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Nataly [62]
3 years ago
11

Water enters a tank of diameter Dt steadily at a mass flow rate of m& . An orifice at the bottom with a diameter of Do allow

s water to escape. The orifice has a round entrance, so frictional losses are negligible. If the tank is initially empty.
a. Determine the maximum height that the water will reach in the tank.
b. Obtain a relation for water height as a function of time.

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

Answer:

Detailed solution for this tank system is given in the images attached below:

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A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp
cricket20 [7]

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

6 0
3 years ago
A video inspection snake is use
LekaFEV [45]

Answer:

very good thx

Explanation:

5 0
3 years ago
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8
marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0
3 years ago
Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm
Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0
3 years ago
For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____
7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>
  • A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

To learn more about it, refer

to brainly.com/question/25689052

#SPJ4

4 0
2 years ago
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