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Chapter 3
The First law
of thermodynamics
Energy interactions between a system and its surroundings across the boundary in the form of heat and work have been discussed separately in the previous chapter. So far, no attempt has been made to relate these interactions between themselves and with the energy content of the system.
First law of thermodynamics, often called as law of conservation of energy, relating work, heat, and energy content of the system will be discussed in detail in this chapter.
3.1 First Law of Thermodynamics
In its more general form, the first law may be stated as follows
“ When energy is either transferred or transformed, the final total energy present in all forms must precisely equal the original total energy ”.
It is based on the experimental observations and can not be proved mathematically. All the observations made so far, confirm the correctness of this law.
3.2 First Law of Thermodynamics for a Closed System
Undergoing a Process
First law can be written for a closed system in an equation form as
For a system of constant mass, energy can enter or leave the system only in two forms namely work and heat.
Let a closed system of initial energy E 1 receives Q units of net heat and gives out W units of work during a process. If E 2 is energy content at the end of the process as given in Figure 3.1, applying first law we get
Q − W = (E 2 −
E 1 ) ...(3.1)
Where the total energy content
Ε = Internal Energy + Kinetic energy + Potential energy = U + + mgz
The term internal energy usually denoted by the letter U is the energy due to such factors as electron spin and vibrations, molecular motion and chemical bond.
Kinetic energy term is due to the system movement with a velocity C. For stationary systems this term will be zero. The term g c^ is a constant of value 1 in SI unit. It will be dropped here after since SI unit is followed throughout the book.
Potential energy term is due to the location of the system in the gravitational field. It remains constant for a stationary system. The unit of energy in SI is kJ.
3.3 The Thermodynamic Property Enthalpy
Consider a stationary system of fixed mass undergoing a quasi- equilibrium constant pressure process
Applying first law
Q^12 −^1 W^2 =^ E^2 −^ E^1
where E 2 − E 1 = (U 2 − U 1 ) + m(C 22 − C 12 ) + mg(Z 2 − Z 1 )
= U 2 − U 1 since it is a stationary system.
also 1 W 2 = p(V 2 − V 1 )
= p 2 V 2 − p 1 V 1
∴ Q 12 = (p 2 V 2 − p 1 V 1 ) + (U 2 − U 1 )
∴Q 12 = (U 2 + p 2 V 2 ) − (U 1 + p 1 V 1 )
The terms within brackets are all properties depending on the end states. This combination of properties may be regarded as a single property known as enthalpy. It is usually denoted by the letter H.
Energy may cross the control surface not only in the form of heat and work but also by total energy associated with the mass crossing the boundaries. Hence apart from kinetic, potential and internal energies, flow energy should also be taken into account.
Conservation of mass
Conservation of energy
Figure 3.3 First Law of Thermodynamics Applied to a control Volume
As a rate equation, it becomes
...(3.6)
3.6 The Steady-state Flow Process
When a flow process is satisfying the following conditions, it is known as a steady flow process.
- The mass and energy content of the control volume remains constant with time.
- The state and energy of the fluid at inlet, at the exit and at every point within the control volume are time independent.
- The rate of energy transfer in the form of work and heat across the control surface is constant with time.
Therefore for a steady flow process ...(3.7) ...(3.7)
also ...(3.8)
For problem of single inlet stream and single outlet stream
This equation is commonly known as steady flow energy equation (SFEE).
3.7 Application of SFEE
SFEE governs the working of a large number of components used in many engineering practices. In this section a brief analysis of such components working under steady flow conditions are given and the respective governing equations are obtained.
3.7.1. Turbines
Turbines are devices used in hydraulic, steam and gas turbine power plants. As the fluid passes through the turbine, work is done on the blades of the turbine which are attached to a shaft. Due to the work given to the blades, the turbine shaft rotates producing work.
Figure 3.4 Schematic Representation of a Turbine
General Assumptions
Similar to compressors pumps are also work consuming devices. But pumps handle incompressible fluids, whereas compressors deal with compressible fluids.
General Assumptions
- No heat energy is gained or lost by the fluids;
- Changes in kinetic energy of the fluid are negligible.
Governing Equation
As the fluid passes through a pump, enthalpy of the fluid increases, (internal energy of the fluid remains constant) due to the increase in pv (flow energy). Increase in potential energy of fluid is the most important change found in almost all pump applications.
3.7.4 Nozzles
Nozzles are devices which increase the velocity of a fluid at the expense of pressure. A typical nozzle used for fluid flow at subsonic * speeds is shown in Figure 3.7.
General Assumptions
- In nozzles fluids flow at a speed which is high enough to neglect heat lost or gained as it crosses the entire length of the nozzle. Therefore, flow through nozzles can be regarded as adiabatic. That is = 0.
- There is no shaft or any other form of work transfer to the fluid or from the fluid; that is = 0.
- Changes in the potential energy of the fluid are negligible.
Governing Equation
3.7.5 Diffusers
Diffusers are (reverse of nozzles) devices which increase the pressure of a fluid stream by reducing its kinetic energy.
General Assumptions
Similar to nozzles, the following assumptions hold good for diffusers.
1. Heat lost or gained as it crosses the entire length of the nozzle.
Therefore, flow through nozzles can be regarded as adiabatic. That is
- There is no shaft or any other form of work transfer to the fluid or from the fluid; that is = 0.
- Changes in the potential energy of the fluid are negligible
Governing Equation
...(3.14)
A throttling process occurs when a fluid flowing in a line suddenly encounters a restriction in the flow passage. It may be
- a plate with a small hole as shown in Figure 3.10 (a)
- a valve partially closed as shown in Figure 3.10 (b)
- a capillary tube which is normally found in a refrigerator as shown in Figure 3.10 (c)
- a porous plug as shown in Figure 3.10 (d)
General assumptions
- No heat energy is gained or lost by the fluid; ie., = 0
- There is typically some increase in velocity in a throttle, but both inlet and exit kinetic energies are usually small enough to be neglected.
- There is no means for doing work; ie., = 0.
- Changes in potential energy of the fluid is negligible.
Governing Equation h 2 = h 1 ...(3.16)
Therefore, throttling is an isenthalpic process.
3.8 First Law for a Cyclic Process
In a cyclic process the system is taken through a series of processes and finally returned to its original state. The end state of a cyclic process is identical with the state of the system at the beginning of the cycle. This is possible if the energy level at the beginning and end of the cyclic process are also the same. In other words, the net energy change in a cyclic process is zero.
Figure 3.11 First Law for a Cyclic Process
Consider a system undergoing a cycle consisting of two processes A & B as shown in Figure 3.11 Net energy change
ΔE A + ΔE B = 0 ..(3.17)
(Q A − W A) + (QB − W B) = 0 ...(3.18)
Q A + QC = [WA + W C]
Q A − WA = −[Q C − W C]
ΔE A = −ΔE C ...(3.22)
From Equation (3.21) and (3.22) it can be concluded that energy change in path B and path C are equal and hence energy is a point function depending only on the end states.
It has been already shown that all the properties are point functions and hence energy is also a property of the system.
3.10 Specific Heat at Constant Volume and at Constant
Pressure
Specific heat at constant volume of a substance is the amount of heat added to rise the temperature of unit mass of the given substance by 1 degree at constant volume
From first law for a stationary closed system undergoing a process
dQ = pdV + dU or dq = pdv + du
For a constant volume process
dQ = dU or dq = du
∴ or du = Cv^ dT^ ...(3.23)
Similarly specific heat at constant pressure is the quantity of heat added to rise the temperature of unit mass of the given substance by 1 degree at constant pressure
where dQ = pdV + dU dQ = pdV + d(H − PV)
dQ = pdV + dH − Vdp − pdV
dQ = dH − Vdp
For a constant pressure process dp = 0
Hence dQ = dH or dq = dh
∴ or dh = Cp^ dT^ ....(3.24)
Note
- For solids and liquids, constant volume and constant pressure processes are identical and hence, there will be only one **specific heat.
- The difference in specific heats C p − C v** = R = - The ratio of sp. heat γ = C (^) p /C (^) **v
- Since h and u are properties of a system, dh** = Cp dT and du = Cv dT, for all processes.
3.11 Work Interaction in a Reversible Steady Flow Process
In a steady flow process the work interaction per unit mass between an open system and the surroundings can be expressed in differential form as
dq − dw = dh + CdC + gdz
dw = dq − (dh + CdC +gdz)
Also, dq = du + pdv (or) dh − vdp
Therefore, dw = dh − vdp − (dh + CdC + gdz)
= − vdp − (CdC + gdz)
For a stationary system
3.12 First law for an open system under unsteady flow
conditions
Many processes of engineering interest involve unsteady flow, where energy and mass content of the control volume increase or decrease.
Example for such conditions are:
Filling closed tanks with a gas or liquid.
Discharge from closed vessels.
An engine which could provide work transfer continuously without heat transfer is known as perpetual motion machine of first kind. It is impossible to have such an engine as it violates first law of thermodynamics.
Exercises
- Define internal energy.
- Express mathematically first law of thermodynamic for the following.
a. a closed system undergoing a process b. a stationary system of fixed mass undergoing a change of state c. a closed system undergoing a cycle. d. an open system. e. an open system with steady-state flow conditions.
- Define flow energy and enthalpy.
- For a stationary system of fixed mass undergoing a process such that its volume remains constant, Q^12 =^ ΔU(T/F)
- dQ = dh − vdp for closed system undergoing a process (T/F).
- Define specific heat at (a) constant pressure (b) constant volume
- Determine the power of the cycle comprising four processes in which the heat transfers are : 50 kJ/kg, −20 kJ/kg, −7l J/kg and 12 kJ/kg having 100 cycles per minute. [48.3 kW]
- Write the steady flow energy equation and explain the terms involved in it.
- Show that energy is a property of the system.
- What are conditions for steady flow process?
- A piston-cylinder assembly contains 1kg or nitrogen at 100 kPa. The initial volume is 0.5 m 3. Heat is transferred to the substance in an amount necessary to cause a slow expansion at constant temperature. This process is terminated when the final volume is twice the initial volume. [34.7 kJ]
- 2 kg of air enclosed in a rigid container receives 0.2 kJ of paddle wheel work and 0.5 kJ of electrical energy per second. Heat loss from the system is 0.6 kJ/s. If the initial temperature is 25 o C what will be the temperature after 5 minutes? [45.9o^ C]
- A well insulated, frictionless piston-cylinder assembly contains 0.5 kg of air initially at 75 oC and 300 kPa. An electric - resistance heating element inside the cylinder is energized and causes the air temperature to reach 150 o C. The pressure of the air is maintained constant throughout the process. Determine the work for the process and the amount of electrical work.
{Hint Q net − Wnet = ΔU; W net =+Welectric}
[−26.9 kJ ; − 37.7]
- A cylinder contains 168 litres of a gas at a pressure of 1 bar and temperature of 47o C. If this gas is compressed to one-twelfth of its volume, pressure is then 21 bar. Find a. index of compression b. change in internal energy c. heat rejected during compression
Take Cp = 1.089 and Cv = 0.837 both in kJ/kg
[1.225 ; 41.81 kJ ; −14.05 kJ]
- a. A mass of 10 kg is falling from a height of 100 m from the datum. What will be the velocity when it reaches a height of 20 m from the datum? Take the total heat loss from the mass when it falls from 100 m height to 20 m height is 5 kJ. [8.68 m/s]
b. An insulated box containing carbon dioxide gas falls from a balloon 3.5 km above the earths surface. Determine the temperature rise of the carbon dioxide when box hits the ground.
Take C v = 0.6556 kJ/kg [52.37o C]
process form state 3 and to state 1. The heat rejected in process 3-1 is 1560 kJ/kg and the substance is air. Determine
(a) the pressures, temperatures and specific volumes around the cycle (b) the heat transfer in process 1- (c) the heat transfer in process 2- (d) work done in each process and (e) net work done in the cycle [137 kPa ; 0.6515 m^3 /kg^ ;^ 311.0^ K^ ;^1095 kPa^ ; 0.1630 m 3 /kg ; 621.8 K ; 1095 kPa ; 0.6515 m 3 /kg ; 2487.0 K ; 44.44 kJ ; 1872.25 kJ ; −178 kJ ; 534.9 kJ ; 0 ; 356.9 kJ]
- 0.15 m 3 of air at a pressure of 900 kPa and 300 o C is expanded at constant pressure to 3 time its initial volume. It is then expanded polytropically following the law PV 1.5 = C and finally compressed back to initial state isothermally. Calculate (a) heat received (b) heat rejected (c) efficiency of the cycle [944.5kJ ; −224.906 kJ ; 0.291]
- A piston and cylinder device contains 1 kg of air, Initially, v = 0.8 m 3 /kg and T = 298 K. The air is compressed in a slow frictionless process to a specific volume of 0.2 m 3 /kg and a temperature of 580 K according to the equation p V 1 .3 = 0.75 ( p in bar, v in m 3 /kg). If Cv of air is 0.78 kJ/kg determine : (a) work and (b) heat transfer (both in kJ)
[ −137.85 kJ ; 82.11 kJ]
- The internal energy of a closed system is given by U = 100 + 50 T + 0.04 T 2 in Joules, and the heat absorbed by Q = 4000 + 16 T in Joules, where T is in Kelvin. If the system changes from 500 K to 1000 K, what is the work done? [47 kJ]
- One kg of air, volume 0.05 m^3 ,^ pressure^20 bar^ expands^ reversibly according to the law pv1.3 = C until the volume is doubled. It is then cooled at constant pressure to initial volume and further heat at constant volume so that it returns back to initial process. Calculate the network done by air. [21.98 kJ]
28. Air at the rate of 14 kg/s expands from 3 bar, 150° C to 1bar reversibly
and adiabatically. Find the exit temperature and power developed. Neglect the changes in kinetic and potential energy. [ 309 k ; 1.603 kW]
- Specific internal energy of a certain substance can be expressed as follows:
u = 831.0 + 0.617 pv Where u is the specific internal energy in kJ/kg p is the pressure in k Pa v is the specific volume in m 3 /kg One kg of such substance expands from 850 kPa, 0.25 m 3 /kg to 600 kPa, 0.5 m 3 /kg. Find the work done and heat transferred. [ 176.06 kJ ; 230 kJ]
- A cylinder of 8 cm internal diameter is fitted with a piston loaded by a coil spring of stiffness 140 N/cm of compression. The cylinder
contains 0.0005 m 3 of air at 15 °C and 3 bar. Find the amount of
heat which must be supplied for the piston to a distance of 4 cm. Sketch the process on a p-V diagram. [ 0.417 kJ]
- Prove that Q = mC v for a polytropic process of index n.
- An air conditioning system for a computer room in a tower block draws in air on the roof at a height of 100 m with a velocity of 25 m/s. The air is at 28o C. The air is discharged at a height of 10 m with a velocity of 2 m/s at 14 oC. The mass flow rate is 2 kg/s, and a heat transfer of − 40. kW cools the air before it is discharged. Calculate the rate of work for the air passing through the system. Take Cp for air as 1005 J/kgK.
[− 10.23 kW]
- A diffuser reduces the velocity of an air stream from 300 m/s to 30 m/s. If the inlet pressure and temperature are 1.01 bar and 315 o^ C, determine the outlet pressure. Find also the area required for the diffuser to pass a mass flow of 9 kg/s. [4.586 bar, 0.17 m 2 ]
- A centrifugal air compressor operating at steady state has an air intake of 1.2 kg/min. Inlet and exit conditions are as follows:
Properties p (kPa) To^ C^ u kJ/kg^ v m^3 /kg Inlet 100 0 195.14 0. Exit 200 50 230.99 0.
