A pipe inclined at 45o to be horizontal converges over a length l of 22 m from a diameter d1=200 mm to a diameter d1=100 mm Oil of relative density 0.9 flows through the pipe at a mean velocity v1 = 2m/s at the lower end of pipe. A mercury manometer is connected across this length as shown in the diagram. The relative density of mercury is 13.6 and limbs of manometer are filled with oil.

πd124⋅v1=πd224v2 224⋅x2=124⋅v2v2=8m/sP1+12ρ(2)2=P2+12ρ(8)2+ρg(2)P1−P2=12ρ(8)2−12ρ(2)2+ρg(2)=ρ[32−2+20]=0.9×103×50=45kPafor the manometer:P1+ρ0gz1=P2+ρ0gz1+2−h+ρHggh45×103=ρ0g(2−h)+ρHggh45×103=0.9×103×10×2+hρHg−ρ0g27×103=h(13.6−0.9)×103×10h=2712.7×10

A pipe inclined at 45o to be horizontal converges over a length l of 22 m from a diameter d1=200 mm to a diameter d1=100 mm Oil of relative density 0.9 flows through the pipe at a mean velocity v1 = 2m/s at the lower end of pipe. A mercury manometer is connected across this length as shown in the diagram. The relative density of mercury is 13.6 and limbs of manometer are filled with oil.

πd124⋅v1=πd224v2 224⋅x2=124⋅v2v2=8m/sP1+12ρ(2)2=P2+12ρ(8)2+ρg(2)P1−P2=12ρ(8)2−12ρ(2)2+ρg(2)=ρ[32−2+20]=0.9×103×50=45kPafor the manometer:P1+ρ0gz1=P2+ρ0gz1+2−h+ρHggh45×103=ρ0g(2−h)+ρHggh45×103=0.9×103×10×2+hρHg−ρ0g27×103=h(13.6−0.9)×103×10h=2712.7×10

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A pipe inclined at 45^o to be horizontal converges over a length l of $2 \sqrt{2} m$ from a diameter $d_{1} = 200 mm$ to a diameter $d_{1} = 100 mm$ Oil of relative density 0.9 flows through the pipe at a mean velocity v₁ = 2m/s at the lower end of pipe. A mercury manometer is connected across this length as shown in the diagram. The relative density of mercury is 13.6 and limbs of manometer are filled with oil.

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The reading of manometer is approximately h = 0.3m.

The pressure difference across the $2 \sqrt{2}$ length is 40 kPa.

The speed with which oil flows out from the pipe is 8 m/s.

The reading of manometer is approximately h = 0.21 m.

answer is A, C.

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Detailed Solution

$\frac{{πd}_{1}^{2}}{4} \cdot v_{1} = \frac{{πd}_{2}^{2}}{4} v_{2} \begin{array}{l} \frac{2^{2}}{4} \cdot x^{2} = \frac{1^{2}}{4} \cdot v_{2} \\ v_{2} = 8 m / s \\ P_{1} + \frac{1}{2} ρ (2)^{2} = P_{2} + \frac{1}{2} ρ (8)^{2} + ρg (2) \\ (P_{1} - P_{2}) = \frac{1}{2} ρ (8)^{2} - \frac{1}{2} ρ (2)^{2} + ρg (2) \\ = ρ [32 - 2 + 20] \\ = (0 .9 \times 10^{3} \times 50) = 45 kPa \end{array}$
for the manometer:
$\begin{array}{l} P_{1} + ρ_{0} {gz}_{1} = P_{2} + ρ_{0} g (z_{1} + 2 - h) + ρ_{Hg} gh \\ 45 \times 10^{3} = ρ_{0} g (2 - h) + ρ_{Hg} gh \\ 45 \times 10^{3} = 0 .9 \times 10^{3} \times 10 \times 2 + h (ρ_{Hg} - ρ_{0}) g \\ 27 \times 10^{3} = h (13 .6 - 0 .9) \times 10^{3} \times 10 \\ h = \frac{27}{12 .7 \times 10} \end{array}$