PCB1043

Tutorial 1

BASIC CONCEPTS OF THERMODYNAMICS

1.What is the difference between pound-mass and pound-force?

Analysis: The lbf is the force unit in the English system which is a primary dimension,

while the lbm is the unit of mass as a derivative dimension. Both magnitudes, however, is

the same

2. What is the difference between intensive and extensive properties? Separate the list P

(pressure), F (force), u (velocity), v (specific volume), (density), T (temperature), a

(acceleration), m (mass) , L (length), t (time) and V (volume) into intensive, extensive and non-

properties. Is it possible to convert one extensive property to an intensive one?

Analysis: Intensive properties do not depend on the size (extent) of the system but

extensive properties do.

Intensive properties are independent upon mass: P, v, , T

Extensive properties scales with mass: V, m

Non-properties: F, a, L, t, u

If an extensive property divides by the mass or mole of system, it converts to an

intensive property.

3. At the beach, atmospheric pressure is 14.8psi. You dive 15 ft down in the ocean and you later

climb a hill up to 250 ft elevation. Assume the density of water is about 62.4 lbm/ft3 and the density

of air is 0.074 lbm/ft3. What pressure do you feel at each place?

Assumption(s): It is assumed the properties of air and water are constant versus height

and depth respectively.

Analysis: The pressure at each level is calculated by:

P= P0-*h*g/gc

The term,P0, is pressure at the datum depth.

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= ( ) = 14.8 62.4 1144 (15 0) 32.232.2 ( )= 21.3 = ( ) = 14.8 0.074 1144 (255 0) 32.232.2 ( )= 14.69

4. A laboratory room keeps a vacuum of 0.1 psi. What net force does that put on the door of size 6 ft

by 3 ft?

Assumption(s): It is assumed that the pressure

distribution on the door is uniform

Analysis: The net force on the door is the difference

between the forces on the two side applied by pressure difference between inside and outside.

= = = 0.1 ( ) 6 3 = 259.2 5. What is the smallest temperature in degrees Celsuis you can have? Kelvin?

Analysis:The lowest temperature is absolute zero which is at zero Kelvin at which

point the temperature in Celsius scale is negative

TK= 0 K = 273.15 C

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6. The standard acceleration (at sea level and 45 latitude) due to gravity is 9.80665 m/s2.

What is the force needed to hold a mass of 2 kg at rest in this gravitational field? How much

mass can a force of 1 N support?

Assumption(s): The friction force caused by air is negligible.

Analysis: Thereare two forces that are applied to the mass in

opposite directions. For holding the mass, acceleration or total of

forces must be zero.

ma = 0 = F = F - mg

F = mg = 2 9.80665 = 19.613 N

F = mg =>m = F/g = 1 / 9.80665 = 0.102 kg

7. A gasoline line is connected to a pressure gage through a double-U manometer as shown in the

figure. If the reading of the pressure gage is 370 kPa, determine the gage pressure of the gasoline

line.

Assumption(s): The density of each fluid is constant

in each column.

Analysis: The pressure at each level is calculated

by:P= P0- gh

PA+gas.ghgas. + HgghHg - oilghOil + wghw = Pgage PA= Pgage- wg ( hgasSGgas.+ hHgSGHg - hOilSGoil+ hw) PA= 370- 10009.81 (0.22 0.7+ 0.113.6 -0.5 0.79+0.45)/1000 PA=354.6 kPa

8. It is well-known that cold air feels much colder in windy weather than what the thermometer

reading indicates because of the chilling effect of the wind. This effect is due to the increase in

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the convection heat transfer coefficient with increasing air velocities. The equivalent wind chill

temperature in F is given by

= 91.4 (91.4 ) 0.475 0.0203 + 0.304 where V is the wind velocity in mi/h and Tambient is the ambient air temperature in F in calm air,

which is taken to be air with light winds at speeds up to 4 mi/h. Using proper conversion factors,

obtain an equivalent relation in SI units where V is the wind velocity in km/h and Tambient is the

ambient air temperature in C.

Analysis: The required conversion relations are 1 mi/h = 1.609 km/h and T(F) = 1.8T(C) + 32.

(F) = 91.4 (91.4 (F)) 0.475 0.0203 ( ) + 0.304( )

1.8 (C) + 32= 91.4 (91.4 1.8 (C) 32) 0.475 0.0203/1.609 (

) + 0.304/1.609(

) = 33.0 (33.0 ) 0.475 0.0126 + 0.240

9. A tank has two rooms separated by a membrane. Room A has 0.5lbm air and volume 18ft3, room B has 30ft3 air with density 0.05lbm/ft3. The membrane is broken and the air comes to a uniform state. Find the final density of the air.

Assumption(s): There is equilibrium in cylinder (P and T =

Constant)

Analysis:According to definition of = mV, thus we

must find total mass and volume.

Density is mass per unit volume

m = mA + mB = mA + BVB = 0.5 + 0.05 30 = 2lbm

V = VA + VB = 18 + 30 = 48ft3

= m/V = 2/48 = 0.042lbm/ft3

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1. 10. A hydraulic lift has a maximum fluid pressure of 80 psia. What should the piston-cylinder diameter be so it can lift a mass of 1600 lbm?

Assumption(s): The friction force caused by air is

negligible.

Analysis: There are two forces that are applied to the mass

in opposite directions. For holding the mass, acceleration or

total of forces must be zero since F=0 F1=F2.

With the piston at rest the static force balance is:

F = P A = F = mg/gc

A = D2/4

D4 = = 4 = 4 1600 80 1441 2232.232.2 = 0.42