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2 years ago

Who has good grades in school like A's B's dm me

Engineering

2 answers:

serious [3.7K]2 years ago

8 0

Why do you need help with school work

siniylev [52]2 years ago

4 0

Answer:

I do

Explanation:

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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

A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto

vazorg [7]

Answer:

$(M_t)_{rated}=61.11lb-in$

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = $4 \times 0.7457$ =2.9828 KW

service factor(k)= 2.75

now,

$KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }$