**Among the important laws of thermodynamics there are two very important law is the law of conservation of mass and energy. Basically the law states that mass or energy balance is made before and after the system must undergo a process of balance.**

3.1. MASS CONSERVATIONAt a closed system where the mass is constant is the mass conservation law, and does not require a mathematician to explain persaaman. In the open system in which the volume The system is set and there is mass flow equation so it is necessary to explain it.

Control volume in a flow fieldAt time t (initial) surface and set the system boundaries coincide. Δt after the mass within the control volume moves across the surface of the set. This is shown by the limit of the system at time t + Δt. Space (volume) which includes the mass at time t and time t + Δt visualized as parts I, II and III. Since the mass of the system is constant over time:

Interest pertma a mass change in bagia II, at Δt approaches zero section II coincides with volune set. So the change in part II is identical to the changes in the volume set.

The mass of the parts I was reduced or not increased with time. For Δt approaches zero mass change on the part I represent the mass that enters the control volume (). And mass changes on the part III represents the mass that comes out of the control volume (), then:

In the case of steady state:

Mass per unit time (mass flow rate) is the product of velocity by the cross-channel flow: V x A (m3/sc), to make berunit kg / sc masses should be multiplied by Jenia fluidanya. ( kg/m3).

A channel system as shown below the scheme. Steady state water flow through the apparatus. What is the flow rate in channel 1Answer:Note:Channel system with a data rate and channel diameterAsked:Flow velocity in channel 1?Scheme:

Assumptions:A. Incompressible flow2. Uniform flow3. Steady state conditions (steady state)Analysis:The law of conservation of mass gives the equation:, For steady state flow, so that:

, Incompressible

3.2. CONSERVATION OF ENERGYThere are two Shittim thermodynamics, closed systems and open systems. Therefore there are two energy conservation equations. Two equations are generated from the I law of thermodynamics. The law basically states that the energy balance before and after the process must balance.3.2.1. Energy conservation equation for Closed SystemsFor a closed system holds:

12)In this case the heat of entry marked Q + Q,-Q heat marked out. Work entry marked W + W and-W marked out of work.

Process P is constant, V is constant and T is constant and the adiabaticE is the energy of the system:13)U: energy (J)

: Kinetic energy, with m: mass (kg) and V: velocity (m / sc)mgh: potential energy, denagn g: acceleration of gravity (= 9.8 m/sc2) and h: elevation / altitude (m)For a closed system of kinetic and potential energy there is no (zero)or14)Can also be written per unit mass:15)If taken derifatifnya;16)17)3.2.1.1. Constant Pressure ProcessRepresented by the line 1-2 Figure 3.2.

18)The definition of enthalpy is:

, Where dP = 0 for the constant pressure, Referring to the press. 18) then:

For an ideal gas with constant specific heat capacity effect:19)20)For constant pressure reversible process:21)Ideal gas equation is:, With R (kJ / kg K): the gas constant table B622)Of the press. 19, 20 and 22 obtained:

23)Example 2:A pair of piston and cylinder containing air is expanded at constant pressure 30 atm, temperature of 1000 K to 400 K. The air mass 0.5 kg. Get Heat, work, changes in enthalpy and energy changes.Scheme:

Analysis:Closed system, constant pressure process the balance enrgi:

The process of constant pressure Q = ΔH

3.2.1.2. Constant Volume ProcessRepresented by the line 1-3 Figure 3.2., For a constant volume process = 0 so:Example 3A sealed container contains 1 m3 air-pressure-temperature 273 K. 344 800 N/m2 Heat put into the container until the temperature to 600 K. How much heat is added.Scheme:

Analysis:Closed system, the volume of container remains, then:

Ideal gas equation:

3.2.1.3. Constant temperature processRepresented by the line 1-4 Figure 3.2.

Ideal gas: as T1 = T4 or acquired, or

Because T is constant then ΔU = 0, so

Example 4:A compressor to suck the air at a pressure of 101 000 N/m2, temperature 30oC, the initial volume of 0.00078 m3 and a density of 1.23 kg/m3. After compressing the isotermik pressure 10 times its original pressure. What is the volume end of the process and also how much work is needed?Scheme

Analysis:Closed system of constant temperature processes:

, Obtainedor

3.2.1.4. Adiabatic ProcessRepresented by the line 1-5 Figure 3.2., With Q = 0For the adiabatic process persamanan line is:Where k; adiabaitk constants.So that work:

Example 5:A compressor to suck the air at a pressure of 101 000 N/m2, temperature 30oC, the initial volume of 0.00078 m3 and a density of 1.23 kg/m3. After compressing the adiabatic pressure is 10 times the original pressure. What is the final temperature and process work and how many are needed?Scheme:

Analysis:Adiabatic compression process, then:

cp = 1.004 kJ / kg K, cv = 0718 kJ / kg K, then

3.2.2. Energy conservation equation for Open SystemsAt the time tmassa set is the number of parts I and II. At time t + Δt some mass moves to the III, the interaction of heat and mechanical work on the mass set. Energy on the mass balance after set intervals of time Δt is:

W * is the fluid mechanical work done, if the cross-sectional area on the surface of the set A and the fluid moves as far as dx, then: fluid work done is:, A volume A dx = dx = dm / dm = υ

For the case of steady state:

For steady flow (steady flow) in = m * m * = m * it out:

If the total energy substitutable volume set, and replaced an index in the index out to be replaced 2 is obtained:

Enthalpy, the

Example 5:An electricity power plant's steam turbine steam at a pressure receive of 20 MPa at a velocity of 10 m / sc. The steam leaves the turbine at 0.1 MPa, with velocity of 1.5 m / sc. The turbine entrance is 12 m above the exit level. Find the work output if the mass flow rate is 3 tons / day and lost of heat transfer is Q * = 8000 kJ / hr. The steam has tabulated the Following Properties:Term Inlet Exit20 MPa pressure 0.1 MPATemperature 365.8 C 99.6 CVelocity 10 m / sc 1.5 m / scSpecific internal energy 2293.2 kJ / g 417.5 kJ / kgSpecific volume m3/kg t 0.001043 0.005836 mt3/kgWere:Work output?Scheme:

AssumptionsA. Steady state, steady flow2. Constant properties3. Friction is ignoredAnalysis:Open system steady state steady flow, the energy balance gives the equation:

Written per unit mass of the

Example 6:A ship's steam turbine steam at a pressure receive of 6205 kPa at a velocity of 30 m / sc. The steam leaves the turbine at 9899 kPa, with velocity of 3.5 m / sc. The turbine entrance is 12 m above the exit level. Find the work output if the mass flow rate is 15 kg / sc and lost of heat transfer is Q * = 14 kW. The steam has tabulated the Following Properties:Term Inlet ExitPressure 6205 kPa 9899 kPaTemperature 811.1 K 318.8 KVelocity 30 m / 3.5 m sc / scSpecific internal energy 3150.3 kJ / g 2211.8 kJ / kgSpecific volume m3/kg 0.005789 mt3/kg 13:36

Example 7:A nozzle is used to convert kinetic energy into enthalpy. Air enters the nozzle with a pressure 400 psia and at 100 ft / sc and enthalpy 240.98 BTU / LBM. Pressurized air nozzle exit 100 psia, and the enthalpy 162.98 BTU / LBM. Air mass flow rate 600 LBM / hr. (A) If the heat loss by 2 BTU / LBM calculated air velocity nozzle exit. (B) If the nozzle is insulated against heat (heat loss q * = 0, is also undergoing a process called adiabatic) calculated nozzle exit velocity.Note:Nozzle to convert the enthalpy to velocity.Were:a). Speed 2BTU/lbm out if the heat loss?b) The speed of exit if the adiabatic?Scheme:

Assumptions:A. Steady state, steady flow2. Constant properties3. Friction is ignoredAnalysis:Open system steady state, steady flow, the energy balance gives the press:a) The heat loss:Total energy in = energy out of the total.

Description:A. The location of the center point of the inlet and exit together so z1 = z22. For the British system known standard gravity factor gc = 32.17 ft/sc23. Also known as the conversion factor from ft (unit of potential energy) into BTU / LBM is factor

b) Without heat loss:

Example:Pump water for household needs brkapasitas 2.5 l / sc. The emnghisap Ompa water from the wells with a depth of 4 m from the axis of the pump shaft, then raising it above the tank on the roof with a height of 8 m. Water temperature 30 ° C, with a density of 998 kg/m3. Heat loss at the pump at 5 W. Compute power is required if the total efficiency of the pump and the driving motor is 70%.

3.1. MASS CONSERVATIONAt a closed system where the mass is constant is the mass conservation law, and does not require a mathematician to explain persaaman. In the open system in which the volume The system is set and there is mass flow equation so it is necessary to explain it.

Control volume in a flow fieldAt time t (initial) surface and set the system boundaries coincide. Δt after the mass within the control volume moves across the surface of the set. This is shown by the limit of the system at time t + Δt. Space (volume) which includes the mass at time t and time t + Δt visualized as parts I, II and III. Since the mass of the system is constant over time:

Interest pertma a mass change in bagia II, at Δt approaches zero section II coincides with volune set. So the change in part II is identical to the changes in the volume set.

The mass of the parts I was reduced or not increased with time. For Δt approaches zero mass change on the part I represent the mass that enters the control volume (). And mass changes on the part III represents the mass that comes out of the control volume (), then:

In the case of steady state:

Mass per unit time (mass flow rate) is the product of velocity by the cross-channel flow: V x A (m3/sc), to make berunit kg / sc masses should be multiplied by Jenia fluidanya. ( kg/m3).

A channel system as shown below the scheme. Steady state water flow through the apparatus. What is the flow rate in channel 1Answer:Note:Channel system with a data rate and channel diameterAsked:Flow velocity in channel 1?Scheme:

Assumptions:A. Incompressible flow2. Uniform flow3. Steady state conditions (steady state)Analysis:The law of conservation of mass gives the equation:, For steady state flow, so that:

, Incompressible

3.2. CONSERVATION OF ENERGYThere are two Shittim thermodynamics, closed systems and open systems. Therefore there are two energy conservation equations. Two equations are generated from the I law of thermodynamics. The law basically states that the energy balance before and after the process must balance.3.2.1. Energy conservation equation for Closed SystemsFor a closed system holds:

12)In this case the heat of entry marked Q + Q,-Q heat marked out. Work entry marked W + W and-W marked out of work.

Process P is constant, V is constant and T is constant and the adiabaticE is the energy of the system:13)U: energy (J)

: Kinetic energy, with m: mass (kg) and V: velocity (m / sc)mgh: potential energy, denagn g: acceleration of gravity (= 9.8 m/sc2) and h: elevation / altitude (m)For a closed system of kinetic and potential energy there is no (zero)or14)Can also be written per unit mass:15)If taken derifatifnya;16)17)3.2.1.1. Constant Pressure ProcessRepresented by the line 1-2 Figure 3.2.

18)The definition of enthalpy is:

, Where dP = 0 for the constant pressure, Referring to the press. 18) then:

For an ideal gas with constant specific heat capacity effect:19)20)For constant pressure reversible process:21)Ideal gas equation is:, With R (kJ / kg K): the gas constant table B622)Of the press. 19, 20 and 22 obtained:

23)Example 2:A pair of piston and cylinder containing air is expanded at constant pressure 30 atm, temperature of 1000 K to 400 K. The air mass 0.5 kg. Get Heat, work, changes in enthalpy and energy changes.Scheme:

Analysis:Closed system, constant pressure process the balance enrgi:

The process of constant pressure Q = ΔH

3.2.1.2. Constant Volume ProcessRepresented by the line 1-3 Figure 3.2., For a constant volume process = 0 so:Example 3A sealed container contains 1 m3 air-pressure-temperature 273 K. 344 800 N/m2 Heat put into the container until the temperature to 600 K. How much heat is added.Scheme:

Analysis:Closed system, the volume of container remains, then:

Ideal gas equation:

3.2.1.3. Constant temperature processRepresented by the line 1-4 Figure 3.2.

Ideal gas: as T1 = T4 or acquired, or

Because T is constant then ΔU = 0, so

Example 4:A compressor to suck the air at a pressure of 101 000 N/m2, temperature 30oC, the initial volume of 0.00078 m3 and a density of 1.23 kg/m3. After compressing the isotermik pressure 10 times its original pressure. What is the volume end of the process and also how much work is needed?Scheme

Analysis:Closed system of constant temperature processes:

, Obtainedor

3.2.1.4. Adiabatic ProcessRepresented by the line 1-5 Figure 3.2., With Q = 0For the adiabatic process persamanan line is:Where k; adiabaitk constants.So that work:

Example 5:A compressor to suck the air at a pressure of 101 000 N/m2, temperature 30oC, the initial volume of 0.00078 m3 and a density of 1.23 kg/m3. After compressing the adiabatic pressure is 10 times the original pressure. What is the final temperature and process work and how many are needed?Scheme:

Analysis:Adiabatic compression process, then:

cp = 1.004 kJ / kg K, cv = 0718 kJ / kg K, then

3.2.2. Energy conservation equation for Open SystemsAt the time tmassa set is the number of parts I and II. At time t + Δt some mass moves to the III, the interaction of heat and mechanical work on the mass set. Energy on the mass balance after set intervals of time Δt is:

W * is the fluid mechanical work done, if the cross-sectional area on the surface of the set A and the fluid moves as far as dx, then: fluid work done is:, A volume A dx = dx = dm / dm = υ

For the case of steady state:

For steady flow (steady flow) in = m * m * = m * it out:

If the total energy substitutable volume set, and replaced an index in the index out to be replaced 2 is obtained:

Enthalpy, the

Example 5:An electricity power plant's steam turbine steam at a pressure receive of 20 MPa at a velocity of 10 m / sc. The steam leaves the turbine at 0.1 MPa, with velocity of 1.5 m / sc. The turbine entrance is 12 m above the exit level. Find the work output if the mass flow rate is 3 tons / day and lost of heat transfer is Q * = 8000 kJ / hr. The steam has tabulated the Following Properties:Term Inlet Exit20 MPa pressure 0.1 MPATemperature 365.8 C 99.6 CVelocity 10 m / sc 1.5 m / scSpecific internal energy 2293.2 kJ / g 417.5 kJ / kgSpecific volume m3/kg t 0.001043 0.005836 mt3/kgWere:Work output?Scheme:

AssumptionsA. Steady state, steady flow2. Constant properties3. Friction is ignoredAnalysis:Open system steady state steady flow, the energy balance gives the equation:

Written per unit mass of the

Example 6:A ship's steam turbine steam at a pressure receive of 6205 kPa at a velocity of 30 m / sc. The steam leaves the turbine at 9899 kPa, with velocity of 3.5 m / sc. The turbine entrance is 12 m above the exit level. Find the work output if the mass flow rate is 15 kg / sc and lost of heat transfer is Q * = 14 kW. The steam has tabulated the Following Properties:Term Inlet ExitPressure 6205 kPa 9899 kPaTemperature 811.1 K 318.8 KVelocity 30 m / 3.5 m sc / scSpecific internal energy 3150.3 kJ / g 2211.8 kJ / kgSpecific volume m3/kg 0.005789 mt3/kg 13:36

Example 7:A nozzle is used to convert kinetic energy into enthalpy. Air enters the nozzle with a pressure 400 psia and at 100 ft / sc and enthalpy 240.98 BTU / LBM. Pressurized air nozzle exit 100 psia, and the enthalpy 162.98 BTU / LBM. Air mass flow rate 600 LBM / hr. (A) If the heat loss by 2 BTU / LBM calculated air velocity nozzle exit. (B) If the nozzle is insulated against heat (heat loss q * = 0, is also undergoing a process called adiabatic) calculated nozzle exit velocity.Note:Nozzle to convert the enthalpy to velocity.Were:a). Speed 2BTU/lbm out if the heat loss?b) The speed of exit if the adiabatic?Scheme:

Assumptions:A. Steady state, steady flow2. Constant properties3. Friction is ignoredAnalysis:Open system steady state, steady flow, the energy balance gives the press:a) The heat loss:Total energy in = energy out of the total.

Description:A. The location of the center point of the inlet and exit together so z1 = z22. For the British system known standard gravity factor gc = 32.17 ft/sc23. Also known as the conversion factor from ft (unit of potential energy) into BTU / LBM is factor

b) Without heat loss:

Example:Pump water for household needs brkapasitas 2.5 l / sc. The emnghisap Ompa water from the wells with a depth of 4 m from the axis of the pump shaft, then raising it above the tank on the roof with a height of 8 m. Water temperature 30 ° C, with a density of 998 kg/m3. Heat loss at the pump at 5 W. Compute power is required if the total efficiency of the pump and the driving motor is 70%.

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