Application of Momentum Equation

From Eqn. 3.43 we have the momentum Equation-

$$F_{Bs} ~+~F_{Ss}~=~~~{\partial \over \partial t} \int_{CV} V_s \rho d \forall~~+~~ \int_{CS}~V_s\rho \overrightarrow{V_s}.d\overrightarrow{A}$$ (3.45)

Since the flow is steady, the first term on the RHS drops out. We need to evaluate the body forces $F_{Bs}$ and surface forces $F_{Ss}$ acting on the control volume.

Body Force.

The only body force acting is the weight of the fluid within the control volume. We need to consider the component of this in the $s$ direction. Accordingly,

$$F_{Bs}~ =-dW~\sin \theta$$

$$=~ -~\left(m \right)g \sin\theta$$

$$=~-~\left(\rho~d \forall~\right) g \sin\theta$$

$$=~-~\left(\rho~ds~(A+{{dA} \over 2})\right)g \sin\theta$$

$$F_{Bs}=~-\rho~g~\left(A+{{dA} \over 2} \right) dz$$ noting that sin q = dz, we have ,

$$F_{Bs}~ -~\rho~g~A~dz$$ (3.46)

Surface Forces

The surface force is due to pressure acting upon the boundaries of the control surface. There are three terms that contribute - end (1), end (2) and the bounding surface of the stream tube. Force on each of these is given by the product of pressure and area. For the bounding surface we take this force to be the product of an average pressure, ${1 \over 2}({p+p+dp}) = p+{{dp} \over 2}$ multiplied by the effective area, dA. Thus we have for the surface forces,

$$F_{Ss}~ = pA~-~(p+dp)(A+dA)~+~(p+{{dp} \over 2})~dA$$

$$=~p~A-p~A-p~dA-A~dp-dp~dA~+~p~dA~+~{{dp} \over 2}~dA$$

$$F_{Ss}~ =~-A~dp$$ (3.47)

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney