Correct Answer - Option 2 : Flow from A to B and head loss is 0.5 m
Concept:
Bernoulli’s Equations
At Section 1
\(\frac{{{p_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1}\)
At Section 2
\(\frac{{{p_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {z_2}\)
Calculation:
Given:
P1 = 100 kPa, Z1 = 10 m, P2 = 75 kPa, z2 = 12 m, v = 4 m/s, d = 500 mm = 0.5 m
At Section 1
\(\frac{{{p_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1} = \frac{{100 \times {{10}^3}}}{{1000 \times 10}} + \frac{{{4^2}}}{{2 \times 10}} + 10 = 20.8\;m\)
At Section 2
\(\frac{{{p_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {z_2} = \frac{{75 \times {{10}^3}}}{{1000 \times 10}} + \frac{{{4^2}}}{{2 \times 10}} + 12 = 20.3\;m\)
(Total Energy) 1 > (Total Energy) 2
∴ Flow takes place from section 1 to section 2 with a head loss of 0.5 m.