Heat exchange between oil and water. water is flowing at the rate of 1.13 kg/s in a 1-2 shell-and-tube heat exchanger and is heated from 45°c to 85°c by an oil having a heat capacity of 1.95 kj/kg. k. the oil enters at 120°c and leaves at 85°c. calculate the area of the exchanger if the overall heat-transfer coefficient is 300 w/m2.k.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.