THERMO FLUIDS NQF LEVEL 3 OUTCOME 4 OPEN …

© D.J.Dunn freestudy.co.uk 1

THERMO – FLUIDS

NQF LEVEL 3

OUTCOME 4 – OPEN THERMODYNAMIC SYSTEMS

This is set at the British Edexcel National level NQF 3.

Learning Outcome 4 - On completion of this tutorial you should be able to quantify

energy transfer in open thermodynamic systems

You should be able to explain and solve problems involving:

The Steady Flow Energy Equation (SFEE)

Work and Power transfer for various steady flow process

Heat transfer and transfer rate

The change of enthalpy in various steady flow systems

Applications to gas turbines, rotary compressors and coolers

Contents

1. Introduction

2. Steady Flow Systems

The Steady Flow Energy Equation

Gas Turbines

Compressors

Axial Flow or Turbine Compressor

Other Rotary Forms

Reciprocating Compressor

Gas Coolers, Heaters and Heat Exchangers

3. Polytropic Work in Open Systems

Turbine Expansions

Adiabatic Expansion

Polytropic Expansion

Isothermal Compression

Polytropic Compression

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

1 Introduction

This tutorial covers expansions and compressions in open steady flow systems such as

turbines, axial flow compressors, fluid cooler and heaters. The major difference between

this and closed systems is the time element. Since we are dealing with fluid flowing

through the system then all energy has to be considered as applying to the mass per

second not the mass and so energy transfer becomes energy transfer rate. Heat transfer

rate is denoted (Greek capital Phi) and work transfer rate is P (for power) and these are

measured in Watts not Joules.

2. Steady Flow Systems (Open Systems)

The Steady Flow Energy Equation

The laws governing this type of system are as follows.

Fluid enters and leaves through the boundary at a steady rate.

Energy may be transferred into or out of the system.

A good example of this system is a gas turbine. Energy may be transferred out as a rate of

heat transfer or as a rate of work transfer P.

The fluid entering and leaving possesses:

potential energy (PE)

kinetic energy (KE)

Internal energy (U)

Flow Energy (pV)

You should already know that Enthalpy is the sum of the last two H = U + pV

The first law of thermodynamics becomes:

+ P = Nett change in energy of the fluid.

This is called the Steady Flow Energy Equation (S. F. E. E.)


The convention is that energy entering the system is positive and leaving is negative.

Putting in the possible energy forms we have:

Φ + P = (PE)/s + (KE)/s + (U)/s + Δ(pV)/s

Or Φ + P = (PE)/s + (KE)/s + (H)/s

The equation may be applied to any open system.

Note that the term means ‘change of’ and if the inlet is denoted point (1) and the outlet

point (2) the change is the difference between the values at (2) and (1).

For example H means (H2 - H1). Putting in the formulae for the energy forms we have:

Φ + P = g Δz + (Δv2)/2 + (U)/s + Δ(pV)/s

Or Φ + P = g Δz + Δ(v2)/2 + Δh

is the mass flow rate (kg/s).

z is the height of the fluid relative to some datum level (m).

v is velocity (m/s)

h is the specific enthalpy (J/kg)

In the case of gas h = m cp T so the equation becomes

Φ + P = g Δz + Δ(v2)/2 + cp ΔT

cp is the specific heat at constant pressure.

In the following examples, you will find that the change in potential and kinetic energy

are usually too small to matter. Let's see how it may be applied to a turbine and

compressor next.


Gas Turbines

A gas turbine uses hot high pressure gas and expands it (usually to atmosphere). In the

process the gas gets cooler and work is removed. The expansion process follows the

polytropic law pVn = C. The change in potential energy is always negligible. If the

turbine is only designed to rotate a shaft then the change in kinetic energy should be

negligible. In the case of aircraft engines the turbines may be used to produce a jet thrust

with the exhaust. In this case the change in kinetic energy should be considered. Normally

the significant energy change is due to the pressure (flow energy) and temperature

(internal energy). Both together are the enthalpy. The Steady flow energy equation

simplifies to:

Φ + P = ΔKE/s + ΔH/s

The diagrams show the basic construction of an axial flow turbine. The rotor has rows of

aerofoil blades that spin between other rows fixed to the casing. The flow of the gas

makes the rotor spin.

The diagram shows the graphic symbols for turbines with and without a jet thruster. Note

the narrow end is always the high pressure end.

WORKED EXAMPLE No. 1

A gas turbine uses 5 kg/s of hot air. It takes it high pressure air at 900oC and

exhausts it at 450oC. The turbine loses 20 kW of heat from the casing. Calculate the

theoretical power output given that cp = 1.005 kJ/kg K. Changes in kinetic and

potential energy are too small to matter.

SOLUTION

With no potential or kinetic energy change we have Φ + P = ΔH/s = cp ΔT

The heat transfer rate is Φ = -20 kW. (out is negative)

P = cp ΔT - Φ = 5 x 1.005 x (450 - 900) - (-20)

P = -2 261 - (-20) = -2 241 kW

The minus sign indicates that the power is leaving the turbine. Note that if this was a

steam turbine, you would look up the h values in the steam tables.



If the turbine in the last example has a jet thruster on the exhaust that produces a

velocity of 300 m/s, what is the power output from the shaft? Assume P.E. is

negligible and that the inlet velocity is negligible.

SOLUTION

If the K.E. is taken into account then:


Φ + P = ( /2)(v22 - v1

2) + cp ΔT

We need to take care with units as the kinetic energy will give watts not kW. The

heat transfer rate is Φ = -20 kW. (Out is negative)

-20 + P = (5/2)(3002 - 0)/1000 + 5 x 1.005 x (450 - 900)

-20 + P = 225 – 2261

P = 225 - 2261 + 20 = -2016 kW (Note all energy units are in kJ)

The minus sign indicates that the power is leaving the turbine.

Compressors

Axial Flow or Turbine Compressor

This is the reverse of the turbine. Air is drawn in by the aerofoil blades and compressed

by each row of blades in succession. The air gets warmer as it is compressed and comes

out hotter than at inlet. The compression process follows the polytropic law pVn = C. The

changes in kinetic and potential energy should normally be negligible but on aeroplanes,

the velocity at inlet could be very significant. A compressor requires power to drive it.

The steady flow energy equation reduces to:


The graphic symbol is shown with the inlet at the wide end.



An air compressor sucks in air at 15oC and delivers it 86

oC. Cooling removes 30 kW

from the air. Calculate the power consumption for a unit mass flow. Neglect K.E. and

P.E.

Take cp = 1.005 kJ/kg K and cv = 0.718 kJ/kg K.

SOLUTION

Calculate the change in enthalpy for 1 kg/s

Φ + P = ΔH/s = cp (T2 - T1) = 1 x 1.005 x (86 - 15) = 71.3 kW

P = 71.3 - (-30) = 101.3 kW

Other Rotary Forms

You might read up on these and other forms of compressors. Many like the vane type will

work in reverse as motors. In all cases we can apply the steady flow energy equation.


Reciprocating Compressor

A reciprocating air compressor might be arranged as shown. There are two sets of

cylinders (3) and (5) and coolers (4) and (8) as well as other things like an air receiver

(10). Inside the compressor cylinders, the gas is compressed as in a closed system with

the pressure and volume varying polytropically. Overall, the system is an open steady

flow system when the boundary shown is used. We can do an overall energy balance on it

by applying the steady flow energy equation.


A reciprocating air compressor shown above sucks in 3.5 kg/min at 1 bar and 10oC

and delivers it at 8 bar and 30oC. The total energy loss to the surroundings is 0.5 kW.

This is due to heat generated in the motor and heat removed by the coolers. The

velocity of the air at inlet and outlet is the same so K. E. may be ignored.

Calculate the theoretical power input to the motor. Take cp = 1005 J/kg K.

SOLUTION

First identify this as a steady flow system for which the equation is

+ P = K.E./s + P.E./s + H/s

For lack of further information we assume K.E. and PE to be negligible. The heat

transfer rate is -0.5 kW.

The enthalpy change for a gas is H = cp T

H = (3.5/60) x 1005 x (30 - 10) = 1172.5 W or 1.172 kW

P = H - = 1.172 - (-0.5) = 1.672 kW

This will be the power supplied to the motor as electricity.


SELF ASSESSMENT EXERCISE No. 1

1. A gas turbine is supplied with hot air and 980oC. The exhaust air is at 320

oC. Due to

good insulation the heat loss is negligible. The change in P.E. and K.E. is negligible.

Calculate the power produced for unit mass flow.

Take cp = 1.005 kJ/kg K

(663.3 kW)

If there was 10% loss due to heat radiation from the casing, what would the power

output be?

(596.97 kW)

2. 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

out of 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. (75.35 kW)


Gas Coolers, Heaters and Heat Exchangers

Coolers and heaters are examples of heat

exchanger. The simple cooler shown may be

used to cool gas with water in an air

compressor or to produce hot water from

exhaust gases of engines.

Large gas heat exchangers are used on gas turbine sets to transfer heat from the exhaust

gas to the incoming air.

In heat exchangers there is no work transferred in or out. K.E. and P.E. are not normally a

feature of such systems so the S.F.E.E. reduces to

= H/s

If there is no energy lost to the surroundings then the same heat transfer rate must occur

out of one fluid and into the other.


0.4 kg/s of air passing into the combustion chamber of a gas turbine is pre-heated

from 120oC to 290

oC in an exhaust gas heat exchanger.

Calculate the heat transfer rate into the air given cp = 1.005 kJ/kg K. Ignore kinetic

and potential energy and assume constant pressure.

SOLUTION

Since no work is done the steady flow energy equation gives us:

Φ = ΔH/s = cp ΔT = 0.4 x 1.005 x (290 - 120) = 68.34 kW


In an air compressor, 0.05 kg/s is cooled from 110oC to 55

oC in an intercooler.

Calculate the heat transfer rate from the air given cp = 1.005 kJ/kg K. Ignore kinetic

and potential energy and assume constant pressure.

SOLUTION

Since no work is done the steady flow energy equation gives us:

Φ = ΔH/s = cp ΔT = 0.05 x 1.005 x (55 - 110) = -2.764 kW (out of air)



1. 0.2 kg/s of gas is heated at constant pressure in a steady flow system from 10oC to

180oC. Calculate the heat transfer rate . (37.4 kW)

Cp = 1.1 kJ/kg K

2. Carbon Dioxide gas is to be heated from -20oC to 15

oC in a heater with an electric

element that passes 2.5 kW into it. Calculate the flow rate of gas.

Take cp = 0.815 kJ/kg K. (0.0876 kg/s)

Next we will take a deeper look at the turbines and compressors and see how to use the

polytropic law.


3. Polytropic Work in Open Systems

When gas flows through a turbine or steady flow

compressor, the overall change of volume and pressure

obeys the polytropic law pVn = C. The work laws derived for

closed systems do not apply to a steady flow system.

Without explanation, it can be shown that the work done is

the area between the curve and the pressure axis of the p - V

diagram.

For a polytropic change pVn = C so:

Substitute into:

Substitute:

Between the limits of p2 and p1 this becomes:

Note this does not work when n = 1 and this is covered in the next section.

For gas the pV = mRT so in terms of the mass we have:

This the work done for every kg of gas expanded or compressed. Remember that the

value of the polytropic index 'n' depends on the degree of cooling and heating taking

place during the process.


Turbine Expansions

Adiabatic Expansion

This is when there is no heat transfer and n is replaced with the adiabatic index γ. It

follows that:

From the steady flow energy equation that:

is the mass flow rate. From this it follows that:

Since it was shown that R = cp = cv then once again we have:

Polytropic Expansion

In practical turbine expansions there may be some heat loss and in this case the final

temperature is reduced and this happens when n > γ. The only way to solve the power

output is by using the work formula.


A gas turbine is supplied with hot air at 1 100oC and 6 bar. The air is expanded to 1

bar adiabatically. Calculate the power produced for a unit mass flow.


SOLUTION

First calculate the adiabatic index. γ = cp/cv = 1.005/0.718 = 1.4

Next calculate the temperature after expansion. It was shown previously that the

simplest way is with the combined gas and polytropic formula:

Note the temperatures must be absolute. Assume the pressures are absolute unless told

otherwise. In this case we know the pressures so:

Now calculate the change in enthalpy for 1 kg/s

ΔH/s = cp (T2 - T1) = 1 x 1.005 x (823 - 1373) = -552.75 kW

Because the process is adiabatic, there is no heat loss so P = 552.75 kW out of the

system.



Repeat the last example but this time the expansion has some heat loss so the

expansion has a polytropic index n = 1.5

SOLUTION

First calculate the temperature after expansion.


ΔH/s = m cp (T2 - T1) = 1 x 1.005 x (755.6 - 1373) = -620.5 (out of the system)

This time we need the work formula to find the power output.

R = cp - cv = 0.287 kJ/kg K

Now apply the steady flow energy equation.

Φ + P = ΔH/s = m cp (T2 - T1) = -620.5 kW

Φ = -620.5 + P = -620.5 -(- 531.6)

Φ = -620.5 + 531.6 = -88.9 kW (out of the system)

Note the power has been reduced by the heat loss.


Isothermal Compression

This is the very best we can do when we cool a gas as it is compressed and represents an

ideal compression. In reality we cannot achieve this degree of cooling.

If we managed to produce isothermal compression then n = 1. Try putting this in the work

formula and it makes no sense. This is because the derivation only applied to cases when

n ≠ 1. We had:

But V = C/p and substituting:

This is the special case in integration. Between the limits of p2 and p1 this becomes:

For gas the pV = RT so in terms of the mass we have:

T is the isothermal temperature.

Since the temperature is constant ΔH = 0 and it follows that Φ = -P

Polytropic Compression

In reality the compression is accompanied by some cooling so the process is polytropic

with an index n>1.


A centrifugal air compressor sucks in air at 1 bar and 15oC and delivers it 3 bar. The

polytropic index n is 1.25. Calculate the power consumption for a unit mass flow.


SOLUTION

First calculate the temperature after expansion.


ΔH/s = cp (T2 - T1) = 1 x 1.005 x (358.8 - 288) = 71.1 kW



R = cp- cv = 0.287 kJ/kg K


Φ + P = ΔH/s = m cp (T2 - T1) = 71.1 kW

Φ = 71.1 - P = 71.1 - 101.6

Φ = -30.4 kW (out of the system)


Repeat the last problem but this time it has isothermal compression.


SOLUTION

The temperature is constant so ΔH = 0


R = cp- cv = 0.287 kJ/kg K


Φ + P = ΔH/s = 0

Φ = -P

Φ = -90.8 kW (out of the system)



1. A turbine is supplied with hot gas at 980oC and 4.5 bar. It expands the gas to 0.9 bar

by the polytropic law pV1.5

= C. Calculate the following:

i. The final temperature. (732.8 K)

ii. The change in enthalpy. (-572.3 kW)

iii. The power produced for a unit mass flow. (-546.3 kW)

iv. The heat transfer rate for a unit mass flow. (-26 kW)

v. The mass flow rate required to produce an output power of 1 MW. (1.831kg/s)

Take cp = 1.1 kJ/kg K and cv = 0.75 kJ/kg K

2. Repeat the last problem but assume adiabatic expansion.

(750.8 K, -552.4 kW, 0 kW and 1.81 kg/s)

3. A rotary air compressor sucks in air at 0.95 bar and -5oC and compresses it to 5 bar by

the polytropic law pV1.2

= C. Calculate the following:

i. The final temperature. (353.5 K)

ii. The change in enthalpy. (85.8 kW)

iii. The power required for a unit mass flow. (147.2 kW)

iv. The heat transfer rate for a unit mass flow. (-61.3 kW)

v. The mass flow rate produced when the power input is 1 MW. (6.8kg/s)

Take cp = 1.005 kJ/kg K and cv = 0.718 kJ/kg K

4. Repeat the last problem but this time with isothermal compression.

(268 K, 0 kW, 127.7 kW, -127.7 kW and 7.829 kg/s)

THERMO FLUIDS NQF LEVEL 3 OUTCOME 4 OPEN …

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