Question

If the cells operate with 18% efficiency, and there is no mismatch loss among the cells, calculate the overall module efficiency if the round cells are used. Assume a PV Module is to have dimensions of 37.5 X 63.2 in. Also assume that 8-in. round cells are available

Answer 1

Ans. Calculate 1/[S] and 1/Vo. Make LB Plot using excel or other similar software.

# Determination of Vmax and Km using LB Plot”

Lineweaver-Burk plot gives an equation in from of y = mx + c

where, y = 1/ Vo, x = 1/ [S],

Intercept, c = 1/ Vmax ,

Slope, m = Km/ Vmax

#A. Enzyme kinetics at I = 0.0 mM            [y = 0.4836x + 0.1951]:

Vmax = 1 / c = 1 / 0.1951 = 5.1256

Hence, Vmax = 5.1256 mmol L^-1 min^-1

Now,

Km = m x Vmax = 0.4836 x 5.1256 = 2.4787

Hence, Km = 2.4787 mM

#B. Vmax,app and Km,aapp in presence of inhibitor is calculated similarly, the result is tabulated in excel sheet.

At [I] = 3.0 mM;        Vmax,app = 5.1335 mmol L^-1min^-1

At [I] = 5.0 mM;        Vmax,app = 4.9727 mmol L^-1min^-1

#C. At [I] = 3.0 mM; Km,app = 3.8773 mM

At [I] = 5.0 mM;        Km,app = 4.9632 mM

#D. Type of Inhibitor: Note that the Vmax remains almost constant whereas Km increases in presence of inhibitor. It is the characteristic of a competitive inhibitor. So, the inhibitor is a competitive inhibitor.

# At [I] = 3.0 mM

KI of completive inhibition is given by-

            KI = [I] / [(Km,app / Km) – 1]                              - equation 1

Putting the values in equation 1-

            KI = 3 mM / [(3.8773 mM / 2.4787 mM) - 1]

            Or, KI = 3 mM / (1.564 - 1)

            Or, KI = 3 mM / 0.564

            Hence, KI = 5.319 mM

# At [I] = 5.0 mM

Putting the values in equation 1-

            KI = 5 mM / [(4.9632 mM / 2.4787 mM) - 1]

            Or, KI = 5 mM / 1.00

            Hence, KI = 5.00 mM