Which of these is shown in the animation below?

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

notsponge [240]

<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 />

<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.

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

8 0

2 years ago

A sphere is assumed to have the properties of water and has an initial heat generation 46480 W/m^3 How much should the heat gene

Verdich [7]

Answer:

Resulting heat generation, Q = 77.638 kcal/h

Given:

Initial heat generation of the sphere, $Q_{Gi} = 46480 W/m^{3}$

Maximum temperature, $T_{m} = 360 K$

Radius of the sphere, r = 0.1 m

Ambient air temperature, $T = 25^{\circ}C$ = 298 K

Solution:

Now, maximum heat generation, $Q_{m}$ is given by:

$T_{m} = \frac{Q_{m}r^{2}}{6K} + T$ (1)

where

K = Thermal conductivity of water at $T_{m} = 360 K = 0.67 W/m^{\circ}C$

Now, using eqn (1):

$360 = \frac{Q_{m}\times 0.1^{2}}{6\times 0.67} + 298$

$Q_{m} = 24924 W/m^{3}$

max. heat generation at maintained max. temperature of 360 K is 24924 $W/m^{3}$

For excess heat generation, Q:

$Q = (Q_{Gi} - Q_{m})\times volume of sphere, V$

where

$V = \frac{4}{3}\pi r^{3}$

$Q = (46480 - 24924)\times \frac{4}{3}\pi\0.1^{3} = 21556\times \frac{4}{3}\pi\0.1^{3} W/m^{3}$

$Q = 90.294 W$

Now, 1 kcal/h = 1.163 W

Therefore,

$Q = \frac{90.294}{1.163} = 77.638 kcal/h$

3 0

3 years ago

(1) Prompt the user for the number of cups of lemon juice, water, and agave nectar needed to make lemonade. Prompt the user to s

Lady bird [3.3K]

Answer:

The program to this question as follows:

Program:

Lemon= float(input('Enter lemon juice value in cups: ')) #defining float variable and input value by user

Water= float(input('Enter water value in cups: ')) #defining float variable and input value by user

Agave= float(input('Enter agave nectar value in cups: ')) #defining float variable and input value by user

Serve= float(input('enter serving value: ')) #defining float variable and input value by user

print ('Lemonade ingredients - yields',Serve,'servings',) #print value

print (Lemon,'cups in lemon juice',) #print value

print (Water,'cups in water',) #print value

print (Agave,'cups in agave nectar',) #print value

Output:

Enter lemon juice value in cups: 2

Enter water value in cups: 3

Enter agave nectar value in cups: 2

enter serving value: 2

Lemonade ingredients - yields 2.0 servings

2.0 cups in lemon juice

3.0 cups in water

2.0 cups in agave nectar

Explanation:

In the above python code, four float variable "Lemon, Water, Agave, and Serve" is defined, which is the input function is used.

The input function message is written that passes in function parameter, that accepts value in the above variables.
In the next step, the print function is used, which prints the above user input values.

6 0

3 years ago

A layer of viscous fluid of constant thickness (no velocity perpenducilar to plate) flows steadily down an infinite, inclined pl

mixer [17]

Answer:

q = (ρg/μ)(sin θ)(h³/3)

Explanation:

I've attached an image of a figure showing the coordinate system.

In this system: the velocity components v and w are equal to zero.

From continuity equation, we know that δu/δx = 0

Now,from the x-component of the navier stokes equation, we have;

-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)

Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0

Then our eq 1 is now;

ρg(sin θ) + μ(δ²u/δy²) = 0

μ(δ²u/δy²) = -ρg(sin θ)

Divide both sides by μ to get;

(δ²u/δy²) = -(ρg/μ)(sin θ)

Integrating both sides gives;

δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)

Now, the shear stress is given by the formula;

τ_yx = μ[δu/δy + δv/δx]

From the diagram, at the free surface,τ_yx = 0 and y = h

This means that δu/δy = 0

Thus, putting 0 for δu/δy in eq 2, we have;

0 = -(ρg/μ)(sin θ)h + b1

b1 = h(ρg/μ)(sin θ)

So, eq 2 is now;

δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)

Integrating both sides gives;

u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3

Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.

Thus, we now have:

u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y

Factorizing like terms, we have;

u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)

The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;

∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]

q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0

q = (ρg/μ)(sin θ)[h³/2 - h³/6]

q = (ρg/μ)(sin θ)(h³/3)

7 0

3 years ago

If the speed of a particle is decreased by a factor of three, by what factor does its kinetic energy change?

lidiya [134]

Answer:

decreases by 1/9

Explanation:

the equation of kinetic energy is $KE = \frac{mv^2}{2}\\$

here, we see that Kinetic Energy and v (velocity) are on opposide sides of the equation, this means that they are directly proportional (when one increases, the other one must increase (assuming mass (m) is constant)): KE∝ $v^2$

since V is decreasing by factor of 1/3, simply plug it in for v^2 to get (1/3)^2,

this equals (1/9). So KE will become 1/9 of original value

6 0

2 years ago