Set 6 2008
5.1*
d (u + gz ) = 0
(a)
(b)
Btu 1.29 u lbm o T = = = 11 F CV ,steel Btu 0.12 o lbm F Btu 1.29 u lbm o T = = = 1.3 F CV ,water Btu 1.0 o lbm F
5.3*
(a)
Note
as
; we employ the results of equation 5.7
ChE 218 Spring 2008 Exam 1 Name: _ Part 1 (30 points) CLOSED Book/Notes This part should take not longer then 10 minutes. Do any 5 of the six problems. Mark an `X' through the number you do not wish graded. When complete, turn this page over and com
2.7- 6
Energy Balance on a Flow System with a Pump and Heat Exchanger. Water stored in a large, well-insulated storage tank at 21.0 C and atmospheric pressure is being pumped at steady state from this tank by a pump at the rate of 40 m3 /hr. The mot
Problem 6.81 - rework Example 6.19 for particle settling in water at 68F . Dp := 10
-6
cP := 10
-2
poise
m
p := 2000
kg m
3
f := 1000
kg m
3
f := 1.002cP
Assuming that stokes law applies, velocity may be directly calculated [Equation 6.5
Set 07
6.13
r02 - r 2 P Recall the velocity profile for Poisueille Flow: V (r ) = - (eqn 6.8) 4 x and maximum velocity occurs at r = 0: Vmax = V (0) =
r02 P D02 P - = - 4 x 16 x Vmax D 2 P = 0 - which, 2 32 x
Subsequent work
Set 4 2008
4.2 (a)
(b)
Relative to the gun barrel, 2 ft 2000 mV 2 lbf s 2 s 3 K.E. = = 0.02 lbm = 1.242 10 ft lbf = 1.684 kJ 2 2 32.2lbm ft ft V0 2000 s V = V0 - gt; V = 0 when t = = = 62.11s g 32.2 ft s2 1 ft 1 ft 2 z = Vdt = V0 t - gt 2
Set 03 2.58* The pressure at the bottom of the dip tube is h (a) That is the depth at the end of the dip tube. The total depth is 4.807 + 0.5 =5.3074 ft. (b) If we use 2.000 psig, the calculated depth is 4.800 ft. The difference is 0.0074 ft or 0.15%
2.30* See Example 2.9.
lbf 1 ft in 2 t= = 0.05 ft = 0.60 in = 1.52 cm lbf 2 10,000 2 in 1000 From the larger table in Perry's equivalent to Appendix A.3 we see that a 12 inch diameter, schedule 80 pipe has a wall thickness of 0.687 inches. It would p
Set 1 2008
1.3*
= mass
volume:
For 100lbm
= 100 lbm
102
50lbm 50lbm lbm =102 3 + 3 3 ft 4.49 62.3lbm ft 62.3lbm ft
lbm ft 3 453.6g 1.63g = 3 3 lbm ft 28310cm cm 3 Discussion; this assumes no volume change on mixing. That is a good
kJ := 10 J Problem 1 Given/Directly Calculated D1 := 4.026in from A.2, page 598) := 6cp D2 := 2in D2 := = 0.497 D1 Part a) Assuming a discharge coefficient: (This will need to be checked later) Using Equation 5.18 (page 149) V2 := Cv
3
kPa := 1
Set 5 2008
4.6* System: the water path through the dam, from inlet to outlet., steady flow The initial and final velocities are essentially 0. Assume also the enthalpy does not change. Then, the only energy to be converted to work is potential energ