PHYSICS COLLEGE

A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

Answers

Answer 1

Answer:

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3

D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3

D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0)=88 BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1 + m2 = m3

0.376m1 + 1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for 2" Sch 40=2.067in=0.17225ft

$A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2$

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6.use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

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Answers

Answer:

Explanation:

E= −L ΔI / Δt.

L = E Δt / ΔI

Hence the unit of inductance may be V s A⁻¹

or volt s per ampere .

In the given case

change in current ΔI = - 2.5 A

change in time = .015 s

L = .56 H

E = − L ΔI / Δt.

= .56 x 2.5 / .015

= 93.33 V .

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B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.

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Answers

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

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and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

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I hope this is usefull for you

regards

How does the mass of an object affect its motion through the air?

Answers

The motion of an object through the air does not affect by its mass. The rate of fall of objects does not depend upon the mass.

What are free fall and air resistance?

Free fall is a motion of a body in which gravity is the only force acting upon it. An object moving upwards might not be considered to be falling. But if the object is under the effect of the force of gravity, it is said to be in free fall.

Free fall is a type of motion in which the force acting upon an object is only gravity. Objects are not encountering a significant force of airresistance as they are only falling under the sole influence of gravity. All objects under such conditions will fall with the same rate of acceleration, regardless of their masses.

As an object falls through the air, have gone through some degree of air resistance. Air resistance is the collisions of the object's leading surface with molecules present in the air. The two most common factors that have a direct effect on the amount of air resistance are the cross-sectional area of the object and the speed of the object.

Learn more about free-fall motion, here:

brainly.com/question/13297394

#SPJ2

Mass does not affect the speed of falling objects assuming there is only gravity acting on it. For example Both Bullets will strike the ground at the same time. The horizontal force applied does not affect the downward motion of the bullets — only gravity and friction (air resistance), which is the same for both bullets
(Sorry if u don’t get what I mean)

Which of the following statements is true about the variation of pressure in function of the depth? O Pressure decreases exponentially with the depth O Pressure increases exponentially with the depth O Pressure decreases linearly with the depth o Pressure increases linearly with the depth O None of the above

Answers

Answer:

The answer is: Pressure increases linearly with the depth

Explanation:

In this case, the definition of pressure is:

$P = (F)/(A)$

where F = mg is the weight of the fluid over the body, and A is the area of the surface to which the force is exerted. If we consider $\rho = m/V$ , then

$P = (mg)/(A) = (\rho Vg)/(A)$ .

Volume can be expressed as V = A*h, where A is the cross section of the column of the fluid over the body and h is the height of the column, in other words, the depth.

$P = (A\rho gh)/(A)= \rho g h$ ,

which means that pressure increases linearly with the depth in a factor of $\rho g$ .

A frictionless pendulum is made with a bob of mass 12.6 kg. The bob is held at height = 0.650 meter above the bottom of its trajectory, and then pushedforward with an initial speed of 4.22 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?

Answers

The answer to your question is 55

A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.

Answers

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

$\Sigma F = F - m\cdot g = m\cdot a$ (1)