Exam2 Solution

Exam 2 Key 1. p1= 60 psig= 413.7 kPa p2= 0 psig ρ= 1,000 kg/m3 V1=10 ft/s=3.048 ft/s V2=13 ft/s=3.962 ft/s z1=2 ft=0.6096 ft z2=0

p1 − p2

a. F=

ρ

b. ∆T =

V12 − V22 + + g ( z1 − z2 ) =416.46 J/kg=139.33 lbf.ft/lbm= 0.179 Btu/lbm= 4482.7 ft2/s2 2

416.46 F = =0.0995 oC=0.179 oF 4184 CV

2. ρ= 800 kg/m3 µ=0.5 cp= 0.0005 Pa.s L=5,000 ft= 1524 m D= 5.5 inch=0.1397 m q=10,000 bbl/day=0.0184 m3/s p2=100 psig= 689.5 kPa

V=

Re =

0.0184 q = A π 0.1397 2

=1.2 m/s

4

ρVD =268,338 µ

So the flow is turbulent. Using moody chart, or solving the equation for ε/d=0.002/5.5, we obtain f= 0.0044.

L V2 ∆p = 4fρ =110.4 kPa D 2 p1=∆p+ p2=800.0 kPa=116.0 psig

3. ρ= 1,000 kg/m3 D=0.2 m L=500 m q=0.05 m3/s n=0.5 K= 3Pa.s0.5

a. Vavg=

0.2 q = A π 0.52

Re Power law =

=1.59 m/s

4 ( 2− n ) n 8 ρVavg D

K  2 ( 3n + 1) n 

n

=757.2

b. Re < 2,000, so (according to the Power Law plot attached) the flow is laminar. c. f = 16/Re=0.0211

dp 1 V2 =4fρ =535.2 Pa/m dx D 2 m = ρ.q= 50 kg/s dp W m= = ρ L 13.38 kW dx d. 13.38 kW/0.65= 20.59 kW =27.595 hp

4.

= F m (Vin − Vout ) =150 kg/s (30-(-30))= 9,000 N m=F/g= 917.7 kg= 2,023.3 lbm

5.

g fv 2 dL + dZ + = 0 2 g c Di ρ gc

dP

P=Pressure, lbf/ft2 (Please note that psi is not acceptable!!!) ρ= Density, lbm/ft3 g= 32.2 ft/s2 gc=gravitational conversion factor, 32.17 lbm.ft/lbf.s2 Z=vertical depth, ft f= Darcy friction factor (4×Fanning friction factor) v= velocity, ft/s L=measured depth, ft Di=diameter of well, ft

v 2 dL v 2 dL Since we have Darcy friction factor instead of Fanning, the friction term, 4 f , becomes f . (In 2 g c Di 2 g c Di the book the 2 was left out by mistake.)