```ENSC 388 Week #2, Tutorial #1– Dimensions and Units
Problem 1: Water flows through a pipe with diameter =2 in. If the average
velocity of water is 1 m/s, find mass flow rate of water in (lbm/s) and (kg/s).
d2
Consider density of water 62.1 lbm/ft and use m   V .
4
3
Solution
Step 1: Write out what you are required to solve for (this is so you don’t
Find:
m – Mass flow rate of water in (lbm/s) and (kg/s)
Step 2: Prepare a data table
Data
Value
d
2

62.1
V
1
Unit
in
 lbm 
 ft 3 


m
 s 
Step 3: Calculations
Part a) English Unit
2
(Eq1)
 lbm 
 1 ft 
 m   1 ft 
 lbm 
2
2
m  (62.1) 
  (2) in  
  (1)  s    0.3048 m   4.44  s 
3
4
12
in
  


 ft 




Part b) SI Unit
Using conversion factors the mass flow rate can be written in SI units.
M. Bahrami
ENSC388
Tutorial #1
1
 lbm   0.4536 kg 
 kg 
m  4.44 
2.014
0


 s 
 s   lbm 
(Eq2)
Step 4: Concluding Statement
 lbm 
 kg 
or
2.0140
The mass flow rate was found to be 4.44 
 s  .
 s 
Problem 2: A car goes with average velocity of 100 km/h. Find kinetic
energy of the car in [Btu] and [J].
100 km/h
2800 lbm
Solution
Step 1: Write out what you are required to solve for (this is so you don’t
Find:
KE: kinetic energy of the car in [Btu] and [J]
Step 2: Prepare a data table
Data
m
Value
2800
Unit
lbm
100
 km 
 h 
V
Step 3: Calculations
1
KE  mV
2
M. Bahrami
ENSC388
(Eq1)
2
Tutorial #1
2
Part a) English Unit
2
2
1
 km  1000 m   1 ft 
KE  (2800)lbm (100) 2    
 
 
2
 h   1 km   0.3048 m 
2
(Eq2)
2
 1 h   1 slug 
 slug . ft 2 


361400
 3600 s   32.174 lbm 
 s 2   361400 lbf . ft 



 

 ft 
Note: 1[lbf ]  1[ slug ]  1  2 
s 
 1 Btu 
KE  361400 [lbf . ft ]  
  465 Btu 
 778 lbf . ft 
(Eq3)
Part b) SI Unit
Using conversion factors the kinetic energy can be written in SI units.
(Eq4)
1054 J 
3


KE  465 Btu  

490

10
J
 1Btu 
Step 4: Concluding Statement
The kinetic energy was found to be 465 Btu  or 490  10 3 J .
Problem 3: Calculate power required to lift a 1ton mass to 30 yards above
the ground in 10 minutes. Express your result in [hp] and [kW].
Solution
Step 1: Write out what you are required to solve for (this is so you don’t
Find:
W : power required to lift a 1 ton mass to 30 yards elevation in [hp] and
[kW]
M. Bahrami
ENSC388
Tutorial #1
3
Step 2: Prepare a data table
Data
Value
m
1
z
t
30
10
Step 3: Calculations
Unit
[ton]
 yards
[min]
mgz
W 
t
(Eq1)
Part a) English Unit
 2000 lbm 
1
m
W  1ton (9.8) 2  2   30 yards 


10 [min]  1ton 
s 
(Eq2)
 1 slug   1 ft   3 ft 
 lbf . ft 

17988


 32.174 lbm   0.3048 m  1 yard 
 min 

 
 

 1 min 
1 hp
 lbf . ft  
W  17988 


 
  0.545 hp 
 min   550 lbf . ft / s   60 s 
(Eq3)
Part b) SI Unit
Using conversion factors the power can be written in SI units.
 0.746 kW 
W  0.545 hp  
  0.407 kW 
1
hp


(Eq4)
Step 4: Concluding Statement
The power was found to be 0.545 hp  or 0.407 kW  .
M. Bahrami
ENSC388
Tutorial #1
4
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