Week#2 , Tutorial#1–Dimensions and Units

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). Consider density of water 62.1 lbm/ft3 and use m d 2 V .

4

Solution

Step 1: Write out what you are required to solve for (this is so you don't forget to answer everything the question is asking for)

Find: m ? Mass flow rate of water in (lbm/s) and (kg/s)

Step 2: Prepare a data table

Data

d

V

Value 2

62.1

1

Unit

in

lbm ft 3 m s

Step 3: Calculations

Part a) English Unit

m

4

(62.1)

lbftm3

(2)2

in2

1 ft 12 in

2

(1)

m s

1 ft 0.3048

m

4.44

lbm s

(Eq1)

Part b) SI Unit

Using conversion factors the mass flow rate can be written in SI units.

M. Bahrami

ENSC388

Tutorial #1

1

m

4.44

lbm s

0.4536 lbm

kg

2.0140

kg s

Step 4: Concluding Statement

The

mass

flow

rate

was

found

to

be

4.44

lbm s

or 2.0140

kg s

.

(Eq2)

Problem 2: A car goes with average velocity of 100 km/h. Find kinetic energy of the car in [Btu] and [J].

2800 lbm

100 km/h

Solution

Step 1: Write out what you are required to solve for (this is so you don't forget to answer everything the question is asking for)

Find: KE: kinetic energy of the car in [Btu] and [J]

Step 2: Prepare a data table

Data

m

V

Value 2800

100

Unit

lbm

km h

Step 3: Calculations

KE 1 mV 2 2

(Eq1)

M. Bahrami

ENSC388

Tutorial #1

2

Part a) English Unit

KE

1 2

(2800)lbm

(100)2

km 2 h

1000 m 2 1km

1 ft 0.3048

2 m

1h 3600

s

2

1 slug 32.174 lbm

361400

slug. ft 2 s2

361400

lbf . ft

(Eq2)

Note:

1[lbf

]

1[

slug

]

1

ft s2

KE

361400

[lbf

.

ft ]

1Btu 778 lbf .

ft

465

Btu

(Eq3)

Part b) SI Unit

Using conversion factors the kinetic energy can be written in SI units.

KE

465Btu

1054J 1Btu

490

103

J

(Eq4)

Step 4: Concluding Statement

The kinetic energy was found to be 465 Btu or 490 103 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 forget to answer everything the question is asking for)

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

m

z t

Value 1

30 10

Unit

[ton ]

yards

[min]

Step 3: Calculations

W mgz t

Part a) English Unit

W

1ton

(9.8)

2

m s2

30

yards

10

1 [min]

2000 lbm 1ton

1 slug 32.174 lbm

1 ft 0.3048

m

1

3 ft yard

17988

lbf . ft min

W

17988

lbf . ft min

550

1hp lbf .

ft

/

s

1min 60 s

0.545

hp

Part b) SI Unit Using conversion factors the power can be written in SI units.

W

0.545hp

0.746 kW

1hp

0.407

kW

Step 4: Concluding Statement

The power was found to be 0.545 hp or 0.407 kW .

(Eq1) (Eq2)

(Eq3) (Eq4)

M. Bahrami

ENSC388

Tutorial #1

4

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