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Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi

PolarNik [594]

Answer:

a) P₂ = 3219.11 lbf / ft² , b) P₂ = 721.91 lbf / ft² , c) P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

The area of a circle is

A = π r² = π/4 d²

v₂ = v₁ (d₁ / d₂)²

Let's calculate

v₂ = 12 (7/3)²

v₂ = 65 feet / s

Now let's use Bernoulli's equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

y₁ = y₂

P₁ - P₂ = ½ ρ (v₂² –v₁²)

P₂ = P₁ - ½ ρ (v₂² –v₁²)

ρ = 62.43 lbf / ft³

P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

P₂ = 7200 - 3980.89

P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

y₂ –y₁ = 40 ft

P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

P₂ = 7200 - 2497.2 - 3980.89

P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

y₂ –y₁ = -40

P₂ = 7200 + 2497.2 - 3980.89

P₂ = 5707.31 lbf / ft²

3 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

Decreasing unnecessary and wasteful purchases, decreasing the volume of waste, and decreasing

Feliz [49]

The answer is A
Hope this helps 8)

6 0

3 years ago

Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r

irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000 × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0

3 years ago

Calculate the unit length of a roof using common rafters if the unit run is 7 inches. Round your answer to two decimal places.

elena-14-01-66 [18.8K]

Answer:

Explanation:

https://www.pole-barn.info/roof-rafter-calculations.html

4 0

3 years ago

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