The system in Fig. P 2.48 is open to 1 atm on the right side. (a) If L=120 cm, what is the air p

step 1 Okay, I've transcribed our diagram here and our problem asks us. Hang on a second. I've got to d

The system in Fig. P 2.48 is open to 1 atm on the right...

Key Concepts

Hydrostatic Pressure

Hydrostatic pressure is the pressure in a fluid at rest that arises from the weight of the fluid above. It is directly proportional to the fluid density, the gravitational acceleration, and the vertical depth. This concept helps determine how pressure changes with height in any fluid system and is central when analyzing pressure differences between connected fluid containers.

Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of the air above a given surface and is a common reference or boundary condition in problems involving open containers. When one side of a system is open to the atmosphere, its pressure is fixed at the atmospheric value, which allows for the determination of pressure differences and equilibrium states in the system.

Pressure-Height Relationship

The pressure-height relationship in a fluid, typically represented by the equation p = p0 + ?gh, expresses how pressure increases with depth. This relationship is utilized in systems where differences in fluid column heights correspond to differences in pressure, enabling the calculation of unknown pressures or fluid column lengths based on equilibrium considerations.

Transcript

00:01 Okay, i've transcribed our diagram here and our problem asks us.

00:08 Hang on a second.

00:10 I've got to do a couple things.

00:12 It asks us if the system is open to the atmosphere on the right side, if r equals 120 centimeters, or if l equals 120 centimeters, what's the air pressure in a? okay.

00:31 I think we can figure this out here.

00:36 Let's begin.

00:42 Okay, we're going to neglect air pressure, assume all liquids at 20.

00:47 So for a, let's get started on an a.

00:51 We can find from table 8 .3 at 20 degrees c, which is what we're going to assume.

01:00 We can assume the density of water is 998 kilograms per cubic meter.

01:10 We can assume the density of mercury is going to be 13 ,550.

01:17 Kilograms per meter cubed and then from the figure we can say 15 centimeters and 32 centimeters at point x and y and 18 and to get the air pressure okay so now let's get the pressure let me go look at my diagram one more time so we're going to find our pressure at 18 centimeters of water and that'll p p z will equal our p have half atmosphere plus the density of water times g times l times the cosine of 35 minus 18 times 10 to the minus 2.

02:24 So this will equal 101 ,325 plus we said 998 times 9 .81 times 120 times the cosine of 17 times 10 to the minus 2.

02:48 And this is going to equal 109, 187 pascals, which will be equal to 109 .2 kilopascals.

03:02 Let's see what this looks like.

03:04 And then we know that our py has to equal pz.

03:09 So at 15 centimeters for mercury, we'll have that'll equal py plus the density of mercury times g times 32 minus 15 times 10 to the minus 2.

03:32 And this will be 109, 187 plus, and then this was 13550, make sure that's a comma, times 9 .81, times 9 .81, times.

03:51 This will be not cosine here.

03:54 This will just be 0 .17.

03:58 And this will equal 13174 pascals, and that'll be 131 .8 kilopascals.

04:13 And then to get the length, okay, there's our first answer.

04:18 131 .8 kilopascals.

04:22 That's the answer to a.

04:26 B asks us another question...

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