Question Video: Calculating the Cross-Sectional Area of a Pipe from Mass Flow Rate | Nagwa Question Video: Calculating the Cross-Sectional Area of a Pipe from Mass Flow Rate | Nagwa

# Question Video: Calculating the Cross-Sectional Area of a Pipe from Mass Flow Rate Physics • Second Year of Secondary School

45 kg of a liquid with a constant density of 1,055 kg/mΒ³ flows smoothly through a 2.5 m long pipe each second. What is the cross-sectionalarea of the pipe?

03:59

### Video Transcript

45 kilograms of a liquid with a constant density of 1055 kilograms per meter cubed flows smoothly through a 2.5-meter-long pipe each second. What is the cross-sectional area of the pipe?

Letβs start by drawing a diagram. We have a pipe that has a length that we will call πΏ and a cross-sectional area that we will call π΄. We are told that there is a liquid flowing through the pipe that has a constant density that we will call π. Now, if we were to place some sort of container at the end of the pipe to catch the liquid and we were to observe it for a certain amount of time, after a time that we will call capital π, a certain mass of liquid will have collected here in our container. And we will call this mass π.

The question tells us that the pipe is 2.5 meters long. So πΏ is equal to 2.5 meters. The question also tells us that the liquid has a constant density of 1055 kilograms per meter cubed. So π is equal to 1055 kilograms per meter cubed. And we are also told that 45 kilograms of the liquid flows through the pipe each second. So π is equal to 45 kilograms and π is equal to one second. The question asks us to calculate the cross-sectional area of the pipe. So we must calculate π΄.

We will start with the liquidβs mass flow rate, which is the amount of mass of the liquid that has flowed through the pipe per unit of time. And this is equal to the density of the liquid multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the liquidβs speed. However, we arenβt told the liquidβs speed, so we must calculate it. We know that speed is equal to distance traveled divided by the time taken to travel that distance. And in this case we are told that the liquid flows through the entire pipe each second. So π is equal to πΏ divided by π. Substituting this back into our equation, we get π divided by π is equal to ππ΄ multiplied by πΏ divided by π.

We can simplify this by multiplying both sides of the equation by π. And then we see that the πβs on the left and also the right cancel. Now, we will rearrange this equation to get an expression for π΄, the cross-sectional area of the pipe. We will divide both sides of the equation by π multiplied by πΏ. And we see that these πβs on the right cancel and the same for the πΏβs, leaving us with our expression for π΄. Writing this a bit more neatly, π΄ is equal to π divided by π multiplied by πΏ.

Now, we can carry on and substitute our known values of π, π, and πΏ into this equation. And we notice that these are all in SI units. So we donβt need to convert any of these before continuing. So π΄ is equal to 45 kilograms divided by 1055 kilograms per meter cubed multiplied by 2.5 meters. Evaluating this expression gives π΄ is equal to 0.017 meters squared to three decimal places.

The cross-sectional area of the pipe is equal to 0.017 meters squared.

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