unit 47: engineering plant technology technology/out2t2.pdfa hot water boiler produces 0. 4 kg/s of...

© D.J.Dunn freestudy.co.uk 1

UNIT 47: Engineering Plant Technology

Unit code: F/601/1433 QCF level: 5 Credit value: 15

OUTCOME 2

TUTORIAL 2 – STEADY FLOW PLANT

2 Be able to apply the steady flow energy equation (SFEE) to plant and equipment

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; pressure-flow energy; internal energy; enthalpy

Application of SFEE to plant: assumptions made in specific applications; energy transfer

and efficiency calculations for specific items of plant e.g. economisers, boilers, super-

heaters, 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/

http://www.freestudy.co.uk/

http://www.spiraxsarco.com/learn/


© D.J.Dunn www.freestudy.co.uk 2

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.


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. Φ + 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: = ΔH/s = m (Δh)

Δh is the change in specific enthalpy of the fluid. Φ 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 + hfg

For superheated steam h values

For wet steam with dryness fraction x h = hf + xhfg

For low pressure water h = cw θ where cw is the specific heat (about 4.2 kJ/kg K) and θ 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 Δh = cp Δθ 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

Temp oC Pressure in bar

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:

Φ = mass/s x calorific value - solid fuels

Φ =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/m3

Other 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 18

oC.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 100

oC.

The boiler burns 5.3 m3/min of natural gas with a calorific value of 38 MJ/ m

3. 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/m

3) = 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 10

oC. 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 120

oC. The

boiler burns 17 m3/min of natural gas with a calorific value of 38 MJ/ m

3. Calculate the

thermal efficiency of the boiler. (80.8%)

ECONOMISER

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 10°C

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 100°C 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.



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 = mwΔh = 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%


Measurements for the boiler and economiser described in example 1 show that the flue

gas is cooled from 250oC to 150

oC 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 mgΔh = mg cpΔT = 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


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 Δh

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.


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

Φ = mΔh = 34(3158 - 2726.85) = 14659 kW or 14.659 MW


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.

Φ = mg cp Δθ = mg 1.1(700) = 770 mg kW

Φ = 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:

Φ + 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 = ΔH = m(Δh)

P is the power given up by the steam and Δh 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.


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 = msΔh = 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:

Φ + 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 = ΔH = m (Δh)

Φ is the heat transfer rate from the steam and water. Δh is the change in enthalpy of the

steam and water.


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(Δh) = 55(1928 - 163) = 97.1 MW

If all this is transferred to the cooling water then the we can equate :

mw cw ΔT = mw 4.2 (20) = 97100 kW

kg/s 1156(4.2)(20)

97100mw


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

Φ + 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 = ΔH = m (Δh)

Remember that Δh = Δu + Δ(pV). If the water is incompressible, the temperature is not

raised and the volume does not change so Δu = 0 and Δ(pV) = VΔp. The work (power)

transferred from the fluid is then:

P = V Δp

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 Actual

Power Idealη


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 Δp

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



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 Δp

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 θ = 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:

Φ + P = ΔH/s = mΔh and Φ = 0 so P = mΔh

640.588= 55 Δh Hence Δh = 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:

θ = 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/m

3.

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:

Φ + P = (PE)/s + (KE)/s + (H)/s

We can normally neglect Potential Energy and Kinetic Energy so this reduces to:

Φ + P = (H)/s = m cp Δ(θ)

The Heat loss Φ 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.



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

Φ = mwcwΔθ = 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 ΔH = macpΔθ

The mass flow rate of air = volume/density = 3.41/1.205 = 2.83 kg/min or 0.0471 kg/s

ΔH = macpΔθ = (0.0471)(1.005)(80) = 3.79 kW

Apply the SFEE Φ + P = (H)/s = m cp Δ(θ)

-1.764 + P = 3.79 kW

P = 5.55 kW This is the theoretical power input.

Efficiency η = 5.55/7.1 = 0.78 or 78%


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

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:

Φ + 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 Δh/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 + p2V2

THROTTLING A LIQUID

If the liquid does not evaporate the volume V is near constant so U1 - U2 = V Δp

The internal energy of a liquid is m c θ where c is the specific heat and θ the temperature.

m c (θ1 – θ2) = V Δp

c ρ

Δp

c m

ΔpV θθ 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).


THROTTLING A PERFECT GAS

If the fluid is a gas, then h2 – h1 = mcpΔθ = 0

It follows that Δθ = 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.


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.


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.

Φ + P = (PE)/s + (KE)/s + (H)/s

Φ is the heat input from the fuel if no other source or losses exist.

P = Net Power output

ΔH = enthalpy change of the air.

Neglecting PE and KE we have:

Φ + P = (H)/s

Note that P = -2000 kW because the power is removed from the system.

In KW units Φ - 2 000 = mcpΔθ = (11.84)(1.01)(400 – 10) = 4663.8

Heat Input Φ = 6663.8 kW

Efficiency η = P/Φ = 2000/6663.8 = 0.3 or 30%


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 500

oC. 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)

unit 47: engineering plant technology technology/out2t2.pdfa hot water boiler produces 0. 4 kg/s of...

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