MIDTERM REVIEW
I. Chapter 1: Fluid Mechanics
1.
Density
-
Unit: (kg/m3 )
EXAMPLES:
2.
Pressure: The ratio of normal force to area
- Atmospheric pressure (po): pressure caused by air.
- Absolute pressure (p): All the pressure that exerted on the subject (can
be understood as all the weight of the fluid above the object)
-
Unit: 1 atm = 1.01 x 105 Pa = 1.01 x 105 N/m2
1 atm = 760 torr
Gauge pressure (ρ𝑔ℎ) is the difference between the absolute pressure and
the atmospheric pressure.
Formula to compute absolute pressure:
EXAMPLES:
1.
A diver is currently located at the depth of 50m in the ocean.
a) What is the gauge pressure at this point?
b) What is the absolute pressure? (the density of the sea water is 1025 kg/m3 , the
atmospheric pressure is 101.3 kPa)
The gauge pressure at the point of the diver:
pg ,= ρgh = 1025 x 9.8 x 50 = 502250 (pa)
The absolute pressure at the point of the diver:
pT = pair + pg = 101.3 + 502,250 = 603,55 (kPa)
2.
A vertical tub, open at the top to the atmosphere, contains 10 cm of oil floating on
20 cm of water. What is the gauge pressure (pressure in excess of atmospheric) at the
bottom of the tube? (ρoil = 0.6ρwater; ρwater = 1000 kg/m3).
April 2019
The gauge pressure at the bottom of the tube:
∆ρ= ρoilghoil + ρwaterg hwater
= 0.6 x 1000 x 9.8 x 0.1 + 1000 x 9.8 x 0.2 = 2548 (pa)
3.
Pascal’s principle
EXAMPLES:
1.
What force must be extended on the master cylinder of a hydraulic lift to support
the weight of a 2000kg car (a large car) resting on a slave cylinder? The master
cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter.
April 2013
4.
Archimedes’s Principle
Where
Fb: the buoyant force acting on the submerged part of the object (N)
ρ𝑓𝑙𝑢𝑖𝑑: density of the fluid (kg/m3)
V: volume of the fluid which is displaced by the object (m3)
→ If the object is fully submerged in water
V = V object
EXAMPLES:
1. A block of wood floats in water with one-third of its volume submerged. Determine
the density of the wood if the density of water is 1000 kg/m3
Since the block floats in the water:
Fnet = Fb - Fg = 0
→ Fb = Fg
⇔ ρwater gVsubmerged = mwoodg
⇔ ρwater g*⅓ Vwood = ρwoodVwoodg
⇔ ρwood = ⅓ ρwater = ⅓ * 1000 = 333.3 (kg/m3)
5.
The equation of continuity
6.
Bernoulli’s equation
EXAMPLES:
The horizontal pipe shown in Figure 1 has a cross-sectional area
of 40.0 cm2 at the wider portion and 10.0 cm2 at the constriction,
Water is flowing in the pipe and the volume flow rate from the pipe
is 6.00 L/s. Mass densities of water and mercury are 1 kg/L and
13,6 kg/s respectively. Find:
a) The flow speech at the wide and narrow portions
b) The height difference h of the 2 mercury columns
November 2018
II. Chapter 2: Heat, temperature and the first law of Thermodynamics
1.
Temperature
TC = TK - 273.15o (0oC = 32oF)
TF = 9/5 TC + 32o (5oC = 9oF)
2.
Thermal expansion
EXAMPLES:
The increase in the volume of the aluminum cup at 28oC:
VAl = Vo ∆T 3α= 100 x (28 -22) x 3 x 23 x 10-6 = 0.0414 (cm3)
The increase in the volume of the glycerin at 28oC:
Vglycerin = Vo ∆T β= 100 x (28 -22) x 5.1 x 10-4 = 0.306 (cm3)
The volume of spilled glycerin:
∆V = 0.306 - 0.0414 = 0.2646 (cm3)
A pendulum consists of a brass sphere with a diameter of 35.0 cm suspended from a steel cable
10.5 m long (both measurements are made at 20℃. The swinging sphere clears the floor by a
distance of only 2.00 mm when the temperature is 20.0℃. At what temperature will the sphere
begin to brush the floor?
April 2017
3.
Heat
The heat capacity C of an object is the amount of energy needed to raise the
temperature of the object by 1 degree .
Specific heat: The heat capacity per unit mass
EXAMPLES:
4.
The first law of thermodynamics
EXAMPLES:
A gas within a closed chamber undergoes a cycle shown in Figure 2. For the
cycle, calculate:
a) The change in internal energy of the gas
b) The net work done by the gas
c) The net heat transferred out of the gas
April 2019
5.
Heat transfer mechanisms
EXAMPLES: