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Law Incorporation [45]
2 years ago
12

120 litres of water is discharge from container in 25 seconds. Find the rate of discharge in cumecs.if the discharge took place

from pipe of diameter 50 mm. find the velocity flow​
Engineering
1 answer:
notsponge [240]2 years ago
8 0
<h2>Answer:</h2>

Rate of discharge in cumecs: <u>0.0048m³/s</u>.

Velocity flow: <u>24m/s</u>.

<h2>Explanation:</h2>

<h3>1. Find the rate of discharge in cumecs.</h3>

a. Convert from litres to m³.
120L*1000= 120000mL

120000mL=120000cm³

120000cm³/100³=0.12 m³.

b. Rate of discharge.

<em>If  0.12 m³ where discharged in 25 seconds, the rate of discharge is:</em>

0.12m³/25s = 0.0048m³/s.

<em />

<em />

<h3>2. Find the velocity flow.</h3>

Let's refer to the fluid mechanics equation that relates volume flow, area and velocity. This is the formula:

\frac{dV}{dt}=Av; where the expression \frac{dV}{dt} is the volume flow rate (in m³/s); A is the cross-sectional area of the pipe (in m²), and v is the velocity flow (in m/s).

a. Solve the equation for v.

\frac{dV}{dt}=Av\\ \\(\frac{dV}{dt})/A=v\\ \\v=(\frac{dV}{dt})/A

b. Calculate the cross-sectional area of the pipe.

<em>The cross-sectional area of the pipe is a circle. Hence, the formula of this area is:</em>

A=\pi r^{2}

<em>We'll have to convert the diameter to meters, because the formula for flow velocity needs the area in m². Let's go ahead and do that.</em>

<em />50mm/1000=0.05m.

<em>We were given the diameter, and the formula uses the radius, but the radius is just half of the diameter, therefore, we can substitute in toe formula like this:</em>

A=\pi (\frac{0.05}{2} )^{2}=0.0020m^{2}

c. Substitute in the new expression for velocity flow and calculate.

v=(\frac{dV}{dt})/A\\ \\v=(\frac{0.048m^{3} }{1s})/(0.0020m^{2} )\\\\ v= 24m/s

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Answer:

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Explanation:

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The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

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Now we can assume that the input and output areas are equal

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\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

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3 years ago
Which of these fuel injection systems operates with fuel injectors located only in the intake manifold near each intake
Stella [2.4K]

Answer:

C. Multipoint fuel injection

Explanation:

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Explanation:

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A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)
Leni [432]

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

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To calculate for the weight of octane(W):

W=g*m

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b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

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Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

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