Thermodynamic Properties of Steam & Steam Tables MCQ Quiz - Objective Question with Answer for Thermodynamic Properties of Steam & Steam Tables - Download Free PDF

Explanation:

The graphical representation of the transformation of 1 kg of ice into 1 kg of superheated steam at constant pressure is depicted using a t-h diagram (temperature-enthalpy diagram). This kind of graph is particularly useful for visualizing and understanding the various phase changes and energy exchanges involved in the process.

When considering the transformation of ice to superheated steam, several important stages can be identified on a t-h diagram:

• **Heating of ice:** As heat is added to ice at a constant pressure, its temperature increases until it reaches the melting point. This is represented by a sloping line on the t-h diagram.
• **Melting of ice:** At the melting point, the temperature remains constant while the ice absorbs latent heat of fusion and transitions into liquid water. This stage is depicted as a horizontal line on the t-h diagram.
• **Heating of water:** Once all the ice has melted, the temperature of the liquid water begins to increase as more heat is added. This phase is represented by another sloping line on the t-h diagram.
• **Boiling of water:** At the boiling point, the temperature again remains constant while the water absorbs latent heat of vaporization and transitions into steam. This is shown as another horizontal line on the t-h diagram.
• **Superheating of steam:** After all the water has evaporated, the temperature of the steam increases further as additional heat is added. This final stage is represented by a sloping line on the t-h diagram.

The t-h diagram provides a clear and comprehensive visualization of these phase changes and the corresponding energy exchanges involved in the transformation from ice to superheated steam. The temperature (t) is plotted on the y-axis, and the enthalpy (h) is plotted on the x-axis. The diagram effectively illustrates how the temperature remains constant during the phase changes (melting and boiling) and how it increases during the heating of each phase (ice, water, and steam).

Correct Option Analysis:

The correct option is:

Option 4: t-h diagram

This option correctly identifies the graphical representation used to illustrate the transformation of 1 kg of ice into 1 kg of superheated steam at constant pressure. The t-h diagram effectively shows the temperature and enthalpy changes during this process.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: u-v diagram

The u-v diagram is a plot of internal energy (u) versus specific volume (v). While this diagram can be useful in thermodynamic analysis, it is not typically used for representing the transformation of ice to superheated steam. The t-h diagram is more appropriate for visualizing the temperature and enthalpy changes during phase transitions.

Option 2: h-s diagram

The h-s diagram, also known as the Mollier diagram, plots enthalpy (h) against entropy (s). This diagram is commonly used in thermodynamics to analyze processes involving heat and work, particularly in the context of steam turbines and refrigeration cycles. However, it is not specifically used for illustrating the transformation of ice to superheated steam at constant pressure.

Option 3: p-v diagram

The p-v diagram plots pressure (p) against specific volume (v). This diagram is often used to analyze the behavior of gases and the performance of engines and compressors. While the p-v diagram is valuable for understanding the relationships between pressure and volume, it is not the best choice for representing the transformation of ice to superheated steam, which involves temperature and enthalpy changes.

Conclusion:

Understanding the transformation of ice to superheated steam at constant pressure involves recognizing the various phase changes and energy exchanges that occur. The t-h diagram is the most suitable graphical representation for this process, as it effectively illustrates the temperature and enthalpy changes during each phase transition. While other diagrams like the u-v, h-s, and p-v diagrams have their own applications in thermodynamic analysis, they are not as appropriate for visualizing the specific transformation of ice to superheated steam.

Explanation:

Statement A: At a given pressure, the capacity of superheated steam to do work will be comparatively higher.

• Superheated steam is steam that has been heated beyond its saturation temperature, meaning it exists at a temperature higher than the boiling point at a given pressure. This additional heat increases the steam's energy content without increasing its pressure. As a result, superheated steam can do more work compared to saturated steam at the same pressure because it has more thermal energy available for conversion to mechanical work. Therefore, statement A is true.

Dryness fraction:

• Dryness fraction is a measure of the quality of steam, indicating the proportion of steam in a mixture of steam and water. It is defined as the mass of dry steam divided by the total mass of the mixture (dry steam and water). A dryness fraction of 1 indicates completely dry saturated steam, while a lower value indicates the presence of water. Therefore, statement B is true.

Statement C: Wet steam does not exist.

• Wet steam is a mixture of saturated steam and water droplets. It is a common phase of steam in many practical applications, especially in steam engines and boilers where steam is generated and used. Therefore, statement C is false because wet steam does exist.

Explanation:

Superheater:

• A superheater is a device used to convert saturated steam or wet steam into superheated steam or dry steam
• Superheated steam is used in steam turbines for electricity generation, steam engines, and in processes such as steam reforming
• While the temperature of the steam in the superheater rises, the pressure remains the same as that of the boiler

Economizer:

• It is also known as a feedwater heater. It is a device in which the waste heat of the flue gases is utilized for heating the feed water.
• In an economizer, feed water is preheated by using flue gases to improve overall efficiency and only sensible heat transfer is taking place so feed water is heated without converting it into steam. Therefore, the economizer is placed after the superheater and located in the feeding water circuit.

Functions of economizer:

• Reduce fuel consumption
• Preheating a fluid (feed-water in case of steam boiler)
• Increases the efficiency of the power plant

Concept:

When dealing with the properties of superheated steam using steam tables, it is crucial to understand the method of interpolation and how steam tables are structured. Steam tables typically provide data for saturated steam and superheated steam at various pressures and temperatures. The correct approach to determine the properties of superheated steam involves interpolation between known data points.

Explanation:

To find the properties of superheated steam at a specific pressure and temperature, you generally follow these steps:

1. Locate the superheated steam tables in the steam table resource.
2. Identify the tables corresponding to the pressure closest to your given pressure.
3. Within that pressure-specific table, identify the temperatures that bracket your given temperature.
4. Interpolate between these temperatures to find the desired property at the given temperature.

Given the options provided in the question, let's analyze each one:

• Option 1: Use the saturated steam tables and adjust for superheat.

This approach is not correct. Saturated steam tables are for steam at the saturation point (where liquid and vapor coexist). Superheated steam is at a higher temperature than the saturation temperature for a given pressure, and its properties differ significantly from those of saturated steam. Therefore, using saturated steam tables and adjusting for superheat is not a valid method.

• Option 2: Interpolate between two temperatures at constant pressure.

This is the correct approach. Superheated steam tables provide data at specific pressures and a range of temperatures for each pressure. To find the properties at a specific temperature that is not directly listed, you interpolate between the two closest temperatures at that constant pressure. This method ensures accuracy in determining the properties of superheated steam.

• Option 3: Interpolate between two pressures at constant temperature.

This approach is not standard practice. While it might be theoretically possible, steam tables are typically organized to facilitate interpolation between temperatures at a constant pressure rather than between pressures at a constant temperature. This method is less common and can be more complex.

• Option 4: Directly read the values without any interpolation.

This is only possible if the exact pressure and temperature values are listed in the steam tables. However, in practice, it is rare to find tables with values for every possible combination of pressure and temperature. Therefore, interpolation is usually necessary.

Solution Statement:

The correct approach to determine the properties of superheated steam at a given pressure and temperature using steam tables is to interpolate between two temperatures at constant pressure. This method ensures that you accurately find the properties of superheated steam at the desired conditions by using the data provided in the steam tables.

Thus, the correct option is:

Option 2: Interpolate between two temperatures at constant pressure.

Explanation:

Mollier Diagram: 

• Dr. Mollier, in 1904, conceived the idea of plotting total heat against entropy, and his diagram is more widely used than any other entropy diagram since the work done on vapour cycles can be scaled from this diagram directly as a length; whereas on T-s diagram it is represented by an area.
• Mollier diagram is enthalpy (h) versus entropy (s) plot.

• It consists of a family of constant pressure lines, constant temperature lines and constant volume lines plotted on enthalpy versus entropy coordinates.
• From the T-ds equation:

 Tds = dh – vdp

 In the two-phase region, the constant pressure and constant temperature lines coincide.

\({\left( {\frac{{\partial h}}{{\partial s}}} \right)_p} = T\)

• The slope of an isobar on the h-s coordinates is equal to the absolute temperature. If the temperature remains constant the slope will remain constant.
• If the temperature increases the slope of the isobar will increase.
• For the saturated liquid and saturated vapour i.e. within the dome the temperature and pressure remains constant.

                               Figure: Mollier diagram

• A vertical line in the Mollier diagram represents the isentropic process (s = C).
• A horizontal line in the Mollier diagram represents the isenthalpic process (h = C).
   

Therefore, statements A and B are true, while statement C is false.

Explanation:

• Adiabatic process: It is the thermodynamic process in which there is no exchange of heat and mass between the system and its surrounding during the process.
• The adiabatic process can be either reversible or irreversible.
• The essential conditions for the adiabatic process to take place:
  1. ​​The system must be perfectly insulated from the surrounding.
  2. The process must be carried out quickly so that there is no sufficient amount of time for heat and mass transfer to take place.
• ​​The adiabatic process equation:

​⇒ PVγ = constant  

The dryness fraction of wet steam undergoing adiabatic expansion can increase, decrease, or remain constant depending on several factors.

Factors affecting dryness fraction:

• Initial dryness fraction: The starting dryness fraction of the wet steam significantly impacts the outcome. If the initial dryness fraction is high (closer to 1), it's less likely to decrease significantly during expansion.
• Temperature and pressure changes: During adiabatic expansion, the temperature and pressure of the steam change. The specific direction and magnitude of these changes influence the phase change behavior of the water in the steam. Depending on the temperature and pressure drop, some of the liquid water might evaporate, increasing the dryness fraction, or some of the vapor might condense, decreasing the dryness fraction.
• Friction: While adiabatic expansion implies no heat transfer, internal friction within the steam can generate heat locally. This heat can influence the phase change behavior, potentially promoting some evaporation and increasing the dryness fraction.
• Specific heat capacities: The specific heat capacities of steam and water also play a role. If the temperature drop during expansion is more prominent for the liquid water component than the vapor component, it can result in less condensation and a potential increase in dryness fraction.

Explanation:

Critical point:

At a critical point, the liquid is directly converted into vapor without having a two-phase transition.

For water at the critical point: 

Pcr = 220.6 bar = 22.06 MPa

Tcr = 373.95oC

vcr = 0.005155 m3/kg

Enthalpy of vaporization at a critical point is zero. 

The figure below represents the P-V diagram for a pure substance (water).

Explanation:

Concept:

The specific volume of steam is given by

v = v_f + xv_fg

v = vf + x(v_g - v_f)

Where v is the volume at any point in the wet region, v_f is the volume at the saturated liquid, v_g is the volume at saturated vapor, and x is the quality of steam.

If the steam lies within the vapor dome (saturated liquid-vapor region) then it is called wet steam.

Calculation:

Given:

x = 80 % = 0.8, vf = 0.011 m3/kg and vg = 0.3071 m3/kg.

v = vf + x (vg - vf)

v = 0.011 + 0.8 (0.3071 - 0.011)

v = 0.248

v = 0.25 m3/kg. 

Additional Information

Enthalpy in wet region is calculated by

h = h_f + xh_fg

h = hf  + x( h_g - h_f)    

Where h is enthalpy at any point in the wet region, h f is enthalpy at the saturated liquid, hg is enthalpy at saturated vapor and x is the quality of steam.

Entropy in wet region is calculated by

s = s_f + xs_fg

s = s_f + x(s_g - s_f)

Where s is entropy at any point in the wet region, s f is entropy at the saturated liquid, s_g is entropy at saturated vapor and x is quality of steam.

Explanation:

Degree of Super Heat:

It is the difference between the temperature of superheated steam and saturation temperature resultant to the given pressure.

So, the degree of superheated = tsup - ts

where, tsup  = Temperature of superheated steam

ts  =  Saturation temperature corresponding to the given pressure of steam generation

Super Heat:

It is the quantity of heat needed to transform unit mass of dry saturated steam to unit mass of superheated steam at constant pressure.

Super heat = Cp  × (tsup  - ts) kJ/kg

When the vessel is at 25°C the saturation pressure of the vapour inside the vessel 3.169 kPa is much lesser than the atmospheric pressure.

When this vessel is kept inside the isothermal oven then at 80°C the saturation pressure of vapour at 80°C 47.39 kPa is again lesser than the atmospheric pressure.

Since the pressure inside the vessel is lesser than the atmospheric pressure, then if the valve is open inside the oven which is at atmospheric pressure the hot air from the oven will go inside the vessel through the Valve.

In stage 1:

The pressure will be less than the atmospheric pressure because of the presence of water vapour. 

In stage 2:

The water vapour receives heat and gets superheated since the vessel is placed in an oven at 80°C.

Also, we need to remember that in stage 2 the heating of water vapour takes place at constant volume since the oven has a fixed boundary. 

When volume is constant then pressure is directly proportional to temperature i.e

P ∝ T

where P is the pressure and T is the temperature

∴ \(\frac{{{{\rm{P}}_1}}}{{{{\rm{P}}_2}}} = \frac{{{{\rm{T}}_1}}}{{{{\rm{T}}_2}}}\)...(i)

Where P₁ = Saturated vapour pressure of the mixture at 25°C = 3.17 kPa (Standard value)

P₂ = Pressure of the mixture at 80°C, T₁ = Initial temperature of the mixture at 25°C, T₂ = Final temperature of the mixture 

T₁ = 25 + 273 = 298 K, T₂ = 80 + 273 = 353 K 

Substituting in equation (i) and simplifying we will get

\(\frac{{3.17}}{{{{\rm{P}}_2}}} = \frac{{298}}{{353}}\)

P₂ = 3.75 kPa

Note:

P₂ is the pressure of the mixture at 80°C after thermal equilibrium is attained.

Therefore, Since the pressure inside the vessel after heating is less than the atmospheric pressure (101.32 kPa) the hot air goes inside the vessel.

Explanation:

Superheated steam:

• The state of steam lies beyond the saturated vapour line or the state of steam at which temperature is more than the saturated steam. 
• A superheated region is a single-phase region (vapour only), temperature and pressure are no longer dependent. 
• If the steam exists in only one phase (superheated steam), it is necessary to specify two independent variables, pressure and temperature, for the complete specification of the state. In the superheated steam tables, the properties- v, u, h, and s- are tabulated from the saturation temperature to some temperature for a given pressure.​

Additional Information 

Wet steam:

• The state of steam which contains moisture or droplet of water particles in suspension is called Wet steam.
• The moisture or water particles lead to a loss in efficiency and damage of mechanical parts of the turbine by the erosion of blades.

Dry saturated steam:

• The state of steam that lies on the saturated vapour line is called dry saturated steam. It is free from any moisture or water droplets.

Calculation:

Given:

Pressure, P = 0.5 MPa, Enthalpy, h₂ = 2890 kJ/kg

T1 = 200°C, h1 = 2855 kJ/kg

T3 = 250°C, h3 = 2960 kJ/kg

The given enthalpy h₂ lie between h₁ (enthalpy at 200°C) and h₃ (enthalpy at 250°C)

By using the method of Interpolation, we can calculate the temperature as follows:

\(\frac{T_2 - T_1}{T_3 - T_2} = \frac{h_2 - h_1}{h_3 - h_2}\)

Put all the known values in the above equation:

\(\frac{T_2 - 200}{250 - T_2} = \frac{2890 - 2855}{2960 - 2890}\)

T₂ - 200 = 0.5(250 - T₂)

2T₂ - 400 = 250 - T₂

3T2 = 650

T₂ = 216.67°C

The temperature of water at a pressure of 0.5 MPa having the specific enthalpy (h) of 2890 kJ/kg is 216.67°C.

Explanation:

Steam table:

It is the thermodynamic data that contain the properties of water or steam.

In the saturated steam table, the data always refer to the steam at saturation or boiling point. 

If saturated steam is further heated after its boiling point at a given constant pressure then it will be converted into superheated Steam.

Additional InformationCritical pressure:

It is the vapour pressure of the fluid at a critical temperature above which the distinct liquid and gas phases do not exist. 

Explanation:

Saturation state:

• A saturation state is a thermodynamic state where a substance is in equilibrium between its liquid and vapour phases at a given temperature and pressure. At this state, the substance exists as a mixture of both liquid and vapour phases, and the concentrations of the substance in the liquid and vapour phases are constant. The term "saturation" refers to the fact that the substance is at its maximum capacity to hold the vapor phase at that given temperature and pressure.
• At a saturation state, any additional heat or mass transfer to the system would cause a phase change without changing the temperature or pressure. For example, if we heat a liquid at a constant pressure, the temperature of the liquid would increase until it reaches the saturation temperature, at which point some of the liquid would start vaporizing to maintain the equilibrium between the liquid and vapor phases. The temperature would remain constant during this process of vaporization, and any further heat transfer would cause more vaporization until all the liquid has vaporized.
• Similarly, if we add more substance to a saturated liquid at a constant temperature and pressure, the liquid would start vaporizing until the concentration of the substance in the liquid and vapour phases reaches a constant value.

Explanation:

Dry steam, or saturated steam, is characterized by the vapor quality or dryness fraction(x), which is equal to unity.

When the vapor quality is equal to 0, it is referred to as the saturated liquid state (single-phase).

On the other hand, when the vapor quality is equal to 1, it is referred to as the saturated vapor state or dry steam (single-phase).

Between these two states, vapor-liquid mixture or wet steam (two-phase mixture) exist. At constant pressure, the addition of energy does not change the temperature of the mixture, but the vapor quality and specific volume change.

In the case of dry steam (100% quality), it contains 100% of the latent heat available at that pressure. Saturated liquid water, which has no latent heat and therefore 0% quality, will therefore only contain sensible heat.

Dryness fraction is defined as the ratio of the mass of dry steam and the combined mass of dry steam & mass of water in the mixture. It is denoted by x.

\(Dryness\;fraction\;\left( x \right) = \frac{{mass\;of\;vapour\;\left( {{m_v}} \right)}}{{mass\;of\;vapour\;\left( {{m_v}} \right) + mass\;of\;liquid\;\left( {{m_l}} \right)}}\)

For saturated liquid x = 0

For saturated vapour = 1

The value of the dryness fraction lies between 0 and 1.

Note:

The term quality is used in the wet vapor zone and is of no significance in superheated or compressed liquid regions.

The difference between the temperature of the superheated vapor and the saturation temperature at that pressure is called the superheat or the degree of superheat.

If the temperature of a fluid is greater than its saturation temperature corresponding to its pressure, the fluid is called superheated vapor. The difference between the temperature of the superheated vapor and the saturation temperature corresponding to its pressure is called the degree of superheat.