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Kay [80]
3 years ago
9

A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradu

al contraction to a width of 4.5 m.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions made.

Engineering
1 answer:
Trava [24]3 years ago
7 0

Answer:

Depth in the contracted section = 2.896m

Velocity in the contracted section = 2.072m/s

Explanation:

Please see that attachment for the solving.

Assumptions:

1. Negligible head losses

2. Horizontal channel bottom

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MrRa [10]

Answer:

  C.  Welded contacts on the thermostat

Explanation:

Any fault that keeps the heating element heating when it should not is a fault that will cause the symptom described. The details <em>depend on the design of the brewer</em> (not given).

"A short at the terminals" depends on what terminals are being referenced. The device on-off switch terminals are normally connected together when the brewer is turned on, so a short there may not be observable.

"Welded contacts on the thermostat" will have the observed effect if the thermostat is the primary means of ending the brewing cycle. If the thermostat of interest is an overheat protective device not normally involved in ending the brewing cycle, then that fault may not cause the observed symptom.

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If the heating element is open-circuit, no heating will occur. A gasket leak may cause a puddle, but may have nothing to do with the end of the brewing cycle. (Loss of water can be expected to end boiling, rather than prolong it.)

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The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 2 mm thick at a rate of 15
KATRIN_1 [288]

Answer:

attached below

Explanation:

7 0
3 years ago
A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob
scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

                                             $ \sim 72 \ kN/m^2 $

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 = $ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

                      = 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                        = 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                         = 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

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8 0
2 years ago
An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).
Phoenix [80]

Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

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G_ A = 2.631

W_ B = 25%

G_B = 2.331

W_C = 20%

G_C = 2.609

Now,

On applying the formula, we get

⇒ G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}

On substituting the values, we get

⇒          =\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          = \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          =2.545

8 0
3 years ago
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