The Equation of Continuity — Question 2

Section 14-8 · Physics pages 2194–2195

Question 2 Conservation can determine an unknown branch direction At a steady water junction, branch A brings in 8.0 L/s and branch B carries out 3.0 L/s. In branch C the mean speed magnitude is 2.0 m/s through area 40 cm2, but its direction is unknown. A fourth branch D is also present. No storage occurs. A B C D Known arrows are shown; C and D directions must be inferred. (a) Find the flow in D for each possible direction in C. (b) If an indicator shows D is an outlet, which direction in C is physically consistent? Find D's speed if its area is 50 cm2.
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Question 2 - Solution Conservation can determine an unknown branch direction A B C D (a) Derivation and calculation |QC| = AC|vC| = (40 × 10−4)(2.0) = .00800 m3/s = 8.0 L/s. Take inward flow positive: QA + QB + QC + QD = 0, QA = 8, quadQB = −3. C out : QD = −[8 −3 −8] = +3 L/s ; C in : QD = −[8 −3 + 8] = −13 L/s . (b) Result and interpretation An outlet D selects inward C. Then |vD| = |QD|/AD = .0130/.00500 = 2.60 m/s . Qin = 8 + 8 = 16, Qout = 3 + 13 = 16 L/s. Assign directions before combining flow rates; magnitudes alone do not conserve signed flux. Physical check Both possible branch assignments satisfy conservation, but only one matches the outlet indicator.

Question 2

Conservation can determine an unknown branch direction

At a steady water junction, branch A brings in 8.0L/s8.0\,\mathrm{L/s} and branch B carries out 3.0L/s3.0\,\mathrm{L/s}. In branch C the mean speed magnitude is 2.0m/s2.0\,\mathrm{m/s} through area 40cm240\,\mathrm{cm^2}, but its direction is unknown. A fourth branch D is also present. No storage occurs.

Diagram shown in the original SVG worksheet.

  1. Find the flow in D for each possible direction in C.

  2. If an indicator shows D is an outlet, which direction in C is physically consistent? Find D's speed if its area is 50cm250\,\mathrm{cm^2}.

Question 2 - Solution

Conservation can determine an unknown branch direction

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

|QC|=AC|vC|=(40×10−4)(2.0)=.00800m3/s=8.0L/s.|Q_C|=A_C|v_C|=(40\times10^{-4})(2.0)=.00800\,\mathrm{m^3/s}=8.0\,\mathrm{L/s}.Take inward flow positive:QA+QB+QC+QD=0,QA=8,quadQB=−3.Q_A+Q_B+Q_C+Q_D=0,\quad Q_A=8,quad Q_B=-3.Cout:QD=−[8−3−8]=+3L/s;C\ \text{out}:\ Q_D=-[8-3-8]=\boxed{+3\,\mathrm{L/s}};Cin:QD=−[8−3+8]=−13L/s.C\ \text{in}:\ Q_D=-[8-3+8]=\boxed{-13\,\mathrm{L/s}}.

(b) Result and interpretation

An outlet D selects inward C. Then|vD|=|QD|/AD=.0130/.00500=2.60m/s.\boxed{|v_D|=|Q_D|/A_D=.0130/.00500=2.60\,\mathrm{m/s}}.Qin=8+8=16,Qout=3+13=16L/s.Q_{\mathrm{in}}=8+8=16,\quad Q_{\mathrm{out}}=3+13=16\,\mathrm{L/s}.Assign directions before combining flow rates; magnitudes alone do not conserve signed flux.

Physical check

Both possible branch assignments satisfy conservation, but only one matches the outlet indicator.