Mathematical Methods of Conceptual Engineering Examples

Mathematical Methods 1
MATHEMATICAL METHODS OF CONCEPTUAL ENGINEERING EXAMPLES
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Mathematical Methods 2
Mathematical Methods of Conceptual Engineering Examples
Task 1A
i) Prove the equation for speed of sound in any fluid
Bulk modulus, K, is given by K= −dP
( dV
v )
Where dP = change in pressure, V = initial volume and dV = change in volume (Greshnyakov &
Belenkov, 2014).
According to conservation of mass principle, density x volume = constant i.e. ρv = c
→ d(ρv) = 0; differentiating this by parts gives
ρdv + vdρ = 0
ρdV = - vdρ; collecting like terms gives
−dv
v = dρ
ρ ; substituting −dv
v in the equation for bulk modulus gives
dP
dρ = K
ρ
But dP
dρ = K
ρ
But velocity of sound, Vs= √ dP
dρ = √ K
ρ
In dimensional analysis, the units for K is N/m2 and the units for ρ is kg/m3

Mathematical Methods 3
In dimensional analysis, N = force = ML
T ² (where M = mass in kg, L = length in m and T = time
in seconds)
Thus Vs= √ K
ρ = kg∗m∗m³
s2∗m2∗kg = √ m ²
s ² =m/s
Thus Vs is i
ii) Maximum permissible flow of water
Calculate the maximum permissible flow of water in the pipeline if maximum pressure in the
pipeline is 94 x 105 N/m2 using the equation ΔP = ρVvs
units for maximum permissible flow of water through a pipe
Calculate the maximum permissible flow of water in the pipeline if maximum pressure in the
pipeline using the equation ΔP = ρVvs
According to Bernoulli equation, velocity s determined as follows:
Vs= √ 2 ΔP
ρ = √ 2 x 94 x 105 N /m2
1000 kg/m3 = 137.11 m/s
Flow rate = Vs x cross-sectional area of the pipe
Cross-sectional area of pipe = πr2 = π x (0.075m)2 = 0.01767m2
Flow rate = 137.11 m/s x 0.01767 m2 = 2.423 m3/s
how to convert m3/s into ft/s using dimensional analysis