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CHAPTER 2 The First Law and Other Basic Concepts ERT 206/4 Thermodynamics Miss. Rahimah Bt. Othman

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Presentation on theme: "CHAPTER 2 The First Law and Other Basic Concepts ERT 206/4 Thermodynamics Miss. Rahimah Bt. Othman"— Presentation transcript:

1 CHAPTER 2 The First Law and Other Basic Concepts ERT 206/4 Thermodynamics Miss. Rahimah Bt. Othman

2 COURSE OUTCOME 1 CO1) 1.Chapter 1: Introduction to Thermodynamics 2. Chapter 2: The First Law and Other Basic Concepts Define, discuss, apply and analyze internal energy, first law, energy balance-closed system, thermodynamic state and state function, equilibrium, the Phase Rule, reversible process, constant-V and constant-P processes, enthalpy and heat capacity. 3. Chapter 3: Volumetric properties of pure fluids 4. Chapter 4: Heat effects 5. Chapter 5: Second law of thermodynamics 6. Chapter 6: Thermodynamics properties of fluids

3 W meanwhile can be replaced by; Will be explained by Carnot group next week (2.6) (1.2) Combined both equations; (2.8) Equation 2.8 is a general equation for n moles of homogeneous fluid in a closed system experienced a mechanically reversible process For n moles of homogeneous fluid in a closed system; 2.9 CONSTANT-V and CONSTANT-P PROCESSES

4 W meanwhile can be replaced by; (2.6) (2.9) For the constant-V, closed system process, the heat transfer is equal to internal energy change of the system. 0 The integration of 2.9 yield (2.10) For constant-V process, the work is equal to zero CONSTANT-V PROCESS

5 Solved for dQ; (2.8) The term (U+PV) is known as enthalpy and has only mathematical definition, which is; Since P now is a constant, therefore (2.11) For constant-P process, CONSTANT-P PROCESS

6 Example kg of water Constant T = 100 O C Constant P = kPa vapor V water = m 3 /kg V vapor = m 3 /kg For this process, heat in the amount of 2, kJ is added to the water Q=2,256.9 kJ Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the constant temperature and pressure. Enthalpy

7 Since the process is a constant-pressure process; Q = nΔH So, ΔH = 2,256.9 kJ For ΔU; from eq 2.11, H = U+PV and ΔH = ΔU+ Δ(PV) ΔH = ΔU+ Δ(PV) ΔU = ΔH - P ΔV = 2,256.9 kJ –[ kPa ( )m 3 ] = 2,256.9 kJ – kPa m 3 = 2,087.5 kJ Aware of the units Solution

8 2.11 HEAT CAPACITY Let say we have two blocks of different metals. A is a copper and B is an aluminum. Mass for both blocks are 1 kg. Now, we supply 5000 J of heat to the blocks. What will happen??? - We are sure that temperature will rise, since the heat will stimulate the atomic energy in the metals. - The question are; a) how high the rise in temperature will occur? b) does the rise of temperature will be the same for both metals blocks?

9 C = dQ/dT Heat capacity is the amount of heat needed to produce a specified temperature change (∆T) in a system

10 Heat Capacity – Constant Volume This definition is true for both molar heat capacity and specific heat capacity. For constant volume process, This can be written as;

11 Comparing with Eq 2.10 Q = nΔU, thus Eq 2.18 become;

12 Heat Capacity – Constant Pressure This definition is true for both molar heat capacity and specific heat capacity. For constant pressure process, This can be written as;

13 Comparing with Eq 2.13 Q = nΔH, thus Eq 2.22 become; Try examples 2.9 and 2.10

14 2.12 Mass and Energy Balances for Open System Since most of the bioprocess eng problems are in open system situation, it is compulsory for those students to understand the mass and energy balances for open system. Measures of Flow

15 The measure of flow are interrelated; Importantly, and relate to velocity, u Try examples 2.11

16 Mass Balance for Open System Control volume – a space identified for analysis in open system Control surface – surface where it is separated from surrounding Fluid within the control volume is where thermodynamics analysis will be done.

17 The flow process known as steady state – condition within the control volume do not change with time. Therefore 0 Because specific volume is reciprocal of density 2.26 Mass Balance for Open System- cont’

18 General Energy Balances for Open System. The rate of change of energy within the control volume equals the net rate of energy transfer into the control volume. Stream flowing in and out of the control volume are associated with energy. Each stream will have total energy of; Each stream transport energy at rate;

19 Net energy transport into the system by flowing streams is The total rate of energy in the control volume including this quantity and heat transfer rate and work rate. work rate – may include work of several forms and has a set of average properties of P, V, U, H, etc.

20 Piston, supply the constant pressure done the work, PV, and work rate is; The net work done on the system is Another form of work is the shaft at the rate =

21 The equation now can be written as;

22 Usually, for many applications of bioprocess engineers, kinetic and potential energy are negligible, and simplify equation 2.28 to Try examples 2.12, 2.13 and 2.14

23 Energy Balances for Steady-State Flow Processes At steady state; Thus, equation 2.28 becomes Assume system has one entrance and one system

24 So far, all the energy unit are presume as Joule. For the English unit, the kinetic and potential energy unit have to be divided by the constant, g c. Therefore, Eq 2.32a becomes Usual unit for ΔH and Q is Btu. Kinetic, potential energy, work are in (ft lb f ). Try examples 2.15, 2.16 and 2.17

25 Thank you


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