Question

The saturation concentration of dissolved oxygen in freshwater can be calculated with the

equation,

ln Osf = −139.34411 + 1.575701 × 105

Ta

−

6.642308 × 107

T2

a

+

1.243800 × 1010

T3

a

−

8.621949 × 1011

T4

where Osf is the saturation concentration of dissolved oxygen in freshwater at 1 atm (mg/L) and Ta

is the absolute temperature in Kelvin (K), where Ta = T + 273.15, where T is the temperature in

Celsius (oC). According to this equation, saturation decreases with increasing temperature. For typical

natural waters in temperate climates, the equation can be used to determine that oxygen concentration

ranges from 14.621 mg/L at 0 oC (273 K) to 6.949 mg/L at 35 oC (308 K). Given a target value of oxygen

concentration Osf = 10 mg/L, it is required to solve for temperature in (K).

(c) Is it possible to solve the equation as a polynomial? If yes, determine all the roots and decide on

the feasible ones (Hint: use roots).

Answer 1

ANSWER:

EXPLANATION:

AS per question

☆ This question is very difficult to find it analytically so we will solve by using matlab.

☆ Formula used in, is in the form of logarithm type. By taking antilog to both sides we get

Code for Q.1)

>> dt = 0.01;

>> x = 273:dt:308;

>> y = exp(-139.34411 + 1.5757501*10.^5*x.^-1 - 6.642308*10.^7*x.^-2 + 1.243800*10.^10*x.^-3 - 8.621949*10.^11*x.^-4);

>> plot(x,y)

>> xlabel('Absolute Temperature (K) ')

>> ylabel('osf (mg/liter)')

>> grid on

☆☆ The graph we will get is

☆ Now we have to obtain a value Temperature at which osf = 10 mg/mites

For this code is

>> [difference, index_At_y_Equals_10] = min(abs(y-10))

>> x10 = x(index_At_y_Equals_10)

☆ The value of x will be 288.54 K.

☆ remember we had taken temperature as 'x' and osf as 'y' in matlab.

Temperature = 288.5400 K

☆ For Q.2)

We will plot the graph by increasing the range. We take the value of temperature from 273 to 800K

>> dt = 0.01;

>> x = 273:dt:800;

>> y = exp(-139.34411 + 1.5757501*10.^5*x.^-1 - 6.642308*10.^7*x.^-2 + 1.243800*10.^10*x.^-3 - 8.621949*10.^11*x.^-4);

>> plot(x,y)

>> xlabel('Absolute Temperature (K) ')

>> ylabel('osf (mg/liter)')

>> grid on  

☆ We can see the graph

☆ In graph, the function will never cut the x-axis at all, so we can say that no matter how much will be the magnitude of Temperatue the dissolved oxygen will never 0.

or we can check by MATLAB

add this line in addition to above code.

>> zero = yzero(y, 1000)

☆ The result will be

'undefined function'

☆ Here we write the 1000 because it had asked for initial guess, as we can see the graph the function y will cut the x-axis at infinity.

☆ So, we had choose x = 1000.