A useful theoretical equation for computing the relation between pressure, velocity, and altitude in a steady flow of a nearly inviscid, nearly incompressible fluid with negligible heat transfer and shaft work^5 is the Bernoulli relation, named after Daniel Bernoulli, who published a hydrodynamics

Question

A useful theoretical equation for computing the relation between pressure, velocity, and altitude in a steady flow of a nearly inviscid, nearly incompressible fluid with negligible heat transfer and shaft [latex]work^5[/latex] is the Bernoulli relation, named after Daniel Bernoulli, who published a hydrodynamics textbook in 1738:

[latex]p_0 = p + \frac{1}{2} \rho V^2 + \rho gZ[/latex] (1)

where [latex]p_0[/latex] = stagnation pressure
p = pressure in moving fluid
V = velocity
ρ = density
Z = altitude
g = gravitational acceleration

[latex]^5[/latex]That’s an awful lot of assumptions, which need further study in Chap. 3.

(a) Show that Eq. (1) satisfies the principle of dimensional homogeneity, which states that all additive terms in a physical equation must have the same dimensions. (b) Show that consistent units result without additional conversion factors in SI units. (c) Repeat (b) for BG units.

Accepted Answer

Part (a)

We can express Eq. (1) dimensionally, using braces, by entering the dimensions of each term from Table 1.2:
[latex]\{ML^{-1}T^{-2}\} = \{ML^{-1}T^{-2}\} + \{ML^{-3}\} \{L^2T^{-2}\} + \{ML^{-3}\}\{LT^{-2}\}\{L\} = \{ML^{-1}T^{-2}\}[/latex] for all terms Ans. (a)

Table 1.2 Secondary Dimensions in Fluid Mechanics
Secondary dimension	SI unit	BG unit	Conversion factor
Area {[latex]L^2[/latex]}	[latex]m^2[/latex]	[latex]ft^2[/latex]	1 [latex]m^2 = 10.764 ft^2[/latex]
Volume {[latex]L^3[/latex]}	[latex]m^3[/latex]	[latex]ft^3[/latex]	1 [latex]m^3 = 35.315 ft^3[/latex]
Velocity {L[latex]T^{-1}[/latex]}	m/s	ft/s	1 ft/s = 0.3048 m/s
Acceleration {L[latex]T^{-2}[/latex]}	m/[latex]s^2[/latex]	ft/[latex]s^2[/latex]	1 ft/[latex]s^2[/latex] = 0.3048 m/[latex]s^2[/latex]
Pressure or stress {ML[latex]^{-1}T^{-2}[/latex]}	Pa = N/[latex]m^2[/latex]	lbf/[latex]ft^2[/latex]	1 lbf/[latex]ft^2[/latex] = 47.88 Pa
Angular velocity {T[latex]^{-1}[/latex]}	[latex]s^{-1}[/latex]	[latex]s^{-1}[/latex]	1 [latex]s^{-1}[/latex] = 1 [latex]s^{-1}[/latex]
Energy, heat, work {ML[latex]^2T^{-2}[/latex]}	J = N · m	ft · lbf	1 ft · lbf = 1.3558 J
Power {ML[latex]^2T^{-3}[/latex]}	W = J/s	ft · lbf/s	1 ft · lbf/s = 1.3558 W
Density {ML[latex]^{-3}[/latex]}	kg/[latex]m^3[/latex]	slugs/[latex]ft^3[/latex]	1 slug/[latex]ft^3[/latex] = 515.4 kg/[latex]m^3[/latex]
Viscosity {ML[latex]^{-1}T^{-1}[/latex]}	kg/(m · s)	slugs/(ft · s)	1 slug/(ft · s) = 47.88 kg/(m · s)
Specific heat {L[latex]^2T^{-2}\Theta^{-1}[/latex]}	[latex]m^2/(s^2 \cdot K)[/latex]	[latex]ft^2/(s^2 \cdot \circ R)[/latex]	1 [latex]m^2/(s^2 \cdot K)[/latex] = 5.980 [latex]ft^2/(s^2 \cdot \circ R)[/latex]

Part (b)

Enter the SI units for each quantity from Table 1.2:
[latex]\{N/m^2\} = \{N/m^2\} + \{kg/m^3\}\{m^2/s^2\} + \{kg/m^3\}\{m/s^2\}\{m\} = \{N/m^2\} + \{kg/(m \cdot s^2)\}[/latex]
The right-hand side looks bad until we remember from Eq. (1.3) that 1 kg = 1 N · [latex]s^2[/latex]/m.

[latex]F = \frac{ma}{g_c}, where g_c = 32.174 \frac{ft \cdot lbm}{lbf \cdot s^2}[/latex] (1.3)

[latex]\{kg/(m \cdot s^2)\} = \frac{\{N \cdot s^2/m\}}{\{m \cdot s^2\}} = \{N/m^2\}[/latex] Ans. (b)

Thus all terms in Bernoulli’s equation will have units of pascals, or newtons per square meter, when SI units are used. No conversion factors are needed, which is true of all theoretical equations in fluid mechanics.

Part (c)

Introducing BG units for each term, we have
[latex]\{lbf/ft^2\} = \{lbf/ft^2\} + \{slugs/ft^3\}\{ft^2/s^2\} + \{slugs/ft^3\}\{ft/s^2\}\{ft\} = \{lbf/ft^2\} + \{slugs/(ft \cdot s^2)\}[/latex]
But, from Eq. (1.3), 1 slug = 1 lbf · [latex]s^2[/latex]/ft, so that

[latex]\{slugs/(ft \cdot s^2)\} = \frac{\{lbf \cdot s^2/ft\}} {\{ft \cdot s^2\}} = \{lbf/ft^2\}[/latex] Ans. (c)

All terms have the unit of pounds-force per square foot. No conversion factors are needed in the BG system either.

Table 1.2 Secondary Dimensions in Fluid Mechanics
Secondary dimension	SI unit	BG unit	Conversion factor
Area { $L^2$ }	$m^2$	$ft^2$	1 $m^2 = 10.764 ft^2$
Volume { $L^3$ }	$m^3$	$ft^3$	1 $m^3 = 35.315 ft^3$
Velocity {L $T^{-1}$ }	m/s	ft/s	1 ft/s = 0.3048 m/s
Acceleration {L $T^{-2}$ }	m/ $s^2$	ft/ $s^2$	1 ft/ $s^2$ = 0.3048 m/ $s^2$
Pressure or stress {ML $^{-1}T^{-2}$ }	Pa = N/ $m^2$	lbf/ $ft^2$	1 lbf/ $ft^2$ = 47.88 Pa
Angular velocity {T $^{-1}$ }	$s^{-1}$	$s^{-1}$	1 $s^{-1}$ = 1 $s^{-1}$
Energy, heat, work {ML $^2T^{-2}$ }	J = N · m	ft · lbf	1 ft · lbf = 1.3558 J
Power {ML $^2T^{-3}$ }	W = J/s	ft · lbf/s	1 ft · lbf/s = 1.3558 W
Density {ML $^{-3}$ }	kg/ $m^3$	slugs/ $ft^3$	1 slug/ $ft^3$ = 515.4 kg/ $m^3$
Viscosity {ML $^{-1}T^{-1}$ }	kg/(m · s)	slugs/(ft · s)	1 slug/(ft · s) = 47.88 kg/(m · s)
Specific heat {L $^2T^{-2}\Theta^{-1}$ }	$m^2/(s^2 \cdot K)$	$ft^2/(s^2 \cdot \circ R)$	1 $m^2/(s^2 \cdot K)$ = 5.980 $ft^2/(s^2 \cdot \circ R)$

Question 1.3: A useful theoretical equation for computing the relation bet...

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