In a system with a pump delivering head Hp(Q)=50 - 0.5 Q and a pipe with head loss Hf(Q)=10 + 0.3 Q^2, determine the Q where the system is in balance.

• A. 0.3 Q^2 + 0.5 Q - 40 = 0
• B. 50 - 0.25 Q = 10 + 0.3 Q^2
• C. 50 - 0.5 Q = 10 + 0.3 Q^2
• D. Q ≈ 5.0 m^3/s

Answer: (C)

In this kind of system, balance occurs when the head added by the pump equals the head lost in the pipe. So set the pump head Hp(Q) equal to the pipe head loss Hf(Q).

That gives 50 - 0.5 Q = 10 + 0.3 Q^2. This direct equality is the condition for balance, and it’s the simplest way to express the requirement that there is no net head around the loop.

If you rearrange to collect terms on one side, you get 0.3 Q^2 + 0.5 Q - 40 = 0. Solving this quadratic yields a positive root of about Q ≈ 10.7 m^3/s. So the system balances at roughly 10.7 m^3/s.

The numeric guess of 5.0 m^3/s doesn’t satisfy the equality—substituting Q = 5 gives Hp ≈ 47.5 and Hf ≈ 17.5, which are not equal. The expanded form is just another way to write the same balance condition.