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Eva8 [605]
3 years ago
5

The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

ear with depth what is the surface velocity of flow in the channel?
Engineering
1 answer:
IceJOKER [234]3 years ago
4 0

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

\dot{m}=\rho AV

Putting all the value to get the velocity of the flow

\frac{\dot{m}}{\rho A} = V

V = \frac{4000}{1000*4*2}

V = 0.5 m/s

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Answer:

Object-Oriented Software Engineering Using UML, Patterns, and Java, 3e, shows readers how to use both the principles of software engineering and the practices of various object-oriented tools, processes, and products.

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2 years ago
¿Cuándo se formarán cristales en una mezcla que se está evaporando?
Georgia [21]
Answer - La cristalización ye un procesu químicu pol cual a partir d'un gas, un líquidu o una disolución, los iones, átomos o molécules establecen enllaces hasta formar una rede cristalina, la unidá básica d'un cristal. La cristalización emplegar con bastante frecuencia en química para purificar una sustancia sólida.
5 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
3 years ago
When a retaining structure moves towards the soil backfill, the stress condition is called:__________.
Alecsey [184]

Answer:

(C) passive state.

Explanation:

The earth pressure is the pressure exerted by the soil on the shoring system. They are three types of earth pressure which are:

a) Rest state: In this state, the retaining wall is stationary, this makes the lateral stress to be zero.

b) Active state: In this state, the wall moves away from the back fill, this leads to an internal resistance. Hence the active earth pressure is less than earth pressure at rest

c) Passive state: In this state the wall is pushed towards the back fill, this leads to shearing resistance. Hence, the passive earth pressure is greater than earth pressure at rest

6 0
2 years ago
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