Calculating flow rate of an infusion pump

[Osama Mohammed]

Hello, My project is to calculate the flow rate of a medical infusion pump, that is , when the first drop falls there is a infrared pair transmitter and receiver ,when each drop falls the light between the ir pair is interrupted ,and a pulse generated is used to initialize the interrupt service routine ,inside the Interrupt service routine there is the instruction that runs the timer to start measuring the time between the first drop and the second drop, when the second drop falls down another interrupt is occurred ,and there are instructions that stops the timer and display the flow rate according to some calculations that I have made earlier. 

 

 

This is the code: 

 

#include <reg51.h>

msDelay(unsigned int);

cmd(unsigned char);

dat(unsigned char );

 

 

 

unsigned char a[]="Welcome";

unsigned char b[]="Flow Rate:25ml/hr";

unsigned char c[]="Flow Rate:50ml/hr";

unsigned char d[]="Flow Rate:250ml/hr";

unsigned char e[]="Flow Rate:100ml/hr";

 

 

sbit regs=P3^0; 

sbit wr=P3^1;   

sbit enab=P3^2;   

 

 

void LCD_write_string(unsigned char *str)  

{

    int i = 0;

    while (str != 0)

    {

        dat(str);    

        msDelay(100);

                i++;

    }

    return;

}

 

unsigned int i,j,k;

void timer0() interrupt 1

{  

  k–;

 

}

 

void timer1() interrupt 2

{

  unsigned int k;

  i–;

  TR0=1;

  if(i==1)

  {

  P1=0xAA;

   }

  

   if(i==0)

   {

   TR0=0;

msDelay(5000);

   cmd(0x01);

  if(k>=139 || k<=141)

  {

 

  LCD_write_string(b);

 

 

 

   

  }

  else if(k>=100 || k<139 )

  {

 

  LCD_write_string(c);

 

  }

 

  else if(k<=100 || k>=20)

  {

 

  LCD_write_string(d);

 

  }

 

  else 

 

  {

  LCD_write_string(e);

 

  

  } 

   k=0;

  i=2;

  k=141;

  msDelay(10000);  

  

    TL0=0x00;

   TH0=0x00;

  

 

  

  }

  

  }

 

 

 

 

 main ()

{

unsigned int k;

// Initialize the LCD 

cmd(0x38);

msDelay(1000);

cmd(0x0C);

msDelay(1000);

cmd(0x01);

msDelay(1000);

// *********************

//

---

Welcome Message 

cmd(0x84);

msDelay(2000);

LCD_write_string(a);

//******************

msDelay(60000);

cmd(0x01);

TMOD=0x01;

IE=0x86;

 

TL0=0x00;

TH0=0x00; 

TCON=0x04;

k=141;

i=4;

 

while(1);

 

}

 

cmd(unsigned char value)

{

 P2=value;

 regs=0;

 wr=0;

 enab=0;

 msDelay(500);

 enab=1;

}

 

dat(unsigned char info)

{

 P2=info;

  regs=1;

  wr=0;

 enab=0;

 msDelay(500);

 enab=1;

}

 

msDelay(unsigned int t)

{

unsigned int i,j;

for(i=0;i<=t;i++)

for(j=0;j<=100;j++);

}

 

 

 

This is a picture illustrating the idea of my project:

 

 

 

 

The problem is that as I think  ( if conditions ) are not run ,that is ,all the time the flow rate displayed on the LCD is 25ml/h and is not changed even if I change the time between the drops.

 

I am using Proteus simulation software.

 

[Ashutosh Bhatt]

r u getting pulses when there is a drop?

if yes then anything else is very simple

you just count number of pulses (that means drops) per unit time (1 sec, 10 sec etc) 

do some calibration like in 10 sec 100 drops falls that means 50 ml liquid flowed like wise. then calculate flow rate as ml / sec or ml / minute etc.

[Osama Mohammed]

Thanks, 

The problem is solved 

but now I have another problem, 

I don’t know how to display decimal numbers on LCD , for example 244 how if I want to display on LCD ?

Also how to display a number preserved inside a variable ,for instance if k is a variable and k=200; , how to display this value of k ????

[AJISH ALFRED]

Hi Mohammed,

 

It is very simple. Tell you the logic!

Take another variable, say ‘m’,

 

step : 1; m = k % 10, you will get the unit place value of k in m,

step : 2; now m = m + ‘0’, you will get the ascii value of that unit place, which you can display on the LCD.

step : 3; display m on LCD

step : 4; Now k = k / 10 and, continue these 4 steps until k becomes zero at step 4.

[venkateshwaran]

Hi mohammed,

this is the code for you.

 

void convert (void)
{
if(sourse<=9) { b = sourse;
c = 0;
d = 0;}
else if(( sourse>= 10) && (sourse<20) ) {
b = sourse – 10 ;
c = 1 ;
d = 0 ; }
else if( (sourse>= 20) && (sourse<30) ) {
b = sourse – 20 ;
c = 2 ;
d = 0 ; }
else if(( sourse>= 30) && (sourse<40) ) {
b = sourse – 30 ;
c = 3 ;
d = 0 ; }

else if(( sourse>= 40) &&( sourse<50 )) {
b = sourse – 40 ;
c = 4 ;
d = 0 ; }
else if( (sourse>= 50) &&(sourse<60) ) {
b = sourse – 50 ;
c = 5 ;
d = 0 ; }
else if( (sourse>= 60) && (sourse<70 )) {
b = sourse – 60 ;
c = 6 ;
d = 0 ; }
else if(( sourse>= 70) &&(sourse<80) ) {
b = sourse – 10 ;
c = 7 ;
d = 0 ; }
else if( sourse>= 80 && sourse<90 ) {
b = sourse – 10 ;
c = 8 ;
d = 0 ; }
else { b = sourse – 10 ;
c = 9 ;
d = 0 ; }
b=0x30|b;
c=0x30|c;
d=0x30|d;}

b is first digit c and d are second and third digits,

this code works only  sourse value is below 100, you can edit this code for

more values. 

[SHAH DISHANT H.]

Hi,

 

This function will help you. In this whatever the number you want to display, pass it as a parameter and it will be displayed on LCD.

 

void DisplayDigit(unsigned int dat)

{

unsigned int temp,i=0, buf[10];

 

while(dat>=10)

{

buf=dat%10;

dat=dat/10;

i++;

} 

 

buf=dat;

 

for(temp=0;temp<=i;temp++)

{

LCDWriteData(buf[i-temp]+’0′);

}

 

}

 

 

All the best…!!!

 

[Ashutosh Bhatt]

congrets for getting solution.

pls write in detail how ur system works now. how problem is solved?

this will help many other students, hobbysists etc 

[Osama Mohammed]

Hi,

 

 

Thank you all,

 

the problem is solved

 

the if statement argument was wrong ,that is all !! 

 

 

 

 

 

SHAH DISHANT H. thank you for posting this function I will try it 

[Osama Mohammed]

Hi,

 

 

Thank you all,

 

the problem is solved

 

the if statement argument was wrong ,that is all !! 

 

 

 

 

 

Dishant 

 thank you for posting this function I will try it 

[Osama Mohammed]

Please I hgave another question 

 

my hex code file is 9 kilo bytes in size

 

can it be used with 89C52 ?

[Ashutosh Bhatt]

try to limit the code size less then 8 KB.

[Author]

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