--- title: "Specific force and its minimum at critical flow" subject: "Hydraulics" chapter: 7 key: "semester_4/hydraulics:specific-force-expression-minimum-critical" source: https://www.bctengineer.com/q/semester_4/hydraulics/specific-force-expression-minimum-critical site: BCT Engineering --- # Specific force and its minimum at critical flow Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow - Asked in: 2080 Chaitra, 5 marks; 2073 Magh, 5 marks; 2071 Bhadra, 4+2 marks (asked 3 times) - Video: [Lab 4 CE234--Specific Force and Momentum](https://www.youtube.com/watch?v=Xuip2FSXDWY) — Robert Holmes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/specific-force-expression-minimum-critical - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7 ## Question Find the expression for specific force and prove that when the specific force is minimum the flow is critical. (Explain also the use of this concept in open channel flow.) ## Short answer Specific force F = Q²/(gA) + Aȳ, with centroid depth ȳ. dF/dy = −Q²T/(gA²) + A = 0 gives Q²T/(gA³)=1 (Fr=1), so F is minimum at critical flow. This momentum function is used to find conjugate depths of a hydraulic jump and to compute forces on structures. ## Answer #### Specific force (momentum function) Specific force $F$ (or specific momentum $M$) is the sum of the momentum flux and the hydrostatic pressure force per unit weight of water at a section of the channel: $$ F=\frac{Q^2}{gA}+A\bar y $$ where $A$ is the flow area and $\bar y$ is the depth of the centroid of the area below the free surface. Its unit is m³. #### Derivation (momentum equation) Apply the momentum equation between two sections 1 and 2 over a short horizontal reach of a prismatic channel (friction and the weight component neglected, as in a hydraulic jump): $$ \gamma A_1\bar y_1-\gamma A_2\bar y_2=\rho Q(V_2-V_1)=\rho Q^2\left(\frac1{A_2}-\frac1{A_1}\right) $$ Dividing by $\gamma$ and rearranging: $$ \frac{Q^2}{gA_1}+A_1\bar y_1=\frac{Q^2}{gA_2}+A_2\bar y_2\;\Rightarrow\;F_1=F_2 $$ #### Proof that $F$ is minimum at critical flow For a given discharge, differentiate $F$ with respect to depth $y$: $$ \frac{dF}{dy}=-\frac{Q^2}{gA^2}\frac{dA}{dy}+\frac{d(A\bar y)}{dy} $$ With top width $T$, $dA=T\,dy$. Also $\dfrac{d(A\bar y)}{dy}=A$ (when the depth increases by $dy$, the first moment of area about the new surface increases by $A\,dy$). Therefore $$ \frac{dF}{dy}=-\frac{Q^2T}{gA^2}+A $$ For minimum $F$, $\dfrac{dF}{dy}=0$: $$ \frac{Q^2T}{gA^3}=1\;\Rightarrow\;\frac{Q^2}{g}=\frac{A^3}{T} $$ This is exactly the condition for critical flow ($Fr=V/\sqrt{gA/T}=1$). The second derivative $\dfrac{d^2F}{dy^2}=\dfrac{2Q^2T^2}{gA^3}-\dfrac{Q^2}{gA^2}\dfrac{dT}{dy}+T$ is positive for usual channel sections, so $F$ is a minimum there. **Hence, for a given discharge, the specific force is minimum when the flow is critical.** ```text y ^ | . upper limb: subcritical y2 |----------------* conjugate depth y2 | . | . yc |-------* <- F_min (critical depth) | . | . y1 |----------------* conjugate depth y1 | . lower limb: supercritical +-------+--------+-------> F F_min F1 = F2 ``` For each $F>F_{min}$ there are two depths (the conjugate depths) with the same specific force. #### Uses of the concept 1. **Hydraulic jump:** the conjugate (sequent) depths $y_1$ and $y_2$ have equal specific force, so $F_1=F_2$ gives the depth after the jump. 2. Calculation of the force on structures (sluice gate, weir, baffle blocks, stilling basin) from the difference in specific force. 3. Locating the jump position in a stilling basin. 4. Analysis of flows where the energy loss is not known (jump, abrupt expansion, flow below gates). 5. Checking the flow type: the lower limb of the F-curve is supercritical and the upper limb is subcritical. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Specific force at a section is: A. F = Q^2/(gA) + A ybar, momentum flux plus hydrostatic force per unit weight B. F = y + Q^2/(2gA^2) C. F = Q^2/(2gA) + y D. F = Q A ybar / g Answer: A. F combines momentum flux Q^2/(gA) and the hydrostatic term A ybar, with ybar the centroid depth. (Key point: Expression for specific force) 2. Why do the two conjugate sections of a jump have equal F? A. Energy is conserved B. Applying momentum with negligible friction and weight gives F1 = F2 C. Velocities are equal D. Depths are equal Answer: B. Net pressure force equals rate of change of momentum, so F1 = F2. (Key point: Origin of equal F) 3. Setting dF/dy = -Q^2 T/(gA^2) + A = 0 gives: A. Q^2 = g A y B. Q^2/g = A^3/T, the critical flow condition (Fr = 1) C. Q = A T D. V^2 = 2gy Answer: B. Rearranging gives Q^2 T/(g A^3) = 1, which is Fr = 1. (Key point: Proof of minimum at critical) 4. In open-channel flow, specific force is chiefly used to: A. Find energy loss in uniform flow B. Compute Manning's n C. Design channel slope D. Find conjugate depths of a hydraulic jump and forces on structures such as gates and baffles Answer: D. F1 = F2 gives the sequent depths, and differences of F give forces on structures. 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