# Chapter 8 Steam Condensers – Thermal Engineering

## Steam Condensers

##### 8.1 ❐ DEFINITION

A condenser is a closed vessel in which steam is condensed by abstracting the heat and where the pressure is maintained below atmospheric pressure.

##### 8.2 ❐ FUNCTIONS OF A CONDENSER

There are two main functions of a condenser, which are explained below:

1. To reduce the back pressure considerably upon the prime mover, and thereby increasing the work output of steam during expansion.

Consider the ideal pv diagram of a non-condensing steam engine, as shown in Fig. 8.1(a), in which steam is supplied at pressure p1 and the exhaust takes place at atmospheric pressure p2. The work output is given by the area 1–2–3–4–5–1. If we reduce the back pressure p2 to p′2, which is below the atmospheric pressure, keeping p1 same, then the work output will be given by the area 1–2–3–4′–5′–1, as shown in Fig. 8.1(b). This work output is obviously greater, than the previous work output. The additional work output is 4–4′–5′–5–4.

2. The condensate condensed from the exhaust steam in the condenser is free from impurities and can be directly used as feed water for the boiler without any treatment. This recovery of feed water is an incidental advantage.

Figure 8.1 Effect of back pressure on work output of a steam engine

##### 8.3 ❐ ELEMENTS OF STEAM CONDENSING PLANT

A steam power plant using a condenser is shown in Fig. 8.2. The various components of a steam condensing plant are described below.

1. Condenser: It is a closed vessel heat exchanger in which steam coming from the turbine is condensed by abstracting heat by cooling water where the pressure is maintained below atmospheric pressure.
2. Condenser cooling water pump: The cooled water from the cooling tower is circulated through the condenser by this pump. This pump is not required if water from the cooling tower is not re-circulated.
3. Make up water pump: This pump circulates the required quantity of water through the condenser.
4. Condensate extraction pump: The function of the condensate pump is to extract the condensate from the condenser where the pressure is below atmospheric pressure and feed it to the hot well, where the pressure is at atmospheric level.
5. Hot well: It is a sump between the condenser and the boiler where the condensate coming from the condenser is collected.
6. Air-extraction pump: The function of the air-extraction pump is to extract the air which has leaked into the condenser by various paths and to maintain the required vacuum inside.
7. Boiler feed pump: Its function is to pump the condensate from hot well into the boiler by increasing the pressure of condensate above the boiler pressure.
8. Cooling tower: It is essential when large amount of cooling water required is not available from the river throughout the year. Its function is to cool the hot water coming out of the condenser instead of discharging it to waste. The hot water is cooled by evaporative cooling.

Figure 8.2 Steam power plant using a condenser

##### 8.4 ❐ TYPES OF STEAM CONDENSERS

Steam condensers may be broadly classified into two categories.

1. Jet condensers: They are contact-type condensers in which the steam to be condensed mixes with the cooling water and the temperature of the condensate and the cooling water is the same when leaving the condenser. The condensate cannot be recovered for use as feed water to the boiler. Heat transfer is by direct conduction between steam and water.

These condensers may be further classified as follows:

1. Low-level, counter flow type condenser
2. Low-level, parallel flow type condenser
3. High-level jet condenser
4. Ejector condenser
2. Surface condensers: There is no direct contact between steam to be condensed and the circulating cooling water in surface condensers. Heat is transferred through a wall interposed between steam and cooling water. The temperature of condensate may be higher than the cooling water. The condensate can be used as feed water to the boiler.

These condensers may be further classified as follows:

1. Cooling water in tubes and steam surrounding it
2. Central flow or regenerative condenser
3. Evaporative condenser

#### 8.4.1 Jet Condensers

1. Low level counter-flow jet condenser: This type of condenser is shown in Fig. 8.3. The cold water is drawn up in the condenser shell from the cooling pond due to the vacuum head created in the shell. The level of the cooling pond should not be much below the delivery pipe to the shell as no pump is used. The shell is arranged with two or three water trays with perforations to break up water into small jets. The exhaust steam and any mixed air enters at lower portion of the shell and tries to ascend up through the falling sprays. Thus, steam gets condensed and air ascends up and cools down. The air is removed by a separate suction pump at the top. The mixture of condensate and cooling water descends down through a vertical pipe to the extraction pump and pumped to the hot well. From the hot well, the boiler feed pump delivers water to the boiler and surplus water overflows to the cooling pond. Since the boiler feed and the cool injection water mix, this type of condenser may be used only where cheap pure water is available. The non-condensable air and gas are continuously removed by air pump suction to maintain vacuum.
2. Low level parallel flow Jet condenser: It is similar to the counter flow jet condenser except that steam enters the condenser shell at the top and just below it the cooling water is delivered. It is less efficient than the counter flow type.
3. High level jet condenser: This is also called Barometric condenser and is shown in Fig. 8.4. The shell is placed at a height greater than 10.363 m, the barometric height of water column. If a pipe of height more than 10.363 m is held vertical with one end immersed in water vessel open to atmosphere and the other end subjected to suction pressure, the atmospheric pressure will hold the column of water in the pipe equal to the suction pressure. This fact is made use of in this condenser by making the tail pipe more than 10.363 m in height and thus making it impossible for any vacuum in the condenser to cause the water to rise high in the tail pipe of the water leg and flood the engine.

Figure 8.3 Low level counter flow jet condenser

Figure 8.4 High-level jet condenser

Figure 8.5 Ejector condenser

The height of the shell much above the barometric height requires a separate pump for injecting cool water. The condensate and water will fall under gravity to the hot well and maintain a column of water in the water leg, depending on the vacuum in the condenser. A separate air pump to remove air is employed.

4. Ejector condenser: It is a low-level type of condenser (as shown in Fig. 8.5). This condenser works on the principle that by discharging a smooth jet of cold water under a head of about 6 m through a series of converging guide cones, the steam and the associated air are drawn in through the hollow truncated cones and led to the diverging cones. In the converging cones, the pressure energy is partly converted to kinetic energy. In the diverging cones, the kinetic energy is again partly converted to pressure energy so as to obtain pressure greater than the atmospheric to enable the condensate and water mixture to be discharged to the hot well exposed to atmospheric pressure.

A non-return valve is fitted on the exhaust steam inlet to condenser so that water from the hot well does not rush back to the engine in case of cold injected water failure.

#### Water Tube Surface Condenser

In this type of surface condenser, water flows in the tubes and steam surrounds it. Thus, the condensate is directly available as an ideal boiler feed. A surface condenser produces more vacuum and any type of cooling water can be used. However, it is bulky and the initial capital cost is higher than jet condensers. But, this is justified by the saving in the running cost due to high efficiency.

This type of condensers may be further classified as follows:

1. Single pass, two pass condensers
2. Down flow: central flow regenerative condensers

A longitudinal section of a two pass, down flow condenser is shown in Fig. 8.6. It consists of a cast iron shell, cylindrical in shape, with the two ends covered by cover plates. A nest of brass tubes is fixed in the two tube plates at the end. They are fixed by brass ferrules so that they can be replaced easily.The space between the tube plates and the cover plates is known as water boxes. One of the water boxes has baffle partitioning the box into two sections—one upper half and the other lower half.

The cold water is sent through the lower half section tubes and comes out through the upper half section tubes. Exhaust steam enters the shell at the top and flows down surrounding the cold water tubes. The condensate is removed at the bottom by an extraction pump.

The surface condensers may work on the wet vacuum system or the dry vacuum system. For the dry vacuum system, three pumps are required—one for the circulating cooling water, the other for removing the condensate, and the third, the dry air pump, for removing air. The cross-section of such a dry vacuum system is shown in Fig. 8.7. The air exit is shielded from the down flow of steam by means of a baffle.

The main requirements of a surface condenser are as follows:

1. Air should be removed at a cooler section and it should be as far as possible dry.
2. The pressure drop of steam in the condenser should be minimum.

Figure 8.6 Water tube surface condenser

Figure 8.7 Down flow surface condenser

#### Central Flow or Regenerative Condenser

In such a condenser as shown in Fig. 8.8, the steam passages are all around the periphery of the shell. The exhaust steam converges to the centre of the shell where there are no tubes. The air is removed by the air pump suction from the centre. Some of the exhaust steam also passes to the centre and gets condensed. The condensate is reheated to nearly the temperature of the exhaust steam. The pressure drop of steam is also reduced due to large flow passages.

Figure 8.8 Central flow surface condenser

#### Evaporative Condenser

In evaporative condenser, steam circulates in pipes surrounded by water spray, as shown in Fig. 8.9. It is used when there is scarcity of cooling water. The cooling water is allowed to evaporate under a small partial pressure. It consists of gilled piping bent to form many rows and is placed vertically. Steam passes through these pipes. Pumped water is sprayed from the top and descends down forming a thin film over the pipes. A natural or forced air draught causes the evaporation of the water film. Steam while circulating inside the pipes, latent heat of evaporation is removed from steam and thereby the steam gets condensed. This mode of heat transfer reduces the cooling water requirement of condenser to 10% of the requirement for surface condensers. The water particles carried with air due to high velocity of air are removed with the help of eliminator. This type of condenser works better in dry weather.

Figure 8.9 Evaporative condenser

##### 8.5 ❐ REQUIREMENTS OF MODERN SURFACE CONDENSERS
1. The steam should be evenly distributed over the entire cooling surface of the condenser with minimum pressure loss.
2. There should be no undercooling of the condensate.
3. Water should be passed through the tubes and steam must surround the tubes from outside.
4. There should be no air leakage into the condenser to maintain proper vacuum.
5. Air must be cooled to the maximum possible extent before extraction.
6. For better thermal efficiency, the rise in temperature of the cooling water should be limited to 10°C.

#### 8.6.1 Jet Condensers

1. Mixing of steam and cooling water is more intimate.
2. Less quantity of circulating water for condensing exhaust steam is required.
3. Condensing plant is simple and less costly.
4. Spare parts required are less.
5. Condensate extraction pump is not needed in barometric jet and ejector condensers.

1. Condensate is wasted unless cooling water is pure.
2. In the low-level jet condenser, if the condensate extraction pump fails, then the engine would get flooded unless safe guarded.
3. Barometric condenser cost is high due to long piping.
4. There is vacuum loss in barometric condenser due to leakage in long piping.
5. There are limitations of vacuum of up to 650 mm of Hg due to liberation of dissolved gases in cooling water.
6. The air extraction pump requires comparatively high power.

#### 8.6.2 Surface Condensers

1. Vacuum obtained is comparatively very high, resulting in high efficiency.
2. Condensate can be used directly as boiler feed.
3. There is no need of water treatment plant.
4. It is more suitable for high capacity plants.

1. It requires more space.
2. Maintenance cost is more.
##### 8.7 ❐ VACUUM MEASUREMENT

A vacuum gauge is used to measure the vacuum in the condenser. Figure 8.10 shows a vacuum gauge. It consists of a glass tube containing mercury to indicate the barometric pressure and another mercury tube connected to the condenser and vacuum gauge. The vacuum gauge has readings marked in cm of Hg. It represents the height at which a column of mercury, the upper surface of which is in communication with the cylinder, will stand when supported by barometric pressure. The vacuum is dependent upon the barometric pressure and the absolute pressure in the condenser.

Figure 8.10 Vacuum gauge

Absolute pressure in the condenser = barometric pressure − vacuum gauge pressure

Usually, the vacuum gauge readings are corrected to standard barometer reading of 76 cm of Hg.

Thus,

Corrected vacuum in cm of Hg = 76 − absolute pressure in cm of Hg

= 76 − (actual barometric height − vacuum)

Let   Hb = actual barometric height in cm of Hg

Hg = actual vacuum gauge reading in cm of Hg

Hb Hg = absolute pressure in condenser in cm of Hg

p = absolute condenser pressure in bar

Now, 76 cm of Hg = 1.01325 bar

##### 8.8 ❐ DALTON’S LAW OF PARTIAL PRESSURES

Dalton’s law states that

1. The total pressure in a container having mixture of gas and vapour is the sum of the partial pressure of the vapour at the common temperature and the partial pressure of the gas.

If the degree of saturation of air at any point inside the condenser is 100%, the partial pressure of vapour is the same as the saturation pressure.

Let t°C = temperature of mixture in a container.

pa = absolute pressure of any gas or air at C.

ps = saturation pressure of any vapour corresponding to C.

pt = total pressure in the container

2. Each constituent of the mixture in the container occupies the whole volume of the container and exerts its own partial pressure.

Let  V = volume of a container, m3

ma = mass of air in the container, kg

mg = mass of water vapour in the container, kg

m = total mass of mixture in the container, kg

vg = specific volume of saturated vapour at C and pa in m3/kg

va = specific volume of air at C in m3/kg

#### Example 8.1

A vacuum of 67 cm of Hg was obtained with the barometer reading of 75 cm of Hg. The condensate temperature is 20°C. Correct the vacuum to a standard barometer of 76 cm and hence determine the partial pressure of air and steam. Also find the mass of air present with 1 kg of steam.

Solution

Corrected vacuum = 76 − (actual barometric reading − actual vacuum)

= 76 − (75 − 67) = 68 cm of Hg

Pressure of dry saturated steam at 20°C from steam tables is 0.023385 bar

and specific volume is v = 57.79 m3/kg

Total absolute pressure in the condenser = 75 − 67 = 8 cm of Hg

Partial pressure of steam, pg = 1.754 cm of Hg

∴ Partial pressure of air, pa = 8 − 1.754 = 6.246 cm of Hg

According to Dalton’s law the volume of air present per kg of steam

= 57.79 m3/kg at 20°C

Also for air paV = maRT

##### 8.9 ❐ MASS OF COOLING WATER REQUIRED IN A CONDENSER

Let   w = mass flow rate of circulating water, kg/h

s = mass flow rate of steam condensed, kg/h

t = temperature of wet exhaust steam, °C

tc = temperature of condensate, °C

hf = enthalpy of water at t°C, kJ/kg

hfg = enthalpy of evaporation of steam at t°C, kJ/kg

x = dryness fraction of exhaust steam

tiw = inlet temperature of cooling water, °C

tow = outlet temperature of cooling water, °C

hc = enthalpy of condensate, kJ/kg = cpw tc

cpw = specific heat of cooling water, 4.1868 kJ/kg.K

Now Heat given by steam = Heat absorbed by water

For jet condenser,

tow = tc

For surface condenser,

ms (hf + xhfghc) = mw cpw (towtiw)

#### Example 8.2

A 200 kW steam engine has a steam consumption of 10 kg/kWh. The back pressure of the engine is equal to the condenser pressure of 0.15 bar. The condensate temperature is 32°C. The cooling water temperature at inlet and outlet are 20°C and 30°C, respectively. Calculate the mass of cooling water required per hour if the exhaust steam is dry saturated.

Solution

At 0.15 bar, hf = 225.91 kJ/kg, hfg = 2373.1 kJ/kg, tc = 32°C, tow = 30°C, tiw = 20°C

Total steam used per hour = 200 × 10 = 2000 kg

Total circulating water used = 58.88 × 2000 = 117760 kg/h

#### 8.10.1 Sources of Air Infiltration in Condenser

1. Air may pass into the condenser with the steam.
2. Air leakage may happen through the packings.
3. The cooling water contains dissolved gases which are released under vacuum.

The amount of air leakage should not be more than 5−15 kg/10000 kg of steam.

#### 8.10.2 Effects of Air Infiltration in Condensers

1. The leaked air in the condenser increases the back pressure on the prime mover, and consequently, the thermal efficiency of the plant is lowered.
2. The infiltrated air lowers the partial pressure of steam which means a lower saturation temperature and increased latent heat. Hence, it will require greater amount of cooling water.
3. Because of poor thermal conductivity of air, the rate of heat transfer from the vapour is reduced which requires increased surface area of tubes for a given condenser duty.
##### 8.11 ❐ AIR PUMP

The function of the air pump is to maintain a desired vacuum in the condenser by removing air and condensed steam. Edward’s air pump is commonly used for this purpose.

#### 8.11.1 Edward’s Air Pump

This pump is shown in Fig. 8.11. It consists of a plunger or piston D having a conical head E which slides inside a barrel or cylinder liner C having a cover B which has a number of delivery valves A. The passage G is connected to the condenser. The barrel has a number of parts, F pierced in the cylinder liner C. These parts communicate to the air pump suction pipe. H is the water weir where the discharged condensate collects and overflows to the hot well. This retention of sufficient water above the cylinder head acts as a water seal against leakage of air. J is the relief valve.

On the downward stroke of the piston, partial vacuum is created above the piston because the head valves are closed and sealed by water. As soon as the ports are uncovered, the water vapour and air push into the space above the piston. The condensate collected at the piston is displaced by further motion of the piston with conical head and pushed rapidly through the ports to the top of the piston. On the upward stroke, the water, air, and water vapour above the piston are compressed to pressure above the atmospheric pressure and discharged outside through the head valves.

Capacity of the pump,

where D = piston diameter, L = stroke length, N = rpm, ηv = volumetric efficiency.

Figure 8.11 Edward’s air pump

##### 8.12 ❐ VACUUM EFFICIENCY

The lowest pressure which can exist in a condenser is the saturation pressure of steam corresponding to the temperature of water entering the condenser. However, the actual pressure is always greater than the ideal pressure by an amount equal to the partial pressure of air present in the condenser.

Let   ps = saturation pressure of steam corresponding to the temperature of water entering the condenser.

pa = partial pressure of air in the condenser.

pb = barometric pressure

pt = pa + ps

= total pressure of air and steam in the condenser

Then ideal vacuum possible without air leakage = pbps

Actual vacuum existing in condenser due to air leakage

= pb pt = pb (pa + ps)

The “vacuum efficiency” is defined as the ratio of actual vacuum to ideal vacuum.

The factors affecting the vacuum efficiency are as follows:

1. Air leakage increases pa, and hence, it decreases vacuum efficiency.
2. The vacuum efficiency decreases with increase in barometric pressure, keeping pa and ps the same.
3. With insufficient cooling water, the pressure in the condenser increases and subsequently, reduces the vacuum efficiency of the condenser.
##### 8.13 ❐ CONDENSER EFFICIENCY

It is defined as the ratio of the difference between the outlet and inlet temperatures of the cooling water to the difference between the temperature ts corresponding to the vacuum in the condenser and the inlet temperature of cooling water.

where tiw, tow = temperature at inlet and outlet respectively of cooling water, ts = temperature of steam corresponding to the actual absolute pressure in the condenser.

##### 8.14 ❐ COOLING TOWER

There is a need to recycle the cooling water through the condenser when cooling water supply is limited. Cooling tower is an artificial device used to cool hot cooling water coming out from the condenser.

There are different types of cooling towers. The induced draft type cooling towers are commonly used in high capacity power plants. The schematic diagram of an induced draft cooling tower is shown in Fig. 8.12. The hot water coming out from the condenser is sprayed at the top of the tower and air is induced to flow through the tower with the help of induced draft fans mounted at the top of the tower.

Figure 8.12 duced draft cooling tower

The amount of water supply lost in such a tower ranges from 1% − 2% by evaporation and 0.5% − 2% by drift losses. To compensate these losses, make-up water is supplied from external sources. The cooled water is collected at the bottom.

#### Example 8.3

The following readings were taken during a test on a surface condenser:

Vacuum in condenser = 71 cm of Hg

Barometer reading = 76 cm of Hg

Temperature in condenser = 33°C

Hot well temperature = 30°C

Cooling water circulated = 48,000 kg/h

Inlet temperature of cooling water = 15°C

Outlet temperature of cooling water = 28°C

Condensate = 1500 kg/h

Calculate:

1. the mass of air in kg/m3 of condensate volume,
2. dryness fraction of steam entering the condenser, and
3. the vacuum efficiency

Solution

1. Absolute pressure in condenser,

pt = (76 − 71) × 0.01333 = 0.0667 bar

At t = 33°C, ps = 0.05106 bar from steam tables = 3.83 cm of Hg

Partial pressure of air, pa = pt ps= 0.0667 − 0.05106 = 0.01564 bar

2. At p = 0.0667 bar, hf = 159.47 kJ/kg, hfg = 2411.4 kJ/kg

3. Vacuum efficiency = 0.9837 or 98.37%

#### Example 8.4

A steam turbine discharges 6000 kg of steam per hour at 45°C and 0.82 dryness. The estimated air leakage is 15 kg/h. The temperature at the suction of air pump is 30°C and temperature of condensate is 33°C. Find (a) the vacuum gauge reading, (b) capacity of air pump, (c) loss of condensate per hour, and (d) the quantity of cooling water required by limiting its temperature rise by 10°C.

Solution

1. Total pressure in the condenser, pt = pa + ps

At 45°C,  ps = 0.095934 bar from steam tables

vg = 15.258 m3/kg

Volume of 6000 kg steam, V = 6000 ×xvg = 6000 × 0.82 × 15.258 = 75069.36 m3/h

= volume of air

Partial pressure of air,

pt = 18.236× 10-5 + 0.09582 = 0.096 bar

Vacuum in condenser

2. Partial pressure of steam at air pump suction at 30°C = 0.04242 bar

pa = pt ps = 0.096 − 0.042461 = 0.0535 bar

Volume of air at 30°C and 0.0535 bar,

Air pump capacity = 243.82 m3/h

3. Specific volume of steam at 30°C, vg = 32.893 m3/kg
4. At 45°C, hf = 188.42 kJ/kg, hfg = 2394.8 kJ/kg

Mass of cooling water,

= 288.62 tonnes/h

#### Example 8.5

The air leakage into a surface condenser operating with a steam turbine is estimated as 80 kg/h. The vacuum near the inlet of air pump is 70 cm of Hg when barometer reads 76 cm of Hg. The temperature at the inlet of vacuum pump is 20°C. Calculate (a) the minimum capacity of the air pump, (b) dimensions of the reciprocating air pump to remove air if it runs at 200 rpm. Take L:D:: 3:2 and volume efficiency = 98%, (c) mass of vapours extracted per hour.

Solution

1. Total pressure at inlet of air pump at 1–1, (Fig. 8.13)

Partial pressure of water vapour at 1–1 corresponding to saturation temperature of 20°C, from steam tables,

ps = 0.023385 bar

Partial pressure of air at 1–1,

pa = pt ps = 0.08 − 0.023385 = 0.0566 bar

Volume of 80 kg/h of air at 20°C and 0.0566 bar,

Capacity of air pump = 1188.6 m3/h

2. Now   V = D2LNηv

D3 = 0.0858

D = 0.441 m or 46.1 cm

L = 66.2 cm

Figure 8.13 Surface condenser

3. Mass of water vapour going with air in the air pump,

#### Example 8.6

The quality of steam entering a jet condenser is 0.92. It is condensed by using cooling water at 15°C. The mass of air in the condenser is 35% of the mixture. Assuming that only latent heat of steam is absorbed by the cooling water, calculate (a) the temperature of mixture of condensate and water leaving the condenser, and (b) the mass of water required per kg of steam condensed.

Take for air, R = 287 J/kg.K and for steam, R = 463 J/kg.K. Condenser vacuum = 61 cm of Hg, Barometer reading = 76 cm of Hg.

Solution

1. Given that pa = 0.35 pt

pt = (76 − 61) × 0.0133 = 0.2 bar

pa = 0.35 × 0.2 = 0.07 bar

ps = pt pa = 0.2 − 0.07 = 0.13 bar

Saturation temperature of steam at 0.13 bar, from steam tables,

ts = 50.71°C

As only latent heat of steam is to be absorbed, therefore, outlet temperature of cooling water,

tow = ts = 50.71°C

Since steam and cooling water mix together in the jet condenser,

Mixture temperature coming out of condenser = 50.71°C

2. Now

At ps = 0.13 bar, hf = 212.27 kJ/kg, hfg = 2380.98 kJ/kg from steam tables

Minimum quantity of cooling water required per kg of steam condensed

= 14.65 kg

#### Example 8.7

In a trial on a surface condenser, the vacuum gauge reading was 660 mm of Hg when barometer read 760 mm of Hg. The temperature in the condenser was 45°C.

1. Determine the alteration in vacuum if the quantity of air entering the condenser is reduced by 100 kg/h.
2. If the mass of condensate is 960 kg/h, calculate the quantities of air and vapour which the pump has to handle.

Solution

Absolute condenser pressure = 760 − 660 = 100 mm of Hg

pt = × 1.01325 = 0.1333 bar

At 45°C, from steam tables,

Partial pressure of steam, ps = 0.095934 bar

Specific volume of steam, vsg = 15.258 m3/kg

Partial pressure of air, pa = 0.1333 − 0.095934 = 0.037366 bar

Volume of steam per hour, Vs = 960 × 15.258 = 14647.68 m3

= Volume of air, Va

If the mass of air is reduced to 100 kg/h, then partial pressure of air,

New absolute pressure of condenser, p′t = 0.095934 + 0.00623 = 0.102 bar

New vacuum reading = 760 − 76.52 = 683.48 mm Hg

Alteration in vacuum = 683.48 − 660 = 23.48 mm Hg

Capacity of pump = 960 + 100 = 1060 kg/h

#### Example 8.8

The following observations were recorded during a condenser test;

Vacuum reading = 700 mm of Hg; Barometer reading = 760 mm of Hg;

Condensate temperature = 34°C

Find: (a) partial pressure of air and (b) mass of air per m3 of condenser volume.

Take R for air = 287 J/kg K

Solution

Given that Vacuum reading = 700 mm of Hg: Barometer reading = 760 mm of Hg; T = 34°C = 34 + 273 = 307 K

1. Pressure in the condenser

= 760 − 700 = 60 mm of Hg

= 60 × 0.00133 = 0.0798 bar (∵ 1 mm of Hg = 0.001 33 bar)

From steam tables, corresponding to a temperature of 34°C, we find that pressure of steam,

ps = 0.0535 bar

∴ Partial pressure of air,

pa = pcps = 0.0798 − 0.0535 = 0.0263 bar
2. Mass of air per m3 of condenser volume.

#### Example 8.9

A vacuum gauge fitted to a condenser reads 680 mm of Hg, when the barometer reads 750 mm of Hg. Determine the corrected vacuum in terms of mm of Hg and bar.

Solution

Given that Vacuum gauge reading = 680 mm of Hg; Barometer reading = 750 mm of Hg:

Pressure in the condenser = 750 − 680 = 70 mm of Hg

Corrected vacuum = 760 − 70 = 690 mm of Hg

= 690 × 0.00133 = 0.918 bar

#### Example 8.10

Calculate the vacuum efficiency from the following data:

Vacuum at steam inlet to condenser = 700 mm of Hg

Barometer reading = 760 mm of Hg
Hot well temperature = 30°C

Solution

Given that Vacuum reading or actual vacuum = 700 mm of Hg; Barometer reading = 760 mm of Hg; t = 30°C

Pressure in the condenser = 760 − 700 = 60 mm of Hg

From steam tables, corresponding to a temperature of 30°C, we find that ideal pressure of steam,

Ideal vacuum = Barometer reading − Ideal pressure

= 760 − 31.93 = 728.07 mm of Hg

Vacuum efficiency, ηvac = = 0.9614 or 96.14%

#### Example 8.11

The vacuum efficiency of a condenser is 96%. The temperature of condensate is 40°C. If the barometer reads 752 mm of Hg, find the vacuum gauge reading of the condenser.

Solution

Given that ηv = 96% = 0.96%; t = 40°C;

We find that ideal pressure of steam, = 0.073837 bar = = 55.52 mm of Hg

∴ Ideal vacuum = Barometer reading − Ideal pressure

= 752 − 55.52 = 696.48 mm of Hg

∴ Actual vacuum or vacuum gauge reading of the condenser

= 0.96 × 696.48 = 668.62 mm of Hg

#### Example 8.12

In a surface condenser, the vacuum maintained is 700 mm of Hg. The barometer reads 754 mm. If the temperature of condensate is 18°C, determine: (a) mass of air per kg of steam and (b) vacuum efficiency.

Solution

Given that Actual vacuum = 700 mm of Hg; Barometer reading = 754 mm of Hg; T = 18°C = 18 + 273 = 291 K

Pressure in the condenser, pc = 754 − 700 = 54 mm of Hg

From steam tables, corresponding to 18°C, we find that absolute or ideal pressure of steam,

Specific volume of steam, vs = 65.844 m3/kg

1. Pressure of air (as per Dalton’s law)

pa = pcps = 54 − 15.68 = 38.32 mm of Hg

= 38.32 × 0.00133 = 0.051 bar = 0.051 × 105 N/m2

Mass of air per kg of steam,

2. Ideal vacuum = Barometer reading − Ideal pressure
= 754 − 15.68 = 738.32 mm of Hg

#### Example 8.13

A surface condenser, fitted with separate air and water extraction pumps, has a portion of the tubes near the air pump suction screened off from the steam so that the air is cooled below the condensate temperature. The steam enters the condenser at 38°C and the condensate is removed at 37°C. The air removed has a temperature of 36°C. (a) If the total air infiltration from all sources together is 5 kg/h, determine the volume of air handled by the air pump per hour. (b) What would be the corresponding value of the air handled if a combined air and condensate pump was employed? Assume uniform pressure in the condenser.

Solution

Given that Ts = 38°C = 38 + 273 = 311 K; Tc = 37°C = 37 + 273 = 310 K; Ta = 36°C = 36 + 273 = 309 K; a = 5 kg/h

1. Since the pressure at entry to the condenser (pc) is equal to the pressure of steam corresponding to 38°C, therefore, from the steam tables,
pc = 0.0668 bar

Pressure of steam at the air pump suction, corresponding to 36°C (from steam tables),

ps = 0.0598 bar

∴ Pressure of air at the air pump suction (as per Dalton’s law),

pa = pcps = 0.0668 − 0.0598 = 0.007 bar
= 0.007 × 105 = 700 N/m2

Volume of air handled by the air pump,

2. From steam tables corresponding to a condensate temperature of 37°C, we find that pressure of steam,
ps = 0.0633 bar

∴ Pressure of air (as per Dalton’s law),

pa = pcps = 0.0668 − 0.0633 = 0.0035 bar
= 0.0035 × 105 = 350 N/m2

Volume of air handled,

#### Example 8.14

The air leakage into a surface condenser operating with a steam turbine is estimated as 84 kg/h. The vacuum near the inlet of air pump is 700 mm of Hg when barometer reads 760 mm of Hg. The temperature at inlet of vacuum pump is 20°C. Calculate the following:

1. The minimum capacity of the air pump in m3/h;
2. The dimensions of reciprocating air pump to remove the air if it runs at 200 rpm. Take L/D ratio = 1.5 and volumetric efficiency = 100%
3. The mass of vapour extracted per minute

Solution

Given that a = 84 kg/h; Vacuum = 700 mm of Hg; Barometer reading = 760 mm of Hg; T = 20°C = 20 + 273 = 293 K

1. Pressure in the condenser,

pc = Barometer reading − Condenser vacuum

= 760 − 700 = 60 mm of Hg

= 60 × 0.00133 = 0.0798 bar

From steam tables, corresponding to a temperature of 20°C, we find that pressure of steam,

ps = 0.023385 bar

∴ Pressure of air (as per Dalton’s law),

pa = pcps = 0.0798 − 0.023385 = 0.056415 bar
= 0.056415 × 105 = 5641.5 N/m2

Minimum capacity of the air pump,

2. Let   D = Diameter of the cylinder in metre,

L = Length of the stroke in metres = 1.5 D

ηv = Volumetric efficiency = 100% = 1

N = Speed of the pump = 200 r.p.m.

Minimum capacity of the air pump (Va),

D3 = 0.0886 or D = 0.446 m

and L = 1.5 D = 1.5 × 0.446 = 0.669 m

3. From steam tables, corresponding to a temperature of 20°C, we find that specific volume of steam,
vg = 57.79 m3/kg

∴ Mass of vapour extracted per minute

#### Example 8.15

The vacuum at the extraction pipe in a condenser is 710 mm of mercury and the temperature is 35°C. The barometer reads 760 mm of mercury. The air leakage into the condenser is 4 kg per 10,000 kg of steam. Determine: (a) the volume of air to be dealt with by the dry air pump per kg of steam entering the condenser, and (b) the mass of water vapour associated with this air. Take R = 287 J/kg K for air.

Solution

Given that vacuum = 710 mm of Hg; T = 35°C = 35 + 273 = 308 K; Barometer reading = 760 mm of Hg; ma = 4 kg per 10000 kg of steam = 0.0004 kg/kg of steam.

1. Let  Va = Volume of air per kg of steam entering the condenser.

Pressure in the condenser

pc = Barometer reading − Condenser vacuum
= 760 − 710 = 50 mm of Hg = 50 × 0.001 33
= 0.0065 bar

From steam tables, corresponding to the temperature of 35°C, we find that the pressure of steam,

ps = 0.05628 bar

∴ Pressure of air (as per Dalton’s law),

pa = pcps = 0.0665 − 0.05628 = 0.01022 bar
= 0.01022 × 105 = 1022 N/m2

Now paVa = ma RT

2. From steam tables, corresponding to a temperature of 35°C, we find that specific volume of steam,
vg = 25.216 m3/kg

∴ Mass of water vapour associated with the air

#### Example 8.16

The amount of steam condensed in a condenser is 6800 kg/h and the air leakage into the condenser is 12 kg/h. The air pump suction is screened off. The exhaust steam temperature is 32°C, the condensate temperature is 30°C and the temperature at the air pump suction is 25°C. Determine the following:

1. The mass condensed in the air cooler section per hour
2. The volume of air in m3 handled by air pump
3. The percentage reduction in air pump capacity due to cooling of the air.

Solution

The saturation pressure corresponding to steam temperature 32°C from steam table is

pa = 0.04799 bar

Specific volume corresponding to 32°C, vs = 29.82 m3/kg

∴ Volume of steam handled, Vs = 6800 × 29.82 = 202776 m3/h

According to Dalton’s law, the volume of air handled per hour is equal to the volume of steam handled per hour. Hence the partial pressure of air,

We observe that the value of partial pressure of air is very small and hence the total condenser pressure may be taken as due to steam alone i.e. 0.04799 bar.

Since the temperature at the air pump suction is 25°C, hence the steam pressure at the air pump suction corresponding to 25°C from steam tables,

ps = 0.031691 bar

Specific volume corresponding to 25°C, vs = 43.359 m3/kg

∴ Partial pressure of air handled,

pa = 0.04799 - 0.031691 = 0.0163 bar

∴ Volume of air handled per hour by the pump,

According to Dalton’s law, the volume of steam will also be equal to 629.64 m3/h.

The temperature may be assumed to be that of the condensate, i.e., ts = 30°C at the point where the saturated steam enters the air cooling section. Corresponding to 30°C, the pressure is:

ps = 0.042461 bar

Specific volume, vs = 32.893 m3/h

∴ Partial pressure of air is given by,

pa = (0.04799 − 0.042461) = 0.00553 bar

Hence, the volume of air handled

According to Dalton’s law, this volume of air is equal to the volume of steam hence the mass of steam associated with the air,

∴ Mass of steam condensed in the cooler section

= 57.37 − 14.52 = 42.85 kg/h

Hence, the percentage reduction in volume handled by the air pump due to cooling section is:

#### Example 8.17

In a condenser test, the following observations were made

Vacuum = 70 cm of Hg

Barometer = 76 cm of Hg

Mean temperature of condensation = 35°C

Hot well temperature = 29°C

Mass of cooling water = 45500 kg/h

Inlet water temperature = 16.5°C

Outlet water temperature = 31°C

Mass of condensate = 1200 kg/h

Find (a) the mass of air present per unit condenser volume, (b) the state of steam entering the condenser, (c) the vacuum efficiency, (d) the condensate undercooling and (e) condenser efficiency.

Solution

1. Corresponding to 35°C, the partial pressure of steam from steam table = 0.05628 bar

Absolute pressure of mixture in condenser = (76 − 70) × = 0.07997 bar

∴ Partial pressure of air = 0.07997 − 0.05628 = 0.02369 bar

Mass of air =

2. Heat absorbed by circulating water = 45500 × 4.18 (31 − 16.5) = 2727755 kJ/h

Heat rejected by steam in condensing at 35°C and in being under cooled from 35°C to 29°C

= mc (hw1 + xL1hwc)

= 1200 (146.66 + 2418.6x − 121.59)

= 1200 (25.07 + 2418.6x)

Now, Heat absorbed by circulating water = Heat rejected by steam

3. Vacuum efficiency
4. Condensate undercooling = 35 − 29 = 6°C
5. Condenser efficiency =

#### Example 8.18

In a surface condenser 2500 kg/h of steam is condensed and the air leakage is 2 kg/h. A separate air and water extraction pumps are fitted which draws the air in air cooler and a portion of the tube near the air pump suction is screened off from the steam so that the air is cooled below the condensate temperature. The steam enters the condenser dry and saturated at 38°C and the condensate is extracted at the lowest point of the condenser at a temperature of 37°C (i.e., the temperature at the entrance of air cooler is 3°C). The temperature at the air pump suction is 31°C. Assuming a constant vacuum throughout the condenser, find the following:

1. The mass of steam condensed/min in the air cooler
2. The volume of air to be dealt with per minute by the dry pump
3. The reduction in the necessary air pump capacity following the cooling of air.

Solution

Partial pressure of steam at 38°C, ps = 0.0668 bar

Specific volume of steam at 38°C, vs = 21.8 m3/kg

This must be the volume of 2 kg/h of air when exerting its partial pressure, i.e., Va = Vs = 908.33 m3/min

Partial pressure of air, pa = = 0.000033 bar

Hence, total pressure in the condenser = ps + pa = 0.0668 + 0.000033 = 0.06683 bar

At condensate extraction:

Partial pressure of steam at 37°C = 0.0633 bar

Specific volume of steam at 37°C, vs = 22.94 m3/kg

∴ Partial pressure of air = 0.06683 − 0.0633 bar = 0.00353 bar

∴ Volume of air, a = = 8.473 m3/min = Vs

∴ The mass of the steam associated with air,

At air pump suction:

Partial pressure of steam of 31°C = 0.04522 bar

∴ Partial pressure of air = 0.06683 - 0.04522 = 0.02161 bar

1. Mass of steam associated with this air and which is partially condensed in the air cooler

2. Mass of steam condensed in the air cooler or saving in condensate by using the separate extraction method
= 0.369 − 0.0429 = 0.3261 kg/min
3. Air capacity without air cooler = 8.473 m3/min

Air pump capacity with air cooler = 1.346 m3/min

∴ Percentage reduction in pump capacity

#### Example 8.19

A barometric jet condenser maintains a vacuum of 64.2 cm of Hg when the barometer reads 76 cm of Hg. The condenser handles 4600 kg of steam/h and 0.97 dryness fraction. The inlet temperature of the cooling water is 15°C. The mixture of condensate and cooling water leaves at 43°C. Calculate (a) the minimum height of the tail pipe above the level of the hot well and (b) the amount of cooling water required. Assume no undercooling.

Solution

1. Absolute condenser pressure of steam = (76 − 64.2) × 0.0133 = 1.6048 m of water

Barometer height = 76 × = 10.336 m of water

Hence the length of the tail pipe = 10.336 − 1.6048 = 8.7312 m

2. The saturation temperature of steam entering the condenser will be 43°C in the absence of any undercooling of the condensate. Corresponding to 43°C,

#### Example 8.20

A condenser is equipped in a steam turbine which handles 14500 kg of steam per hour and develops 2484.3 kW. The initial conditions of steam entering to turbine are 27 bar and 350°C. The exhaust from the turbine is condensed in the condenser and the vacuum maintained is 72.5 cm of Hg while barometer reads 75.8 cm of Hg. The temperature of the circulating water is increased from 8 to 28°C while the condensate is removed from the condenser at a temperature of 29°C. Calculate the following:

1. Dryness fraction of steam entering the condenser
2. Mass of circulating water per hour and the cooling ratio
3. Minimum quantity of cooling water required per kg of steam.

Solution

1. Absolute condenser pressure = (75.8 − 72.5) × 1.013/76 = 0.04398 bar

From the steam tables, corresponding to 0.043398 bar, hf = 327.19 kJ/kg and hfg = 2313.46 kJ/kg

Corresponding to 27 bar and 350°C, h = 3124.84 kJ/kg

If x be the dryness fraction of steam entering the condenser, then drop of enthalpy

x = 0.9414

2. Heat given by steam = Heat received by water

14500[(121.48 + 0.9414 × 2433.1) − 117.31] = mw × 4.18 (28 − 8)

Mass of cooling water = mw = 398002.5 kg/h

3. The maximum possible temperature of cooling water at outlet will be 29.4°C corresponding to condenser pressure of 0.04398 bar.

Hence maximum rise in temperature of cooling water = 29.4 − 8 = 21.4°C

If under this condition mw1 be the quantity of cooling water required per kg of steam, by the heat balance equation, then

[(123.22 + 0.942 × 2432.4 − 121.48)] mw1 × 4.18 × 21.4

mw1 = 25.634 kg of water/kg of steam

#### Example 8.21

The following observations were recorded during a trial on a steam condenser:

Condenser vacuum recorded = 70 cm (0.93325 bar)

Barometer reading = 76.5 cm Hg (1.02 bar)

Mean condenser temperature = 35°C

Hot well temperature = 28°C

Condensate formed/hr. = 1800 kg

Circulating cooling water inlet temperature = 15°C

Circulating cooling water outlet temperature = 27°C

Quantity of cooling water/h = 80000 kg

Determine the following:

1. Vacuum corrected to standard barometer of 76 cm Hg (1.01325 bar)
2. Vacuum efficiency
3. Under cooling of condensate
4. Condenser efficiency
5. State of steam entering condenser
6. Mass of air per cubic metre of condenser volume.
7. Mass of air present per kg of uncondensed steam

Assume R for air as 287 kJ/kgK.

Solution

1. Vacuum corrected to standard barometer
= Standard barometric pressure − (Barometric pressure − Gauge pressure)
= 1.01325 bar − (1.02 bar − 0.93335 bar)
= 1.01325 − 0.8665 = 0.9266 bar
2. From steam table, the pressure corresponding to condenser temperature of 35°C = 0.0562 bar
3. Undercooling of condensate = 35° − 28° = 7°C
4. Absolute condenser pressure = Barometer pressure − Vacuum reading

= 1.02 − 0.93325 = 0.08675 bar

Saturation temperature corresponding to pressure of 0.08665 bar from steam tables is 42.9°C

Thus, cooling water temperature can get raised to 42.9°C

5. From steam tables for pressure of 0.08665 bar
hf = 179.8 kJ/kg, hfg = 2399.6 kJ/kg.

Enthalpy of condensate corresponding to hot well temperature of 28°C is 117 kJ/kg.

Heat absorbed by cooling water = Heat given by steam

mcpw (t2t1) = ms (hf + xhfghc)

6. At condenser temperature of 35°C, the partial pressure of steam,
ps = 0.0562 bar.

Hence, ma = = 1.911 kg per kg of vapour

or mv = = 0.523 kg of vapour per kg of air.

With external cooling from 25°C to 20°C we have, from steam tables, at 20°C

ps = 0.023366 bar = 1.7526 cm Hg, vg = 57.834 m3

pa = 5.2 − 1.7526 = 3.4474 cm Hg

7. where pa = 3.4474 × = 0.04596 bar

Va = vg = 57.834 m3 per kg of vapour

R = 287 J/kg ×K; T = (273 + 20) = 293 K

Hence, ma = = 3.161 kg per kg of vapour

mv = = 0.316 kg of vapour per kg of air

These calculations show that by cooling the mixture form 25°C to 20°C, vapour accompanying each kg of air withdrawn from the condenser is reduced from 0.523 kg to 0.316 kg. This will have a significant effect both on the ejector capacity and work done for air pump suction.

#### Example 8.22

In a condenser test, the following observations were made:

Vacuum = 690 mm of Hg; Barometer reading = 750 mm of Hg; Mean temperature of condensation = 35°C; Hot well temperature = 28°C; Mass of cooling water = 50,000 kg/h; Inlet temperature = 17°C; Outlet temperature = 30°C; Mass of condensate per hour = 1250 kg.

Find: (a) the mass of air present per m3 of condenser volume, (b) the state of steam entering the condenser and (c) the vacuum efficiency.

Take R for air = 287 J/kg K.

Solution

Given that vacuum = 690 mm of Hg; Barometer reading = 750 mm of Hg; tc = 35°C; th = 28°C; w = 50000 kg/h; tf = 17°C; to = 30°C; s = 1250 kg/h; R = 287 J/kg K

1. We know that pressure in the condenser,
pc = 750 − 690 = 60 mm of Hg = 60 × 0.00133= 0.08 bar

From steam tables, corresponding to a condensation temperature of 35°C, we find that the pressure of steam,

ps = 0.05628 bar

∴ Pressure of air (as per Dalton’s law),

pa = pcps = 0.08 − 0.05628 = 0.02372 bar = 0.02372 × 105 = 2372 N/m2

Mass of air per m3 of condenser volume,

2. Let x = Dryness fraction (i.e., state) of steam entering the condenser.

From steam tables, corresponding to a pressure of 0.05628 bar (or 35°C), we find that,

hf = 146.66 kJ/kg and hfg = 2418.6 kJ/kg

Corresponding to a hot well temperature of 28°C,

hf1 = 117.3 kJ/kg

Total heat of entering steam,

h = hf + xhfg = 146.66 + x × 2418.6 kJ/kg

Mass of cooling water (mw).

3. Corresponding to a condensation temperature of 35°C,

Ideal pressure of steam, = 0.05628 bar = = 42.32 mm of Hg

∴ Ideal vacuum = Barometer reading − Ideal pressure

= 750 − 42.32 = 707.68 mm of Hg

Vacuum efficiency, ηvac =

#### Example 8.23

The following observations were recorded during a test on a steam condenser:

Barometer reading = 765 mm of Hg

Condenser vacuum = 710 mm of Hg

Mean condenser temperature = 35°C

Condensate temperature = 28°C

Condensate collected per hour = 2 tonnes

Quantity of cooling water per hour = 60 tonnes

Temperature of cooling water at inlet = 10°C

Temperature of cooling water at outlet = 25°C

Find: (a) vacuum corrected to the standard barometer reading, (b) vacuum efficiency of the condenser, (c) undercooling of the condensate, (d) condenser efficiency, (e) quality of the steam entering of condenser, (f) mass of air per m3 of condenser volume and (g) mass of air per kg of uncondensed steam.

Solution

Given that Barometer reading = 765 mm of Hg; Condenser vacuum = 710 mm of Hg; T = 35°C = 35 + 273 = 308 K; tc = 28°C; s = 2 t/h = 2000 kg/h; w = 60 t/h = 60000 kg/h; ti = 10°C; to = 25°C

1. Absolute pressure in the condenser = Barometer reading − Condenser vacuum
= 765 − 710 = 55 mm of Hg

Vacuum corrected to the standard barometer reading (assuming 760 mm of Hg)

= 760 − 55 = 705 mm of Hg
2. From steam tables, corresponding to the mean condenser temperature of 35°C, we find that ideal pressure of steam,

Ideal vacuum = Barometer pressure − Ideal pressure

= 765 − 42.32 = 722.68 mm of Hg

Vacuum efficiency, ηvac = = 0.9824 or 98.24%

3. Undercooling of the condensate = Mean condenser temperature − Condensate temperature
= 35 − 28 = 7°C
4. Pressure in the condenser, pc = 765 − 710 = 55 mm of Hg
= 55 × 0.00133 = 0.073 bar

From steam tables, corresponding to a pressure of 0.073 bar, we find that vacuum temperature,

5. Let   x = Quality of steam entering the condenser.

From steam tables, corresponding to a pressure of 0.073 bar, we find that,

hf = 166.7 kJ/kg; and hfg = 2407.4 kJ/kg

Corresponding to a condensate temperature of 28°C, heat in condensate,

hf1 = 117.3 kJ/kg

Total heat of entering steam,

h = hf + xhfg = 166.7 + x × 2407.4

Mass of cooling water (mw),

6. Absolute pressure of air (as per Dalton’s law),

pa = pcps = 0.073 − 0.0562 = 0.0168 bar

= 0.0168 × 105 = 1680 N/m2

∴ Mass of air per m3 of condenser volume,

7. From steam tables, corresponding to 35°C (i.e., mean condenser temperature), specific volume of steam,
vg = 25.245 m3/kg

Air associated with 1 kg of steam at 35°C will occupy the same volume, i.e., 25.245 m3

∴ Mass of air per kg of uncondensed steam,

#### Example 8.24

The following observations refer to a surface condenser:

Mass flow rate of condensate = 20 kg/min

Mass flow rate of cooling water = 800 kg/min

Mean temperature of condensation = 35°C

Condenser vacuum = 0.954 bar

Intel cooling water temperature = 20°C

Outlet cooling water temperature = 30°C

Temperature of the hot well = 29°C

Calculate the following:

1. Weight of air per unit volume of condenser
2. Entering condition of steam to the condenser
3. Vacuum efficiency of the condenser.

Properties of saturated steam:

Take R for air = 0.287 kJ/kg. K.

[IAS, 2009]

Solution

Given that s = 20 kg/min, w = 800 kg/min, T = 273 + 35 = 308 K, twi = 20°C, two = 30°C, ph = 1.03248 N/m2, pcu = 0.954 N/m2, thw = 29°C

Absolute pressure in condenser, pt = (phpcu) = 1.03248 − 0.954 = 0.07848 bar

At t = 35°C, ps = 0.05628

Partial pressure of air, pa = ptps = 0.07848 − 0.05628 = 0.0222 bar

1. Mass of air, ma =

2. At pt = 0.07848 bar, hf = 169.15 kJ/kg, hfg = 2405.80 kJ.kg

3. Vacuum efficiency, ηvac =

#### Summary for Quick Revision

1. A condenser is a closed vessel in which steam is condensed by abstracting heat and the pressure is maintained below atmospheric pressure.
2. Steam condensers are basically of two types—jet condensers and surface condensers.
3. In jet condensers, the steam to be condensed mixes with the cooling water.
4. In surface condensers, steam does not mix with the cooling water.
5. Jet condensers can be classified as low-level counter flow or parallel flow type; high-level type, and ejector type.
6. Surface condensers can be classified as water tube type, central flow or regenerative type, and evaporative type.
7. Vacuum is sub-atmospheric pressure. It is measured by a vacuum gauge.
8. Absolute condenser pressure. p = 0.133322 (Hb Hg) bar

where Hb = actual barometric height, cm of Hg

Hg = actual vacuum gauge reading, cm of Hg.

Corrected or standard vacuum = 76 − (HbHg) cm of Hg.

9. According to Delton’s law of partial pressures,

Total pressure in the condenser,

pt = absolute pressure of air, pa + saturation pressure of steam, ps

10. Mass of mixture,

where ma, mg = mass of air and water vapour respectively.

va, vg = specific volume of air and water vapour respectively.

11. Mass of cooling water required, for jet condensers
12. An Edward’s air pump is used to maintain the desired vacuum in the condenser.
13. Vacuum efficiency is defined as the ratio of actual vacuum to ideal vacuum.

where pb = barometric pressure

pa = partial pressure of air

ps = saturation pressure of steam

14. Condenser efficiency,

where tiw, tow = inlet and outlet temperatures of cooling water

ts = steam temperature in the condenser

15. A cooling tower is an artificial device used to cool hot cooling water coming out from the condenser.

#### Multiple-choice Questions

1. In a steam condenser, the partial pressure of steam and air are 0.060 bar and 0.007 bar, respectively. The condenser pressure is
1. 0.067 bar
2. 0.060 bar
3. 0.053 bar
4. 0.007 bar
2. Cooling tower in a steam power plant is a device for
1. condensing steam into water
2. cooling the exhaust gases coming out of the boiler
3. reducing the temperature of superheated steam
4. reducing the temperature of cooling water used in condenser
3. The function of the surface condenser is to
1. lower the engine thermal efficiency
2. lncrease the engine thermal efficiency
3. lncrease the back pressure of the engine
4. cool the exhaust gases
4. In a jet condenser,
1. steam and cooling water mix together
2. steam and cooling water do not mix together
3. steam passes through tubes and cooling water surrounds them
4. cooling water passes through tubes and steam surrounds them
5. A surface condenser is a
1. water tube device
2. steam tube device
3. steam and cooling water mix to give the condensate
4. All of the above
6. The air removal from the surface condenser leads to
1. fall in condenser pressure
2. rise in condenser pressure
3. no change in condenser pressure
4. rise in condenser temperature
7. Edward’s air pump
1. removes only air from the condenser
2. removes air and vapour form the condenser
3. removes only uncondensed vapour from condenser
4. removes air along with vapour and condensed water from the condenser
8. In a surface condenser used in a steam power station, undercooling of condensate is undesirable as this would
1. not absorb the gases in steam.
2. reduce efficiency of the plant.
3. increase the cooling water requirements.
4. increase thermal stresses in the condenser
9. Consider the following statement:

The effect of fouling in a water-cooled steam condenser is that it

1. reduces the heat transfer coefficient of water
2. reduces the overall heat transfer coefficient
3. reduces the area available for heat transfer
4. increases the pressure drop of water

Of these statements:

1. 1, 2, and 4 are correct
2. 2 and 4 are correct
3. 2 and 4 are correct
4. 1 and 3 are correct.

#### Review Questions

1. Define a condenser.
2. What are the functions of a condenser?
3. Name the various elements of a steam condensing plant.
4. What are the various types of steam condensers?
5. Differentiate between jet and surface condensers.
6. What is an evaporative condenser?
7. What are the requirements of a modern surface condenser?
10. State Dalton’s law of partial pressures.
11. How the mass of cooling water can be estimated in a surface condenser?
12. What are the sources of air infiltration in a condenser?
13. What are the effects of air infiltration in condensers?
14. Define vacuum efficiency and condenser efficiency.
15. What is the role of cooling towers in surface condenser?

#### Exercises

8.1 The vacuum in a condenser handling 9000 kg/h of steam is found to be 72 cm of Hg when the barometer reading is 76 cm of Hg and the temperature is 25°C. The air leakage amounts to 2 kg for every 5000 kg of steam. Determine the capacity of a suitable dry air pump in m3/min required for the condenser. Take volumetric efficiency of the pump as 85%.

[Ans. 2.83m3/min]

8.2 A steam condenser has separate air and condensate pumps. The entry to the air pump suction is screened. Steam enters the condenser at 38°C and the condensate is removed at 37°C. The air removed has a temperature of 36°C. If the quantity of air infiltration from various sources is 5 kg/h, determine the volume of air handled by the air pump per hour. Compare this with the quantity that would have to be dealt with by using a combined air and condensate pump. Neglect the pressure due to air at entry of steam and assume uniform pressure in the condenser.

[Ans. 647.3m3/h, 1267.38 m3/h]

8.3 A surface condenser deals with 13625 kg of steam per hour at a pressure of 0.09 bar. The steam enters 0.85 dryness fraction and the temperature at the condensate and air extraction pipes is 36°C. The air leakage amounts to 7.26 kg/hour. Determine the following:

1. The surface area required if the average heat transmission rate is 4 kJ/cm per second.
2. The cylinder diameter for the dry air pump, if it is to be single acting and running at 60 rpm with a stroke-bore ratio of 1.25 and volumetric efficiency of 85%.

[Ans. (a) 1956.57 cm2, (b) 74 cm]

8.4 A surface condenser, having an absolute pressure of 0.10 bar, is supplied with cooling water at the rate of 40kg of steam condensed. The rise in the temperature of cooling water is 14°C. Find the dryness fraction of steam entering the condenser. The condensate leaves at 44°C.

[Ans. 0.98]

8.5 A steam turbine discharges 5000 kg/h of steam at 40°C and 0.85 dry. The air leakage in the condenser is 15 kg/h. The temperature at the suction of air pump is 32°C and temperature of condensate is 35°C. Find (a) vacuum gauge reading, (b) capacity of air pump, (c) loss of condensate in kg/h, and (d) quantity of cooling water if its temperature rise is limited to 10°C.

[Ans.(a) 70.45 cm, (b) 502.5 m3/h, (c) 17 kg/h, (d) 254 × 103 kg/h]

8.6 A steam turbine uses 50,000 kg/h of steam. The exhaust steam with dryness fraction 0.9 enters the condenser fitted with water extraction and air pumps. When the barometer reads 76 cm of Hg, vacuum of air pump suction is 72 cm of Hg and temperature is 32°C. The air leakage is estimated at 500 kg/h. Calculate (a) net capacity of air pump and (b) quantity of water circulated per minute if temperature rise is limited to 15°C.

[Ans. (a) 1291.83 m3.min, (b) 27160 kg/min]

8.7 The temperature in a surface condenser is 40°C and the vacuum is 69 cm of Hg while the barometer reads 75 cm of Hg. Determine the partial pressure of steam and air and the mass of air present per kg of steam.

[Ans. 0.07375 bar, 0.00625 bar, 0.136 kg/kg vapour]

8.8 In a condenser, to check the leakage of air, the following procedure is adopted:

After running the plant to reach the steady conditions, the steam supply to the condenser and also the air and condensate pumps are shut down, thus completely isolating the condenser. The temperature and vacuum readings are noted at shut down and also after a period of 5 minutes. They are 39°C and 68.5 cm Hg; and 28°C and 27 cm Hg respectively. The barometer reads 75 cm Hg. The effective volume of condenser is 1 m3.Calculate (a) the quantity of air leakage into the condenser during the period of observation, and (b) the quantity of water vapour condensed during the period.

[Ans. 0.3523 kg, 0.0214 kg]