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21-Mat-A2 Materials Transport Phenomena · December 2018

Question 7 of 8: Pump Power for a Direct-Chill Casting Mold Water Circuit

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2018 — Met-A2 Metallurgical Rate Phenomena. Three-hour, open-book exam; any Casio or Sharp approved calculator permitted. Candidates answer Question 1 (compulsory) plus any four of Questions 2–8; the five answered count equally (20 marks each). All eight are solved below for completeness. Candidates were told to state any interpretive assumptions where doubt exists.

Reference texts: Geiger, G. H. & Poirier, D. R., Transport Phenomena in Materials Processing (TMS) — the primary reference for the momentum-, heat- and mass-transfer analyses in Questions 3, 5, 6, 7 and 8; Szekely, J., Evans, J. W. & Sohn, H. Y., Rate Phenomena in Process Metallurgy — reactor-network and interfacial mass-transfer modelling (Questions 2, 3); Welty, J. R. et al., Fundamentals of Momentum, Heat and Mass Transfer — dimensional analysis and pipe-friction data (Questions 4, 7); Gaskell, D. R., Introduction to the Thermodynamics of Materials — Fe–C phase relations (Question 1h).

Note on the questions
Question 8 specifies a fill hole of explicit length $L$ and asks whether to increase $L$, and it is solved below from first principles. Question 5 asks for the maximum thickness of slab and adds an explicit instruction to justify the four simplifying assumptions rather than merely state them.

Question 7: Pump Power for a Direct-Chill Casting Mold Water Circuit (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

WaterSupplyPumpPWater-cooled moldIngotwater level (tank), z₁3 m3 mgauge P: 1.22×10⁵ N/m²Q = 3.93×10⁻³ m³/s, D = 30.5 mm
Fig. 7 — tank → pump → mold manifold (gauge P) → ingot water circuit.

Given.

QuantitySymbolValue
Volume flow rate$Q$$3.93\times10^{-3}$ m$^3$/s
Pipe diameter$D$30.5 mm
Total straight pipe length$L$9.14 m
Elbow equivalent-length ratio$(L_e/D)_{elbows}$25
Fanning friction factor$f$0.004
Entrance loss coefficient$e_{s,entry}$0.4
Exit (enlargement) loss coefficient$e_{s,exit}$0.8
Tank surface pressure$P_1$$1.0133\times10^5$ N/m$^2$
Gauge P pressure$P_2$$1.22\times10^5$ N/m$^2$

Find. Theoretical pump power (W) needed to deliver the stated flow from the tank surface to gauge point P.

Approach. Apply the extended (mechanical-energy) Bernoulli equation between the tank free surface (1, atmospheric, negligible velocity) and the gauge point P (2, known pressure and velocity), computing frictional losses over the piping run from the given length, elbow, entrance and exit data, then solve for the pump work per unit mass.

  1. Pipe velocity at P. $A=\pi D^2/4=\pi(0.0305)^2/4=7.31\times10^{-4}$ m$^2$, so $$v=\dfrac{Q}{A}=\dfrac{3.93\times10^{-3}}{7.31\times10^{-4}}=\boxed{5.38\ \text{m/s}}$$
  2. Friction + fitting losses (tank to P), Fanning convention. $$h_f=4f\left(\dfrac{L}{D}+\left(\dfrac{L_e}{D}\right)_{elbows}\right)\dfrac{v^2}{2}+(e_{s,entry}+e_{s,exit})\dfrac{v^2}{2}$$ $$=4(0.004)\!\left(\dfrac{9.14}{0.0305}+25\right)\!\dfrac{(5.38)^2}{2}+(0.4+0.8)\dfrac{(5.38)^2}{2}$$ $$=4(0.004)(299.7+25)(14.5)+(1.2)(14.5)=75.3+17.3=\boxed{92.5\ \text{J/kg}}$$
  3. Elevation change, tank surface to P. Both the tank water level and the pipe's vertical rise are quoted as 3 m from the pump (the low point of the circuit per the layout diagram), so the tank free surface and gauge P sit at the same elevation: $z_2-z_1=0$.
  4. Kinetic-energy term at station 2. The paper states that "the kinetic energy of the water within the manifold portion of the mold is negligible", so $v_2\approx0$ and there is no velocity-head term at station 2. That statement and the supplied $e_{s,exit}=0.8$ are a matched pair describing one physical event: the pipe's velocity head $v^2/2$ is dissipated in the sudden enlargement into the manifold and is already carried inside $h_f$ (Step 2). Adding $v^2/2$ again as a station-2 kinetic energy would charge the same 14.5 J/kg twice.
  5. Extended Bernoulli, solve for pump work per unit mass. $$w_{pump}=\dfrac{P_2-P_1}{\rho}+\dfrac{v_2^2-v_1^2}{2}+g(z_2-z_1)+h_f$$ $$=\dfrac{1.22\times10^5-1.0133\times10^5}{1000}+0+0+92.5=20.7+92.5=\boxed{113.2\ \text{J/kg}}$$
  6. Pump power. Mass flow $\dot m=\rho Q=(1000)(3.93\times10^{-3})=3.93$ kg/s, so $$\boxed{P_{pump}=w_{pump}\,\dot m=(113.2)(3.93)\approx445\ \text{W}\ (\approx0.45\ \text{kW})}$$
QuantityValue
Pipe velocity, $v$5.38 m/s
Friction + fitting head loss, $h_f$92.5 J/kg
Pump specific work, $w_{pump}$113.2 J/kg
Theoretical pump power≈ 445 W
Check
Three assumptions are load-bearing: (i) the station-2 velocity head is omitted, on the paper's own instruction that the manifold kinetic energy is negligible — the pipe's $v^2/2$ leaves the system in the sudden enlargement and is already inside $h_f$ through $e_{s,exit}=0.8$, so carrying both would double-charge it (14.5 J/kg, which would inflate the answer to 127.6 J/kg and 502 W); (ii) $f=0.004$ is taken as a Fanning friction factor (hence the $4f(L/D)$ form) — a Darcy factor of 0.004 would be implausibly low for turbulent flow in a 30.5 mm pipe; (iii) the tank water level and the pipe's vertical rise, both quoted as “3 m,” are read as measured from the same low point (the pump, per the layout diagram), making the tank surface and gauge P co-elevational. The elbow $L_e/D=25$ is applied as a single combined total for the circuit, since no elbow count is given.