Question 4 of 6: Momentum on a 45° pipe elbow — downstream pressure
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — May 2019 · 3 hours · closed book (one aid sheet) · six questions, complete any four (all six solved here) · each question 20 marks, equal-value parts.
Reference texts: Chin, Water-Resources Engineering (Pearson); Chow, Open-Channel Hydraulics (McGraw-Hill); Chaudhry, Open-Channel Flow (Springer); Wylie & Streeter, Fluid Transients in Systems (Prentice-Hall); Munson, Young & Okiishi, Fundamentals of Fluid Mechanics (Wiley); Roberson, Cassidy & Chaudhry, Hydraulic Engineering.
Governing relations supplied on the exam cover sheet. Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI, $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^5}Q^2$. Unless stated, local losses and velocity head are neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
The Question 3 pipe specification used below is reconstructed from Figure 1. Check it against your copy of the paper.
Question 4: Momentum on a 45° pipe elbow — downstream pressure (20 marks)
Given. Water enters horizontally at $P_1$ and turns $45^\circ$ downward through a constant-diameter elbow (Figure 2). The support (anchor) reactions are $R_x=6$ kN acting in the $-x$ sense and $R_y=2$ kN acting in the $-y$ sense.
Given data — Question 4
Quantity
Symbol
Value
Diameter
$D$
0.153 m
Discharge
$Q$
0.200 m³/s
Upstream pressure (gauge)
$P_1$
500 kPa
Turn angle
$\theta$
$45^\circ$
Anchor reactions
$R_x,\ R_y$
6 kN, 2 kN
Find. (a) the $x$- and $y$-momentum statements for the elbow control volume; (b) the downstream pressure $P_2$.
Figure 4-1. 45° elbow (Figure 2 on the exam). Flow enters along $+x$ and leaves $45^\circ$ below horizontal; the anchor reactions $R_x,\,R_y$ (shown $-x$, $-y$) hold the bend against the fluid’s push.
Approach. Take the elbow interior as a control volume. Continuity fixes the (equal) inlet and outlet speeds; the steady momentum theorem in each direction sums the two pressure forces and the anchor force to the momentum-flux change. Solving the $x$-equation for $P_2$ isolates the unknown.
Kinematics. Area $A=\tfrac{\pi}{4}D^2=\tfrac{\pi}{4}(0.153)^2=0.018385\ \text{m}^2$. Constant diameter, so by continuity $V_1=V_2=V=Q/A=0.200/0.018385=10.88\ \text{m/s}$, and mass flow $\dot m=\rho Q=1000(0.200)=200\ \text{kg/s}$.
Momentum statements (part a). With the inlet velocity along $+x$ and the outlet velocity along $(\cos\theta,\,-\sin\theta)$, the steady momentum theorem $\sum \mathbf F=\dot m(\mathbf V_2-\mathbf V_1)$ on the control volume, taking the anchor force on the fluid as $(-R_x,\,-R_y)$ and the outlet pressure acting back into the volume, reads
$$x:\quad P_1A-P_2A\cos\theta-R_x=\dot m\,V(\cos\theta-1),$$
$$y:\quad P_2A\sin\theta-R_y=\dot m\,V(-\sin\theta-0).$$
These are the two control-volume momentum equations requested.
Solve for $P_2$ (part b). Rearranging the $x$-equation for $P_2$ (it contains the given $P_1$ and the axial reaction):
$$P_2=\frac{\dot m\,V(1-\cos\theta)+P_1A-R_x}{A\cos\theta}
=\frac{200(10.88)(1-0.7071)+500000(0.018385)-6000}{0.018385(0.7071)}.$$
The numerator is $637+9193-6000=3830\ \text{N}$, and $A\cos\theta=0.013000\ \text{m}^2$, so
$$\boxed{P_2\approx 2.95\times10^{5}\ \text{Pa}=295\ \text{kPa}.}$$
Physically the bend and the direction change drop the pressure from 500 to about 295 kPa.
The result is physically reasonable: turning the flow costs momentum, and part of the upstream pressure head is spent supplying the force that changes the flow direction, so $P_2$ sits well below $P_1$. Because the diameter is constant the speed is unchanged, so this is not a Bernoulli velocity effect — it is entirely the momentum reaction of the bend registering as a lower downstream pressure. The pressure force on the inlet face, $P_1A=9.2$ kN, is the single largest term in the balance and dwarfs the $0.64$ kN momentum-flux change, which is why the pressure terms and the anchor reaction essentially set the answer between them.
Check — over-specified data
The exam supplies both reaction components for a single unknown $P_2$. Solving the $y$-equation instead, $P_2=(R_y/\sin\theta-\dot m V)/A=(2000/0.7071-2176)/0.018385\approx 35\ \text{kPa}$, which does not equal the $x$-result. With $\theta=45^\circ$, $Q$ and $P_1$ fixed, the two stated reactions are not simultaneously exact (rounded/measured values). The $x$-direction equation is used because it is anchored on the given upstream pressure $P_1$ and the axial (dominant) reaction; $P_2\approx 295$ kPa is reported, and the inconsistency is noted per Exam Note 1.
