(,,)=^3−^3+^3, where is the sphere ^2 + ^2 + ^2=^

The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbac

bazaltina [42]

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

$\bold{Area =B \times D_f}$

$=6\times 5\\\\=30 \ ft^2$

In point b, Calculating the wetter perimeter.

$\bold{P_w =B+2\times D_f}$

$= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft$

In point c, Calculating the hydraulic radius:

$\bold{R=\frac{A}{P_w}}$

$=\frac{30}{16}\\\\= 1.875 \ ft$

In point d, Calculating the value of Reynolds's number.

$\bold{Re =\frac{4VR}{v}}$

$=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\$

$=750,000 V$

Calculating the velocity:

$V= \sqrt{\frac{8gRS}{f}}$

$= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\$

$\sqrt{f}=\frac{3.108}{V}\\\\$

calculating the Cole-brook-White value:

$\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\$

$\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})$

After calculating the value of V it will give:

$V= 25.18 \ \frac{ft}{s^2}\\$

In point a, Calculating the value of Froude:

$F= \frac{V}{\sqrt{gD}}$

$= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\$

$= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98$

The flow is supercritical because the amount of Froude is greater than 1.

Calculating the channel flow rate.

$Q= AV$

$=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\$

4 0

3 years ago

Find the dryness fraction, specific volume and internal energy of steam at 7bar nd enthalpy 2600kj/kh. (0.921,0.2515m³/kg , 2420

Blizzard [7]

Answer:

hdwhifniuewohfoyyhyeiudhwbwuxvgusvfgcvxg

Explanation:

7 0

2 years ago

The wet density of a sand was found to be 1.9 Mg/m3 and the field water content was 10%. In the laboratory, the density of solid

Nookie1986 [14]

Answer:

a) 44.4%

b) 72 mm

Explanation:

See attached pictures.

3 0

3 years ago

A manufacturer makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't

aleksklad [387]

Answer:

probability P = 0.32

Explanation:

this is incomplete question

i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?

solution

we know that Standard error is expess as

Standard error = $\frac{sd}{\sqrt{n}}$

Standard error = $\frac{12}{\sqrt{16}}$

Standard error = 3

so here we get Z value for 3 units away are from mean are

mean = -1 and + 1

so here

probability P will be

probability P = P( z < -1 or z > 1)

probability P = 0.1587 + 0.1587

probability P = 0.3174

probability P = 0.32

7 0

3 years ago

A complete mix of an activated sludge system without primary clarification is used for treatment of municipal wastewater with a

Hitman42 [59]

Answer:

sorry di ko alam

Explanation:

4 0

2 years ago