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3 years ago

Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar.

Engineering

1 answer:

rodikova [14]3 years ago

6 0

This question is incomplete, the complete question is;

Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar.

The 125-mm-diameter smooth pipe is transporting SAE 10W-30 oil with ρ=920 kg/m3 and µ=0.2 N.s/m2 .

Answer: the greatest pressure drop allowed is 14247 Pascals

Explanation:

LAMINAR FLOW INF PIPE

greatest pressure drop

(P1 - P2) = pressure drop

(P1 - P2) = 32uvl / D²

so (P1 - P2) ∝ V

greatest pressure drop is only at very high velocity

now Re (Reynold Number = svd / u

Re ∝ V

so velocity is high when Re is high

Re = 2000 ( Highest for LAMINAR)

Re = 2000 = svd / u

now in the question we were given that

s = 920 kg/m³, u = 0.2 NS/m², d = 125mm = 0.125m, l = 10m

so we substitute;

2000 = (920 × v × 0.125) / 0.2

v = 3.48 m/s

now pressure drop = (P1 - P2) = 32UVL / d²

we substitute

(P1 - P2) = (32(0.2) (3.48)(10)) / (0.125)²

(P1 - P2) = 14247 Pascals

therefore the greatest pressure drop allowed is 14247 Pascals

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4 years ago

A helicopter moves horizontally in the x direction at a speed of 120 mi/h. Knowing that the main blades rotate clockwise when vi

devlian [24]

Answer:

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4 0

3 years ago

A water tank is completely filled with liquid waterat 20°C.The tank material is such that it can withstand tensioncaused by a vo

Xelga [282]

Answer:

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Explanation:

From literature, the coefficient of volume expansion of water between 20°C and 50°C = β = (0.377 × 10⁻³) K⁻¹

Volume expansivity is given by

ΔV = V β ΔT

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ΔT = Change in temperature = ?

It is given in the question that maximum volume increase the tank can withstand is

(ΔV/V) × 100% = 0.8%

(ΔV/V) = 0.008

V β ΔT = ΔV

β ΔT = (ΔV/V)

β ΔT = 0.008

ΔT = (0.008/β)

ΔT = (0.008/0.000377)

ΔT = 21.22°C

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Hope this Helps

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4 years ago

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton

SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

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We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

$F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}$

Applying value of the constant we get

$F_{1}=\frac{62.208}{0.7^{2}}$

Thus $F_{2}=126.955Newtons$

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3 years ago