Head loss and flow direction in tapering oil pipe

Most of the data are unreadable in the scan, so the standard textbook values for this problem are assumed: $d_A = 200$ mm, $d_B = 500$ mm, $p_A = 9.81$ N/cm² (98.1 kPa), $p_B = 5.886$ N/cm² (58.86 kPa), $Q = 200$ L/s, oil S = 0.90, B is 5 m above A. The method is the same for any values.

Principle: flow always goes from the section of higher total energy to the section of lower total energy, and the difference is the head loss.

                         B  (d = 500 mm, z = 5 m)
                      ___/
          ________---
 A ___---            flow ?
 (d = 200 mm, z = 0)

Velocities

$$\begin{aligned} V_A &= \frac{0.2}{\frac{\pi}{4}(0.2)^2} = 6.366\ \text{m/s} \\ V_B &= \frac{0.2}{\frac{\pi}{4}(0.5)^2} = 1.019\ \text{m/s} \end{aligned}$$

Total energy (take datum through A, $\rho g = 0.9 \times 9810 = 8829$ N/m³)

$$\begin{aligned} E_A &= \frac{p_A}{\rho g} + \frac{V_A^2}{2g} + z_A = \frac{98100}{8829} + \frac{6.366^2}{19.62} + 0 \\ &= 11.111 + 2.066 + 0 = 13.177\ \text{m} \\ E_B &= \frac{58860}{8829} + \frac{1.019^2}{19.62} + 5 \\ &= 6.667 + 0.053 + 5 = 11.720\ \text{m} \end{aligned}$$

Result

Since $E_A > E_B$, flow is from A to B (upwards, from the smaller to the larger end).

$$h_L = E_A - E_B = 13.177 - 11.720 = 1.457\ \text{m of oil}$$

Answer (with assumed data): Flow from A to B; loss of head ≈ 1.46 m of oil. If $E_B$ had come out larger, the flow would be from B to A.