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bernoulli's equation.pptx
Deriving of Bernoulli’s Equation
Consider a tube with a varying cross sectional and different heights at both ends.
Show liquid flow
Volume in
Volume out
1
2
𝐴 1 𝐴𝐴 𝐴 1 1 𝐴 1
𝐴 2 𝐴𝐴 𝐴 2 2 𝐴 2
ℎ 1 ℎ ℎ 1 1 ℎ 1
ℎ 2 ℎ ℎ 2 2 ℎ 2
The amount of water coming in must be equal to the amount of water coming out per time.
Work needs to be done to move the liquid from point 1 to point 2.
𝐿 1 𝐿𝐿 𝐿 1 1 𝐿 1
𝐿 2 𝐿𝐿 𝐿 2 2 𝐿 2
Total work done is equal to the total change of energy.
Volume in
Volume out
1
2
𝐴 1 𝐴𝐴 𝐴 1 1 𝐴 1
𝐴 2 𝐴𝐴 𝐴 2 2 𝐴 2
ℎ 1 ℎ ℎ 1 1 ℎ 1
ℎ 2 ℎ ℎ 2 2 ℎ 2
𝐿 1 𝐿𝐿 𝐿 1 1 𝐿 1
𝐿 2 𝐿𝐿 𝐿 2 2 𝐿 2
∆𝑊𝑊=∆ 𝑃𝐸 𝑔 𝑃𝑃𝐸𝐸 𝑃𝐸 𝑔 𝑔𝑔 𝑃𝐸 𝑔 +∆𝐾𝐾𝐸𝐸
𝑊 2 − 𝑊 1 = 𝑃𝐸 𝑔2 − 𝑃𝐸 𝑔1 + 𝐾𝐸 2 − 𝐾𝐸 1
𝑊=𝐹𝑑
𝐹=𝑃𝐴
𝑑=𝐿
𝑊=𝑃𝐴𝐿=𝑃𝑉
𝑃𝐸 𝑔 =𝑚𝑔ℎ
𝐾𝐸= 1 2 𝑚 𝑣 2
Volume in
Volume out
1
2
𝐴 1 𝐴𝐴 𝐴 1 1 𝐴 1
𝐴 2 𝐴𝐴 𝐴 2 2 𝐴 2
ℎ 1 ℎ ℎ 1 1 ℎ 1
ℎ 2 ℎ ℎ 2 2 ℎ 2
𝐿 1 𝐿𝐿 𝐿 1 1 𝐿 1
𝐿 2 𝐿𝐿 𝐿 2 2 𝐿 2
∆𝑊𝑊=∆ 𝑃𝐸 𝑔 𝑃𝑃𝐸𝐸 𝑃𝐸 𝑔 𝑔𝑔 𝑃𝐸 𝑔 +∆𝐾𝐾𝐸𝐸
𝑊 2 − 𝑊 1 = 𝑃𝐸 𝑔2 − 𝑃𝐸 𝑔1 + 𝐾𝐸 2 − 𝐾𝐸 1
𝑃 2 𝑃𝑃 𝑃 2 2 𝑃 2 𝑉𝑉− 𝑃 1 𝑃𝑃 𝑃 1 1 𝑃 1 𝑉𝑉= 𝑚𝑔 ℎ 2 − 𝑚𝑔ℎ 1 𝑚𝑚𝑔𝑔 ℎ 2 ℎ ℎ 2 2 ℎ 2 − 𝑚𝑔ℎ 1 𝑚𝑚𝑔𝑔ℎ 𝑚𝑔ℎ 1 1 𝑚𝑔ℎ 1 𝑚𝑔 ℎ 2 − 𝑚𝑔ℎ 1 + 1 2 𝑚 𝑣 2 2 − 1 2 𝑚 𝑣 1 2 1 2 1 1 2 2 1 2 𝑚𝑚 𝑣 2 2 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 𝑣 2 2 2 𝑣 2 2 − 1 2 1 1 2 2 1 2 𝑚𝑚 𝑣 1 2 𝑣 1 𝑣𝑣 𝑣 1 1 𝑣 1 𝑣 1 2 2 𝑣 1 2 1 2 𝑚 𝑣 2 2 − 1 2 𝑚 𝑣 1 2
𝑃 1 𝑃𝑃 𝑃 1 1 𝑃 1 𝑉𝑉+ 𝑚𝑔ℎ 1 𝑚𝑚𝑔𝑔ℎ 𝑚𝑔ℎ 1 1 𝑚𝑔ℎ 1 + 1 2 1 1 2 2 1 2 𝑚𝑚 𝑣 1 2 𝑣 1 𝑣𝑣 𝑣 1 1 𝑣 1 𝑣 1 2 2 𝑣 1 2 = 𝑃 2 𝑃𝑃 𝑃 2 2 𝑃 2 𝑉𝑉+𝑚𝑚𝑔𝑔 ℎ 2 ℎ ℎ 2 2 ℎ 2 + 1 2 1 1 2 2 1 2 𝑚𝑚 𝑣 2 2 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 𝑣 2 2 2 𝑣 2 2
𝑃 1 𝑃𝑃 𝑃 1 1 𝑃 1 + 𝑚 𝑉 𝑚𝑚 𝑚 𝑉 𝑉𝑉 𝑚 𝑉 𝑔ℎ 1 𝑔𝑔ℎ 𝑔ℎ 1 1 𝑔ℎ 1 + 1 2 1 1 2 2 1 2 𝑚 𝑉 𝑚𝑚 𝑚 𝑉 𝑉𝑉 𝑚 𝑉 𝑣 1 2 𝑣 1 𝑣𝑣 𝑣 1 1 𝑣 1 𝑣 1 2 2 𝑣 1 2 = 𝑃 2 𝑃𝑃 𝑃 2 2 𝑃 2 + 𝑚 𝑉 𝑚𝑚 𝑚 𝑉 𝑉𝑉 𝑚 𝑉 𝑔𝑔 ℎ 2 ℎ ℎ 2 2 ℎ 2 + 1 2 1 1 2 2 1 2 𝑚 𝑉 𝑚𝑚 𝑚 𝑉 𝑉𝑉 𝑚 𝑉 𝑣 2 2 𝑣 2 𝑣𝑣 𝑣 2 2 𝑣 2 𝑣 2 2 2 𝑣 2 2
𝑷 𝟏 +𝝆 𝒈𝒉 𝟏 + 𝟏 𝟐 𝝆 𝒗 𝟏 𝟐 = 𝑷 𝟐 +𝝆𝒈 𝒉 𝟐 + 𝟏 𝟐 𝝆 𝒗 𝟐 𝟐
𝑷+𝝆𝒈𝒉+ 𝟏 𝟐 𝝆 𝒗 𝟐 =𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑷 𝟏 +𝝆 𝒈𝒉 𝟏 + 𝟏 𝟐 𝝆 𝒗 𝟏 𝟐 = 𝑷 𝟐 +𝝆𝒈 𝒉 𝟐 + 𝟏 𝟐 𝝆 𝒗 𝟐 𝟐
𝑷 𝟏 +𝟎+ 𝟏 𝟐 𝝆 𝒗 𝟏 𝟐 = 𝑷 𝟐 +𝟎+ 𝟏 𝟐 𝝆 𝒗 𝟐 𝟐
Pressure 1
Pressure 2
Velocity 1
Velocity 2
𝑷 𝟏 + 𝟏 𝟐 𝝆 𝒗 𝟏 𝟐 = 𝑷 𝟐 + 𝟏 𝟐 𝝆 𝒗 𝟐 𝟐
Pressure 1
Pressure 2
Velocity 1
Velocity 2
𝑷 𝟏 − 𝑷 𝟐 = 𝟏 𝟐 𝝆 𝒗 𝟐 𝟐 − 𝟏 𝟐 𝝆 𝒗 𝟏 𝟐
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