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atroni [7]
3 years ago
9

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Express your answer in terms of quantities given in the problem introduction and g, the magnitude of the acceleration due to gravity.

Physics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

(a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

P_{1}=\rho g h_{1}

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

P_{1}=\rho gh_{1}.....(I)

Pressure for second pipe,

P_{2}=\rho gh_{2}.....(II)

From equation (I) and (II)

P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

Put the value of P₁ and P₂

\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)

2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2....(III)

We know that,

The continuity equation

v_{1}A_{1}=v_{2}A_{2}

v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})

Put the value of v₂ in equation (III)

2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2

2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2

Here, \dfrac{A_{1}}{A_{2}}=\gamma

So, 2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)

v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Hence, (a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
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Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

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distance the bucket is lifted = height = 11m

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Let W = work

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= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

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Force = 9.8 * 10 = 98N

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Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

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Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

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