44 eBook PSP | First Law of Thermodynamics 2. Differences between Steady-Flow and Unsteady-Flow Process Table 3.1 : The differences between steady-flow and unsteady-flow process Steady-Flow Process Unsteady-Flow Process The mass of the control volume remains constant during the process. The mass within the system boundaries does not remain constant during the process. Fixed in space, size and shape The shape and size of a control volume may change during an unsteady-flow process. The total energy content remains constant with time The energy content of a control volume changes with time 3.3 Energy Analysis of Steady-Flow Systems The mass balance for a general steady-flow system is given as ∑̇ = ∑̇ Therefore, for a single stream which is one-inlet and one outlet, the mass balance of a steadyflow is given as; or 111 = 222 The above equation is known as Continuity Equation, where the subscripts 1 and 2 refer to the inlet and the outlet states, respectively. While the various terms in the above equation is ̇1 = ̇ 2 11 1 = 22 2
First Law of Thermodynamics | eBook PSP 45 referred as follows: ̇ = mass flow rate (kg/s). It defines as the amount of mass flowing through a cross section per unit time. = The average flow velocity (m/s) = the cross-sectional area of pipe (m2 ) = specific volume of fluid (m3 /kg) = fluid density (kg/m3 ) During the steady-flow process, the total energy content of a control volume remains constant (Ecv = constant). Thus, the amount of energy entering an open system is equal to the amount of energy leaving it. ̇ = ̇ Or ̇ − ̇ = 0 By combining all forms of energy involved in a control volume (heat, work, enthalpy, kinetic energy and potential energy) into the above equation, then a Steady-Flow Energy Equation can be expressed as; ̇ + ∑̇ [ℎ + 2 2 + ] = ̇ + ∑̇ [ℎ + 2 2 + ] If the fluid enter a control volume at state 1 and exit at state 2 , thus the Steady-Flow Energy Equation can be written as; ̇ + ̇1 [ℎ1 + 1 2 2 + 1] = ̇ + ̇ 2 [ℎ2 + 2 2 2 + 2] Rearrange the equation, then; (Note that ̇1 = ̇ 2 = ̇) ̇ − ̇ = ̇ [(ℎ2 − ℎ1 ) + ቆ 2 2 − 1 2 2 ቇ + (2 − 1 )]
46 eBook PSP | First Law of Thermodynamics Or for a unit mass; Where: ̇ = rate of heat transfer (kJ/s or kW) = heat transfer per unit mass (kJ/kg) ̇ = Power (kJ/s or kW) = work done per unit mass (kJ/kg) ̇ = mass flow rate (kg/s) ℎ2 − ℎ1 = Change of specific enthalpy (kJ/kg) ( 2 2−1 2 2 ) = Change in kinetic energy (m2 /s2 or J/kg). By dividing the change in kinetic energy by 1000 then the unit will convert to kJ/kg. Note that C represents fluid velocity in m/s. (2 − 1 ) = Change in potential energy (m2 /s2 or J/kg) which is z represents an elevation of inlet or outlet channel from the datum. By dividing the change in potential energy by 1000 then the unit will convert to kJ/kg. The value of gravitational acceleration, g is 9.81 m/s2 . According to the Steady-Flow Energy Equation, the relationship between ̇ , and ̇ , are as follows; ̇ = ̇ × And ̇ = ̇ × In ddition, the equation in defining the change of specific enthalphy can be expressed as below; ℎ = + If the fluid enters a control volume at state 1 and leaves it at state 2, then − = (ℎ2 − ℎ1 ) + ቆ 2 2 − 1 2 2 ቇ + (2 − 1 ) (ℎ2 − ℎ1 ) = (2 − 1 ) + (22 − 21)
First Law of Thermodynamics | eBook PSP 47 Where: (ℎ2 − ℎ1) = Change of specific enthalpy (kJ/kg) (2 − 1) = Change of Specific Internal Energy (kJ/kg) (22 − 11) = Change of flow work (kJ/kg) = The fluid pressure (kN/m2 ) v = Specific volume (m3 /kg) For an ideal gas, the specific enthalpy change of the fluid can also be determined through the following equation; Where; = Specific heat capacity at constant pressure (kJ/kg.K) (2 − 1) = Temperature difference (Kelvin) 3.4 Application of A Steady-Flow Energy Equation Table 3.2 : Steady-flow engineering devices DEVICES WORKING PRINCIPLE Turbine • A device used to produce a mechanical work. • As the high-velocity gas or liquid is flowed through the turbine blades, work is done against the blades. This will cause the shaft which are attached to the turbine blade will rotate and as the result the turbine produce work output. TURBINE Wout (ℎ2 − ℎ1 ) = (2 − 1 )
48 eBook PSP | First Law of Thermodynamics Compressor And Pump • Compressor A device used to increase the pressure of a fluid in the form of gas or air. For this purpose, a certain amount of work needs to be provided. Typically, an electric motor is used to supply work to the compressor. • Pump A device used to increase the flow rate of liquid flowing through it. The working fluid involved is liquid instead of gases. Boiler • A closed vessel that is used to heat a fluid, usually water. An integral component of thermal power plants are boilers, more especially steam boilers. • Hot gases are created in a furnace when fuel, usually coal, is burned. When these heated gases meet a water vessel, the heat from the gases transfers to the water, which causes the boiler to produce steam. Nozzle • A device used to increase the velocity or kinetic energy of a fluid. Therefore, the velocity at the exit of the nozzle is higher than the velocity at the entrance. • Does not involve any work transfer. • In mostly case, the distance between the inlet and outlet pipe is very small. As a COMPRESSOR / PUMP Wout Water In Steam out Fluid In Fluid Out
First Law of Thermodynamics | eBook PSP 49 result, the total heat transfer and the change in potential energy is small and may be negligible. Diffuser • A device used to increase the pressure of a fluid by decrease the velocity of a fluid. • Involve no work transfer. • The distance between the inlet and outlet pipe is very small. Therefore, the total heat transfer and the change in potential energy is small and may be neglected. Heat Exchanger • A device where two moving fluid streams exchange heat without mixing since, it flows through a different tube and space. • The exchange heat takes place in the device while the pressure is constant. • The example of heat exchanger in engineering practices is boiler and condenser in steam power plant. Fluid A In Fluid A Out Fluid B Out Fluid B In Fluid In Fluid Out
50 eBook PSP | First Law of Thermodynamics Example of Final Examination Questions May 2014 Session QUESTION Define the mass flow rates in an open system. ANSWER Mass flow rate is defined as the amount of mass flowing through a cross section per unit time. December 2014 Session QUESTION i. Define the steady flow process. ii. List TWO (2) conditions that fulfil the steady flow process. ANSWER i. Steady flow process A process where a fluid steadily passes through a control volume. Although the fluid properties may differ from point to point throughout the control volume, but they do not change with time. ii. TWO (2) conditions that fulfil the steady flow process. • The total mass inside the control volume does not change with time. • The properties of fluid must remain constant over time at any given point in the system. • A steady flow system’s interaction with its surroundings in terms of work energy and heat must remain constant over time. (Choose any two correct answer)
First Law of Thermodynamics | eBook PSP 51 May 2016 Session QUESTION State the correct quantity and unit for each equation’s symbol based on the steady flow energy equation below: ̇ − ̇ = ̇ [(ℎ2 − ℎ1 ) + ቆ 2 2 − 1 2 2 ቇ + (2 − 1 )] ANSWER Symbol Quantity Unit ̇ Heat received or rejected kJ/s or J/s ̇ Work Done kJ/s or J/s ̇ Mass flow rate kg/s ℎ2 − ℎ1 Change of specific enthalpy kJ/kg or J/kg 2 2 − 1 2 2 Change in Kinetic energy m2 /s2 (2 − 1 ) Change in potential energy m2 /s2 December 2016 Session QUESTION: List down SIX (6) equipment that applied the Steady Flow Energy Equations. ANSWER: Six equipment that applied the Steady Flow Energy Equations are: i. Boiler ii. Heat exchanger iii. Diffuser iv. Nozzle v. Turbine vi. Compressor
52 eBook PSP | First Law of Thermodynamics December 2018 Session QUESTION Draw the schematic diagram of the nozzle and the diffuser with their working mechanism. ANSWER Devices Nozzle Diffuser Schematic Diagram Working Principle A device used to increase the velocity or kinetic energy of a fluid. A device used to increase the pressure of a fluid by decrease the velocity of a fluid. 2: 2022/2023 Session QUESTION Compare TWO (2) differences between steady and unsteady flow process. ANSWER Steady-Flow Process Unsteady-Flow Process The mass of the control volume remains constant during the process. The mass within the system boundaries does not remain constant during the process. Fixed in space, size and shape The shape and size of a control volume may change during an unsteady-flow process. The total energy content remains constant with time The energy content of a control volume changes with time (Choose any two correct answer) In Out In Out
First Law of Thermodynamics | eBook PSP 53 May 2016 Session QUESTION A rotary air pump is required to deliver 900 kg of air per hour. The enthalpy at the inlet and exit of the pump are 300 kJ/kg and 500 kJ/kg respectively. The air velocity at the entrance and exit are 10 m/s and 15 m/s respectively. The rate of heat loss from the pump is 2500 W. Determine the power required to drive the pump. ANSWER Given: ̇ = 900 ℎ = 900 ℎ × 1ℎ 3600 = 0.25 ℎ1 = 300 ℎ2 = 500 1 = 10 2 = 15 ̇ = −2500 = −2.5 The power, ̇ required to drive the pump ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+( 2 2−1 2 2 )+ (2 − 1 )] Assumptions: P.E = 0 ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ] −2.5 − ̇ = 0.25 [(500 − 300)+ ቆ 152 −102 2 × 1000ቇ] −2.5 − ̇ = 50.0156 ̇ = −.
54 eBook PSP | First Law of Thermodynamics May 2014 Session QUESTION Steam flows through a turbine at 40 kg/min. During the process, the steam undergoes a drop in enthalpy of 550 kJ/kg and loss 4500 kJ/min of heat energy. Assume the changes in potential energy is neglected, calculate the power produced by the turbine. ANSWER Given: ̇ = 40 = 40 × 1 60 = 0.67 ∆ℎ = ℎ2 − ℎ1 = −500 ̇ = −4500 = −4500 × 1 60 = −75 The power, ̇ required to drive the pump ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: P.E = 0 ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ] −75 − ̇ = 0.67 [(−550) +ቆ 452 − 602 2 × 1000ቇ] −75 − ̇ = −369.028 ̇ = −.
First Law of Thermodynamics | eBook PSP 55 May 2016 Session QUESTION In a steady flow system, a substance flows at the rate of 240 kg/min. It enters the system at a pressure of 550 kPa, a velocity of 0.25 km/s, specific internal energy of 2500 kJ/kg and specific volume of 0.55 m3 /kg. It leaves the system at a pressure of 1.5 bar, a velocity of 140 m/s, specific internal energy of 1.55 MJ/kg and specific volume of 1.5 m3 /kg. When it passes through the system, 45 kJ/kg of heat loss to the surrounding. If the change of potential energy is neglected, calculate the power of the system in kilowatts. ANSWER Given: ̇ = 240 = 240 × 1 60 = 4 1 = 550 = 550 2 1 = 0.25 = 0.25 × 1000 1 = 250 1 = 2500 1 = 0.55 3 2 = 1.5 = 1.5 × 102 2 = 150 2 2 = 140 2 = 1.55 = 1.55 × 106 1 × 1 103 = 1550 2 = 1.5 3 = −45 Heat transfer in kJ/s ̇ = ̇ × ̇ = 4 × (−45) ̇ = −180
56 eBook PSP | First Law of Thermodynamics Change in specific enthalpy ℎ2 − ℎ1 = (2 − 1 )+ (22 −11 ) ℎ2 − ℎ1 = (1550 −2500) + [(150 × 1.5)− (550 × 0.55)] ℎ2 − ℎ1 = −1027.5 / The power, ̇ of the system in kW ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: P.E = 0 ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ] −180 − ̇ = 4 [(−1027.5) +ቆ 1402 − 2502 2 × 1000 ቇ] −180 − ̇ = −4195.8 ̇ = . June 2016 Session QUESTION In a steady flow system, air flows through a compressor at the rate of 90 kg/m3 . It enters at a pressure of 1.2 bar, a velocity of 15 m/s, specific internal energy 1700 kJ/kg and specific volume of 0.37 m3 /kg. It leaves the system at the pressure of 3.8 bar, a velocity of 35 m/s, specific internal energy 3100 kJ/kg and specific volume 1.2 m3 /kg. During its passage through the system the air has a loss heat of 30 kJ/s to the surroundings. Assuming the changes in potential energy are negligible, determine: i. Change of enthalpy ii. Power of the system in kilowatts iii. Area of compressor outlet
First Law of Thermodynamics | eBook PSP 57 ANSWER Given: ̇ = 90 = 90 × 1 60 = 1.5 1 = 1.2 = 1.2 × 102 2 = 120 2 1 = 15 1 = 1700 1 = 0.37 3 2 = 3.8 = 3.8 × 102 2 = 380 2 2 = 35 2 = 3100 2 = 1.2 3 ̇ = −30 i. Change of enthalpy ℎ2 − ℎ1 = (2 − 1 )+ (22 −11 ) ℎ2 − ℎ1 = (3100 −1700) + [(380 × 1.2)− (120 × 0.37)] − = . / ii. Power of the system in kW(s) ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: P.E = 0 ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ] −30 − ̇ = 1.5 [(1811.6) +ቆ 352 − 152 2 × 1000ቇ] −30 − ̇ = 2718.15 ̇ = −.
58 eBook PSP | First Law of Thermodynamics iii. Area of compressor outlet ̇ = 22 2 2 = ̇ × 2 2 2 = 1.5 × 1.2 35 = . June 2014 Session QUESTION Water flow through a boiler with the velocity of 6 m/s, pressure of 0.98 bar and specific volume of 0.1 m3 /kg. Superheated steam produced has velocity of 50 m/s and pressure of 7 bar with specific volume of 0.2 m3 /kg. The specific internal energy in the outlet section is 100 kJ/kg greater than the inlet. If the heat transfer to the boiler is 40 kW, calculate the mass flow rate produced in kg/hr. ANSWER Given: 1 = 6 1 = 0.98 = 0.98 × 102 2 = 98 2 1 = 0.1 3 2 = 50 2 = 7 = 7 × 102 2 = 700 2 2 = 0.2 3 2 > 1 = 100 ; therefore 2 − 1 = 100 ̇ = 40
First Law of Thermodynamics | eBook PSP 59 Change in specific enthalpy ℎ2 − ℎ1 = (2 − 1 )+ (22 −11 ) ℎ2 − ℎ1 = 100 + [(700 × 0.2)− (98 × 0.1)] ℎ2 − ℎ1 = 230.2 The mass flow rate, ̇ produced in kg/hr ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: P.E = 0 =̇ 0 ̇ = ̇ [(ℎ2 − ℎ1 ) +ቆ 2 2 − 1 2 2 ቇ] 40 = ̇ [230.2+ ቆ 502 −6 2 2 × 1000ቇ] 40 = ̇ (231.432) ̇ = 0.1728 × 3600 1ℎ ̇ = . January 2009 Session QUESTION A steam turbine receives a steam flow of 4850 kg/hr and the power output is 500 kW. Calculate the change of specific enthalpy across the turbine when the velocity at the entrance is 55 m/s and the velocity at the exit is 360 m/s. The inlet pipe is 3 meter above the exhaust pipe and the heat loss from the turbine is negligible. ANSWER Given: ̇ = 4850 ℎ = 4850 ℎ × 1ℎ 3600 = 1.35
60 eBook PSP | First Law of Thermodynamics ̇ = 500 1 = 55 2 = 360 1 − 2 = 3 ; Therefore 2 − 1 = −3 The change of specific enthalpy across the turbine ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: ̇ = 0 (heat loss from the turbine is negligible) −̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] −500 = 1.35 [(ℎ2 − ℎ1 )+ቆ 3602 − 552 2 × 1000 ቇ+ (−3 × 9.81) 1000 ] −500 1.35 = (ℎ2 − ℎ1 )+ ቆ 3602 − 552 2 × 1000 ቇ + (−3 × 9.81) 1000 −370.3704 = (ℎ2 − ℎ1 )+ 63.2875 + (−0.02943) ( − ) = −. June 2017 Session QUESTION Fluid with a specific enthalpy of 4100 kJ/kg enters a horizontal nozzle with negligible velocity at the rate of 79200 kg/hr. At the outlet, the specific enthalpy and specific volume of the fluid were 3050 kJ/kg and 1.45 m3 /kg, respectively. Assuming the flow is an adiabatic flow process, find the: i. Velocity outlet. ii. Required outlet area of the nozzle.
First Law of Thermodynamics | eBook PSP 61 ANSWER Given: ℎ1 = 4100 ̇ = 79200 ℎ = 79200 ℎ × 1ℎ 3600 = 22 ℎ2 = 3050 2 = 1.45 3 i. Velocity Outlet ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: 1 = 0 P.E = 0 (horizontal nozzle) ̇ = 0 (adiabatic flow process) ̇ = 0 0 = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 2 ቇ] 0 = 22 [(3050 −4100) + ቆ 2 2 2 × 1000ቇ] 0 22 = (−1050)+ ቆ 2 2 2000ቇ 0 +1050 = 2 2 2000 1050 × 2000 = 2 2 2.1 × 106 = 2 2 2 = √2.1 × 106 = . ii. Required outlet area of the nozzle ̇ = 22 2 2 = ̇ × 2 2
62 eBook PSP | First Law of Thermodynamics 2 = 22 × 1.45 1449.138 = . May 2014 Session QUESTION Air enters steadily in an adiabatic nozzle at the pressure, temperature and velocity of 300 kPa, 200C and 30 m/s and leaves at pressure of 100 kPa with velocity of 180 m/s. If the inlet area of the nozzle is 80 cm2 , determine: i. Mass flow rate through the nozzle. ii. Exit temperature of the air. iii. Nozzle’s exit area ANSWER Given: 1 = 300 = 300 2 1 = 200℃ + 273 = 473 1 = 30 2 = 100 = 100 2 2 = 180 i. Mass flow rate through the nozzle. For air, Rair = 0.287 kJ/kg.K 11 = 1 1 = 1 1 1 = 0.287 × 473 300 1 = 0.453 3
First Law of Thermodynamics | eBook PSP 63 ̇1 = 11 1 ̇1 = 8 × 10−3 × 30 0.453 ̇ = . ii. Exit temperature of the air, T2. 2 1 = ( 2 1 ) ( −1 ) 2 = 1 ( 2 1 ) ( −1 ) 2 = 473 ( 100 300) ( 1.4−1 1.4 ) = . iii. Nozzle’s exit area, A2 ̇ 1 = ̇ 2 = 0.53 Find V2 22 = 2 2 = 2 2 2 = 0.287 × 345.57 100 2 = 0.992 3 ̇ 2 = 22 2 2 = ̇ 22 2 2 = 0.53 × 0.992 180 = . × −
64 eBook PSP | First Law of Thermodynamics 2: 2022/2023 Session QUESTION The power output of an adiabatic steam turbine is 5 MW, with the inlet and the outlet values of the steam are as indicated in figure below. If the given value of g = 9.81 m/s2 , calculate: i. The specific enthalpy of the turbine outlet. ii. The mass flow rate of the steam. ANSWER Given: ̇ = 5 = 5000 = 9.81 2 ℎ1 = 3248 1 = 50 1 = 10 2 = 16 2 = 956 2 = 0.356 3 2 = 180 STEAM TURBINE h1 = 3248 kJ/kg C1 = 50 m/s Z1 = 10 m P2 = 16 kPa u2 = 956 kJ/kg 2 = 0.356 m3 /kg C2 = 180 m/s Z2 = 6 m Wout = 5 MW
First Law of Thermodynamics | eBook PSP 65 2 = 6 i. The specific enthalpy of the turbine outlet. ℎ2 = 2 +22 ℎ2 = 956 + (16 × 0.356) = . ii. The mass flow rate of the steam. ̇ − ̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] Assumptions: ̇ = 0 (Adiabatic steam turbine) −̇ = ̇ [(ℎ2 − ℎ1 )+ ቆ 2 2 − 1 2 2 ቇ+ (2 − 1 )] −5000 = ̇ [(961.696 −3248) + ቆ 1802 − 502 2 × 1000 ቇ+ ቆ (6 − 10) × 9.81 1000 ቇ] −5000 = ̇ (−2271.3932) −5000 −2271.3932 = ̇ ̇ = .
66 eBook PSP | First Law of Thermodynamics References Cengel, Y. and Boles, M.A (2015), Thermodynamics: An Engineering Approach (SI Unit). 8th Edition, United States of America: Mc Graw Hill. ISBN 978-0-07-339817-4 Md. Razali Ayob (1997), Pengenalan Termodinamik, Skudai, Johor: Penerbit Universiti Teknologi Malaysia. Mohd Kamal Ariffin (2005), Termodinamik Asas, Skudai, Johor: Penerbit Universiti Teknologi Malaysia. Muller, I. (2007) A History of Thermodynamics. Springer. Berlin-Heidelberg Rajput, R.K (2007), Engineering Thermodynamics. 3 rd Edition, New Delhi: Laxmi Publications (P) Ltd. BYJU’S Application Program (2023, Sept 19) Website. https://byjus.com/physics/energyconversion/ Electrical 4U (2023, Nov 10), Website. https://www.electrical4u.com/steam-boiler-workingprinciple-and-types-of-boiler/ Learn with Kassia (2023, Sept 27), Website. mskuksclass.weebly.com Lumen Learning (2023, Sept 15) Website. https://courses.lumenlearning.com/sunyphysics/chapter/15-1-the-first-law-of-thermodynamics/ ScienceDirect (2023, Oct 4), Website. https://www.sciencedirect.com/topics/engineering
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