Sd282 Asg2

FLUID MECHANICS I Solution 2 Question 1: Example 2.3 The solution is given in the textbook p74.

Question 2: Example 2.4 The solution is given in the textbook p76.

Question 3: Problem P2.31 Assumming that the fluids in the tubing between pipes A and B are at rest: p1 − pA p2 − p1 p3 − p2 p4 − p3 pB − p4

= = = = =

ρbenzene g ∗ (hA − 0) ρHg g ∗ (0 − h2 ) ρkerosene g ∗ (h2 − h3 ) ρH2 O g ∗ (h3 − h4 ) ρair g ∗ (h4 − hB ) ≈ 0

where the vertical variation of pressure in at air is neglected. Adding the set of pressure differences for each leg together gives: pB −pA = ρbenzene g∗(hA −0)+ρHg g∗(0−h2 )+ρkerosene g∗(h2 −h3 )+ρH2 O g∗(h3 −h4 ) From Table A.3, ρbenzene = 881[kg/m3 ], ρHg = 13550[kg/m3 ], ρkerosene = 804[kg/m3 ], and ρH2 O = 998[kg/m3 ]. From the problem specification, hA = 0.20[m], h2 = 0.08[m], h3 = 0.4[m], and h4 = 0.14[m]. Final Answer: pB − pA = −8881[P a]. This pressure difference is approximately 9% of standard atmospheric pressure. Notice the dominant role played by the heavy fluid, mercury.

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Figure 1: Sketch of pipe layout.

Question 4: Problem P2.32 Start from the hydrostatic relation dp = −ρ gz. See sketch (Figure 2) p1 − pA = −ρw g(z1 − zA ) = −ρw g ∗ H p2 − p1 = −ρoil g(z2 − z1 ) = −ρoil g ∗ (0.18) pB − p2 = −ρHg g(zB − z2 ) = +ρHg g ∗ (0.35 + H + 0.18) Substitute to get pB − pA and then H. pB − pA = ρHg g ∗ (0.35 + 0.18) − ρoil g ∗ 0.18 + (ρHg − ρw ) g ∗ H Use Table A.3: ρHg = 13550 kg.m−3 , ρw = 998 kg.m−3 and ρoil = 0.827∗998 = 825.346 kg.m−3 . H = 22.6 cm 2

z

z2 = (35 + H + 18) cm

2 Meriam red oil z1 = (35 + H) cm

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water A

Mercury

zA = +35 cm

B

zB = 0 cm

datum

Figure 2: Sketch of the manometer.

Question 5: Problem P2.35 -see Figure 3 Assuming that the fluids in the manometer are at rest and that the fluid in the pipe is not accelerating in the vertical direction: p3 − p1 p4 − p3 p2 − p4

= = =

ρH2 O g ∗ (h1 − 0) ρHg g ∗ (h3 − h4 ) = −ρHg gh ρH2 O g ∗ (h4 − h2 )

Adding the above gives: p2 − p1 = ρH2 O g ∗ (h1 + h4 − h2 ) − ρHg gh From geometry h2 = h1 + Ltan θ so the above reduces to: p2 − p1 = −ρH2 O gLtan θ + (ρH2 O − ρHg )gh From Table A.3, ρHg = 13550[kg/m3 ] and ρH2 O = 998[kg/m3 ]. From the problem specification, h = 0.12[m], L = 2.0[m], and θ = 30◦ . Final Answer: p2 − p1 = −26100[P a]

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Figure 3: Sketch of pipe-manometer layout.

Question 6: Problem P2.36 pA = patm + ρw g ∗ 0.5 + ρoil g ∗ 0.5 p3 − pA = −ρw g h = −ρw g Lsinθ patm + SGoil ∗ ρw ∗ g ∗ 0.5 + ρw g ∗ 0.50 = patm + ρw gLsinθ sinθ =

ρw ∗ 0.5 + ρoil ∗ 0.5 = 0.4225 ρw ∗ L θ ≈ 25◦

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patm

oil

50 cm

patm

3 L 50 cm

water

θ

A Figure 4: Sketch of the inclined manometer.

Question 7: P2.34 In the original configuration: pB − pA

= ρH2 O g(h + L) − ρoil g(h + H) = 0

In the displaced configuration: pB − pA

= ρH2 O g(h − ∆h + L − ∆L) − ρoil g(h − ∆h + H + ∆H)

From the original configuration this can be reduced to: pB − pA

=

−ρH2 O g(∆h + ∆L) − ρoil g(∆H − ∆h)

If the oil level in the tube rises by ∆h, the oil in the reservoir must rise by d2 ∆H = ∆h D 2 since the volume of oil that leaves the tube must end up in the d2 reservoir. Similarly, the water level in the reservoir must sink by ∆L = ∆h D 2. Substituting these geometric relationships gives: pB − pA

= (ρoil − ρH2 O )g∆h − (ρoil + ρH2 O )g

d2 ∆h D2

From Table A.3, ρoil = 891[kg/m3 ] (this is for SAE 30W oil) and ρH2 O = 998[kg/m3 ]. Final Answer: If d/D ⇒ 0 then pB − pA = −1050∆h[P a] and if d = 0.15D then pB − pA = −1470∆h[P a].In the latter case, there is a 29% error in neglecting the elevation change in the reservoirs.

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Figure 5: Original and displaced layouts of the oil-water manometer system.

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