Heat and Temperature
Heat is a form of energy. Temperature is the degree
of hotness of a body.
A healthy human body temperature is 37 0C or 98 0F
Specific Heat Capacities
Consider the following specific heat capacities:
Metal
lithium
gallium
Specific Heat (J/g K)
3.560
0.372
If 1000 joules of heat is added to 100 g samples of each of the metals which are all at the
same initial temperature, which metal will have the highest temperature?
Use Q=m.c.Ө ,
Say, lithium, so 1000 joules = (100 g)x(3.560)x(Ө)
Ө = (100 g x 3.560)/1000
Ө = 0.356 0C
So, for gallium, 1000 joules = (100 g)x(0.372)x(Ө)
Ө = (100 g x 0.372)/1000
Ө = 0.037 0C
Lithium has the highest temperature change. Hence lithium has the highest temperature.
Diagram on the right shows the common braking system
where pressure is transmitted equally in a closed system
through a fluid (oil).
Pascal’s Law, P=
F
A
The diagram shows a simple hydraulic system. A1 and A2 are cross sectional area of the
piston. Given areas A1 = 5 cm2 and A2 = 10 cm2. The minimum force F1 which can lift a load
F2 of 100 kg is:
𝐹
𝐴
(Use g = 10 m/s2 and 𝑃 = )
P=
F
A
=
F1
A1
=
F2
A2
F1 =
F1 =
F2
A2
x A1
100 kg ×10 m/s2
10
x 5 = 500 N
pg. 1
Machine and Engine Power
A machine is is a device that can change magnitude or line of action, or both magnitude and
line of action of a force ie. reduces human effort.
Engine power can be measured with a dynamometer/ dyno. The actual engine power that
actually reaches the output shaft or flywheel of an engine is also known as brake power.
Formula for engine power:
Indicated mean effective pressure,
imep= (effective area of indicator diagram/base length of the diagram)×constant
*constant is given in test/exam.
Indicated power (ip) is the power that is developed inside the engine cylinders
Example calculation:
In a test, a single-cylinder 4-stroke engine develops a mean effective pressure of 5 bar at a
speed of 3000 rev/min. The length of the engine stroke is 0.12 m and the cross-sectional
area of the cylinder bore is 0.008 m2. Calculate the indicated power of the engine in kW.
The engine is single-cylinder 4-stroke, so there is one working stroke for every two
revolutions.
N, the number of working strokes per second
N = 3000÷60/2 = 25; P = 5 bar = 500 000 N/m2,
l = 0.12 m; a = 0.008 m2.
Substituting these values in the formula gives
indicated power, ip = 500 000 × 0.12 × 0.008× 25/1000 kW = 12 kW
Heat, Specific Heat Capacity and Latent Heat
You may use Q=m.c. Ө and Q=m.Lf
1. Calculate the temperature change, ɵ, of 3 kg of aluminium when supplied with 5000 J of
heat? Given the specific heat capacity of aluminium is 950 J kg-1 0C -1.
Solution:
Use Q=m.c.Ө. So
5000 = 3 x 950 x Ө
Ө = 5000/(3x950)
Ө = 1.75 0C
2. Calculate the amount of heat, Q required to change 10 kg of ice into liquid water at 0°C,
given the specific latent heat of fusion is 335,000 J/kg.
Solution:
Q=m.Lf
Q = 10 x 335000
Q = 335,000 J
Q = 3.35 x 105 J
pg. 2
3. Calculate the amount of heat required to convert 5.0 kg of water at 20°C to steam at
100°C.
The specific heat capacity of water in this temperature range is 4200 J kg-1 0C-1 and the
specific latent heat at 100°C for liquid to vapour is 2,257,000 J/kg.
Solution:
Total heat, Qtotal = m.c.Ө + m.Lf
Qtotal = [5x4200x(100 – 20)] + [ 5 x 2,257,000]
Qtotal = 1,680,000 J + 11,285,000 J
Qtotal = 12,965,000 J = 12,965 kJ = 1.2 x 107 J
Machine
Terms: Mechanical advantage; Velocity ratio/ movement ratio; Machine efficiency
A simple machine raises a load of 160 kg through a distance of 1.6 m. The effort applied to
the machine is 200N and moves through a distance of 16 m. Taking g as 9.8 m/s2,
determine:
a) the force ratio,
b) movement ratio/ velocity ratio, and
c) machine efficiency (in percentage)
Solutions:
a) Force ratio = Load/effort
= 160kg/200 N
= (160 x 9.8) N/200 N
= 7.84
b) Velocity ratio (movement ratio) = distance moved by effort/distance moved by the load
= 16 m/1.6 m
= 10
c) Efficiency = (force ratio/ velocity ratio) x 100%
= (7.84/10) x 100
= 78.4%
Gears in engine
Use this ratio:
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒𝑟
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑟
=
𝑡𝑒𝑒𝑡ℎ 𝑜𝑛 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑟
𝑡𝑒𝑒𝑡ℎ 𝑜𝑛 𝑑𝑟𝑖𝑣𝑒𝑟
A driver gear on a shaft of a motor has 35 teeth and meshes with a follower
having 98 teeth. If the speed of the motor is 1400 revolutions per minute, find the
speed of rotation of the follower.
Solution:
Speed of driver =
speed of follower
teeth on follower
teeth on driver
1400 / speed of follower = 98/35
Speed of follower = (1400 x 35)/98
= 500 rev/min
pg. 3
A compound gear train similar to that shown in last figure
consists of a driver gear A, having 40 teeth, engaging with
gear B, having 160 teeth. Attached to the same shaft as B,
gear C has 48 teeth and meshes with gear D on the output
shaft, having 96 teeth. Determine;
(a) the movement ratio of this gear system
(b) the efficiency ratio when give, the force ratio is 6
(a) The speed of D = speed of A x TA/TB x TC/TD
movement ratio = NA/ND = TB/TA x TD/TC
= 160/40 x 96/48
=8
(b) The efficiency = (force ratio/movement ratio) x 100%
= 6/8 x 100
= 75%
ENGINE POWER
Use these equations:
1. Indicated Mean Effective Pressure, IMEP =
effective area of indicator diagram
base length of the diagram
× 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2. Indicated power (IP) = (P ×l×a×N)/1000 (in kW)
where P = mean effective pressure in N/m2,
l = length of engine stroke in m,
a = cross sectional area of cylinder bore in m2,
N = number of working strokes per second.
An indicator diagram taken from a single-cylinder 4-stroke engine has an effective area of
600 mm2. If the base length of the indicator diagram is 60 mm and the constant is
80 kPa/mm, calculate the indicated mean effective pressure.
Solution:
Indicated mean effective pressure or
(imep)= (effective area of indicator diagram/base length of the diagram)×constant
imep = (600/60)×80 kPa/mm
imep = 800 kPa = 8 bar
In a test, a single-cylinder 4-stroke engine develops a mean effective pressure of 5 bar at a
speed of 3000 rev/min. The length of the engine stroke is 0.12 m and the cross-sectional
area of the cylinder bore is 0.008 m2. Calculate the indicated power of the engine in kW.
Solution:
The engine is single-cylinder 4-stroke, so there is one working stroke for every two
revolutions.
N, the number of working strokes per second,
N = (no. of cylinders/2)x(no.of revolutions/second)= (4/2)x[(3000 revs/minute)x(1
minute/60 seconds)] = 100;
P = 5 bar = 500 000 N/m2, l = 0.12 m; a = 0.008 m2
Substituting these values in the formula gives
ip = (500 000) × (0.12) × (0.008) × (100) W = 48000 W = 48 kW
pg. 4