Physics I, Lesson 7: Bernoulli's Equation

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Lesson 7: Bernoulli's Equation

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What is Bernoulli's equation in terms of pressure, height, and velocity of a fluid at two points in a pipe?

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

P1 = Pressure at Point 1 (Pa)
P2 = Pressure at Point 2 (Pa)
ρ = Density of the Fluid (kg/m^3)
v1 = Velocity at Point 1 (m/s)
v2 = Velocity at Point 2 (m/s)
g = Acceleration due to Gravity (9.8m/s^2)
h1 = Height at Point 1 (m)
h2 = Height at Point 2 (m)

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<p>P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2<br><br>P1 = Pressure at Point 1 (Pa)<br>P2 = Pressure at Point 2 (Pa)<br>ρ = Density of the Fluid (kg/m^3)<br>v1 = Velocity at Point 1 (m/s)<br>v2 = Velocity at Point 2 (m/s)<br>g = Acceleration due to Gravity (9.8m/s^2)<br>h1 = Height at Point 1 (m)<br>h2 = Height at Point 2 (m)<br><br><b>Struggling to keep your MCAT equations straight? Simply conquer the 100 most important equations using Andrew's 100 Most Essential Equations Mastery Course @ https://mcatselfprep.com/course/andrews-equation-mastery-course/</b></p>

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Water is flowing through a pipe of varying cross sectional area down a hill with a velocity of 10 m/s at the top, and 20 m/s at the bottom. Over this distance the pipe goes from a height of 10 m to a height of 0 m. The water is at a pressure of 1,000,000 Pascals at the top. What is the pressure at the bottom (use 10 m/s^2 for g for simplicity)?

(A) 860,000
(B) 990,000
(C) 1,240,000
(D) 1,560,000

(A) 860,000

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
(1000000) + (.5)(1000)(10^2) + (1000)(10)(10) = P2 + (.5)(1000)(20^2) + (1000)(10)(0)
1060000
P2 = 860,000 Pa

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Water is flowing through a level pipe. One end of the pipe has an area of 4 m^2, while the other end has an area of 1/4 m^2. The pressure at the entrance is 200,000 Pa, while the pressure at the exit is 100,000 Pa. What is the Flow Rate in the pipe (m^3/s)?

(A) 37.86
(B) 53.33
(C) 87.69
(D) 112.47

(B) 53.33

v1A1 = v2A2 = Flow Rate = R
v1 = R/4
v2 = R/(.25)

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
(200,000) + (.5)(1000)(R/4) + (1000)(9.8)(h) = (100,000) + (.5)(1000)(4R) +(1000)(9.8)(h)
Subtract (1000)(9.8)(h) from each side
(200,000) + (.5)(1000)(R/4) = (100,000) + (.5)(1000)(4R)
100,000 + 125R = 2000R
100,000 = 1875R
R = approx. 50 m^3/s (53.33 m^3/s)

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<p>(B) 53.33<br><br>v1A1 = v2A2 = Flow Rate = R<br>v1 = R/4<br>v2 = R/(.25)<br><br>P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2<br>(200,000) + (.5)(1000)(R/4) + (1000)(9.8)(h) = (100,000) + (.5)(1000)(4R) +(1000)(9.8)(h)<br><b>Subtract (1000)(9.8)(h) from each side</b><br>(200,000) + (.5)(1000)(R/4) = (100,000) + (.5)(1000)(4R)<br>100,000 + 125R = 2000R<br>100,000 = 1875R<br>R = approx. 50 m^3/s (53.33 m^3/s)<br><br><b>Need help with MCAT math? Become an MCAT math wizard using Andrew's High-speed Math Mastery Course @ https://mcatselfprep.com/course/andrews-high-speed-math-mastery-course/</b></p>

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