Unit - 2

Available energy

Q1) Explain the Available energy

A1) Consider a fully reversible process with no dissipative effects – that is all work is transferred without loss and all heat is transferred using an ideal Carnot process to generate additional work.

The resulting maximum work in Secondary system given by

The resulting maximum work in Secondary system given by

 ΔB= change in availability

Clearly, the availability B is a state function in the strictest mathematical sense so the maximum (or minimum) work associated with any steady state process is also independent of the path. Availability: • Yields the maximum work producing potential or the minimum work requirement of a process •Allows evaluation and quantitative comparison of options in a sustainability context

Q2) What is the Quality of energy?

A2) “The potential to convert energy into work”. Obviously, the meaning of this quantity is limited to kinds of energy for which the potential to produce work does not equal the energy content, that is kinds of energy that cannot be converted fully into work. From the kinds of energy mentioned in section 2.1 the kinetic energy, potential energy and electrical energy can be converted completely into work by applying reversible conversion systems. This means that the exergy of these kinds of energy is the same as the energy content. So, the quantity exergy is mainly important for heat, internal energy and chemical energy. Heat, work and flows of matter have to be considered for the analysis of chemical plants and energy conversion plants. An exergy analysis of these types of plants will require exergy values for heat and flows of matter. In this chapter equations will be derived that can be used to determine these exergy values.

Q3) Explain the irreversibility process

A3) An irreversible process is one in which heat is transferred through a finite temperature

Example:             

1. Relative motion with friction

2. Combustion                                                

3.  Diffusion

An irreversible process is usually represented by a dotted line joining the end states to indicate that the intermediate.

Two types of irreversibility.

External irreversibilities: These are associated with dissipating effects outside the working fluid.

Internal irreversibility: these are associated with dissipating effects within the working fluid.

Q4) A reversible heat engine receives 3000 KJ of heat from a constant temperature source at 650 K. If the surroundings are at 295 K, determine

ii)  Unavailable heat.

A4)

Change in entropy

The availability of heat energy,

Unavailable heat (U.A)=

Availability of heat energy= 2064.85 KJ

Unavailable heat = 935.15 KJ

Q5) Explain the second law of efficiency

A5) The second law states that, due to the increase in entropy, heat cannot be converted to work without creating some waste heat. There due important efficiency equations with respect to this concept:

The temperatures here should be in Kelvin  K = ºC + 273.15 or Rankin = 460 + ºF.

2.     Second Law Efficiency Second Law efficiency is a measure of how much of the theoretical maximum (Carnot) you achieve, or in other words, a comparison of the system’s thermal efficiency to the maximum possible efficiency. The Second Law efficiency will always be between the Carnot and First Law efficiencies

Q6) Explain the Clapeyron equation

A6) A substance can exist in three states; solid, liquid and gas. Out of these three, only two states can generally coexist in equilibrium. Whenever there is a change of state, either solid changes into liquid or liquid changes into vapour, the temperature remains constant during the change of state. This temperature, although depends upon pressure, is characteristic of each substance. When the change is from solid to liquid state, the characteristic temperature is called the melting point of the solid and when it is from liquid to vapour state, the temperature is called the boiling point of the liquid. The melting point or boiling point have got a specific value at a specific pressure and vice versa. It is possible to obtain a relation showing how the melting and boiling points vary with pressure by applying the second law of thermodynamics.

The idea is to run a Carnot engine between temperatures and T-dT for a two-phase medium and to let it undergo a change in phase. We can then derive an important relation known as the Clausius-Clapeyron equation, which gives the slope of the vapor pressure curve. We could then measure the vapor pressure curve for various substances and compare the measured slope to the Clausius-Clapeyron equation.

Consider the infinitesimal Carnot cycle abcd shown in figure (a) (b). Heat is absorbed between states a and b. To vaporize an arbitrary amount of mass m, the amount of heat.

must be supplied to the system. From the 1st and 2nd laws of thermodynamics the thermal efficiency for a Carnot cycle can be written as

Hence, for the infinitesimal cycle considered above

The work along bc and da nearly cancel such that the network is the difference between the work along and ab and cd and dW can be viewed as the area enclosed by the rectangle abcd.

Substituting Equations (1) and (3) into (2) we obtain

Rearranging terms yields the Clausius-Clapeyron equation, which defines the slope of the vapour pressure curve:

The beauty is that we have found a general relation between experimentally measurable quantities from first principles (1st and 2nd laws of thermodynamics). The Clapeyron’s latent heat equation holds for both the changes of state, i.e., from liquid to vapour and solid to liquid. In the later case hfg will represent the latent of fusion

Q7) Explain the Isothermal compressibility and volume expansivity?

A7)

The isothermal compressibility is defined by the fractional differential change in volume due to a change in pressure.

The negative sign is important in order to keep the value of kT positive, since an increase in pressure will lead to a decrease in volume. The 1/V term is needed to make the property intensive so that it can be tabulated in a useful manner.

Q8) Explain the Joule-Thomson coefficient

A8) There is a pressure drop associated with flow through a restriction like valves, capillary tube, porous plug etc.

 • The enthalpy of the fluid remains a constant.

 • The temperature of a fluid may increase, decrease, or remain constant during a throttling process.

 • The behaviour of fluids in such flows is described by the Joule-Thomson coefficient. The Joule-Thomson coefficient is defined as:

The Joule-Thomson coefficient is a measure of the change in temperature with pressure during a constant-enthalpy process.

Some h= constant lines on the T-P diagram pass through a point of zero slope or zero Joule-Thomson coefficient

•The line that passes through these points is called the inversion line, and the temperature at a point where a constant-enthalpy line intersects the inversion line is called the inversion temperature.

 • The slopes of the h=constant lines are negative (μ0) to the left of the inversion line.

Q9) 5000 J of heat are added to two moles of an ideal monatomic gas, initially at a temperature of 500 K, while the gas performs 7500 J of work. What is the final temperature of the gas?

A9)

Comment: the gas does more work than it takes in as heat, so it must use 2500J of its internal energy.

Q10) Compute the internal energy change and temperature change for the two processes involving 1 mole of an ideal monatomic gas. (a) 1500 J of heat are added to the gas and the gas does no work and no work is done on the gas (b) 1500 J of work are done on the gas and the gas does no work and no heat is added or taken away from the gas

A10)

(a)

(b)

Q11) A cyclic heat engine operates between a source temperature of 8000C and a sink temperature of 300C. What is the least rate of heat rejection per kW net output of the engine?

A11) For a reversible engine, the rate of heat rejection will be minimum

This is the least rate of heat rejection.

Q12) Explain the TDS relations

A12) On a P-v diagram, the area under the process curve is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. On a T-s diagram, the area under an internally reversible process curve is equal, in magnitude, to the heat transferred between the system and its surroundings. That is,

Note the area has no meaning for irreversible processes.

The T-s diagram of a Carnot cycle is shown on the left. The area under process curve 1-2 (area 1-2-B-A-1) equals the heat input from a source (QH). The area under process curve 3-4 (area 4-3-B-A-4) equals the heat rejected to a sink (QL). The area enclosed by the 4 processes (area 1-2-3-4-1) equals the net heat gained during the cycle, which is also the net work output.