How to understand the increase in local resistance in a pipeline when a valve is closed?
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In the textbook on chemical engineering principles from Tianjin University, in the chapter on the regulation of centrifugal pumps, it is stated that when the valve is closed, the local resistance in the pipeline increases, and the characteristic curve of the pipeline becomes steeper. How to understand why the local resistance in the pipeline increases when the valve is closed? As I understand it, when the fluid is in a fully turbulent state, the resistance in the straight pipe is proportional to the square of the flow velocity. So when the valve is closed, the resistance in the straight pipe decreases. Then how does the total resistance (resistance in the straight pipe + local resistance) change? So confused, please help solve it
I’m still very grateful to the original poster for sharing it~
When the valve is fully closed, the resistance is greatest.
In fact, you can analyze the problem in an extreme manner; under extreme conditions, the valve is completely closed, and the fluid cannot pass through at all. In other words, the valve is closed to an infinitesimally small degree – something that can be addressed using mathematics； Or, thinking of reducing the valve opening as a decrease in the pipe diameter means that the resistance increases naturally; there is a formula for this, as resistance is inversely proportional to the pipe diameter.
Thank you, I understand now, but my confusion remains: when the valve is closed, the flow rate decreases. When the fluid is in a fully turbulent state, the resistance in a straight pipe is proportional to the square of the flow rate – so shouldn’t the resistance in a straight pipe decrease as well?
Why does the flow rate decrease when the valve is closed?
When there is no acceleration, the total resistance equals the driving force (such as potential difference, pressure difference, etc.). When the driving force remains constant, the total resistance also remains constant! Total resistance = local resistance + straight-line pipe resistance ； When the valve is closed, the flow rate decreases; as a result, the resistance in the straight pipe reduces, while the local resistance increases!
In the formula for calculating pressure loss, there is a term that is the square of the velocity. When the valve is closed, the flow area decreases, which causes the velocity to increase. Is this understanding correct?
When the valve is closed, the resistance increases because the coefficient of resistance rises, and this increase is far greater than the effect of the reduced velocity