MEC151

Thermodynamics 1
Lecture 4

General Energy Analysis

Ahmed Saleh
Faculty of Engineering
Galala University
 Learning objectives
• Introduce the first law of thermodynamics.
• Energy balances and mechanisms of energy transfer to or
from a system.
• Define energy conversion efficiencies.

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2.6 The first law of thermodynamics
• While so far, we have independently considered various forms of energy (E, Q, W,
U), the 1st law of thermodynamics (conservation of energy principle) provides a
sound basis for linking them to each other during a process.
• The 1st law (in thermodynamics terms): For all adiabatic processes between two
specified states of a closed system, the net work done is the same regardless of the
nature of the closed system and the details of the process. Simply put, it states that:

The change in the total energy during an adiabatic process must be

equal to the net work done.

The decrease in The increase in the

potential energy total energy of a
of the falling rock potato in an oven is
exactly equals the equal to the amount
increase in kinetic of heat transferred
energy (when the to it (when moisture
air resistance is loss from the potato
negligible). is disregarded).

Energy can be neither created nor Heat transfer but no work

destroyed; it can only change forms. interactions.
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2.6 The first law of thermodynamics
→ Implicit in the first law statement is the conservation of energy.

W=0

Q=0

Q=0

Q=0

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 Energy Balance (Conservation of Energy)
• The net change (increase or decrease) in the total energy of the system
during a process is equal to the difference between the total energy
entering and the total energy leaving the system during that process.

Total energy Total energy Change in the total

entering the system leaving the system energy of the system

Ein – Eout = ΔEsystem (kJ)

→ In other words, the energy change of a system

during a process is equal to the difference between
the energy of the system at the beginning and at the
end of the process:

ΔEsystem = Efinal – Einitial = E2 – E1

→ For stationary closed systems (ΔKE = ΔPE = 0):

ΔE = ΔU 5
 Mechanisms of Energy Transfer
• Energy can be generally transferred to or from
a system in three forms:
1. Heat (Q) transfer to a system (Qin) increases its
internal energy (U), whereas heat transfer from a
system (Qout) decreases U.
2. Work (W) transfer to (done on) a system (Win)
increases its energy, whereas work transfer from
(done by) a system (Wout) decreases it.
3. Mass flow (m) into a system (min) increases its
energy, whereas mass flow out of a system (mout)
decreases it.

→ The energy balance can be written more explicitly as:

ΔEsystem = Ein – Eout = (Qin – Qout) + (Win – Wout) + (Emass, in – Emass, out)

Adiabatic process → Q = 0 No work interactions → W = 0 Closed system → Emass = 0

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 Mechanisms of Energy Transfer
• The energy balance can be expressed in the rate form as:
𝑬ሶ in – 𝑬ሶ out = dEsystem /dt (kW)

• The energy balance can also be expressed per unit mass as:
ein – eout = Δesystem (kJ/kg)

→ For a closed system undergoing a cycle, the initial

and final states are identical (ΔEsystem = E2 – E1 = 0),
thus:
Qin + Win = Qout + Wout

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Example 1: Cooling of a hot fluid in a tank
• A rigid tank contains a hot fluid that is cooled while being stirred by a
paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During
the cooling process, the fluid loses 500 kJ of heat and the paddle wheel
does 100 kJ of work on the fluid. Determine the final internal energy of
the fluid (neglect the energy stored in the paddle wheel).

→ Hint: Apply the energy balance: Ein – Eout = ΔEsystem

Solution:
Assume the tank is stationary → ΔKE = ΔPE = 0 → ΔE = ΔU
→ Applying the energy balance on the given closed system:
Ein – Eout = ΔEsystem
Wshaft, in – Qout = ΔU = U2 – U1
100 – 500 = U2 – 800
U2 = 400 kJ → Final internal energy
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Example 2: Annual Lighting Cost of a Classroom
• The lighting needs of a classroom are met by 30 fluorescent lamps, each
consuming 80 W of electricity. The lights in the classroom are kept on for
12 hours a day and 250 days a year. For a unit electricity cost of LE 0.8
per kWh, determine annual energy cost of lighting for this classroom.
Also, discuss the effect of lighting on the heating and air-conditioning
requirements of the room.
Solution:
→ The power consumed by the lamps when all are on & the operating hours/year:
Lighting power = Power consumed per lamp × No. of lamps
= 80 × 30 = 2400 W = 2.4 kW
The power consumed by the
Operating hours = 12 × 250 = 3000 h/year lamps becomes part of thermal
→ The amount and cost of electricity per year: energy of the room. Therefore,
the lighting system reduces the
Lighting energy = Lighting power × Operating hours heating requirements by 2.4
kW, but increases the air-
= 2.4 × 3000 = 7200 kWh/year
conditioning load by 2.4 kW.
Lighting cost = Lighting energy × Unit cost
= 7200 × 0.8 = LE 5760/year 9
2.7 Energy conversion efficiencies
• Efficiency is one of the most frequently used terms in thermodynamics &
it indicates how well an energy conversion process is accomplished.

Efficiency = Desired output/Required input

→ Efficiencies of mechanical & electrical devices: Fluid systems usually involve

increasing the pressure, velocity, and/or elevation of a fluid by supplying
mechanical energy by a pump (driven by a motor), OR extracting mechanical energy
from a fluid by a turbine & producing mechanical power (that can drive a
generator).

→ The combined or overall efficiency of the pump–motor & turbine–generator:

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Example 3: Power generation potential
• Consider a river flowing toward a lake at an average velocity of 3 m/s at a
rate of 500 m3/s at a location 90 m above the lake surface. Determine the
total mechanical energy of the river water per unit mass and the power
generation potential of the entire river at that location (ρ = 1000 kg/m3).
Solution:
∵ The change in the flow energy (P/ρ) is zero (P = Patm at
both the river and lake surfaces).
→ The mechanical energy of the river water per unit mass:
emech = pe + ke = gz + v2/2 emech = P/𝞀 + v2/2 + gz

emech = (9.81 × 90) + ((32)/2) = 887.4 J/kg 𝑚ሶ = 𝜌𝑉ሶ

The mass flow rate: 𝑊ሶ max = 𝐸ሶ mech = 𝑚ሶ emech

𝒎ሶ = 𝜌𝑉ሶ = 1000 × 500 = 500,000 kg/s

→ The power generation potential of the river water:
𝑬ሶ mech = 𝒎ሶ emech = 500,000 × 887.4 = 443.7 MW

→ Solve again ignoring the ke contribution & comment on the result. 11

Example 4: Power Generation from a Hydroelectric Plant
• Electric power is to be generated by installing a hydraulic turbine–
generator at a site 70 m below the free surface of a large water reservoir
that can supply water steadily at a rate of 1500 kg/s. If the mechanical
power output of the turbine is 800 kW and the electric power generation
is 750 kW, determine the turbine efficiency and the combined turbine–
generator efficiency of this plant. Neglect losses in the pipes.

Solution:
∵ The change in the flow energy (P/ρ) is zero and the change
in kinetic energy is negligible.
→ The mechanical energy of the river water per unit mass:
emech = pe = gz = 9.81 × 70 = 686.7 J/kg
→ The power generation potential of the water reservoir:
emech = P/𝞀 + V2/2 + gz
𝑬ሶ mech = 𝒎ሶ emech = 1500 × 686.7 = 1030.05 kW
𝑊ሶ max = 𝐸ሶ mech = 𝑚ሶ emech
→ The turbine efficiency: ηturbine = 800/1030.05 = 77.7 %

→ The combined turbine–generator efficiency = 750/1030.05 = 72.8 % 12

Reading assignment
• Read the following section in Chapter 2 of the textbook:
2-6 The first law of thermodynamics

• Review section 2-7 Energy conversion efficiencies.

→ Solve examples 2-10, 2-12, 2-13, 2-14 and 2-15.

→ Watch the following videos on pumps & hydropower generation:

https://www.youtube.com/watch?v=lmjIQqo8mX4
https://www.youtube.com/watch?v=PvJHjnELVSM

→ Check the following websites on Joule’s experiment:

https://www.geogebra.org/m/JPtWKxqe
https://www.thephysicsaviary.com/Physics/Programs/Labs/MechanicalEquivalentOfHeatLab/
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Concept questions
• The length of a spring can be changed by (a) applying a force to it or (b)
changing its temperature (i.e., thermal expansion). What type of energy
interaction between the system (spring) and surroundings is required to
change the length of the spring in these two ways?

• For a cycle, is the net work necessarily zero? For what kind of systems
will this be the case?

• Can the combined pump–motor efficiency be greater than either the

pump efficiency or the motor efficiency? Explain.

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