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Likurg_2 [28]
2 years ago
14

A hypodermic syringe is used to apply a vaccine. If the plunger is moved forward at the steady rate of 20 mm/s and if vaccine le

aks past the plunger at 0.1 of the volume flowrate out the needle opening, calculate the average velocity of the needle exit flow. The inside diameters of the syringe and the needle are 20 mm and 0.7 mm.
Engineering
1 answer:
svlad2 [7]2 years ago
3 0

Answer:

The average velocity of the needle exit flow is 163.27 m/s

Explanation:

Given;

inside diameter of the syringe, d₁ = 20 mm

inside diameter of the needle, d₂ = 0.7 mm

steady flow rate of the plunger, v = 20 mm/s

Apply continuity equation;

Q_{in} = Q_{out}

Considering 0.1 of the volume flowrate out the needle opening, we will have the following equations.

0.1 *Q_{out \ of \ needle} = Q_{in\ syringe}\\\\0.1(A_2v_2) = A_1v_1\\\\v_2 = \frac{A_1v_1}{0.1A_2}\\\\ v_2 = (\frac{A_1}{A_2} )(\frac{v_1}{0.1})\\\\  v_2 = (\frac{d_1^2}{d_2^2} )(\frac{v_1}{0.1})\\\\ v_2 = (\frac{d_1}{d_2} )^2(\frac{v_1}{0.1})\\\\v_2 =  (\frac{20}{0.7} )^2(\frac{20*10^{-3}}{0.1})\\\\v_2 = 163.27 \ m/s

Therefore, the average velocity of the needle exit flow is 163.27 m/s

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An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs
creativ13 [48]

Answer:

ΔKE=-347.278 kJ

ΔPE= 441.45 kJ

Explanation:

given:

mass m=900 kg

the gravitational acceleration g=9.81 m/s^2

the initial velocity V_{1}=100 km/h-->100*10^3/3600=27.78 m/s

height above the highway h=50 m

h1=0m

the final velocity V_{F}=0 m/s

<u>To find:</u>

the change in kinetic energy ΔKE

the change in potential energy ΔPE

<u>assumption:</u>

We take the highway as a datum

<u>solution:</u>

ΔKE=5*m*(V_{F}^2-V_{1}^2)

      =-347.278 kJ

ΔPE=m*g*(h-h1)

      = 441.45 kJ

5 0
2 years ago
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

3 0
2 years ago
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
yanalaym [24]

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
2 years ago
At his review last year, Lucas was promised a 20 percent raise if he met his production goals. Raises were included in today’s p
Murrr4er [49]

Answer: Instrumentality;low

Explanation: Instrumentality is the impact a person have or will be able to render to a given activity or his or her job. The Instrumentality of a person has been found to be proportional to the what outcome of the person's efforts. Especially if the person's explanations are meant.

When a person's expectations are not meant it will cause the person's Instrumentality to be low.

4 0
2 years ago
To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Kno
azamat

Answer:

a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

Explanation:

The stack of plywood has a certain mass. The weight will depend on that mass.

w = m * g

There will be a normal reaction between the stack and the bed of the truck, this will be:

nr = -m * g * cos(θ)

Being θ the tilt angle of the bed.

The static friction force will be:

ffs = - m * g * cos(θ) * fJK

The dynamic friction force will be:

ffd = - m * g * cos(θ) * fik

These forces would produce accelerations

affs = -g * cos(θ) * fJK

affd = -g * cos(θ) * fik

affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)

affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)

These accelerations oppose to movement and must be overcome by another acceleration to move the stack.

The acceleration of the truck is horizontal, the horizontal component of these friction forces is:

affs = -3.92 * cos(θ)^2

affd = -2.94 * cos(θ)^2

The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

Assuming the bed has a lenght L.

The horizontal movement will be over a distance cos(θ) * L because L is tilted.

Movement under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.

If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.

L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2

2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)

a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

6 0
2 years ago
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