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matrenka [14]
3 years ago
10

A wastewater is to be disinfected using ultraviolet (UV) light. Batch experiments find that the bacterial concentration decays u

nder UV light with first-order kinetics, with a half of the initial bacteria killed after 6.2 minutes of illumination. You need to design a reactor than can kill 99% of bacteria in a stream having a flow rate of 140, 000 gal/d. Assume that regardless of the reactor type it will be possible to have UV illumination conditions identical to those of the batch experiment. a. Based on the batch experiments, what is the first-order rate constant k for bacterial inactivation
Engineering
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

k = 0.1118 per min

Explanation:

Assume;

Initial number of bacteria = N0

Number of bacteria IN 'T' time = Nt

So,

Nt=N0e^{-kt}\\\\in\ 6.2 min\\\\\\frac{N0}{2}= N0e^{-k(6.2)}\\\\ln\frac{1}{2} = -k[6.2]

k = 0.1118 per min

You might be interested in
Compare the output of full-wave rectifier with and without filter
lara31 [8.8K]

Answer:

Full wave rectification flips the negative half cycle of the sine wave to positive so the result is two positive half cycles.

Explanation:

hope it helps a lil

4 0
2 years ago
1. The system must be able to manage multiple students (max of 15) and their grades for assignments from three different assignm
mr_godi [17]

Answer:

#include<iostream>

#include<string.h>

using namespace std;

class gradebook;

class student

{

friend class gradebook;

char name[20]; // std name

int homework_grades[6], quiz_grades[6], exam_grades[4]; //grades array

float homework_avg, quiz_avg, exam_avg, final_avg;

};

class gradebook

{

student list[16];

public:

char gb_name[20]; // gradebook name

int max, total;

// Constructor

gradebook()

{

strcpy(gb_name,"noname");

max = 5;

total = 0;

for(int i=1;i<=15;i++)

{

strcpy(list[i].name,"noname");

for(int j=1;j<=5;j++)

{

list[i].homework_grades[j] = -1;

list[i].quiz_grades[j] = -1;

}

list[i].exam_grades[1] = list[i].exam_grades[2] = list[i].exam_grades[3] = -1;

}

}

// Member functions

void display_std_name();

void display_grades(int, int);

int add_std(char *new_name);

int remove(int);

//int ch_grade(int, int);

};

// Define all functions

// Remove a student from list and adjust remaining

int gradebook::remove(int std_id)

{

if(total==0 || std_id>total || std_id<=0)

return 0;

if(std_id==total)

{

cout<<"\n“"<<list[std_id].name<<"” has been successfully removed! (New class size: 2)";

strcpy(list[total--].name,"noname");

return 1;

}

cout<<"\n“"<<list[std_id].name<<"” has been successfully removed! (New class size: 2)";

for(int i=std_id; i<total; i++)

{

list[i]=list[i+1];

}

total--;

return 1;

}

// Add student to gradebook

int gradebook::add_std(char *new_name)

{

if(total<15)

{

strcpy(list[++total].name, new_name);

return 1;

}

return 0;

}

// Display student name list

void gradebook::display_std_name()

{

for(int i=1;i<=total;i++)

{

cout<<"\n"<<i<<" "<<list[i].name;

}

}

/* Display grades of students

// (grade_id, student_number_in_list)

homework(1), quiz(2), exam(3)

*/

void gradebook::display_grades(int g_id, int std_id)

{

int i;

if(std_id<=total)

{

if(g_id==1)

{

for(i=1;i<6;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].homework_grades[i]==-1) cout<<"<ungraded>";

}

}

else if(g_id==2)

{

for(i=1;i<6;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].quiz_grades[i]==-1) cout<<"<ungraded>";

}

}

else if(g_id==3)

{

for(i=1;i<4;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].exam_grades[i]==-1) cout<<"<ungraded>";

}

}

}

else

cout<<"\nNo such std exists";

}

/****************************** MAIN ****************************/

int main()

{

int main_choice, grade_choice, std_id, new_grade;

char g_name[20], s_name[20];

float a,b,c;

gradebook g;

do

{

cout<<"\n\n-=| MAIN MENU |=-";

cout<<"\n1. Add a student\n2. Remove a student\n3. Change a student’s grade\n4. Change group weights\n5. Change the gradebook name\n6. Display class averages\n7. Display full report\n0. QUIT\n";

cout<<"\nEnter an action: ";

cin>>main_choice;

switch(main_choice)

{

case 0:

break;

case 1:

cout<<"\n-=| ADDING STUDENT |=-";

cout<<"\nPlease enter the student’s name: ";

cin>>s_name;

if(g.add_std(s_name))

cout<<"\n"<<s_name<<" was successfully added to the gradebook!";

else

cout<<"\nStudents cannot be added because the gradebook is full!";

break;

case 2:

cout<<"\n-=| REMOVING STUDENT |=-";

if(g.total!=0)

{

g.display_std_name();

cout<<"\n Enter student to remove: ";

cin>>std_id;

if(!g.remove(std_id))

cout<<"\nCannot remove students";

}

else

cout<<"\nStudents cannot be added because the gradebook is full!";

break;

case 3:

cout<<"\n-=| CHANGING GRADE |=-\n1. Change a homework grade\n2. Change a quiz grade\n3. Change an exam grade";

cout<<"\nWhat type of grade would you like to change: ";

cin>>grade_choice;

g.display_std_name();

cout<<"\nWhich student’s grade would you like to change? ";

cin>>std_id;

//ch_grade(grade_choice, std_id);

g.display_grades(grade_choice, std_id);

break;

/*

case 4:

cout<<"\n-=| CHANGING WEIGHTS |=-";

while(1)

{

cout<<"\nEnter the weights, separated by spaces, in the order of homework, quizzes, and exams (total must add up to 1.0):";

cin>>a>>b>>c;

if(a+b+c==1.00)

break;

else

cout<<"\n Weights do not add up to 1.0, try again...";

}

break;

case 5:

cout<<"\n-=| CHANGING NAME |=-\nPlease enter the new name for the gradebook: ";

cin>>g_name;

cout<<"\n Gradebook name changed from “ ” to "<<g_name;

break;

case 6:

cout<<"\n-=| CLASS AVERAGES |=-";

cout<<"\nHomework average of class: ";

cout<<"\nQuiz average of class: ";

cout<<"\nExam average of class: ";

cout<<"\nFinal average of class: ";

break;

case 7:

break;

*/

default:

cout<<"\n\nInvalid input\n\n";

}

}while(main_choice!=0);

return 0;

}

Explanation:

6 0
3 years ago
If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no
Brut [27]

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

7 0
4 years ago
An escalator in a shopping center is designed to move 50 people, 75 kg each, at a constant speed of 0.6 m/s at 45° slope. Determ
kramer

Answer:

\dot E = 15602.842\,W

Explanation:

By an adequate application of the Principle of Energy Conservation, the escalator need energy to elevate from to the bottom to the top. Hence:

E = m_{total}\cdot g \cdot y

An expression for power needed is found by deriving the equation with respect to time:

\dot E = m_{total}\cdot g \cdot v_{y}

The minimum power is found by substituting known inputs:

\dot E = (50)\cdot (75\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.6\,\frac{m}{s} )\cdot \sin 45^{\textdegree}

\dot E = 15602.842\,W

6 0
3 years ago
Assume we have one road section which has 3 lanes in both directions. If the Sf for both direction is 75 mph, and Dj for both di
givi [52]

Answer:

i) 3750 veh/hr/ln

ii) 100 veh/mi/In

iii) 37.5 mph

Explanation:

number of lanes = 3

sf for both directions = 75 mph  ( free mean speed )

Dj for both directions = 200 veh/mi/In

<u>Calculate the value of  S0, D0 (veh/mi/ln) and maximum Vm (veh/hr)</u>

For either direction we will consider the total volume = 3 lanes

value of Dj = 3 lanes * 200 = 600 veh/mi/

i) value of SO

=  ( Dj * sf ) / 4 = ( 600 * 75 ) / 4 = 11250 veh/hr  = 3750 veh/hr/lane

ii) Value of DO

DO = Dj / 2 = 200 /2 = 100 veh/mi/In

iii) Value of Vm

= sf /2 = 75 / 2 = 37.5 mph

5 0
3 years ago
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