chapter 7: energy & energy balances · chapter 7 energy & energy balances. heat: energy...

14
Chapter 7 Energy & Energy Balances

Upload: others

Post on 20-Oct-2019

31 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Chapter 7

Energy & Energy

Balances

Page 2: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Heat: Energy that flows as a result of temperature difference between system & surroundings.

- Direction of Heat: from high temperature tolower one.

- Heat is positive when it is transferred from surroundings to the system.

Work: Energy that flows as a response to any driving force other than temperature difference such as torque or voltage. Work is positive when it is done by the system on the surrounding

Adiabatic: Means no heat is transferred from or to the system.

Important Definitions

Page 3: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Specific Enthalpy ( 𝐻 ): 𝐻 = 𝑈 + P 𝑉

𝑈 = specific internal energy.P = total pressure. 𝑉 = specific volume.

• Absolute values of 𝐻 & 𝑈 are not known, but the change Δ 𝐻 or Δ 𝑈can be calculated.

• Therefore, tabulated values of 𝐻 & 𝑈 must have a reference state.

Reference State: is to choose a temperature, pressure & a state of aggregation as a reference, where 𝐻 or 𝑈 is equal to zero.

- Then, the rest of the values of 𝐻 & 𝑉 in that table will be the change in 𝐻 or 𝑈 from the tabulated state to the reference state.

Important Definitions

Page 4: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Steam Tables: (Tables B.5, B.6, & B.7)Reference state for all steam tables is liquid water at triple point ( T = 0oC, P ≈ 0 bar )

Check tables:B.5: Saturated steam (temp)B.6: Saturated steam (pressure)B.7: Super heated steam & saturated water (need both temp. & pressure).

Saturated liquid: It means that the liquid has reached the maximum temperature that it can reach under certain pressure before the first bubble is formed. At this pressure & at a temperature below the maximum temperature, then the liquid is called sub-cooled liquid.

sameNeedOnly one Info.

Tables of thermodynamic data (Example: Water)

Page 5: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

• Saturated steam (vapor): it means the vapor that reaches its

minimum temperature at certain pressure without condensing the

first drop of liquid. At this pressure & temperatures greater than

this minimum temperature, the vapor is called superheated.

Tables of thermodynamic data (Example: Water)

Page 6: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction
Page 7: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction
Page 8: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction
Page 9: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Example 1: Saturated water @ P = 10 bar, find 𝑈 , 𝐻 and 𝑉.

Temperature of water = 179.9oC (Table B.6) 𝑈 = 761.5 kJ/kg 𝐻 = 762.6 kJ/kg 𝑉 = 0.001127 m3/kg

Example 2:What is the change in specific enthalpy of water (Δ 𝐻) when its conditions are changed from saturated water @ 10 bars to 300oC and 5 bar.

Saturated water (10 bar) 𝐻 = 762.6 kJ/kg (Table B.6)H2O, T = 300oC, P = 5 bar 𝐻 = 3065 kJ/kg (Table B.7)

∴ Δ 𝐻 = 3065 – 762.6 = 2302.4 kJ/kg

Page 10: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

𝑄 - 𝑊𝑠 = Δ 𝐻 + Δ 𝐸𝑘 + Δ 𝐸𝑝

𝐸𝑘 = 𝑚𝑢2

2: u = velocity

𝐸𝑝 = 𝑚 g z : z = elevation

∴ 𝑄 − 𝑊𝑠 = Δ 𝐻

Δ 𝐻 = 𝑜𝑢𝑡𝑚𝑖 𝐻𝑖 − 𝑖𝑛𝑚𝑖

𝐻𝑖

i = # of components

SystemInputStreams

OutputStreams

𝑄 𝑊𝑠

For chemical processesΔ 𝐸𝑘 ≈ 0Δ 𝐸𝑝 ≈ 0

Heat shaft change change changework in in in

enthalpy kinetic Energy Potential Energy

Energy Balance on Open Systems:

Page 11: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Example 3:

If the turbine is adiabatic, calculate maximum power that can be generated

from the turbine.

𝑄 𝑊𝑠

SteamTurbine

𝑚 = 2000 kg/hrP = 10 barT = 370oC

Saturated SteamP = 1 bar

Page 12: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Example 4:

Calculate the required heat input to the boiler?

Boiler

𝑄 = ?

𝑚3

SteamP = 17 barSaturated

𝑚1 = 120 kg/min H2OLiquid, T = 30oC

𝑚2 = 175 kg/minLiquid, T = 65oC

Page 13: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Example 5:

Calculate the required 𝑄 for the following cooler?

CoolerP = 1 atm

𝑄 𝑊𝑠

10 mol CO2/sec20 mol H2O/secT = 600oC T = 200oC

Page 14: Chapter 7: Energy & Energy Balances · Chapter 7 Energy & Energy Balances. Heat: Energy that flows as a result of temperature difference between system & surroundings. - Direction

Example 6:

Find 𝑚2 & 𝑚3

AdiabaticMixer

𝑚3 kg H2O/hr300oC, 1 atm

𝑚1 = 1150 kg H2O /hrSaturated vaporP = 1 atm

𝑚2 = kg H2O/hr1 atm, T = 400oC