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kompoz [17]
2 years ago
11

The pressure drop across a valve through which air flows is expected to be 10 kPa. If this differential were applied to the two

legs of a U-tube manometer filled with mercury, estimate the manometer deflection. What is the deflection if a 30-degree inclined tube manometer were used instead? SHq - 13.6
Engineering
1 answer:
ozzi2 years ago
6 0

Answer:hit that soulja boy

Explanation:

You might be interested in
Suppose we store a relation R (x,y) in a grid file. Both attributes have a range of values from 0 to 1000. The partitions of thi
leva [86]

Answer:

For (a) The total number of buckets from the given query for the relation is 25 buckets (b) the nearest neighboring query is (80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

Explanation:

From the question stated, we need to define what a Grid file is

Grid File it is a structure of data that are used to divide the total space into a grid non-periodic, where set of point (small) are defined by more than one cells of the grid.

(a)Finding buckets for the query

The relation is divided into two parts which ranges from 0 to 1000, the first part is partitioned in every 20 units, at 20, 40, 60 etc; a second part is partitioned into every 50 units at 50, 100, 150 etc.

The total number of buckets from the given query for the relation is 25 buckets

(b)Finding the closest point or nearest point

The closest point discovered in the distance is little above 15

These points are are the points closer to the point target (110, 205) which can be found in five neighboring rectangles with left corners lower is stated as follows:

(80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

3 0
2 years ago
Derive an equation for the work of a mechanically reversible, isothermal compression of 1 mol of a gas from an initial pressure
Lyrx [107]

Answer:

The long derivation for work of a mechanically reversible, isothermal compression was done with detailed steps as shown in the attachment.

Explanation:

what is applied here is a long derivation from Work done in an isobaric process, the expression for the compressibility factor (Z) and the equation of state that was given. The requisite knowledge of Differentiation and Integration was used.

The detailed derivation from firs principle is as shown in the attachment.

5 0
2 years ago
When should you exercise extreme caution around power lines?
Elis [28]

<em>You should take note and exercise extreme precautions when you are near power lines and consider the following: </em>

<em> </em>

<em>1. Make sure that you have a good distance away from the lines. The minimum distance you can get is 10 feet away from the lines. Be cautious as well when you see broken lines as they could still harm you and electrified you. </em>

<em>2. Do not make ladders, equipments and things around you touch the power lines as it may harm you as well. </em>

<em>3. Clear everything and ensure that no things are near you before you lift your hands and other tools.</em>

6 0
2 years ago
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

3 0
2 years ago
The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use
Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

q = \frac{k * A * (th - tc)}{d}

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

\frac{d * q}{k * A} = th - tc

th = \frac{d * q}{k * A} + tc

The surface area is:

A = \frac{\pi * d^2}{4}

A = \frac{\pi * 0.22^2}{4} = 0.038 m^2

If the pan is aluminum:

th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C

If the pan is copper:

th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C

7 0
2 years ago
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