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tester [92]
4 years ago
6

The flow through a closed, circular sectioned pipe may be metered by measuring the speed of rotation of a propeller having its a

xis along the pipe central line. Using dimensional analysis, derive a set of non-dimensional parameters describing the relationship between the volume flow rate and relevant parameters of the system.
Engineering
1 answer:
spin [16.1K]4 years ago
7 0

Answer:

\frac{Q}{ND_{pi}^{3}}=(\frac{Vis}{pND_{pi}^{2}} )(\frac{D_{pr}}{D_{pi}} )

Explanation:

To solve this problem we have to make a dimensional analysis:

First, we have to write the variables involved and their dimensions:

1. Volume flow rate = Q

2. Speed of rotation= N

3. Density =ρ

4. Viscosity = Vis

5. Propeller diameter= D_{pr}

6. Pipe diameter= D_{pi}

Second, we have to write the fundamental dimensions:

Lenght = L

Mass= M

Time =T

Third, we must express the variable we want to know as a product of the other variables and to each variable we have to assign a respectic exponent:

Q=(N^{a})(p^{b})(Vis^{c})(D_{pi} ^{d})(D_{pr} ^{e})

We have to express the variables with the fundamental dimensions:

(L^{-3}T^{-1})=(T^{-1} )^{a}(ML^{-3} )^{b}(ML^{-1}T^{-1} )^{c}(L)^{d}(L)^{e}

Fourth, developing and agrupating the similar terms, we have:

L^{3}=L^{(-3b-c+d+e)}

T^{-1}=T^{(-a-c)}

0=M^{(b+c)}

From the previous equations we deduce:

a=1-c

b=-c

d=3-2c-e

Now, we have to substitute the found exponents into the first equation that we wrote:  

Q=(N^{a})(p^{b})(Vis^{c})(D_{pi} ^{d})(D_{pr} ^{e})

Q=(N^{1-c})(p^{-c})(Vis^{c})(D_{pi} ^{3-2c-e})(D_{pr} ^{e})

Developing and Agrupating the terms with the same exponent we get:

Q=(\frac{Vis}{pND_{pi}^2} )^c(ND_{pi}^3)(\frac{D_{pr}}{D_{pi}} )^e

Finally, the three non-dimensional group terms which describe the volume flow rate in terms of the relevant parameters of the system are:

\frac{Q}{(ND_{pi}^3)} =(\frac{Vis}{pND_{pi}^2} )^c(\frac{D_{pr}}{D_{pi}} )^e

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maxonik [38]

Answer:

\dot m_{2} = 0.199\,\frac{kg}{s}

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

1 + y = z

Then, the expression is simplified afterwards:

\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0

\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

h = 2994.3\,\frac{kJ}{kg}

State 2 (Saturated liquid)

h = 1008.3\,\frac{kJ}{kg}

State 3 (Liquid-Vapor mixture)

h = 2444.22\,\frac{kJ}{kg}

The ratio of the stream at state 2 to the stream at state 1 is:

y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}

y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }

y = 0.397

The mass flow rate of the saturated liquid is:

\dot m_{2} = y\cdot \dot m_{1}

\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )

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The acceleration of a particle as it moves along a straight line is given
BARSIC [14]

Answer:

V_t=6 = 32 m/s

Explanation:

The origin is at point 0 with the positive motion to the right  

The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:  

a=dv/dt

From here we can express dv as:

dv = a dt

Replace a by 2t — 1

dv = (2t — 1) dt

Integrate both sides of equation  

\int\limits^v_a  {2t-1} \, dv

v=t

a=t_0

putting these value in integral

<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>

We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values

v — 2 = (t^2 — t) — (0^2 — 0)

From here we can write the expression for v as:  

v_t=6=6^2-6+2                             (1)  

So the velocity at t = 6 s is:

v_t=6 = 32 m/s

V_t=6 = 32 m/s

In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero  

To do that, we'll just replace v by 0 in expression (1)

0 = t^2 — t + 2

The roots of the quadratic equation are:

t_1/2=1±  √(1^2-4*2*1)/2

Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right  

Now that we have all the details, we can correctly draw the path of the particle  

We can see from the sketch that the total distance traveled is:  

s^T=Δs_0-1

s^T=| s_1 - s_0 |

Replace s_0 by its value  

s^T=| s_1 - 1 |                                        (2)  

In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time  

Since we have already wrote expression for v as function of time (step 2), we'll use expression  to get the expression for s

v= ds/dt  

Multiply both sides of equation by dt

v dt = ds

Replace v by expression (1)

(t^2 — t + 2) dt = ds

Integrate both sides of equation  

\int\limits^t_b {x} \, dx

t=s

b=(s=0)

x=(t^2 — t + 2)

dx=ds

putting these value in integral

(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0

Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6  

(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1

4 0
4 years ago
For automobile engines, identify the best choice of motor oil.
Vlad1618 [11]

Answer:

C. Low viscosity for Maine in the winter and high viscosity for Florida in the summer.

Explanation:

The choice of motor oil or engine oil depends on the type of engine and also on the type of climate on which the engine runs.

In winter climates, we should use oil with low viscosity as at lower temperatures, the it flows more readily if its viscosity is low.

But the summer season needs a relatively thicker oil so as to reduce the thinning effect of the oil out of the engine.

Thus for Maine in the winter season, we need a low viscosity oil and for Florida a high viscosity oil in the summer season.

Therefore, option (c) is correct.

6 0
3 years ago
Design complementary static CMOS circuits with minimized number of transistors to realize the following Boolean functions (hint:
Pie

Answer:

as pull up network. the metteing point of pull down and pull up is the point where we take the output

note 1: if two n-mos are connected in series it gives logical AND and p-mos paralle gives logical-AND

note 2: if two n-mos are connected in parallel it gives logical OR and p-mos series gives logical-OR

note 3: output is always complement of what we implement

example Y= (AB)'

image attached

A) F = (ABC + D(A+B) )'

pulldown:

this can be realize by takeing three n-mos in series which gives ABC ,two n-mos are parallel which in series with another n-mos whic gives D(A+B), now connect ABC and D(A+B) in parallel

pull up

this can be realize by takeing three p-mos in parallel which gives ABC ,two p-mos are series which is in serires with

another p-mos whic gives D(A+B), now connect ABC and D(A+B) in series

the out put will be (ABC + D(A+B) )'

so we require total 6-mos and 6-pmos total 12mos transistors

B) F = AC + BD

pull down

this can be realize by takeing two n-mos in series which gives AB ,two n-mos are in series

which whic gives BD, now connect AC and BD in parallel

pull up

this can be realize by takeing two p-mos in parallel which gives Ac ,two p-mos are in parallel

which whic gives BD, now connect AC and BD in series

the output is (AC+BD)'

to avoid the complement we have to connect the output to c-mos inverter then we get AC+BD

so we require 5-nmos, 5-pmos total 10 mos transistors

8 0
3 years ago
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