Consider the x component of momentum equation given in the form of Equation (3.15). The first term in the left hand side of this equation can be expanded as

(3.18)

and second term of the equation can be expanded as,

(3.19)

Substituting Equations (3.18) and (3.19) into (3.15), we obtain

or

(3.20)

Examine the two terms inside the square brackets in the above equation. They represent the continuity equation, (2.11). Hence the sum inside the square bracket is zero. Therefore, equation (3.20) becomes

or

(3.21)

Examine the two terms inside the parentheses of above equation. They represent the substantial derivative of x-component of velocity, u. Hence, Equation (3.21) becomes

(3.22)

In a similar manner, y and z momentun equations are given respectively as below,

(3.23)

and

(3.24)

Equations (3.22 to 3.24) are the x, y and z components of the momentum equation written in terms of the substantial derivative. The momentum equation in the vector form can be written as

(3.25)