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solniwko [45]
3 years ago
13

Consider a junction that connects three pipes A, B and C. What can we say about the mass flow rates in each pipe for steady flow

?

Engineering
1 answer:
Elis [28]3 years ago
6 0

Answer:

The statement regarding the mass rate of flow is mathematically represented as follows \Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}

Explanation:

A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure

As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.

The above statement can be mathematically written as

Mass_{Junction}=Constant\\\\\Rightarrow Mass_{in}=Mass_{out}

this is known as equation of conservation of mass / Equation of continuity.

Now we know that in a time 't' the volume that enter's the Junction 'O' is

1) From pipe 1 = V_{1}=Q_{1}\times t

1) From pipe 2 = V_{2}=Q_{2}\times t

Mass leaving the junction 'O' in the same time equals

From pipe 3 = V_{3}=Q_{3}\times t

From the basic relation of density, volume and mass we have

\rho =\frac{mass}{Volume}

Using the above relations in our basic equation of continuity we obtain

\rho \times V_{3}=\rho \times V_{1}+\rho \times V_{2}\\\\Q_{3}\times t=Q_{1}\times t+Q_{2}\times t\\\\\Rightarrow Q_{3}=Q_{1}+Q_{2}

Thus the mass flow rate equation becomes \Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}

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For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time
max2010maxim [7]

Answer with Explanation:

The general equation of simple harmonic motion is

x(t)=Asin(\omega t+\phi)

where,

A is the amplitude of motion

\omega is the angular frequency of the motion

\phi is known as initial phase

part 1)

Now by definition of velocity we have

v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )

part 2)

Now by definition of acceleration we have

a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )

part 3)

The angular frequency is related to Time period 'T' asT =\frac{2\pi }{\omega }

where

\omega is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is A\omega ^{2} and A\omega respectively.

4 0
3 years ago
Air at p=1 atm enters a long tube of length 2.5 m and diameter of 12 mm at an inlet temperature of Tm,i=100oC and mass flowrate
Annette [7]

Answer:

The heat transfer q = 18.32W

Explanation:

In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K

Please check attachment for complete solution and step by step explanation

6 0
3 years ago
A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi
raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation =\frac{N_1 +N_2}{2}

Nm = \frac{500+750}{2} = 625 rpm

w =\frac{2\pi Nm}{60}

  =65.44 rad/sec

F_T = mw^2 e

F_T = mew^2

       = 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio =\frac{300}{428.36} = 0.7

also

transmission ratio = \frac{1}{[\frac{w}{w_n}]^{2} -1}

0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}

SOLVING FOR Wn

Wn = 42 rad/sec

Wn = \sqrt {\frac{k}{m}

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0
3 years ago
Two flat plates, separated by a space of 4 mm, are moving relative to each other at a velocity of 5 m/sec. The space between the
xenn [34]

Answer:

0.008

Explanation:

From the question, the parameters given are:

Velocity V = 5 m/s

Pressure = 10 pa

But pressure = F/A

10 = F/A

F = 10A

Substitute all the parameters into the formula below

Coefficient of viscosity (η) = F × r /[AV]

Where

F = tangential force,

r = distance between layers,

A = Area, and

V = velocity

(η) = 10A × 0.004 /[A × 5]

The A will cancel out

(η) = 10 × 0.004 /[5]

(η) = 0.04 /5

(η) = 0.008

Therefore, the coefficient of viscosity of the fluid is 0.008

5 0
3 years ago
An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compress
pickupchik [31]

Answer:

<em>The temperature will be greater than 25°C</em>

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>

8 0
3 years ago
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