Properties of Steam
2.1 ❐ PURE SUBSTANCE
- Homogeneous in composition: It means that the composition of every part of the system is same throughout. By composition, we mean the relative proportions of the chemical elements into which the sample of the system can be analysed.
- Homogeneous in chemical aggregation: It means that the chemical elements must be combined chemically in the same way in all parts of the system.
- Invariable in chemical aggregation: It means that the state of chemical combination of the system does not change with time and position.
Examples of pure substance are atmospheric air, mixture of steam-water, combustion products of a fuel, etc.
2.2 ❐ CONSTANT PRESSURE FORMATION OF STEAM
Consider 1 kg of ice under atmospheric pressure and at a temperature of −10°C contained in a vessel. If we heat the ice gradually keeping the pressure constant, the following changes will take place, as shown in Fig. 2.1:
Figure 2.1 Constant pressure temperature-heat added diagram for steam and water
- The temperature of ice will gradually increase till it just reaches the freezing temperature (represented by point ‘b’), which is 0°C at atmospheric pressure. This is shown by the straight line ab.
- On addition of more heat at point b, the ice starts melting, and with no rise in temperature till the whole ice has melted, up to point c, the ice is converted into water. The heat added during the process bc is called the latent heat of fusion of ice or simply latent heat of ice.
- Gradual addition of more heat beyond the point c leads to gradual increase in temperature till the water reaches the vapourisation or boiling point ‘d’. Under atmospheric pressure of 1.01325 bar, it is 100°C.
- On further heating beyond point d, the water will gradually turn into steam, with no rise in temperature. This steam contains some water particles and is called the wet steam. With further addition of heat, all the water particles in the steam disappear and this steam is called dry saturated steam or simply dry steam. This corresponds to point ‘e’. The heat added during de is called the latent heat of vapourisation of steam or simply latent heat of steam.
- If the dry saturated steam is further heated beyond point e, keeping the pressure constant, the process is called superheating. The steam so obtained is called the superheated steam. Let the steam be superheated to temperature ts°C, represented by point ‘f’. Then (ts − t)°C is called the degree of superheat. The heat added during process ef is called the heat of superheat.
2.3 ❐ PROPERTIES OF STEAM
- Enthalpy or total heat of water: It is defined as the quantity of heat required to raise the temperature of 1 kg of water from 0°C to its boiling point or saturation temperature corresponding to the pressure applied. It is also called enthalpy of saturated water or liquid heat and is represented by hf.
hf = specific heat of water, cpw × rise in temperature= 4.187 × Δt, kJ/kg
Note that enthalpy of saturated water can also be found from the steam table based on saturation temperature (Appendix A.1.1).
- Heat of steam or latent heat of vapourisation: Latent heat of steam at a particular pressure may be defined as the quantity of heat required to convert 1 kg of water at its boiling point into dry saturated steam at the same pressure. It is denoted by hfg.
- Dryness fraction or quality: Dryness fraction of a liquid−vapour mixture is defined as the ratio of mass of dry steam actually present to the mass of the total mixture. It is denoted by x.
Let ms = mass of dry steam (saturated vapour), kg
mw = mass of water vapour in steam (saturated liquid), kg
x = dryness fraction of the sample
For dry steam, mw = 0 and x = 1.
The quality of steam is the dryness fraction expressed as a percentage.
Wetness fraction = 1 − x
- Specific volume of wet steam: The specific volume of steam is the volume of 1 kg of dry steam.
Total volume of liquid and vapour can be expressed in terms of their specific volume asVf = mwvfVg = msvg
where vf and vg are the specific volumes of saturated liquid and saturated vapour, respectively.
Specific volume of the mixture is then= (1 − x)v f + xvg= v f + x(vg − v f)
Now vf << vg
- Enthalpy of wet steam: It may be defined as the quantity of heat required to convert 1 kg of water at 0°C, at constant pressure, into wet steam. Similar to the analysis of specific volume, one can write
where hf is the specific enthalpy of saturated water and hfg is the change in specific enthalpy during vapourisation. When steam is dry saturated, then x = 1
- Total enthalpy of superheated steam:
Let cps = specific heat of superheated steam
ts = temperature of formation of steam, °C
tsup = temperature of superheated steam, °C
Then heat of superheat = cps (tsup − ts)
Note that enthalpy of superheated steam can also be found from the steam table (Appendix A.1.3).
- Internal energy of steam: The internal energy of steam may be defined as the actual heat energy stored in steam above the freezing point of water. It is the difference between the total heat of steam and the external work of evaporation. Similar to the analysis of specific volume, one can write for 1 kg of wet steam,
where uf is the specific internal energy of saturated water and ufg is the change in specific internal energy during evaporation. From the known value of enthalpy, internal energy can be found as follows:uwet = h − pv1
For superheated steam,hsup = hg + cps (tsup − ts)usup = hsup − pvsup
- Entropy of water: Consider 1 kg of water to be heated from a temperature Ts to T at a constant pressure.
Then, change of entropy =
where cpw = specific heat of water
The entropy of water is always reckoned above the freezing point of water.
Therefore, Ts = saturation temperature of formation of steamT = absolute temperature of freezing water = 273 K
Entropy of saturated water,
- Entropy of wet steam: Consider 1 kg of wet steam at an absolute pressure p bar.
Let x = dryness fraction of steam
Ts = temperature of formation of steam
sf = the specific enthalpy of saturated water
sfg = change in entropy during evaporation
The entropy of wet steam can be calculated as
When steam is dry saturated, then x = 1
- Entropy of superheated steam: Consider 1 kg of dry saturated steam at Ts temperature of formation, to be superheated to a temperature Tsup.
Let cps = specific heat of steam at constant pressure during superheating
Total entropy of superheated steam
ssup = Entropy of dry saturated steam + Change of entropy during superheating
2.4 ❐ STEAM TABLES
In actual practice, it is quite cumbersome to calculate the relation between various quantities such as pressure, temperature, volume, enthalpy, latent heat and entropy, etc., of steam. Such quantities have been determined experimentally and recorded in the form of tables known as steam tables. Steam table is a complete data book that contains various properties, like volume, internal energy, enthalpy and entropy for 1 kg of water in different phases for a given pressure or temperature. Steam table is given in the Appendix A. The properties of steam are given in three different tables—saturated steam table based on temperature (Appendix A.1.1), saturated steam table based on pressure (Appendix A.1.2), and superheated steam table (Appendix A.1.3). When the temperature is selected as independent variable and the properties of steam are tabulated, the steam table is referred to as the saturated steam table based on temperature. When the pressure is chosen as independent variable, it is referred to saturated steam table based on pressure. For superheated steam, there is a separate table for different amount of degree of superheat.
Values for unknown pressure or temperature can be calculated by linear interpolation between two consecutive values. Saturated steam table based on pressure for a particular pressure (= 1 bar) is shown in Table 2.1.
Table 2.1 Saturated steam table based on pressure
2.5 ❐ TEMPERATURE−ENTROPY DIAGRAM FOR WATER AND STEAM
It is the plot of the saturation temperature of water and steam corresponding to various absolute pressure vs entropies at those saturation temperatures, as shown in Fig. 2.2. The curve on the left marked water line (saturated liquid line) shows the relationship between entropy and temperature before steam is formed. The dry steam line (saturated vapour line) shows the relationship between entropy and temperature when all the water has been converted into steam. The area in between two regions is called the wet region. The area to the right of the dry steam lines is called the region of superheat. The lines of constant dryness, constant volume, and constant total heat are also drawn on the diagram. The temperature at which the distinction between the entropy of water and that of dry steam disappears is called the critical temperature (374.15°C). The pressure corresponding to this temperature is called the critical pressure (221.2 bar).
2.6 ❐ ENTHALPY−ENTROPY OR MOLLIER DIAGRAM OF STEAM
In Mollier diagram (Fig. 2.3), the vertical ordinate represents the enthalpy and the horizontal ordinate represents the entropy. The regions above and below the saturation line represent the superheated and wet conditions of steam, respectively. The lines of constant dryness fraction are shown in the wet steam region, whereas the lines of constant temperature and pressure in the region of both wet and superheat are shown.
From the thermodynamic property relationship, one can write,
For constant pressure process, we have
From Eq. (2.13), it is clear that the slope of the constant pressure lines is equal to the corresponding saturation temperatures. It may be noted that as the pressure increases, the saturation temperature also increases. This is the reason for which constant pressure lines (isobar) are divergent on the h−s diagram. The saturated liquid line and saturated vapour line meet at critical point.
2.7 ❐ VARIOUS PROCESSES FOR STEAM
2.7.1 Constant Volume Process
The constant volume heating process is shown on the p−v and T−s diagrams in Fig. 2.4. Consider 1 kg of wet steam at a certain pressure p1, and dryness fraction x1 is heated to state 2 in the wet region and then to state 3 in the superheated region.
Initial volume of steam v1 = vf 1 + x1(vg1 − vf1)
Final volume of steam in wet region v2 = vf2 + x2(vg2 − vf2)
Since the volume remains constant,
Figure 2.4 Constant volume heating: (a) p−v diagram, (b) T−s diagram
From steam table, internal energy of steam can be found as follows:
Initial internal energy of steam, u1 = uf1 + x1 ufg1
Final internal energy of steam in the wet region, u2 = uf2 + x2 ufg2
Since dv = 0, therefore δW = 0
For constant volume heating process, v2 = v3. After knowing the value of v3, the condition of steam can be found from superheated steam table. If pressure of state 3 is given (=p3), then from known values of p3 and v3, temperature of state 3 can be obtained. Similarly, if temperature at state 3 (=T3) is given, pressure can be found from known values of T3 and v3.
2.7.2 Constant Pressure Process
The constant pressure heating process is shown in Fig. 2.5. In this case, p1 = p2 = p (say).
For wet steam, during heating process 1–2,
Work done per kg of steam, w = pdv = p(v2 − v1)
where v1 = vf + x1(vg − vf) and v2 = vf + x2(vg − vf)
The specific enthalpy of steam, h1 = hf + x1 hfg and h2 = hf + x2 hfg
From first law of thermodynamics, heat transfer per kg of steam during constant pressure heating
For superheated steam, specific volume and specific enthalpy of steam can be found from the superheated steam table. Then, during heating process 2-3,
Work done per kg of steam, w = pdv = p(v3 − v2) = p(vsup − v2)
and heat transfer per kg of steam
Figure 2.5 Constant pressure expansion
2.7.3 Isothermal Process
The isothermal heating process is shown in Fig. 2.6. The isothermal process for wet steam will be same as the constant pressure process. However, in the superheated region, the steam will behave as a gas and shall follow Boyle’s law, pv = const. Therefore, the process will be hyperbolic.
In the wet region,
In the superheated region,
Figure 2.6 Isothermal process
2.7.4 Hyperbolic Process
During a hyperbolic process, pv = const. If the superheated vapour behaves as an ideal gas, then the process during which pv = const. may be regarded isothermal, as shown in Fig. 2.7.
For 1 kg of steam in the wet region,
Initial volume of steam v1 = vf1 + x1(vg1 − vf1)
Final volume of steam v2 = vf2 + x2(vg2 − vf2)
As pv = const., therefore
Figure 2.7 Hyperbolic expansion
and hs2 = hg2 + cp (Tsup − T2) for superheated steam
Change is internal energy,
where v1 and v2 are the specific volumes of steam before and after heating.
Total heat supplied, q = w + (u2 − u1)
2.7.5 Reversible Adiabatic or Isentropic Process
Since s1 = s2, therefore
For initially dry steam (Fig. 2.8(b)), s1 = sg1
Figure 2.8 Isentropic expansion: (a) Expanded steam wet, (b) Expanded steam dry saturated, (c) Expanded steam superheated
Then, sg1 = s f2 + x2s fg2
For initially superheated steam at temperature Tsup, s1 can be calculated from superheated steam table. After expansion, state 2 may be superheated (Fig. 2.8(c)), dry or wet. If the state after expansion is superheated, then the condition of steam can be calculated from the relationship s1 = s2. Note that if pressure is specified, then temperature can be calculated or otherwise if temperature is specified, then pressure can be calculated.
Now q = w + (u2 − u1)
where u1 = u f1 + x1u fg1 for wet steam
= ug1 for dry steam
= usup for superheated steam
Similarly, u2 can be found depending on the condition of steam after expansion.
But q = 0
2.7.6 Polytropic Process
Let the steam expand from condition 1 to 2 according to the law pvn = constant. Then for wet steam,
Initial volume of steam, v1 = vf1 + x1(vg1 − vf1)
Final volume of steam v2 = vf2 + x2(vg2 − vf2 )
Also u1 = uf1 + x1ufg1
n = 1.13 for wet steam and 1.3 for superheated steam
For superheated steam on heating from wet steam, v2 can be calculated from superheated steam table for the known pressure and temperature conditions. Similarly, the condition of steam can also be found after knowing the value of v2 from Eq. (2.21).
2.7.7 Throttling Process
The throttling process occurs when steam is expanded through a small aperture as in the case of throat of a nozzle. During this process, no work is done and the enthalpy remains constant. The throttling process is shown on T−s and h−s diagrams in Fig. 2.9.
Now h1 = h2
For wet steam after expansion
For superheated steam after expansion
Figure 2.9 Throttling process: (a) T−s diagram, (b) h−s diagram
2.8 ❐ DETERMINATION OF DRYNESS FRACTION OF STEAM
The different methods commonly used for the measurement of dryness fraction of steam are as follows:
- Barrel Calorimeter
- Separating Calorimeter
- Throttling Calorimeter
- Combined Separating and Throttling Calorimeter.
The working principle of all the calorimeters is to bring the state of the substance from the two-phase region to the single phase region (either compressed liquid or superheated vapour region). This is because in the two-phase region, pressure and temperature are not independent variable, one is independent and the other is dependent. On the other hand, both in compressed liquid and superheated vapour region, both pressure and temperature are independent variables.
2.8.1 Barrel Calorimeter
The barrel calorimeter, shown in Fig. 2.10, consists of a copper barrel placed on wooden block and covered by a wooden cover. The barrel contains a known quantity of cold water and is surrounded by an outer vessel with air space in between, which acts as an insulator. The temperature is measured by a thermometer that passes through one of the holes in the wooden cover. The whole assembly is placed on the platform of a weighing bridge. The steam from the main steam pipe enters through the sampling tube via the control valve and flows into the cold water through fine exit holes of the ring provided at the end of the pipe. The condensation of steam takes place as it comes in contact with cold water, and as a result, the temperature of water rises. The quantity of steam condensed can be known from the difference of readings of the weighing bridge before and after the condensation of steam. The pressure of entering steam is indicated by the pressure gauge.
Figure 2.10 Barrel calorimeter
Let ms = mass of steam condensed
x = dryness fraction of steam
hfg = latent heat of steam
ts = temperature of formation of steam (corresponding to pressure indicated by the pressure gauge)
mw = mass of cold water in the calorimeter in the beginning
mc = water equivalent of calorimeter
t1 = initial temperature of cold water in the calorimeter
t2 = final temperature of the mixture after steam condensation
Heat lost by steam = ms[x × hfg + cps(ts − t2)]
Heat gained by cold water and calorimeter = (mw + mc) × cpw × (t2 − t1)
This method gives only approximate results. The value of x calculated is always lower than the actual value as heat losses due to convection and radiation are not accounted for.
2.8.2 Separating Calorimeter
This calorimeter, shown in Fig. 2.11, consists of two concentric chambers that communicate with each other through an opening at the top. The steam to be tested flows from the main steam pipe through a sampling tube to the calorimeter pipe through the valve which must be fully open when testing. The metal basket has a large number of perforations through which the steam discharges. The water particles due to their heavier momentum get separated from the steam and get collected in the inner chamber. The quantity of water collected is indicated by the level in the gauge glass and the pointer that moves on a scale. The dry steam in the inner chamber moves up and then down again through the annular space between the two chambers. The gauge fitted has two scales: the inner one indicates the pressure, whereas the outer one indicates the rate of discharge of dry saturated steam during a predefined time interval. Then the steam then flows through the calibrated orifice to the bucket calorimeter which is placed on the platform balance by which the most of steam discharged can be further confirmed.
Figure 2.11 Separating calorimeter
mw = mass of suspended moisture collected
It is not possible to remove all the water particles from steam, and the dryness fraction calculated by this method is always greater than the actual. However, it is a quick method and is used for the measurement of dryness fraction for very wet steam.
2.8.3 Throttling Calorimeter
The throttling calorimeter, shown in Fig. 2.12, consists of the sampling tube through which the steam from the main steam pipe flows through the throttling valve and becomes superheated. The steam then flows into the inner chamber, flows down, and rises up again to enter the annular space. The loss of heat by radiation from the inner chamber is minimised by the hot steam around the outside of the inner chamber. The temperature of throttled steam is measured by the thermometer placed in the pocket filled with cylinder oil. To obtain good results, the steam issuing from the throttle valve must be superheated. To ensure this, a manometer is attached to measure the pressure of steam. Corresponding to this pressure, the saturation temperature of steam must be lower than the temperature indicated by the thermometer. The steam finally escapes through the exhaust. This calorimeter is suitable for steam having high dryness fraction.
Let p1 = initial pressure of wet steam before expansion
x1 = dryness fraction of wet steam
hfg1 = latent heat of vapourization of wet steam
tsup = temperature of superheated steam after expansion
hfg2 = latent heat of vapourization of steam after expansion
ts2 = saturation temperature of steam after expansion
cps = specific heat of superheated steam at constant pressure
Enthalpy of steam before throttling = hf 1 + x1 hfg1
Figure 2.12 Throttling calorimeter
Enthalpy before throttling = Enthalpy after throttling
2.8.4 Combined Separating and Throttling Calorimeter
The combined separating and throttling calorimeter is shown in Fig. 2.13. The steam is first passed through the separating calorimeter where it losses most of its moisture and becomes comparatively drier. It is then passed through the throttling calorimeter where superheating takes place without change of enthalpy. The temperature and pressure of steam after throttling are measured by using a thermometer and pressure gauge, respectively.
Let p1 = pressure of wet steam entering the throttling calorimeter
x1 = dryness fraction of steam
hfg1 = latent heat of entering steam
mw = mass of suspended moisture collected
ms = mass of steam leaving the separating calorimeter and entering the throttling calorimeter
p2 = pressure of steam after throttling
hfg2 = latent heat of steam at pressure p2
Enthalpy of steam entering throttling calorimeter = Enthalpy of steam leaving throttling calorimeter
Figure 2.13 Combined separating and throttling calorimeter
and x = dryness fraction of steam entering the separating calorimeter
Mass of dry steam entering the whole apparatus = ms x1
This calorimeter gives quite satisfactory results when the steam is considerably wet.
Determine the condition of steam in the following cases:
- At a pressure of 10 bar and temperature 200°C.
- At a pressure of 10 bar and specific volume 0.175 m3/kg.
Given p = 10 bar; t = 200°C; v = 0.175 m3/kg
- Condition of steam at temperature of 200°C:
From steam table (A.1.2 in the Appendix), corresponding to a pressure of 10 bar, we find the saturation temperature ts = 179.91°C
Since the saturation temperature at 10 bar is (179.91°C) lower than the given temperature of the steam (200°C), therefore the given steam is superheated.
The degree of superheat = 200 − 179.91= 20.09°C
- Condition of steam at a volume of 0.175 m3/kg:
From steam table (A.1.2 in the Appendix), vf = 0.001127 m3/kg, vg = 0.19444 m3/kg.
Since the volume of given steam (0.175 m3/kg) is less than the specific volume of the dry saturated steam (0.19444 m3/kg), therefore the given steam is wet. Therefore, the dryness fraction can be found to be0.175 = 0.001127 + x (0.19444 − 0.001127)
or x = 0.8994
Calculate the quantity of heat required to generate 1 kg of steam at a pressure of 8 bar from water at 30°C (a) when dryness fraction is 0.9, (b) when steam is just dry, and (c) when the temperature of steam is 300°C.
From steam tables (A.1.2 in the appendix) at 8 bar: hf = 721.1 kJ/kg, hfg = 2048.0 kJ/kg
And at 30°C (A.1.1 in the appendix), hf = 125.79 kJ/kg
- h = hf + x hfg = 721.1 + 0.9 × 2048.0 = 2564.3 kJ/kg
Net heat required = h − hf = 2564.3 − 125.79 = 2438.51 kJ/kg
- h = hf + hfg = hg = 2769.1 kJ/kg
Net heat required = h − hf = 2769.1 − 125.79 = 2643.31 kJ/kg
- ts = 170.4°C, tsup = 300°C
Therefore, the steam is in superheated condition. From superheated steam table (A.1.3 in the appendix), we have h = 3056.4 kJ/kg
Net heat required = h − hf = 3056.4 − 125.77 = 2930.63 kJ/kg
A pressure cooker contains 2 kg of steam at 5 bar and 0.9 dry. Find the quantity of heat that must be rejected so that the quality of steam becomes 0.5 dry.
The pressure cooker is a constant volume vessel. Therefore, W = 0
From steam tables (A.1.2 in the appendix) at 5 bar: vf = 0.001093 m3/kg, vg= 0.3749 m3/kg, uf = 639.66 kJ/kg, ufg = 1921.6 kJ/kg,
Specific volume of the steam at 5 bar
From steam tables (A.1.2 in the appendix), the properties of steam at 2.5 and 2.75 bar are given below:
For constant volume of steam, we have
Specific volume of the steam at 2.5 bar
v = vf + x (vg − vf ) = 0.001067 + 0.9 × (0.7187 − 0.001067) = 0.3599 m3/kg
Specific volume of the steam at 2.75 bar
v = vf + x(vg − vf ) = 0.001070 + 0.9 × (0.6573 − 0.001070) = 0.3292 m3/kg
Using linear interpolation, the pressure corresponding to v2 = 0.3375 m3/kg is obtained as
or p2 = 2.68 bar
or u fg2 = 1994.83 kJ/kg
∴ u2 = uf 2 + x2 ufg2 = 544.79 + 0.5 × 1994.83 = 1542.205 kJ/kg
About 5400 m3/h of wet steam with dryness fraction 0.92 and at 10 bar is to be supplied by a boiler for a processing plant. Calculate (a) the mass of steam supplied per hour and (b) the quantity of coal of calorific value 15 MJ/kg to be burnt in the boiler if overall efficiency of the boiler is 30%.
- From steam tables (A.1.2 in the appendix) at 10 bar: vf = 0.001127 m3/kg, vg= 0.19444 m3/kg
Specific volume of the steam v = vf + x (vg − vf ) = 0.001127 + 0.92 × (0.19444 − 0.001127) = 0.179 m3/kg
- At p = 10 bar, hf = 762.79 kJ/kg, hfg = 2015.3 kJ/kg
h = hf + x hfg = 762.79 + 0.92 × 2015.3 = 2616.866 kJ/kg
H = ṁh = 30167.6 × 2616.866 = 78944566 kJ/h
Heat supplied to boiler, Hb = = 263148553 kJ/h
Quantity of coal burnt = = 17.54 tonnes/h
Find the internal energy of 1 kg of superheated steam at a pressure of 10 bar and 300°C. If this steam is expanded to 1.5 bar and 0.9 dryness, then find the change in internal energy.
From steam table (Appendix A.1.3), at p = 10 bar and tsup = 300°C, for superheated steam,
For dry saturated steam, at p = 1.5 bar, we have from table A.1.2 in the Appendix
A rigid vessel is initially divided into two parts A and B by a thin partition. Part A contains 1 kg of steam at 4 bar, dry and saturated, and part B contains 2 kg of steam at 8 bar with dryness fraction of 0.90. The partition is removed and the pressure in the vessel after sometimes is found to be 6 bar. Find (a) the volume of the vessel and the state of steam and (b) the amount of heat transfer from steam to the vessel and the surroundings.
- At pA = 4 bar, from steam table A.1.2 in the Appendix, vgA = 0.4625 m3/kg, uA = 2553.6 kJ/kg
VA = mAvgA = 1 × 0.4625 = 0.4625 m3
At pB = 8 bar, from steam table A.1.2 in the Appendix, vfB = 0.001115 m3/kg, vgB = 0.2404 m3/kgVB = mB (vf B + xBvgB) = 2 × [0.001115 + 0.9 × (0.2404 − 0.001115)] = 0.433 m3
Volume of vessel, V = VA + VB = 0.4625 + 0.433 = 0.8955 m3
At p = 6 bar, from steam table A.1.2 in the Appendix, vf = 0.001101 m3/kg, vg = 0.3157 m3/kg,m = 1 + 2 = 3 kg
Dryness fraction at 6 bar, x can be found to be
or x = 0.945
- Now UA = mAuA = 1 × 2553.6 = 2553.6 kJ
UB = mB (hfB + xBhfgB − pBxBvgB × 102)= 2 (721.1 + 0.9 × 2048.0 − 8 × 0.9 × 0.240 × 102)= 4783.0 kJ
At 6 bar, U = m[hf + x hfg − pxvg × 102]= 3[670.6 + 0.944 × 2086.3 − 6 × 0.944 × 0.316 × 102)= 7383.25 kJQ = U − (UA + UB)= 7383.25 − (2553.3 + 4783.0) = 46.95 kJ
Calculate enthalpy and entropy for 5 kg of steam at 8 bar under the following conditions:
(a) dry and saturated, (b) wet steam having wetness of 40%, and (c) superheated steam at 250°C.
From steam table (Appendix A.1.2), at p = 8 bar, we have
At tsup = 250°C and p = 8 bar, we have from table A.1.3
- H = m hg = 5 × 2769.1 = 13845.5 kJ
S = m sg= 5× 6.6627 = 33.3135 kJ/K
- x = 1 − 0.4 = 0.6
H = m (hf + x hfg) = 5(721.1 + 0.6 × 2048.0) = 9749.5 kJ
S = m (sf + xsfg) = 5(2.0461 + 0.6 × 4.6166) = 24.08 kJ/K
- H = mh = 5 × 2950.0 = 14750.0 kJ
S = ms = 5 × 7.0384 = 35.192 kJ/K
About 1 kg of steam 0.85 dry at a pressure of 1 bar and dryness fraction 0.85 is compressed in a cylinder to a pressure of 2 bar according to pv1.25 = const. Determine the final condition of steam and the heat transfer through the cylinder walls.
From steam table (Appendix A.1.2), at p = 1 bar, vf1 = 0.001043 m3/kg, vg1 = 1.694 m3/kg
Specific volume of steam
Now p1v1n = p2v2n
At p2 = 2 bar, vf 2 = 0.001061 m3/kg, vg2 = 0.8857 m3/kg, uf 2 = 504.47 kJ/kg, ufg2 = 2025 kJ/kg
Now v2 = 0.8271 = vf 2+ x2 (vg2 − vf 2) = 0.001061 + x2 (0.8857− 0.001061)
or x2 = 0.934
Now u1 = uf 1 + x1ufg1 = 417.33 + 0.85 × 2088.7 = 2192.72 kJ/kg
u2 = uf 2 + x2ufg2 = 504.47 + 0.934 × 2025 = 2395.82 kJ/kg
∆u = u2 − u1 = 2395.82 − 2192.72 = 203.1 kJ/kg
Heat transfer through the cylinder walls,
Steam at a pressure of 10 bar and 0.9 dryness fraction expands to 1 bar hyperbolically. Find (a) work done and heat absorbed and (b) internal energy and change in enthalpy.
Take specific heat of steam at constant pressure = 2 kJ/kg.K.
From steam table (Appendix A.1.2), we have
Now, p1v1 = p2v2
Work done, w = p1v1 ln r = 10 × 102 × 0.1751 ln 10 = 403.18 kJ/kg
Now, v2 > vg2. Therefore, the steam is superheated at p2 = 1 bar.
Using linear interpolation, we have
h1 = hf 1 + x1hfg1 = 762.79 + 0.9 × 2015.3 = 2576.56 kJ/kg
Using linear interpolation, we get
h2 = 2699.19 kJ/kg
∆h = h2 − h1 = 2699.19 − 2576.56 = 122.63 kJ/kg
∆u = ∆h for hyperbolic process
q = w + ∆u = 403.18 + 122.63 = 525.81 kJ/kg
Steam, which is initially dry saturated, expands isentropically from pressure of 15 bar to 0.15 bar. Find the index of isentropic expansion.
From steam table (Appendix A.1.2), we have
or x2 = 0.7845
Now, p1v1n = p2v2n
or n = 1.126
Steam at a pressure of 8 bar and 0.95 dry is expanded to a pressure of 1.8 bar. Find the final condition of steam and the heat drop for each of the following methods of expansion:
- Adiabatic expansion
- Throttling expansion
Given that p1 = 8 bar, x1 = 0.95, p2 = 1.75 bar
From steam tables (Appendix A.1.2), at 8 bar, sf1 = 2.0461 kJ/kg.K, sfg1 = 4.6166 kJ/kg.K, hf1 = 721.1 kJ/kg, hfg1 = 2048 kJ/kg, and at 1.75 bar, sf2 = 1.4848 kJ/kg.K, sfg2 = 5.6868 kJ/kg.K, hf 2 =486.97 kJ/kg, hfg2 = 2213.6 kJ/kg
- During an adiabatic expansion, s1 = s2
sf 1 + x1sfg1 = sf 2 + x2sfg22.0461 + 0.95 × 4.6166 = 1.4848 + x2 × 5.6868x2 = 0.8699
- During throttling process, h1 = h2
hf 1 + x1hfg1 = hf 2 + x2hfg2721.1 + 0.95 × 2048.0 = 486.97 + x2 × 2213.6x2 = 0.9847
Steam at 10 bar flows into a barrel containing 150 kg of water at 10°C. The final temperature of water is 25°C and the increase in mass is 4 kg. Calculate the dryness fraction of steam. Take cpw = 4.187 kJ/kg.K.
Given that ms = 4 kg, p = 10 bar, mw = 150 kg, tw1 = 10°C, tw2 = 25°C, and cps = 4.187 kJ/kg.K
From steam table (Appendix A.1.2), at 10 bar, hfg = 2015.3 kJ/kg and ts = 179.91°C
Heat lost by steam = Heat gained by water
Dryness fraction of steam, x = 0.847
In a separating and throttling calorimeter, the steam main pressure is 8 bar absolute and the temperature after throttling is 130°C. The pressure in the throttling calorimeter is 0.001 bar gauge, and the barometer reading is 75 cm. If 0.15 kg of water is trapped in the separator and 1.8 kg of steam is passed through the throttling calorimeter, then determine the dryness fraction of steam in the steam main.
After throttling, p2 = 0.001 + 0.75 × 9.81 × 13600 × 10−5 = 1.01 bar
The saturation temperature for p2 = 1.01 bar, ts2 = 99.62°C
Degree of superheat = tsup − ts2 = 130 − 99.62 = 30.38°C
For superheated steam, from steam tables (Appendix A.1.3), for tsup = 130°C
From steam tables (Appendix A.1.2),
At 8 bar hf1 = 721.1 kJ/kg, hfg1 = 2048 kJ/kg
Now h1 = hsup2
Dryness fraction, x = x1x2
Calculate the enthalpy of 1 kg of steam at a pressure of 8 bar and dryness fraction of 0.8. How much heat would be required to raise 2 kg of this steam from water at 20°C?
Given that ms = 1 kg, p = 8 bar, x = 0.8, and tw = 20°C
Enthalpy of 1 kg of steam:
From steam table (Appendix A.1.2), corresponding to a pressure of 8 bar, we find that,
Enthalpy of 1 kg of wet steam,
Heat required to raise 2 kg of this steam from water at 20°C:
We have calculated above the enthalpy or total heat required to raise 1 kg of steam from water at 0°C. Since the water, in this case, is already at 20°C, therefore,
Heat already in water = 4.2 × 20 = 84 kJ
Heat required per kg of steam = 2359.5 − 84 = 2275.5 kJ
and heat required for 2 kg of steam = 2 × 2275.5 = 4551 kJ
Determine the quantity of heat required to produce 1 kg of steam at a pressure of 6 bar at a temperature of 25°C, under the following conditions:
- When the steam is wet having a dryness fraction 0.9
- When the steam is dry saturated
- When it is superheated at a constant pressure at 250°C, assuming the mean specific heat of superheated steam to be 2.3 kJ/kg.K.
Given that p = 6 bar, tw = 25°C, x = 0.9, tsup = 250°C, and cps = 2.3 kJ/kg.K
From steam table (Appendix A.1.2), corresponding to a pressure of 6 bar, we find that
- When the steam is wet:
Enthalpy or total heat of 1 kg of wet steam,
h = hf + xhfg = 670.54 + 0.9 × 2086.3 = 2548.21 kJ
Since the water is at a temperature of 25°C, therefore from steam tables (Appendix A.1.1)
Enthalpy of water 104.87 kJ
∴ Heat actually required = 2548.21 − 104.87 = 2443.34 kJ
- When the steam is dry saturated:
From steam tables (Appendix A.1.2), enthalpy or total heat of 1 kg of dry saturated steam,hg = 2756.8 kJ
∴ Heat actually required = 2756.8 − 104.87 = 2651.93 kJ
- When the steam is superheated:
From steam tables (Appendix A.1.3), enthalpy or total heat of 1 kg of superheated steam,hsup = 2957.2 kJ
∴ Heat actually required = 2957.2 − 104.87 = 2852.33 kJ
Steam enters an engine at a pressure of 12 bar with a 67°C of superheat. It is exhausted at a pressure of 0.15 bar and 0.95 dry. Find the drop in enthalpy of the steam. Assume specific heat of superheated steam cps = 2.1 kJ/kgK.
Given that p1 = 12 bar, tsup − ts = 67°C, p2 = 0.15 bar, and x = 0.95
From steam table (Appendix A.1.2), corresponding to a pressure of 12 bar, we find that
Enthalpy or total heat of 1 kg of superheated steam,
Similarly, from steam tables (Appendix A.1.2), corresponding to a pressure of 0.15 bar
Enthalpy or total heat of 1 kg of wet steam,
∴ Drop in enthalpy of the steam = hsup − h = 2925.5 − 2480.36 = 445.14 kJ/kg
A steam engine obtains steam from a boiler at a pressure of 15 bar and 0.98 dry. It was observed that the steam loses 21 kJ of heat per kg as it flows through the pipe line and pressure remaining constant. Calculate the dryness fraction of the steam at the engine end of the pipeline.
Given that p = 15 bar, x = 0.98, and heat loss = 21 kJ/kg
From steam table A.1.2 in the Appendix, corresponding to a pressure of 15 bar we find that
hf = 844.87 kJ/kg; hfg = 1947.3 kJ/kg
Enthalpy of wet steam at the boiler end.
Since the steam loses 21 kJ/kg of steam, therefore enthalpy of wet steam at the engine end,
Let x2 = Dryness fraction of steam at the engine end.
Since the pressure remains constant, therefore hf and hfg is same. We know that
or x2 = 0.9692
A coal fired boiler plant consumes 400 kg of coal per hour. The boiler evaporates 3200 kg of water at 45°C into superheated steam at a pressure of 12 bar and 274.5°C. If the calorific value of the fuel is 32600 kJ/kg of coal, then determine the following:
- Equivalent evaporation
- Efficiency of boiler
Assume specific heat of superheated steam as 2.1 kJ/kg.K.
Given that mf = 400 kg/h, ms = 3200 kg/h, feed water temperature, t1 = 45°C, p = 12 bar,
Tsup = 274.5°C, C.V. = 32600 kJ/kg, cps = 2.1 kJ/kg.K
From steam table A.1.2 in the Appendix, at 12 bar, we have
Ts = 187.99°C, hg = 2784.2 kJ/kg
From steam table A.1.3 in the Appendix, using linear interpolation, enthalpy of superheated steam at exit of boiler (12 bar and 274.5°C) is found to be
hsup = 2989.29 kJ/kg
From steam table A.1.1 in the Appendix, enthalpy of water at 45°C,
hf1 = 188.42 kJ/kg
- Equivalent evaporation:
- Boiler efficiency:
During a trial on an oil-fired smoke tube boiler for 1 h, the following data were recorded:
Steam pressure, p = 14 bar
Amount of water evaporated, m1= 5400 kg
Condition of steam, x = 0.92 dryness fraction
Amount of fuel burnt, mf = 540 kg
Calorific value of fuel used, C.V. = 42000 kJ/kg
Temperature of steam leaving the superheater, Tsup = 250°C
Temperature of feed water, t1 = 50°C
Determine the equivalent evaporation from and at 100°C with and without superheater, boiler efficiency, and the percentage of heat utilised in the superheater.
Given that p = 14 bar, m1 = 5400 kg, x = 0.92, mf = 540 kg, C.V. = 42000 kJ/kg, Tsup = 250°C, and t1 = 50°C
Amount of steam generated per kg of fuel,
Enthalpy of wet steam at 15 bar, 0.92 dry,
From steam table A.1.2 in the Appendix, enthalpy of wet steam at 14 bar, 0.92 dry,
From steam table A.1.3 in the Appendix, Enthalpy of superheated steam at 14 bar, 250°C,
Equivalent evaporation with superheater from and at 100°C.
Percentage of heat utilized in the superheater,
Following particulars refer to a steam power plant consisting of a boiler, superheater, and an economiser:
Steam pressure, p = 20 bar
Mass of steam generated, m1 = 10000 kg/h
Mass of coal used, mf = 1300 kg/h
Calorific value of coal, C.V. = 29000 kJ/kg
Temperature of feed water entering the economiser, t1 = 35°C
Temperature of feed water leaving the economiser, t2 = 105°C
Temperature of superheated steam leaving the superheater, tsup = 350°C
Dryness fraction of steam leaving the boiler, x = 0.98
Determine the following:
- Overall efficiency of the plant
- Equivalent evaporation from and at 100°C
- The percentage of heat utilised in boiler, superheater, and economiser
Given that p = 20 bar, m1 = 10000 kg/h, mf = 1300 kg/h, C.V. = 29000 kJ/kg, t1 = 35°C, t2 = 105°C, tsup = 350°C, x = 0.98
From steam table A.1.1 in the Appendix, enthalpy of feed water entering the economiser,
Enthalpy of feed water leaving the economiser at 105°C,
From steam table A.1.2 in the Appendix, enthalpy of steam leaving the boiler,
Mass of steam generated per kg of fuel,
- Overall efficiency of the plant:
- Equivalent evaporation from and at 100°C:
- Percentage of heat utilised in the boiler:
Heat supplied in 1 kg of fuel = 29000 kJ
Percentage of heat utilised in superheater:
Percentage of heat utilised in economiser:
Note: Superheater is considered as a part of the boiler for calculation of equivalent evaporation.
The following readings were recorded during a boiler trial of 6 h duration:
Mean steam pressure = 12 bar
Mass of steam generated = 40000 kg
Mean dryness fraction of steam = 0.85
Mean feed water temperature = 30°C
Coal used = 4000 kg
Calorific value of coal = 33400 kJ/kg
Calculate the following:
- Factor of equivalent evaporation
- Heat rate of boiler in kJ/hr
- Equivalent evaporation from and at 100°C
- Efficiency of the boiler
Given that p = 12 bar, m = 40000 kg, x = 0.85, t1 = 30°C, mf = 4000 kg, C.V. = 33400 kJ/kg, and duration of trial, T = 6 h
From steam table A.1.2 in the Appendix at 12 barhf = 798.64 kJ/kg, hfg = 1986.2 kJ/kg
∴ Enthalpy of steam, h = hf + x.hfg = 798.64 + 0.85 × 1986.2 = 2486.91 kJ/kg
From steam table A.1.1 in the Appendix, enthalpy of feed water at 30°C, hf 1 = 125.77 kJ/kg
- Factor of equivalent evaporation:
- Heat rate of boiler:
- Equivalent evaporation from and at 100°C:
- Boiler efficiency:
During a trial for 8 h on a boiler, the following data were obtained:
Pressure of steam leaving the boiler = 14 bar
Condition of steam = 0.97 dryness fraction
Steam produced = 26700 kg
Temperature of feed water = 50°C
Mass of coal fired = 4260 kg
Calorific value of coal = 28900 kJ/kg
Air fuel ratio = 17
Temperature of flue gases leaving the boiler = 344°C
Boiler house temperature = 21°C
Specific heat of flue gases = 1.1 kJ/kg.K
Determine the following:
- Boiler efficiency
- Equivalent evaporation
- Heat lost to flue gases expressed as percentage of heat supplied
Given that p = 14 bar, x = 0.97, m = 26700 kg, t1 = 50°C, mf = 4260 kg, C.V. = 28900 kJ/kg, air−fuel ratio, = 17, tf = 344°C, t0 = 21°C, Cpg = 1.1 kJ/kg.K, and duration of boiler trial, T = 8 h
- From steam table A.1.2 in the Appendix, at 14 bar, hf = 830.29 kJ/kg and hfg = 1959.7 kJ/kg
Enthalpy of steam, h = (hf + x.hfg) at 14 bar = 830.29 + 0.97 × 1959.7 = 2731.2 kJ/kg
From steam table A.1.1 in the Appendix, enthalpy of water at 50°C, hf 1 = 209.31 kJ/kg
∴ Efficiency of boiler,
- Equivalent evaporation:
- Fuel burnt/hour, 532.5 kg/h
Heat supplied in fuel = ṁf × C.V. = 532.5 × 28900 = 15389250.0 kJ/h
Mass of flue gases formed/hour,
Heat carried away by flue gases,= ṁg × cpg (tf − t0) = 9585 × 1.1 (344 − 21) = 3405550.5 kJ/h
Heat loss of flue gases as per cent of heat supplied
A gas-fired boiler operates at a pressure of 100 bar. The feed water temperature is 255°C. Steam is produced with a dryness fraction of 0.9, and in this condition, it enters a superheater. Superheated steam leaves the superheater at a temperature of 450°C. The boiler generates 1200 tonne of steam per hour with an efficiency of 92%. The gas used has a calorific value of 38 mJ/m3. Determine the following:
- The heat transfer per hour in producing wet steam in the boiler
- The heat transfer per hour in producing superheated steam in the superheater
- The gas used in m3/hour
Given that p = 100 bar, t1 = 256°C, x = 0.9, tsup = 450°C, m = 1200 tonne = 1200 × 103 kg
- From table A.1.1 in the Appendix, enthalpy of feed water: hf 1 = 1109.72 kJ/kg
From table A.1.2 in the Appendix, enthalpy of steam at 100 bar, 0.9 dryh = hf + x.hfg = 1407.53 + 0.9 × 1317.1 = 2592.92 kJ/kghsup = 3240.8 kJ/kg
∴ Heat transfer/h producing wet steam in boiler= m(h − hf1) = 1200 × 103 (2592.92 − 1109.72)= 17.7984 × 108 kJ/h
- Heat transfer/hour in producing superheated steam in superheater:
= m(hsup − h) = 1200 × 103 (3240.8 − 2592.92)= 77.7456 × 107 kJ/h
- Let the gas used in m3/hour be Vf.
Vf = 73149.2 m3/h
Summary for Quick Revision
- A pure substance may be defined as a system which is homogeneous in composition and chemical aggregation and invariable in chemical aggregation.
- Properties of steam are as follows:
- Latent heat of steam, hfg = Quantity of heat required to convert 1 kg of liquid water at its boiling point into dry saturated steam at the same pressure.
- Dryness fraction, x =
Quality of steam = x × 100Wetness fraction = 1 − x
- Enthalpy of wet steam, hwet = hf + x hfg
- Density of wet steam = kg/m3
Density of dry saturated steam,
Volume of wet steam, vwet = (1 − x)vf + xvg = vf + xvfg
- Internal energy of wet steam, uwet = u f + xu fg
- Entropy of wet steam, swet = s f + xs fg
- Steam processes are as follows:
- Constant volume process,
v1 = v2; q = u2 − u1
- Constant pressure process,
w = p (p2 − p1)
q = (x2 − x1) hfg, for wet steam
= hsup − hg, for superheated steam.
- Isothermal process,
In the wet region, q = h2 − h1 and w = p1 (v2 − v1)
- Hyperbolic process,
p1v1 = p2v2
- Isentropic process,
s1 = s2
- Polytropic process,
n = 1.13 for wet steam
= 1.3 for superheated steamp1(v1)n = p2(v2)n
- Throttling process,
h1 = h2h1 = hf1 + x1 hfg1 for wet steamh2 = hf2 + x2 hfg2 for wet steam= hg + cps (Tsup − Ts) for superheated steam.
- Constant volume process,
- Determination of dryness fraction
- Barrel calorimeter,
It gives approximate results which are always lower than the actual value.
- Separating calorimeter,
The value of x calculated is always greater than the actual.
- Throttling calorimeter,
It is suitable for steam with high dryness fraction.
- Combined separating and throttling calorimeter.
It is suitable for considerably wet steam.
- Barrel calorimeter,
- The dryness fraction of 1 kg of steam containing 0.8 kg of dry steam is
- If x1 and x2 be the dryness fractions of steam obtained in the separating and throttling calorimeters, respectively, then the dryness fraction is
- x1 + x2
- x1 − x2
- Superheating of steam is done at constant
- With the increase of pressure, the enthalpy of evaporation of water
- remains same
- changes randomly
- Only throttling calorimeter is used for measuring
- very low dryness fraction up to 0.7
- very high dryness fraction up to 0.98
- dryness fraction of only low pressure steam
- dryness fraction of only high pressure steam
- Constant pressure lines in the superheated region of the Mollier diagram have
- a positive slope
- a negative slope
- zero slope
- both positive and negative slope
- Specific volume of wet steam with dryness fraction x is
- vf + x vfg
- Entropy of wet steam is given by
- sf + x sfg
- x sg
- sg + x sfg
- x sf
- The phase change at constant pressure (or constant temperature) from liquid to vapour is referred to as
- The point that connects the saturated liquid line to the saturated vapour line is called the
- triple point
- critical point
- superheated point
- saturated point
- Define a pure substance.
- Define latent heat of fusion and latent heat of vapourisation.
- Define dryness fraction and wetness of steam.
- What is quality of steam?
- Write expressions for internal energy of steam when it is wet and when it is superheated.
- Write expressions for entropy of steam for wet and superheated.
- What is a Mollier diagram?
- When steam is heated at constant volume and its end condition is still wet, what is heat transfer?
- In the wet region, constant temperature process is also a constant pressure process. Say true or false.
- Hyperbolic process is also an isothermal process in the superheat region. Say true or false.
- What remains constant during isentropic process?
- What is throttling of steam?
- What are the drawbacks of barrel calorimeter?
- Which calorimeter is used for a very wet steam?
2.1 A vessel of 1.35 m3 capacity is filled with steam at 13 bar absolute and 95% dry. The vessel and its contents cool until the pressure is 2 bar absolute. Calculate the mass of the contents in the vessel and the dryness fraction of steam after cooling. Neglect the volume of water present.
[Ans. 9.2 kg, 0.1625]
2.2 Steam at a pressure of 5 bar and temperature 200°C is expanded adiabatically to a pressure of 0.7 bar absolute. Determine the final condition of steam. For superheated steam, cps = 2.2 kJ/kg.K.
2.3 Steam at a pressure of 20 bar absolute and dryness fraction 0.8 is throttled to a pressure of 0.5 bar. Determine the final condition of steam.
2.4 About 2 kg of wet steam at 10 bar and 90% dry is expanded according to the law pv = constant to a pressure of 1 bar. Determine the final condition of steam and the change in internal energy.
[Ans. Superheated to 112°C, 257 kJ]
2.5 While conducting the dryness fraction test with a throttling calorimeter, it was found that the entering steam was at a pressure of 12 bar and a sample after being reduced to 1 bar in the calorimeter was at a temperature of 120°C. Estimate the dryness fraction of steam assuming cps = 2.1 kJ/kg.K.
2.6 In a test with a separating and throttling calorimeter, the following observations were made:
Water separated = 2.04 kg
Steam discharged from throttling calorimeter = 20.6 kg at 150°C
Initial pressure of steam = 12 bar absolute
Final pressure of steam = 12.3 cm of mercury above atmospheric pressure
Barometer = 76 cm of mercury
Calculate the dryness fraction of steam entering the calorimeter.
2.7 Determine the change in internal energy when 1 kg of steam expands from 10 bar and 300°C to 0.5 bar and 0.9 dry. Take cps = 2.1 kJ/kg.K.
[Ans. −669 kJ/kg]
2.8 One kg of steam at 10 bar exists at 200°C. Calculate the enthalpy, specific volume, density, internal energy, and entropy. Take cps = 2.1 kJ/kg.K.
[Ans. 2818 kJ/kg, 0.203 m3/kg, 4.93 kg/m3, 2615 kJ/kg, 6.673 kJ/kg.K]
2.9 A vessel contains one kg of steam which contains one-third liquid and two-third vapour by volume. The temperature of steam is 151.86°C. Calculate the quality, specific volume, and specific enthalpy of the mixture.
[Ans. 0.0576, 0.003245 m3/kg, 650.5 kJ/kg]
2.10 A cylinder fitted with a piston contains 0.5 kg of steam at 4 bar. The initial volume of steam is 0.1 m3. Heat is transferred to steam at constant pressure until the temperature becomes 300°C. Determine the heat transferred and work done during the process.
[Ans. 771 kJ, 91 kJ]
2.11 Steam at 10 bar and 0.9 dry initially occupies 0.35 m3. It is expanded according to the law pv1.25 = constant until the pressure falls to 2 bar. Determine the mass of steam used in the process, the work done, the change in internal energy, and the heat exchange between steam and surroundings.
[Ans. 2 kg, 192 kJ, −896 kJ, −704 kJ]
2.12 One kg of steam at 8.5 bar and 0.95 dry expands adiabatically to a pressure of 1.5 bar. The law of expansion is pv1.2 = constant. Determine the final dryness fraction of steam and the change in internal energy during the expansion.
[Ans. 0.792, −230 kJ/kg]
2.13 A throttling calorimeter is used to measure the dryness fraction of steam in the steam main where the steam is flowing at a pressure of 6 bar. The steam after passing through the calorimeter comes out at 100 kPa pressure and 120°C temperature. Calculate the dryness fraction of steam in the main. Assume cps = 2.09 kJ/kg.K.
2.14 A separating and throttling calorimeter was used to determine the dryness fraction of steam flowing through a steam main at 900 kPa. The pressure and temperature after throttling were 105 kPa and 115°C, respectively. The mass of steam condensed after throttling was 1.8 kg and mass of water collected in the separating calorimeter was 0.16 kg. Determine the dryness fraction of steam flowing through the steam main. Take cps = 2.09 kJ/kg.K.
ANSWERS TO MULTIPLE-CHOICE QUESTIONS