steady flow energy equation ii

EDEXCEL HNC/D PLANT and PROCESS PRINCIPLES

OUTCOME 2

TUTORIAL 2 - STEADY FLOW ENERGY EQUATION

2 Steady flow energy equation (SFEE) SFEE: consideration and applications of continuity of mass; first law of thermodynamics; principle of conservation of energy; work flow; heat transfer; kinetic energy; potential energy; pressureflow energy; internal energy; enthalpy Application of SFEE to plant: assumptions made in specific applications; energy transfer and efficiency calculations for specific items of plant (eg economisers, boilers, superheaters, turbines, pumps, condensers, throttles, compressors); boiler efficiency

It is assumed that the student is already familiar with the use of tables to find the properties of fluids and the use of gas laws. If you are not you will find tutorials on this in the Edexcel module Thermodynamics and in the general area of Thermodynamics (www.freestudy.co.uk) In this tutorial we examine specific plant items as listed above. You will also find good quality learning material at http://www.spiraxsarco.com/learn/ And free tables for finding the properties of steam and water at http://www.spiraxsarco.com/esc/

D.J.Dunn freestudy.co.uk 1

STEADY FLOW PLANT

Steam power plant contains most of the items to be examined but many of them are found elsewhere. The graphic below indicates the basic plant used in simple steam power plant. The thermodynamic cycle is based on the Rankine Cycle and you can find tutorials on this in the thermodynamic module.

Most of the plant items are heat exchangers of one kind or other. The boiler is a series of heat exchangers that evaporates the feed water and then superheats it. The source of heat may be from the combustion of fossil fuel or hot gas from a nuclear reactor. The steam passes to a turbine where it gives up its energy (enthalpy) to mechanical power that drives the alternator and produces electric power. The steam leaving the turbine is condensed in the condenser. This is another heat exchanger that uses cooling water. The cooling water is circulated to a cooling tower (or another source of cooling such as the sea, river or lake). The condensate is pumped back to the boiler but it is heated in another heat exchanger called an economiser. This uses the hot gas (usually flue gas) exiting the boiler.

In a full blown power station, there will be several feed heaters, a reheater and an air pre-heater. There will also be feed water heaters, de-aerators, de-superheaters, hot wells and many other things. The purpose of this module is to examine the energy transfers occurring and the efficiencies of the plant items.

D.J.Dunn www.freestudy.co.uk 2

BOILER There are many types of boilers ranging from small hot water boilers to large power station boilers producing superheated steam. A main power station boiler will have an economiser, an evaporator, a superheater, a reheater and an air preheater (recuperator) all of which improve the efficiency. The diagram shows the arrangement for a medium size boiler using water walls. The walls of the furnace are entirely surrounded by vertical pipes banked so close together that they form a curtain wall. Water from the drum passes down the down comer pipe and is distributed to the water wall with a circulating pump. The steam from the top returns to the steam space inside the drum. Steam is drawn off from the same space and superheated in a bank of heat exchanger tubes. The superheater is placed in the hottest part of the flue. The economiser is in the coolest part of the flue.

MEDIUM SIZE POWER STATION BOILER

Some boilers (once through types) are designed as one long heat exchanger with water entering one end and superheated steam leaving the other and there is no clearly defined evaporator section. Industrial process steam is usually dry or wet so many boilers do not have a superheater. It is not practical to cover the many types of boilers on the market. The pictures show typical package boilers delivered fully constructed to the site.

TYPICAL PACKAGE BOILERS


APPLYING THE STEADY FLOW ENERGY EQUATION All the units described are basically heat exchangers. The main point is that a heat transfer rate is needed into the boiler unit in order to heat up the water, evaporate it and superheat it. The heat transfer to the water or steam is found by applying the Steady Flow Energy Equation. + P = (PE)/s + (KE)/s + (H)/s In the case of any heat exchanger, P = 0 as there is no work done. The velocity of fluids should be low otherwise there will be pressure losses and there should not be much difference between the inlet and outlet so kinetic energy is usually neglected. The inlet and outlets of the heat exchanger may be at different levels so potential energy changes might exist but this is likely to be small compared to the other energy terms so this is normally neglected as well. In this case the SFEE becomes: = H/s = m (h) h is the change in specific enthalpy of the fluid. is the heat transferred into or out of the fluid and m is the mass flow rate. This may be applied to the boiler in its entirety or to any of the units within the boiler. For steam we need to use steam tables to find the enthalpy values. You should already know that: hf is the specific enthalpy of saturated water. hfg is the specific enthalpy of evaporation. hg is the specific enthalpy of dry saturated steam hg = hf + hfgFor superheated steam h values For wet steam with dryness fraction x h = hf + xhfg For low pressure water h = cw where cw is the specific heat (about 4.2 kJ/kg K) and the temperature in oC. Celsius is used because specific enthalpy is arbitrarily taken as zero at 0oC. For high temperatures and pressure the specific heat changes significantly and it is best to use the table below. The enthalpy of steam and water may be found at http://www.spiraxsarco.com/esc/ For gases, the enthalpy change may be found with specific heats so that h = cp where cp is the specific heat at constant pressure and these can be found from tables. Note that these values also vary with temperature and so care should be taken.

TABLE FOR THE ENTHALPY OF HIGH PRESSURE WATER Pressure in bar Temp oC

0 25 50 75 100 125 150 175 200 221 250 0 0 2.5 5 7.5 10 12.6 15 17.5 20 22 25 20 84 86 87 91 93 96 98 100 103 105 107 40 168 170 172 174 176 179 181 183 185 187 190 60 251 253 255 257 259 262 264 266 268 270 272 80 335 337 339 341 343 345 347 349 351 352 355 100 419 421 423 425 427 428 430 432 434 436 439 120 504 505 507 509 511 512 514 516 518 519 521 140 589 591 592 594 595 597 599 600 602 603 605 160 675 677 678 680 681 683 684 686 687 688 690 180 763 764 765 767 767 769 790 772 773 774 776 200 852 853 854 855 856 857 858 859 861 862 863


OVERALL BOILER EFFICIENCY The heat input to a boiler comes from various sources. Fossil fuels are one of the most common sources and the heat input is given by: = mass/s x calorific value - solid fuels =volume/s x calorific value - gas fuels Typical calorific values are: Coal 30 36 MJ/kg Fuel Oils 43 46 MJ/kg Natural Gas 38 MJ/m3Other sources of heat are hot gasses such as the CO2 used on nuclear power stations and hot flue gas used with waste heat boilers. WORKED EXAMPLE No.1 A hot water boiler produces 0.24 kg/s of hot water at 80oC from cold water at 18oC.The

boiler burns fuel oil at a rate of 1.6 g/s with a calorific value of 44 MJ/kg. Calculate the thermal efficiency of the boiler.

SOLUTION The easiest way to find the increase in enthalpy of the water is to use the specific heat

assumed to be 4.186 kJ/kg K. = m c = 0.24 x 4.186 x (80 18) = 62.29 kW Heat released by combustion = mf x C.V. Heat released by combustion = 1.6 x 10-3 (kg/s) x 44 000 (kJ/kg) = 70.4 kW th = (62.29/70.4) x 100 = 88.5% WORKED EXAMPLE No. 2 A steam boiler produces 0.2 kg/s at 50 bar and 400oC from water at 50 bar and 100oC.

The boiler burns 5.3 m3/min of natural gas with a calorific value of 38 MJ/ m3. Calculate the thermal efficiency of the boiler.

SOLUTION The enthalpy of the steam produced is found in tables at 50 bar and 400oC. h2 = 3196 kJ/kg The enthalpy of the water is found from the table at 50 bar and 100oC. h1 = 5 423 kJ/kg Energy given to the water and steam = m (h2 - h1) = 0.2 (3196 - 423) = 3111 kW Energy from burning the fuel = Vol/s x C.V. Energy from burning the fuel = (5.3/60)(m3/s) x 38 000 (kJ/m3) = 3357 kW th = (3111/3357) x 100 = 92.7%


SELF ASSESSMENT EXERCISE No. 1 1. A hot water boiler produces 0. 4 kg/s of hot water at 70oC from cold water at 10oC. The

boiler burns fuel oil at a rate of 3.2 g/s with a calorific value of 44 MJ/kg. The specific heat of water is 4.186 kJ/kg K.

Calculate the thermal efficiency of the boiler. (71.4 %) 2. A steam boiler produces 3 kg/s at 70 bar and 500oC from water at 70 bar and 120oC. The

boiler burns 17 m3/min of natural gas with a calorific value of 38 MJ/ m3. Calculate the thermal efficiency of the boiler. (80.8%)

ECONOMISER

6

The economiser is a heat exchanger that passes heat from the flue gas to the feed water. The feed water temperature is hence hotter than it would otherwise be and less heat is required to produce steam from it. Alternatively more steam can be raised for the same amount of heat. An increase in temperature of 10C can improve the boiler efficiency by as much as 2%. SHELL BOILER WITH ECONOMISER Because the economiser is on the high-pressure side of the feed pump, feed water temperatures in excess of 100C are possible. Note that the flue gas should not be cooled to the dew point to prevent acidic water forming. The diagram shows the layout for a small shell boiler.

The efficiency of an economiser refers only to how well it transfers heat and this is only affected by the heat transfer through the walls of the tubes. They are prone to sooty deposits forming on the flue gas side and this reduces the heat transfer rate. They are designed to be cleaned easily. A finned heat exchanger tube such as shown should have the fins arranged vertically so that soot will fall off them. Plain tubes have scrapers fitted for cleaning them mechanically. Steam blowing is a technique used to blast the soot off the fins with a steam jet but this can cause atmospheric pollution.

D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No. 3 A boiler is supplied with 6 MW of heat and produces 0.5 kg/s of steam at 100 bar from

feed water at 40oC with an efficiency of 65%. If an economiser is fitted that raises the feed water temperature to 180oC, what would be the new boiler efficiency? Assume that the heat input to the boiler is reduced by the amount saved by fitting the economiser. The table for the enthalpy of water should be used.

SOLUTION The heat given to the steam is 65% x 6 MW = 3.9 MW The enthalpy of water at 100 bar and 180oC is 767 kJ/kg The enthalpy of water at 100 bar and 40oC is 176 kJ/kg The heat given to the water = mwh = 0.5(767 176) = 295.5 kW The heat transfer is reduced to 6 0.2955 = 5.7045 MW The efficiency is 3.9/5.7045 = 0.684 or 68.4% WORKED EXAMPLE No. 4 Measurements for the boiler and economiser described in example 1 show that the flue

gas is cooled from 250oC to 150oC over the economiser. The specific heat of the flue gas is 1.1 kJ/kg K. Assuming 100% heat exchange, determine mass flow of the flue gas.

SOLUTION The heat given to the water is 295.5 kW The heat lost from the flue gas is mgh = mg cpT = mg 1.1 x (250 150) = 295.5 kW If the transfer is 100% then we can equate. mg = 295.5/110 = 2.686 kg/s


EVAPORATOR All steam boilers evaporate the feed water. The feed water should be as close to the boiling point as possible to reduce the heat needed to evaporate it. The steam produced should ideally be dry steam but in practice a small amount of water is carried over with it so the steam is typically 98% dry (dryness fraction 0.98). It is not possible to produce superheated steam without a superheater section. In large power station boilers the evaporator is a water wall surrounding the furnace as described earlier. These are also called water tube boilers because the tubes contain the water with hot gas on the outside. The boiler with water walls and heater tube banks as shown earlier fits this category. In smaller boilers the water may be contained inside a shell and heated by hot gas passing through the tubes. These are shell boilers and also called fire tube boilers because the hot gas passes through the tubes.

SHELL BOILER WITH FIRE TUBES

WORKED EXAMPLE No. 5 The evaporator section of a boiler uses 20 kg/s of feed water at 50 bar and 200oC. The

steam produced is 98% dry. Calculate the heat transfer. SOLUTION Heat transfer = mass x h Specific enthalpy of feed water is h1 = 854 kJ/kg (from the table) The specific enthalpy of the steam is h2 = hf + 0.98 (hg) These values must be found from steam tables and we find that at 50 bar: hf = 1155 kJ/kg and hg = 1639 kJ/kg h2 = 1155 + 0.98 (1639) = 2761.2 kJ/kg in = m(h2 - h1) = 20(2761.2- 1639) = 22444 kW or 22.444 MW


SUPERHEATER This is a heater placed in the hottest part of the boiler that raises the temperature of the steam well beyond the saturation temperature. WORKED EXAMPLE No. 6 A superheater produces 34 kg/s of steam at 70 bar and 400oC from steam 97% dry.

Calculate the heat transfer to the steam. SOLUTION From the steam tables (superheat section) the enthalpy of the superheated steam is h2 = 3158 kJ/k K at 70 b and 400oC From the saturated steam section hf = 1267 and hfg = 1505 kJ/kg K at 70 bar. h1 = 1267 + 0.97(1505) = 2726.85 kJ/kg K = mh = 34(3158 - 2726.85) = 14659 kW or 14.659 MW WORKED EXAMPLE No. 7 The same superheater given in the last example cools the flue gas by 700 K. The mean

specific heat is 1.15 kJ/kg K. The air fuel ratio for the furnace is 14/1. Determine the mass flow of the flue gas and the mass of fuel being burned. SOLUTION Assuming the heat lost by the gas all goes into the steam we can equate. = mg cp = mg 1.1(700) = 770 mg kW = 14659 = 770 mg mg= 19.03 kg/s The mass of gas = mass of fuel + mass of air 19.03 = mf + ma 19.03 = mf + 14 mf = 15 mf mf = 19.03/15 = 1.269 kg/s


TURBINE The turbine converts the enthalpy in the steam into mechanical power. Applying the steady flow energy equation gives:

+ P = (PE)/s + (KE)/s + (H)/s In an ideal turbine, the heat loss from the case is reduced to near zero by lagging. The change in potential and kinetic energy are usually negligible so this reduces to :

P = H = m(h) P is the power given up by the steam and h the change in specific enthalpy of the steam. Turbines in real plant are often in several stages and the last stage is specially designed to cope with water droplets in the steam that becomes wet as it gives up its energy. You must use the isentropic expansion theory in order to calculate the dryness fraction and enthalpy of the exhaust steam. The picture shows a large steam turbine stripped for service. There are three sections, high, intermediate and low pressure. The steam enters at the middle of each and flows away in both directions to reduce axial forces on the rotor.

WORKED EXAMPLE No. 8 A steam turbine is supplied at 100 bar and 400oC. The steam exits the turbine at 0.07 bar

and dryness fraction 0.73. The flow rate of steam is 55 kg/s. Calculate the power output from the turbine assuming 100% energy transfer from the steam.

SOLUTION The specific enthalpy of the steam supplied is found from tables for superheated steam. h = 3097 kJ/kg (100 bar and 400oC). The specific enthalpy of wet steam is h = hf + x3 hfg = 163 + 0.733(2409) = 1928 kJ/kg (The values are found in tables for saturated steam). The power out of the turbine is P = msh = 55(3097 - 1928) = 64.3 MW


CONDENSER The condenser removes energy from the steam and converts it back to water. Applying the Steady Flow Energy Equation gives:

+ P = (PE)/s + (KE)/s + (H)/s There is no work (power) extracted from a condenser. Kinetic and potential energy terms are negligible so the equation reduces to = H = m (h) is the heat transfer rate from the steam and water. h is the change in enthalpy of the steam and water. WORKED EXAMPLE No. 9 A steam condenser receives 55 kg/s of wet steam at 0.07 bar and 0.73 dryness fraction.

The steam is completely condensed to saturated water. Assuming 100% heat transfer, determine the heat transfer to the cooling water.

If the cooling water temperature must not rise by more than 20K, what should the flow

rate be? Assume the specific heat of water is 4.2 kJ/kg K. SOLUTION The specific enthalpy of the steam entering the condenser is: h = hf + x hfg = 163 + 0.73(2409) = 1928 kJ/kg (from tables at 0.07 bar) The specific enthalpy of the condensate is simply hf at 0.07 bar = 163 kJ/kg The heat transfer from the steam is = ms(h) = 55(1928 - 163) = 97.1 MW If all this is transferred to the cooling water then the we can equate : mw cw T = mw 4.2 (20) = 97100 kW

kg/s 1156(4.2)(20)

97100mw == SELF ASSESSMENT EXERCISE No. 2 A steam condenser takes in wet steam at 8 kg/s and dryness fraction 0.82. This is

condensed into saturated water at outlet. The working pressure is 0.05 bar. Calculate the heat transfer rate. ( = 15896 kW)


PUMPS Pumps come in many types and sizes. A large boiler will have circulation pumps, feed water pumps, condensate pumps and cooling water circulation pumps.

The energy given to the water by the pump is also found from the Steady Flow Energy Equation as

+ P = (PE)/s + (KE)/s + (H)/s

The ideal pump has no heat loss and again the change in potential and kinetic energies are usually negligible so this reduces to:

P = H = m (h)

Remember that h = u + (pV). If the water is incompressible, the temperature is not raised and the volume does not change so u = 0 and (pV) = Vp. The work (power) transferred from the fluid is then:

P = V p

The actual power input to the pump is greater because of mechanical losses and friction in the impeller. There may also be some heat loss from the casing but this is very small in comparison to the power on large feed pumps. If losses of all kinds are taken into account we define the overall efficiency of the pump as

Power ActualPower Ideal =

WORKED EXAMPLE No. 10 A circulation pump delivers 40 kg/s with a pressure rise of 2 bar. The overall efficiency

is 82%. Calculate the power input required. SOLUTION Neglecting Kinetic and Potential Energy P = V p The volume = mass/density The density of water is nominally 1000 kg/m3 V = 40/1000 = 0.04 m3/s P = 0.04 x 2 x 105 = 8000 W


WORKED EXAMPLE No. 11 A feed pump delivers 55 kg/s of water. The inlet pressure is 1.0 bar and the delivery

pressure is 100 bar. The inlet temperature is 40oC. The efficiency of the pump is 85% and is entirely due to mechanical effects. Heat loss, kinetic and potential energy may be taken as negligible. Assume that the specific volume of water is 0.001 kg/m3 and specific heat is 4.2 kJ/kg K.

Calculate the power input. Calculate the specific enthalpy of the water leaving the pump. Estimate the temperature at exit. SOLUTION To solve this apparently simple problem, we would need very accurate tables and more

advanced theory (involving a property called entropy) to work out the temperature rise over the pump and this would turn out to be quite small. With the information available, we can only assume that the the ideal power is P = V p

V = 55 x 0.001 = 0.055 kg/s P = 0.055(100 - 1) x 105 = 544.5 x 103 W = 544.5 kW Actual Power input = 544.5/85% = 544.5/0.85 = 640.588 kW The specific enthalpy of the water at inlet to the pump is approximately the product of

specific heat x temperature in oc giving h = c = 4.2 x 40 = 168 kJ/kg. The table previous gives 168 kJ/kg so we will take this rounded figure.

Applying the SFEE to the pump and ignoring KE and PE we have: + P = H/s = mh and = 0 so P = mh 640.588= 55 h Hence h = 11.7 kJ/kg The specific enthalpy at outlet is 168 + 11.7 = 179.7 kJ/k The exit temperature could be obtained from accurate tables. A rough estimate is: = h/specific heat = 179.7/4.2 = 42.8oC


COMPRESSORS INSTALLATIONS The purpose of a compressor is to produce a flow of gas and overcome the pressure on the delivery side. An industrial compressor may be fitted with cooling systems. There are many types for many functions. The capacity of the air compressor is usually rated as the volume flow rate at atmospheric conditions and this is called the Free Air Delivery. In other words it is the volume based on 1.013 bar and 20oC which has a density of 1.205 kg/m3.

SCREW TYPE VANE TYPE

CENTRIFUGAL AXIAL VANE

One of the most versatile compressors is the reciprocating type. A typical case is shown below.

Applying the SFEE we have:

+ P = (PE)/s + (KE)/s + (H)/s

We can normally neglect Potential Energy and Kinetic Energy so this reduces to: + P = (H)/s = m cp ()

The Heat loss may be calculated when the compressor is water cooled, otherwise this is difficult. The specific heat is usually taken as the value at the average temperature. For air this is normally around 1.005 kJ/kg K.


WORKED EXAMPLE No. 12 A water-cooled reciprocating air compressor delivers 3.41 m3/min FAD. The cooling

water flows at 0.021 kg/s and its temperature is raised by 20 K. The air temperature is raised by 80 K.

The specific heat of the water and air are respectively 4.2 kJ/kg K and 1.005 kJ/kg K. Calculate the theoretical power input. The actual power input is 7.1 kW. What is the overall efficiency of the compressor? SOLUTION = mwcw = 0.021 x 4.2 x 20 = 1.764 kW this is a heat loss so it is negative. The change in enthalpy for the air is H = macp The mass flow rate of air = volume/density = 3.41/1.205 = 2.83 kg/min or 0.0471 kg/s H = macp = (0.0471)(1.005)(80) = 3.79 kW Apply the SFEE + P = (H)/s = m cp () -1.764 + P = 3.79 kW P = 5.55 kW This is the theoretical power input. Efficiency = 5.55/7.1 = 0.78 or 78% SELF ASSESSMENT EXERCISE No. 3 1. A steady flow air compressor draws in air at 20oC and compresses it to 120oC at outlet.

The mass flow rate is 0.7 kg/s. At the same time, 5 kW of heat is transferred into the system. Take cp as 1005 J/kg K. Calculate the following.

i. The change in enthalpy per second. (70.35 kW) ii. The work transfer rate. (65.35 kW) 2. A steady flow boiler is supplied with water at 15 kg/s, 100 bar pressure and 200oC. The

water is heated and turned into steam. This leaves at 15 kg/s, 100 bar and 500oC. Using your steam tables, find the following.

i. The specific enthalpy of the water entering. (856 kJ/kg) ii. The specific enthalpy of the steam leaving. (3373 kJ/kg) iii. The heat transfer rate. (37.75 kW) 3. A pump delivers 50 dm3/min of water from an inlet pressure of 100 kPa to an outlet

pressure of 3 MPa. There is no measurable rise in temperature. Ignoring K.E. and P.E, calculate the work transfer rate. (2.417 kW)

4. A water pump delivers 130 dm3/minute (0.13 m3/min) drawing it in at 100 kPa and

delivering it at 500 kPa. Assuming that only flow energy changes occur, calculate the power supplied to the pump. (0.86 kW)

5. A steam condenser is supplied with 2 kg/s of steam at 0.07 bar and dryness fraction 0.9.

The steam is condensed into saturated water at outlet. Determine the following. i. The specific enthalpies at inlet and outlet. (2331 and 163 kJ/kg) ii. The heat transfer rate. 4336 kW)


REFRIGERATION PLANT Refrigeration plant contains many of the same features as steam plant but the fluids used are different. A basic circuit is shown below containing a compressor, a cooler/condenser and a heater/evaporator. The other feature shown is a throttle valve and this needs to be studied next.

BASIC REFRIGERATION PLANT STRIPPED TO BARE PIPEWORK.

THROTTLE VALVE

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A throttle is simply a restriction used to drop the pressure of the fluid with high pressure at inlet and low pressure at outlet. The diagram shows a variable type. A fixed type might be as simple as a narrow coiled tube as seen in domestic refrigerators. The affect of dropping the pressure depends on the type and condition of the fluid. In refrigerators this is usually a liquid near to the saturation temperature so that a drop in pressure causes the liquid to boil and instantly evaporate into a wet vapour with an accompanying drop in temperature. Applying the SFEE we have:

+ P = (PE)/s + (KE)/s + (H)/s In this case there is no work (power) and no heat transfer. The change in potential energy is zero negligible. The kinetic energy could be a factor depending on the pipe sizes, especially if the fluid expands into a gas with a large volume but velocities should be kept small to avoid losses. If this is the case then we have the simple result:

(H)/s = 0 and h/s = 0 This means that the enthalpy and hence specific enthalpy has the same value at inlet and outlet. Remember that enthalpy is the sum of two other energies, internal energy (U) and flow energy (pV) so:

U1 + p1V1 = U2 + p2V2THROTTLING A LIQUID If the liquid does not evaporate the volume V is near constant so U1 - U2 = V p The internal energy of a liquid is m c where c is the specific heat and the temperature. m c (1 2) = V p

( )c

pc mpV 21 ==

This assumes the density is the same at inlet and outlet. This allows us to calculate the change in temperature. There would have to be a very large change in pressure to produce a significant temperature change (Typically 40 bar for 1 degree change).

D.J.Dunn www.freestudy.co.uk

THROTTLING A PERFECT GAS If the fluid is a gas, then h2 h1 = mcp = 0 It follows that = 0 and there is no change in temperature. There will be a change in volume THROTTLING A SATURATED LIQUID If a saturated liquid undergoes a drop in pressure, the saturation temperature (boiling point) is also dropped so it cannot exist as a liquid at the lower pressure. Some of the liquid must change to vapour and in the process absorbs latent heat. This causes the temperature to drop. Remember h1 = h2 so if we have access to tables, we can calculate the dryness fraction and determine the temperature. Hot liquids are often vented into a low pressure tank in order to cause rapid cooling and evaporation. WORKED EXAMPLE No. 13 A boiler contains high pressure hot water at 50 bar and 200oC. A valve is opened on the

bottom of the connecting it to atmosphere at 1.013 bar. Determine the condition of the substance being vented.

SOLUTION On the entry side of the valve the water has a specific enthalpy of 854 kJ/kg (from the

water table). h1 = 854 kJ/kg On the atmospheric side of the valve the specific enthalpy is hf + x hfg at 1.103 bar.

These values may be found in steam tables giving h2 = 419.1 + x(2256.7) Equate 854 = 419.1 + x(2256.7) hence x = 0.193 In other words about 81% of the mass is evaporated in the process. The temperature of

the wet steam emerging from the valve is 100oC (the saturation temperature at the lower pressure).


GAS TURBINES A basic gas turbine is illustrated below. Air is compressed and blown into a combustion chamber where it is heated. The hot air expands through the turbines which drive the compressor and produces power output. The hot gas exhausts to atmosphere.

WORKED EXAMPLE No. 14 A gas turbine draws in 11.84 kg/s of air at 1 bar and 10oC. It exhausts at 1 bar and

400oC. The power produced is 2 MW. Neglect the mass of the fuel and assume no heat losses. Neglect Potential and Kinetic Energy. Take the specific heat of the air as 1.01kJ/kg K.

Determine the heat supplied by the fuel and the thermal efficiency of the unit. SOLUTION Taking the unit as a whole we can apply the SFEE. + P = (PE)/s + (KE)/s + (H)/s is the heat input from the fuel if no other source or losses exist. P = Net Power output H = enthalpy change of the air. Neglecting PE and KE we have: + P = (H)/s Note that P = -2000 kW because the power is removed from the system. In KW units - 2 000 = mcp = (11.84)(1.01)(400 10) = 4663.8 Heat Input = 6663.8 kW Efficiency = P/ = 2000/6663.8 = 0.3 or 30% SELF ASSESSMENT EXERCISE No. 4 1. Boiling water at 10 bar is throttled to 1 bar. What is the temperature and dryness fraction

after throttling? (99.6oC and 0.153) 2. A gas turbine set draws in air at 20oC and exhausts at 500oC. The fuel supplies 48 MW

of heat. The power produced is 12 MW. Neglect all other losses and the mass of the fuel burned and take the specific heat as 1.01 kJ/kg K. Determine the thermal efficiency and the mass flow of the air. (25% and 74.25 kg/s)


WORKED EXAMPLE No.1SOLUTIONWORKED EXAMPLE No. 2SOLUTIONSELF ASSESSMENT EXERCISE No. 1