1 SI Units

The International System of Units (SI) is the globally accepted standard for measurement. Established to provide a consistent framework for scientific and technical measurements, SI units facilitate clear communication and data comparison across various fields and countries. The system is based on seven fundamental units: the meter for length, the kilogram for mass, the second for time, the ampere for electric current, the kelvin for temperature, the mole for substance, and the candela for luminous intensity.

Base SI units.
Physical Quantity	SI Base Unit	Symbol
Length	Meter	m
Mass	Kilogram	kg
Time	Second	s
Electric Current	Ampere	A
Temperature	Kelvin	K
Amount of Substance	Mole	mol
Luminous Intensity	Candela	cd

Derived SI units.
Physical Quantity	Derived SI Unit	Symbol
Area	Square meter	m²
Volume	Cubic meter	m³
Speed	Meter per second	m/s
Acceleration	Meter per second squared	m/s²
Force	Newton	N
Pressure	Pascal	Pa
Energy	Joule	J
Power	Watt	W
Electric Charge	Coulomb	C
Electric Potential	Volt	V
Resistance	Ohm	Ω
Capacitance	Farad	F
Frequency	Hertz	Hz
Luminous Flux	Lumen	lm
Illuminance	Lux	lx
Specific Energy	Joule per kilogram	J/kg
Specific Heat Capacity	Joule per kilogram Kelvin	J/(kg·K)

Common multiples and submultiples for SI units.
Factor	Prefix	Symbol
10⁹	giga	G
10⁶	mega	M
10³	kilo	k
10²	hecto	h
10¹	deca	da
10^-1	deci	d
10^-2	centi	c
10^-3	milli	m
10^-6	micro	µ

1.1 Quick Reference

Conversion Reference
Conversion Type	Relationship
Length	1 \(\text{m} = 10^3\) \(\text{mm} = 10^2\) \(\text{cm}\)
Area	1 \(\text{m}^2 = 10^6\) \(\text{mm}^2 = 10^4\) \(\text{cm}^2\)
Volume	1 \(\text{m}^3 = 10^9\) \(\text{mm}^3 = 10^6\) \(\text{cm}^3\)

1.2 Unity Fraction

The unity fraction method, or unit conversion using unity fractions, is a systematic way to convert one unit of measurement into another. This method relies on multiplying by fractions that are equal to one, where the numerator and the denominator represent the same quantity in different units. Since any number multiplied by one remains the same, unity fractions allow for seamless conversion without changing the value.

The principle of unity fractions is based on:

Setting up equal values: Write a fraction where the numerator and denominator are equivalent values in different units, so the fraction equals one. For example, \(\frac{1km}{1000m}\) is a unity fraction because 1 km equals 1000 m.
Multiplying by unity fractions: Multiply the initial quantity by the unity fraction(s) so that the undesired units cancel out, leaving only the desired units.

1.2.1 Classwork

Example 1.1 Suppose we want to convert \(5\) kilometers to meters.

Start with \(5\) kilometers: \[ 5 \, \text{km} \]
Multiply by a unity fraction that cancels kilometers and introduces meters. We use \((\frac{1000 \, \text{m}}{1 \, \text{km}}), since\:1 \, \text{km} = 1000 \, \text{m}\):

\[5 \, \text{km} \times \frac{1000 \, \text{m}}{1 \, \text{km}} = 5000 \, \text{m}\]

The kilometers \(\text{km}\) cancel out, leaving us with meters \(\text{m}\):

\[ 5 \, \text{km} = 5000 \, \text{m} \]

This step-by-step approach illustrates how the unity fraction cancels the undesired units and achieves the correct result in meters.

Unity fractions can be extended by using multiple conversion steps. For example, converting hours to seconds would require two unity fractions: one to convert hours to minutes and another to convert minutes to seconds. This approach ensures accuracy and is widely used in science, engineering, and other fields that require precise unit conversions.

Example 1.2 Convert \(15 \, \text{m/s}\) to \(\text{km/h}\).

Start with \(15 \, \text{m/s}\).
To convert meters to kilometers, multiply by \(\frac{1 \, \text{km}}{1000 \, \text{m}}\).
To convert seconds to hours, multiply by \(\frac{3600 \, \text{s}}{1 \, \text{h}}\).

\[ 15 \, \text{m/s} \times \frac{1 \, \text{km}}{1000 \, \text{m}} \times \frac{3600 \, \text{s}}{1 \, \text{h}} = 54 \, \text{km/h} \]

The meters and seconds cancel out, leaving kilometers per hour: \(54 \, \text{km/h}\).

1.2.2 Self-Test

Instructions:

Use unity fraction to convert between derived SI units.
Show each step of your work to ensure accuracy.
Simplify your answers and include correct units.

Speed
Convert \(72 \, \text{km/h}\) to \(\text{m/s}\).
Force
Convert \(980 \, \text{N}\) (newtons) to \(\text{kg} \cdot \text{m/s}^2\).
Energy
Convert \(2500 \, \text{J}\) (joules) to \(\text{kJ}\).
Power
Convert \(1500 \, \text{W}\) (watts) to \(\text{kW}\).
Pressure
Convert \(101325 \, \text{Pa}\) (pascals) to \(\text{kPa}\).
Volume Flow Rate
Convert \(3 \, \text{m}^3/\text{min}\) to \(\text{L/s}\).
Density
Convert \(1000 \, \text{kg/m}^3\) to \(\text{g/cm}^3\).
Acceleration
Convert \(9.8 \, \text{m/s}^2\) to \(\text{cm/s}^2\).
Torque
Convert \(50 \, \text{N} \cdot \text{m}\) to \(\text{kN} \cdot \text{cm}\).
Frequency
Convert \(500 \, \text{Hz}\) (hertz) to \(\text{kHz}\).
Work to Energy Conversion
A force of \(20 \, \text{N}\) moves an object \(500 \, \text{cm}\). Convert the work done to joules.
Kinetic Energy Conversion
Calculate the kinetic energy in kilojoules of a \(1500 \, \text{kg}\) car moving at \(72 \, \text{km/h}\).
Power to Energy Conversion
A machine operates at \(2 \, \text{kW}\) for \(3\) hours. Convert the energy used to megajoules.
Pressure to Force Conversion
Convert a pressure of \(200 \, \text{kPa}\) applied to an area of \(0.5 \, \text{m}^2\) to force in newtons.
Density to Mass Conversion
Convert \(0.8 \, \text{g/cm}^3\) for an object with a volume of \(250 \, \text{cm}^3\) to mass in grams.

1.2.3 Answer Key

\(72 \, \text{km/h} = 20 \, \text{m/s}\)
\(980 \, \text{N} = 980 \, \text{kg} \cdot \text{m/s}^2\)
\(2500 \, \text{J} = 2.5 \, \text{kJ}\)
\(1500 \, \text{W} = 1.5 \, \text{kW}\)
\(101325 \, \text{Pa} = 101.325 \, \text{kPa}\)
\(3 \, \text{m}^3/\text{min} = 50 \, \text{L/s}\)
\(1000 \, \text{kg/m}^3 = 1 \, \text{g/cm}^3\)
\(9.8 \, \text{m/s}^2 = 980 \, \text{cm/s}^2\)
\(50 \, \text{N} \cdot \text{m} = 5 \, \text{kN} \cdot \text{cm}\)
\(500 \, \text{Hz} = 0.5 \, \text{kHz}\)
\(20 \, \text{N} \times 5 \, \text{m} = 100 \, \text{J}\)
\(\text{Kinetic energy} = 1500 \, \text{kg} \times \left(20 \, \text{m/s}\right)^2 / 2 = 300 \, \text{kJ}\)
\(2 \, \text{kW} \times 3 \, \text{hours} = 21.6 \, \text{MJ}\)
\(200 \, \text{kPa} \times 0.5 \, \text{m}^2 = 100,000 \, \text{N}\)
\(0.8 \, \text{g/cm}^3 \times 250 \, \text{cm}^3 = 200 \, \text{g}\)

1.3 Condenser Vacuum

Condenser vacuum gauge reads 715 mmHg when barometer stands at 757 mmHg. State the absolute pressure in kN/m² and bar.

1.3.1 Given Data

\[ P_{atm} = 757~\text{mmHg}, \quad P_{vac} = 715~\text{mmHg} \]

1.3.2 Absolute Pressure in mmHg

\[ P_{abs} = P_{atm} - P_{vac} = 757 - 715 = 42~\text{mmHg} \]

1.3.3 Convert mmHg → kN/m²

\[ P = \rho g h = 13{,}600 \cdot 9.81 \cdot 0.001 = 133.416~\text{Pa} = 0.133416~\text{kN/m}^2 \]

\[ P_{abs} = 42 \cdot 0.133416 = 5.6034~\text{kN/m²} \]

1.3.4 Convert kN/m² → bar

\[ P_{abs} = \frac{5.6034}{100} = 0.056~\text{bar} \]

1.3.5 Final Answers

\[ \boxed{P_{abs} = 42~\text{mmHg} = 5.6034~\text{kN/m²} = 0.056~\text{bar}} \]

1.3.6 Code

P_atm_mmHg = 757
P_vac_mmHg = 715
MMHG_TO_KN_M2 = 0.133416
KNM2_TO_BAR = 1 / 100
P_abs_mmHg = P_atm_mmHg - P_vac_mmHg
P_abs_kNm2 = P_abs_mmHg * MMHG_TO_KN_M2
P_abs_bar = P_abs_kNm2 * KNM2_TO_BAR
print(f"Absolute Pressure = {P_abs_mmHg:.2f} mmHg")
print(f"Absolute Pressure = {P_abs_kNm2:.3f} kN/m²")
print(f"Absolute Pressure = {P_abs_bar:.4f} bar")

1.4 Oil Flow in Tubes

Oil flows full bore at a velocity of \(V = 2~\text{m/s}\) through 16 tubes of diameter \(d = 30~\text{mm}\). Density of oil: \(\rho = 0.85~\text{g/mL}\). Find volume flow rate (L/s) and mass flow rate (kg/min).

1.4.1 Cross-sectional area of one tube

\[ A = \pi \frac{d^2}{4} = \pi \frac{0.03^2}{4} \approx 7.0686 \times 10^{-4}~\text{m}^2 \]

1.4.2 Total area and volume flow rate

\[ A_\text{total} = 16 \cdot 7.0686 \times 10^{-4} \approx 0.01131~\text{m}^2 \]

\[ \dot{v} = A_\text{total} \cdot V \approx 0.02262~\text{m}^3/\text{s} \approx 22.62~\text{L/s} \]

1.4.3 Mass flow rate

\[ \dot{m} = \rho \cdot \dot{v} = 850 \cdot 0.02262 \approx 19.227~\text{kg/s} \approx 1153.6~\text{kg/min} \]

1.4.4 Final Answers

\[ \text{Volume flow rate: } \dot{v} \approx 22.6~\text{L/s} \] \[ \text{Mass flow rate: } \dot{m} \approx 1154~\text{kg/min} \]

1.4.5 Code

import math
v = 2.0
N = 16
d = 0.03
rho = 0.85 * 1000
A = math.pi * d**2 / 4
A_total = N * A
v_dot_m3_s = A_total * v
v_dot_L_s = v_dot_m3_s * 1000
m_dot_kg_s = rho * v_dot_m3_s
m_dot_kg_min = m_dot_kg_s * 60
print(f"Volume flow rate: {v_dot_L_s:.2f} L/s")
print(f"Mass flow rate: {m_dot_kg_min:.2f} kg/min")

1.5 Gauge Pressure

An oil of specific gravity (relative density) \(\text{SG} = 0.8\) is contained in a vessel to a depth of \(h = 2 \text{ m}\). Find the gauge pressure at this depth in kPa.

1.5.1 Gauge Pressure

\[ P_g = \rho g h \]

where

\(\rho = \text{density of fluid (kg/m³)}\)

g= acceleration due to gravity (9.81 m/s²)

h = depth (m)

1.5.2 Compute the density of oil using specific gravity

Specific gravity is defined as

\[ \text{SG} = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} \]

where \(\rho_{\text{water}} = 1000 \text{ kg/m³}\). Thus,

\[ \rho_{\text{oil}} = \text{SG} \times \rho_{\text{water}} = 0.8 \times 1000 = 800\ \text{kg/m³} \]

1.5.3 Compute the gauge pressure

\[ P_g = \rho g h = 800 \times 9.81 \times 2 \]

\[ P_g = 15696\ \text{Pa} \approx 15.7\ \text{kPa} \]

1.5.4 Answer

The gauge pressure at a depth of 2 m in the oil is:

\[ \boxed{15.7\ \text{kPa}} \]

1.5.5 Code

# Gauge Pressure Calculation for Oil

# Given data
specific_gravity = 0.8  # SG of oil
depth_m = 2.0           # depth in meters
g = 9.81                # acceleration due to gravity in m/s²
rho_water = 1000        # density of water in kg/m³

# Compute density of oil
rho_oil = specific_gravity * rho_water

# Compute gauge pressure (Pa)
P_g_Pa = rho_oil * g * depth_m

# Convert to kPa
P_g_kPa = P_g_Pa / 1000

# Print results
print(f"Density of oil: {rho_oil:.1f} kg/m³")
print(f"Gauge pressure at {depth_m} m depth: {P_g_Pa:.1f} Pa ({P_g_kPa:.2f} kPa)")

1.6 Absolute Pressure from Manometer Reading

A water manometer shows a pressure in a vessel of \(400\ \text{mm}\) below atmospheric pressure. The atmospheric pressure is measured as \(763\ \text{mmHg}\). Determine the absolute pressure in the vessel in kPa.

1.6.1 Relationship between absolute and gauge pressure

\[ P_\text{abs} = P_\text{atm} + P_\text{gauge} \]

Since the manometer shows a pressure below atmospheric, the gauge pressure is negative:

\[ P_\text{gauge} = - \rho_\text{water} g h \]

1.6.2 Convert atmospheric pressure to Pa using

\[ P = \rho g h = 13{,}600 \cdot 9.81 \cdot 0.001 = 133.416~\text{Pa} \]

\[ P_\text{atm} = 763 \times 133.416 \approx 101,801\ \text{Pa} \approx 101.8\ \text{kPa} \]

1.6.3 Compute gauge pressure

Water column height:

\[ h = 400\ \text{mm} = 0.4\ \text{m} \]

Density of water: \(rho_\text{water} = 1000\ \text{kg/m³}\), \(g = 9.81\ \text{m/s²}\)

\[ P_\text{gauge} = - \rho g h = - 1000 \times 9.81 \times 0.4 \]

\[ P_\text{gauge} = -3924\ \text{Pa} \approx -3.92\ \text{kPa} \]

1.6.4 Compute absolute pressure

\[ P_\text{abs} = P_\text{atm} + P_\text{gauge} \approx 101.8 - 3.92 \approx 97.9\ \text{kPa} \]

1.6.5 Answer

The absolute pressure in the vessel is:

\[ \boxed{97.9\ \text{kPa}} \]

1.6.6 Code

# Absolute Pressure Calculation from Water Manometer

# Given data
h_mm = 400               # manometer reading in mm (below atmospheric)
atm_mmHg = 763           # atmospheric pressure in mmHg
rho_water = 1000         # density of water in kg/m³
g = 9.81                 # gravity in m/s²
mmHg_to_Pa = 133.416     # conversion factor

# Convert manometer height to meters
h_m = h_mm / 1000

# Convert atmospheric pressure to Pa
P_atm_Pa = atm_mmHg * mmHg_to_Pa

# Gauge pressure (negative because below atmospheric)
P_gauge_Pa = - rho_water * g * h_m

# Absolute pressure
P_abs_Pa = P_atm_Pa + P_gauge_Pa

# Convert to kPa
P_abs_kPa = P_abs_Pa / 1000

# Print results
print(f"Atmospheric pressure: {P_atm_Pa:.1f} Pa ({P_atm_Pa/1000:.1f} kPa)")
print(f"Gauge pressure: {P_gauge_Pa:.1f} Pa ({P_gauge_Pa/1000:.2f} kPa)")
print(f"Absolute pressure in the vessel: {P_abs_Pa:.1f} Pa ({P_abs_kPa:.2f} kPa)")