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jolli1 [7]
3 years ago
7

Write the heat equation for each of the following cases:

Engineering
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

where;

Q_g = heat generated per unit volume

\alpha = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

where;

The radial directional term = \dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) and the axial directional term is \dfrac{\partial^2 T}{\partial z^2 }

d) The heat equation for a wire going through a furnace is:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

since;

the steady-state is zero, Then:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}

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A team member who has been a good worker for many years has recently been doing poor work. You suspect that he may be tired of h
ohaa [14]

Answer:

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Explanation:

Be specific with facts in hand

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Consider the needs of your employees

Poor performance is not always the result of an employee's carelessness. There can be multiple genuine reasons for lack of performance and it can vary from person to person. The first is to understand the reason and judge whether they are genuine or not. Even if they aren't, don't let the other person know. Focus on your concerns and provide solutions accordingly. For example, if your employees cannot focus on their work due to some personal stress, make appointments for the counseling sessions and make sure they can get back on track.

Focus on feedback

Everyone handles comments differently. Although it is always recommended to be direct and clear in your communication, there may be certain strategies you can adopt to communicate your comments effectively. If your employee has difficulties in achieving his goals, work with him and provide him with all the necessary help to improve his performance. The best way is to provide weekly or monthly comments to your employees, so they know what they must do to achieve their goals.

Provide Performance Support Technology

When faced with existing employees who do not meet expectations, it is a smart decision to offer them training (and / or training) and different resources to help them improve. As an example, you can combine a low-performance worker with someone to act as a mentor or offer you a manual with the procedures to follow. In addition, there are performance improvement tools: WalkMe is a very valuable technology that can help a worker be more efficient and precise, within the workflow (and not take them away from their daily tasks).

Offer rewards and recognition

Whenever you see that your employees have poor performance, it is always better to adopt a carrot and stick approach for instant and consistent improvement. It is a combination of rewards and punishments that you can induce for the best and worst weekly or monthly performance. This has proven to be one of the best ways to combat low performance problems in companies of all kinds for centuries.

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8 0
3 years ago
Write a complete C++ program that is made of functions main() and rShift(). The rShift() function must be a function without ret
erma4kov [3.2K]

Answer:

Explanation:

attached is an uploaded picture to support the answer.

the program is written thus;

#include<iostream>

using namespace std;

// function declaration

void rShift(int&, int&, int&, int&, int&, double&);

int main()

{

   // declare the variables

   int a1, a2, a3, a4;

   int maximum = 0;

   double average = 0;

   // inputting the numbers

   cout << "Enter all the 4 integers seperated by space -> ";

   cin >> a1 >> a2 >> a3 >> a4;

   cout << endl << "Value before shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

        << a3 << ", a4 = " << a4 << endl << endl;

   // calling rSift()

   // passing the actual parameters

   rShift(a1,a2,a3,a4,maximum,average);

   // printing the values

   cout << "Value after shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

           << a3 << ", a4 = " << a4 << endl << endl;

   cout << "Maximum value is: " << maximum << endl;

   cout << "Average is: " << average << endl;

}

// function to right shift the parameters circularly

// and calculate max, average of the numbers

// and return it to main using pass by reference

void rShift(int &n1, int &n2, int &n3, int &n4, int &max, double &avg)

{

   // calculating the max

   max = n1;

   if(n2 > max)

     max = n2;

   if(n3 > max)

     max = n3;

   if(n4 > max)

     max = n4;

   // calculating the average

   avg = (double)(n1+n2+n3+n4)/4;

   // right shifting the numbers circulary

   int temp = n2;

   n2 = n1;

   n1 = n4;

   n4 = n3;

   n3 = temp;

}

8 0
3 years ago
A flow of 12 m/s passes through a 6 m wide, 2 m deep rectangular channel with a bed slope of 0. 001. If the mean velocity of flo
prohojiy [21]

Answer:

manning's coefficient is 0.0357

Explanation:

Given:

Velocity of flow, v = 12 m/s

Width of the channel, b = 6 m

Depth of the channel, d = 2 m

bed slope, s = 0.001

mean velocity of flow, V = 1 m/s

now, the velocity is given as:

V= \frac{1}{n}R^{\frac{2}{3}}S^{\frac{1}{2}}

where,

n is the manning's coefficient

R is hydraulic mean depth

R = (Area of the channel) / (wetted Perimeter of the channel)

now,

R = (2 × 6) / ((2 × 2) + 6)

or

R = 12 / 10 = 1.2 m

now, on substituting the values in the equation for velocity, we get

1= \frac{1}{n}1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}

or

n= 1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}

or

n = 0.0357

hence, the value of manning's coefficient is 0.0357

8 0
3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
This problem has been solved!
lisov135 [29]

Answer: a) 135642 b) 146253

Explanation:

A)

1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.

3- reserving all resources in advance helps would happen most likely if the algorithm has been used.

5- Resource ordering comes first before detection of any deadlock

6- Thread action would be rolled back much easily of Resource ordering precedes.

4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread

2- only option practicable after thread has been killed.

Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.

5 0
3 years ago
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