Appendix C Useful Theoretical Approaches and Formulas
493
Therefore, the stream function 1/J (x, y) finally takes the form:
1/J ( x, y) = Va (y cos () - x sin ()) ,
(C.67)
and the streamlines are given by equations:
Va (y cos () - x sin ()) = K n ,
(C.68)
in which Kn is the constant. The subscript n indicates a value of constant K,
corresponding to the nth streamline. If we rewrite Eq. (C.68) in another form:
Kn
y = tan(} x +
,
Va cos ()
(C.69)
we find that streamlines are straight lines of slope tan (), as should be expected.
Consider now the flow through cross-section A-B in Fig. C.4. The discharge
of flow through this cross-section is:
(C.70)
Assuming that a difference between two adjacent streamlines is constant and
equal to 11K, the discharge for any two adjacent streamlines becomes:
I1Q = Kn - K n- 1 = 11K.
(C.7l)
In a similar way it can be found that the velocity potential function, 1>,
corresponding to the uniform flow shown in Fig. C.4, is:
1> (x, y) = Va (x cos () + y sin ()),
(C.72)
and at equipotential lines where the velocity potential, 1>, is constant, we have:
Va(x cos () + y sin (}) = Pn ,
(C.73)
where Pn is the constant corresponding to the nth equipotential line.
Let us rewrite Eq. (C.73) as:
1
Pn
y = - - - x + - - .
tan ()
Va sin ()
(C.74)
Thus, equipotential lines are also straight lines but of slope (-l/tan ()). Comparing this slope to the slope of the streamline (Eq. C.69), we find that equipotential lines are perpendicular to the streamlines as their slopes are negative
reciprocals.