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The application of Appropriate Technology

The following are the basic equations which can be used to derive the Continuity and Bernoulli Equations:

(a) Velocity = Distance Moved/Time

v=s/t

Metres/Second (m/s)(1)

Example: If I run 100 metres in 10 seconds (in a straight line), I have a velocity of 100/10 = 10 metres per second (m/s).

(b) Acceleration = Velocity Change/Time

A=\left ( V_2-V_1 \right )\setminus t

Metres/Second2 (m/s2)(2)

Example: If at the start of the 100m race I have a velocity of zero (I am in the blocks) and after 2 seconds I have a velocity of 10 m/s then my acceleration is (10-0)/2 = 5 metres per second per second (a = (v2 – v1)/t ). This means I increase my velocity by 5 metres/second every second that passes. This is my average acceleration.

(c) Force = Mass.Acceleration

F=M.a

Kilogramme Metres/Second2 (Kg.m/s2) or Newtons (N)(3)

Example: If I throw a bucket of water at you, you will feel a force when the water hits you because the water is decelerating. Suppose the water had a mass of 10 Kg , was travelling at 5m/s before it hit you and took 1 second to slow down. The deceleration of the water is 5/1 = 5 metres per second per second and so the force applied to you is 5 x 10 = 50 Newtons (F = M.a).

(d) Pressure = Force/Area

P=F/A

Newtons/Meter 2 (N/m2) or Pascals (Pa)(4)

Example: I wear my training shoes and step on your foot. If I then step on your foot wearing my stilettos, you will feel the difference in pressure. This is because the force (my weight) is the same, but the area over which it is applied (the heel of the stiletto) is smaller, so the pressure is higher. If I have a mass of 70 Kg, then my force due to gravity is 70 x 9.81 = 687 Newtons. If the area of my training shoe is 0.0075 m2 (75 cm2) then the pressure due to it on your foot is 687/0.0075 = 91600 N/m2 or Pa. If the area of my stiletto heel is 0.0001 m2 (1cm2) then the pressure due to it on your foot is 687/0.0001 = 6870000 N/m2 or Pa. This is why it hurts !

(e) Density = Mass/Volume

\rho=M/V

Kilogrammes/Meter3 (Kg/m3)(5)

Example: A bucket of water and a bucket of air do not have the same weight. This is because, although the volume of each substance is the same (a bucket full), the water is about 1000 times denser than air. Density (ρ, pronounced ‘ro’) is a measure of how closely the atoms or molecules of a substance are packed together. The density of water is 1000 Kg/m3. Which means that if I have 10 Kg of water it will occupy a volume of 10/1000 = 0.01 m3 (10 litres).

(f) Energy = Force.Distance Moved

E=F.S

Newton Metres (N.m) or Joules (J)(6)

Example: If I pull a bucket of water on a rope to the top floor of my house then I am expending energy (I get tired). This is because I am applying a force (through my arms) to lift the bucket through a distance h (the height of my house) against the force of gravity. If the water has a mass of 10 Kg and knowing that the gravitational acceleration on earth is 9.81 m/s/s, then the force required to lift the water is 10 x 9.81 = 981 Newtons. If the height of the house is 10 metres, then the energy required to lift the bucket is 981 x 10 = 9810 Joules (E = F.S).

(g) Power = Energy/Time

W=E/ t

Newton Metres/Second (N.m/s) or Watts (W)(7)

Example: If I lift the bucket from the previous example to the top of the house in 10 seconds then the power required to carry out the task is 9810/10 = 981 Watts (W = E/t).