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2 years ago

10

Your roommate wants to learn about organizational chart. You can tell her that these charts can tell you all of these about an o

rganization EXCEPT their

1 answer:

mars1129 [50]2 years ago

4 0

Answer:

Control mechanisms

Explanation:

Organizational chart of any company will give details of different aspects of the company such as the major sub-units of the organization with the names of team leaders for different sub-units, it can also give you the span of control and the division of work within the company. However, the chart can't show you control mechanisms of different departments.

You might be interested in

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

If water molecules pass through a membrane with a steady state flux of 220 mole/(m2 day), how long will it take, in hours, for 0

goblinko [34]

Answer:

<em>0.0386 hr</em>

<em></em>

Explanation:

Area = 565 cm^2 = 0.0565 m^2 (1 cm^2 = 0.0001 m^2)

flux state rate = 220 mole/m^2-day

<em>There are 24 hrs in a day,</em> therefore rate in hrs will be

220/24 = 9.17 mole/m^2-hr

mass of water = 0.4 kg

molar mass of water = (1 x 2) + 16 = 18 kg/mole

therefore,

<em>mole of water = mass of water/molar mass of water</em>

mole of water = 0.4/18 = 0.02 mole

<em>mole flux = mole/area</em> = 0.02/0.0565 = 0.354 mol/m^2

<em>time that will be taken will be for water to pass = mole flux/mole flux rate</em>

time = 0.354/9.17 = <em>0.0386 hr</em>

7 0

2 years ago

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0

2 years ago

Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment

aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

}

}

Explanation:

7 0

2 years ago

1. Consider the steady flow in a water pipe joint shown in the diagram. The areas are:

marshall27 [118]

Answer:

-4.5 m/s

Explanation:

Assuming steady and incompressible flow and uniform properties at each section

$V_1A_1+V_2A_2+A_3V_3+Q_4=0$

Here V is velocity of flow and A is area, Q is flow rate out of the leak, subscript 1-4 represent different sections

At the surface, is negative hence the equation above will be

$-V_1A_1+V_2A_2+A_3V_3+Q_4=0$

Making the subject of the formula then

$V_2=\frac {V_1A_1-A_3V_3-Q_4}{A_2}$

Substituting the given values then

$V_2=\frac {(5\times 0.2)-(12\times 0.15)-0.1}{0.2}=-4.5 m/s\\V_2=-4.5m/s$

8 0

2 years ago