
=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, since we know that we must infuse 1 L over 10 h, we can put this total volume over total time to get a rate expressed in volume over time.
_____
= We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.
But we notice that the problem does not give us any other information in L; it does, however, give us a fraction we could use to cancel out mL:
So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction
will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:
_____
= ×Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 10 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____
= ××Now we notice that we want our answer to be in gtt/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation.
(Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other  we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.)
We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××× = ××× = ××× Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 10 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ×××
(Notice that if we want to use the drop factor method, all we have done here is to calculate the drop factor 6. So if we wanted to use the drop factor method we could have simply skipped this step and gone directly to:
_____ = ×÷ 6)
We can divide both 6 and 1000 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ××× We can divide both 10 and 500 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____
= ××× = ××× Now we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____
= ××× = Dividing the top by the bottom and then rounding to the nearest whole drop yields:
= 16.666666666667 = 17 So the correct flow rate is 17
. Calculating Infusion and Completion Times
Sometimes we know the rate an IV is supposed to infuse, but we need to calculate how long the IV will take to infuse and/or exactly at what time it should be finished. In this case, we will be doing almost exactly the same calculations as when we were computing flow rates; we are in a sense going the other direction. We need to know when an IV will finish so that we will know when to return to disconnect the IV, and if another bag must be hung for a continuous IV, to know when we must return with another bag to keep the IV flowing. We may also want to label an IV bag with start, progress and completion times so that anyone who checks on a patient can tell at a glance whether or not their IV is running on schedule, without having to contact the nurse who originally set up the IV.
We may want to calcuation an infusion time, which is a time in hours or minutes that indicates how long it will take for an IV to infuse. We may also want to calculate a completion time, which is a time of day that indicates at what time the IV will finish.
The 24Hour Clock
In hospitals we may see time measured either in the usual 12hour clock which uses am and pm to indicate morning vs. evening times, or we may see time marked using the 24hour clock (sometimes also referred to a military time). If you are not familiar with the 24hour clock, here is how it works:
Time on a 24 hour clock does not start over in the afternoon; rather, the numbers keep increasing until they reach 24 hours. So, for example, with the usual 12hour clock, we start our numbering over in the afternoon: the hour after 12:00 pm (noon) is 1 pm. In the 24hour clock, the hour after 1200 noon is 1300. Numbers writen in the 24hour clock system are also written slightly differently. They:

Numbers are written with leading zeros. So 6:00 am in the 24hour clock would be written 0600.

Times are written without a colon (the symbol ":"). So 9:30 am in the 24hour clock would be written 0930.

No "am" or "pm" is used; this is because it is unnecesary in a 24hour clock. For example, 9:00 am would be written 0900 in the 24hour clock, but 9:00 pm would be written 2100 (because this is 9:00 plus 12 hours).
The general rule is that am times are pretty similar in the 12hour and 24hour clock formats, but pm times must have 12 hours added to them in order to be converted to the 24hour clock format.
Here is a list of every hour of the day as it might be written in both 12hour and 24hour formats:
12hour format 
24hour format 
12:00 am 
0000 (2400) 
1:00 am 
0100 
2:00 am 
0200 
3:00 am 
0300 
.
.
.

.
.
. 
10:00 am 
1000 
11:00 am 
1100 
12:00 pm 
1200 
1:00 pm 
1300 
2:00 pm 
1400 
3:00 pm 
1500 
4:00 pm 
1600 
5:00 pm 
1700 
6:00 pm 
1800 
7:00 pm 
1900 
8:00 pm 
2000 
9:00 pm 
2100 
10:00 pm 
2200 
11:00 pm 
2300 
12:00 am 
2400 (0000) 
Notice that 12:00 am can be written as either 0000 or 2400; however, we can never write any time larger than 2400. So, for example, 12:01 am can be written 0001, but it cannot be written 2401!
Also note that, just like in the 12hour clock, the 24hour clock indicates hours with the first two numbers and minutes with the last two numbers. So we can express times in the 24hour clock which are not whole hours by changing the last two numbers, just like in the 12hour clock system. For example, 5:48 pm would be written 1748 in the 24hour clock system (notice that 17 = 5 + 12).
Calculating Infusion and Completion Times
Let's practice actually calculating infusion and completion times from real drug orders.
Example:
The order is for 900 mL of D2.5W to infuse at 100 mL/h. Start time is 0712
What are the infusion and completion times for this IV order?
We need to find the infusion time, or the total time it will take this IV to infuse, so we want our answer to be a unit of time. So we can choose either hours or minutes, but since the rate we are already given in the problem is
, which contains h, it will be easiest if we start by trying to get our answer in h. It doesn't actually matter whether you choose to get your answer in h or min, so if you want to solve all your problems for h or all your problems for min, feel free to do so. We've chosen hours here, so that's what we will write:
_____ h =
We want to see how long it will take to infuse 900 mL, so we will put this first on the right side of our equation:
_____ h = 900 mL
Putting 900 mL over 1 doesn't change it, so we have:
_____ h =
Now we still need to cancel out mL, and since this is in the top of a fraction on the right side of the equation, we need to multiply by a fraction that is equal to one and which has mL on the bottom.
We have such a fraction:
. So this yields:_____ h =
×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____ h =
× = × = × hNow we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 900 and 100 by 100. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ h =
× hNow we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____ h =
× h = hDividing the top by the bottom and then rounding to the nearest whole drop yields:
h = 9 h = 9 hSo the infusion time is 9 h.
Now, to find the completion time, or time at which the IV will finish, we simply add the infusion time to the start time.
Because this IV has a start time of 0712 today (which is 09/01) and an infusion time of 9 h and 0 min, it will finish at 1612 on 09/01.
So our completion time is 1612 on 09/01.
Example:
The order is for 250 mL of 1/4S to infuse at 12 gtt/min. Set calibration is 15 gtt/mL.Start time is 1555
What are the infusion and completion times for this IV order?
We need to find the infusion time, or the total time it will take this IV to infuse, so we want our answer to be a unit of time. So we can choose either hours or minutes, but since the rate we are already given in the problem is
, which contains min, it will be easiest if we start by trying to get our answer in min. It doesn't actually matter whether you choose to get your answer in h or min, so if you want to solve all your problems for h or all your problems for min, feel free to do so. We've chosen hours here, so that's what we will write:
_____ min =
We want to see how long it will take to infuse 250 mL, so we will put this first on the right side of our equation:
_____ min = 250 mL
Putting 250 mL over 1 doesn't change it, so we have:
_____ min =
Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 15 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____ min =
×Now we still need to cancel out gtt, and since this is in the top of a fraction on the right side of the equation, we need to multiply by a fraction that is equal to one and which has gtt on the bottom.
We have such a fraction:
. So this yields:_____ min =
××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____ min =
×× = ×× = ×× minNow we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 15 and 12 by 3. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ min =
×× minWe can divide both 250 and 4 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ min =
×× minNow we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____ min =
×× min = minDividing the top by the bottom and then rounding to the nearest whole drop yields:
min = 312.5 min = 313 minSo the infusion time is 313 min.
But this answer is not in the right form! Because 313 min is greater than or equal to 60 min, we can convert some of these minutes into hours by dividing by 60.
Then the answer we get when we divide by 60 is the number of hours, and any leftover minutes that don't fit into a whole hour will be the minutes:
Dividing 313 by 60 yields 5, with a remainder of 13. (Notice that the remainder is the number of minutes remaining when you break 313 into hoursized chunks  in other words, the remainder is the number of minutes left over that don't fit into a whole hour.)
So our correctly written infusion time is 5 h 13 min.
Now, to find the completion time, or time at which the IV will finish, we simply add the infusion time to the start time.
Because this IV has a start time of 1555 today (which is 09/01) and an infusion time of 5 h and 13 min, it will finish at 2108 on 09/01.
So our completion time is 2108 on 09/01.
Example:
The order is for 30 mL of an IV antibiotic to infuse at 10 gtt/min. Set calibration is 10 gtt/mL.Start time is 9:48 pm
What are the infusion and completion times for this IV order?
We need to find the infusion time, or the total time it will take this IV to infuse, so we want our answer to be a unit of time. So we can choose either hours or minutes, but since the rate we are already given in the problem is
, which contains min, it will be easiest if we start by trying to get our answer in min. It doesn't actually matter whether you choose to get your answer in h or min, so if you want to solve all your problems for h or all your problems for min, feel free to do so. We've chosen hours here, so that's what we will write:
_____ min =
We want to see how long it will take to infuse 30 mL, so we will put this first on the right side of our equation:
_____ min = 30 mL
Putting 30 mL over 1 doesn't change it, so we have:
_____ min =
Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 10 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____ min =
×Now we still need to cancel out gtt, and since this is in the top of a fraction on the right side of the equation, we need to multiply by a fraction that is equal to one and which has gtt on the bottom.
We have such a fraction:
. So this yields:_____ min =
××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____ min =
×× = ×× = ×× minNow we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 10 and 10 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ min =
×× minNow we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____ min =
×× min = minDividing the top by the bottom and then rounding to the nearest whole drop yields:
min = 30 min = 30 minSo the infusion time is 30 min.
Now, to find the completion time, or time at which the IV will finish, we simply add the infusion time to the start time.
Because this IV has a start time of 9:48 pm today (which is 09/01) and an infusion time of 0 h and 30 min, it will finish at 10:18 pm on 09/01.
So our completion time is 10:18 pm on 09/01.
Example:
The order is for 1 L of D2.5LR to infuse at 25 gtt/min. Set calibration is 20 gtt/mL.Start time is 6:30 am
What are the infusion and completion times for this IV order?
We need to find the infusion time, or the total time it will take this IV to infuse, so we want our answer to be a unit of time. So we can choose either hours or minutes, but since the rate we are already given in the problem is
, which contains min, it will be easiest if we start by trying to get our answer in min. It doesn't actually matter whether you choose to get your answer in h or min, so if you want to solve all your problems for h or all your problems for min, feel free to do so. We've chosen hours here, so that's what we will write:
_____ min =
We want to see how long it will take to infuse 1 L, so we will put this first on the right side of our equation:
_____ min = 1 L
Putting 1 L over 1 doesn't change it, so we have:
_____ min =
We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.
But we notice that the problem does not give us any other information in L; it does, however, give us a fraction we could use to cancel out mL:
So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction
will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:
_____ min =
×Now, since we want to cancel out the mL at the top right of our equation, we need to multiply by a fraction which is equal to one and also has mL in the bottom, to cancel out the mL which appears in the top of the right of our equation.
Because we know from the calibration of our tubing which is given in the problem that 20 gtt = 1 mL, we can use the fraction
, because this fraction is equal to one and has mL in the bottom, which we will need to cancel out the mL in the top of our fraction in the right side of the equation. So our equation becomes:
_____ min =
××Now we still need to cancel out gtt, and since this is in the top of a fraction on the right side of the equation, we need to multiply by a fraction that is equal to one and which has gtt on the bottom.
We have such a fraction:
. So this yields:_____ min =
×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____ min =
××× = ××× = ××× minNow we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 20 and 25 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ min =
××× minWe can divide both 1000 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ min =
××× minNow we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____ min =
××× min = minDividing the top by the bottom and then rounding to the nearest whole drop yields:
min = 800 min = 800 minSo the infusion time is 800 min.
But this answer is not in the right form! Because 800 min is greater than or equal to 60 min, we can convert some of these minutes into hours by dividing by 60.
Then the answer we get when we divide by 60 is the number of hours, and any leftover minutes that don't fit into a whole hour will be the minutes:
Dividing 800 by 60 yields 13, with a remainder of 20. (Notice that the remainder is the number of minutes remaining when you break 800 into hoursized chunks  in other words, the remainder is the number of minutes left over that don't fit into a whole hour.)
So our correctly written infusion time is 13 h 20 min.
Now, to find the completion time, or time at which the IV will finish, we simply add the infusion time to the start time.
Because this IV has a start time of 6:30 am today (which is 09/01) and an infusion time of 13 h and 20 min, it will finish at 7:50 pm on 09/01.
So our completion time is 7:50 pm on 09/01.
Example:
The order is for 1 L of whole blood to infuse at 170 mL/h. Start time is 2300
What are the infusion and completion times for this IV order?
We need to find the infusion time, or the total time it will take this IV to infuse, so we want our answer to be a unit of time. So we can choose either hours or minutes, but since the rate we are already given in the problem is
, which contains h, it will be easiest if we start by trying to get our answer in h. It doesn't actually matter whether you choose to get your answer in h or min, so if you want to solve all your problems for h or all your problems for min, feel free to do so. We've chosen hours here, so that's what we will write:
_____ h =
We want to see how long it will take to infuse 1 L, so we will put this first on the right side of our equation:
_____ h = 1 L
Putting 1 L over 1 doesn't change it, so we have:
_____ h =
We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.
But we notice that the problem does not give us any other information in L; it does, however, give us a fraction we could use to cancel out mL:
So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction
will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:
_____ h =
×Now we still need to cancel out mL, and since this is in the top of a fraction on the right side of the equation, we need to multiply by a fraction that is equal to one and which has mL on the bottom.
We have such a fraction:
. So this yields:_____ h =
××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____ h =
×× = ×× = ×× hNow we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 1000 and 170 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____ h =
×× hNow we have nothing left to simplify, so we multiply across the top and multiply across the bottom to get:
_____ h =
×× h = hDividing the top by the bottom and then rounding to the nearest whole drop yields:
h = 5.8823529411765 h = 6 hSo the infusion time is 5.88 h.
But this answer is not in the right form! Because 5.88 h is an expression of hours as a decimal, we must convert the decimal part of the hours to minutes by multiplying just the part after the decimal by 60.
10 by 60 yields 52.8 min, which when rounded to the neares whole minute is 53 min.
So our correctly written infusion time is 5 h 53 min.
Now, to find the completion time, or time at which the IV will finish, we simply add the infusion time to the start time.
Because this IV has a start time of 2300 today (which is 09/01) and an infusion time of 5 h and 53 min, it will finish at 0453 on 09/02.
So our completion time is 0453 on 09/02.