Parameters for the Steady Flow Energy Equation - Paper Example

To calculate the required parameters for the steady flow energy equation.

To determine the unknown Q in both magnitude and direction which is the heat transfer rate across the boundary in kW.

To identify the importance of kinetic energy, enthalpy and potential energy of the steady flow energy equation and how they contribute to Q as the total heat transfer value.

Total energy flow is found by the application of the steady flow energy equation. During the calculations of energy transfer into or out of the system, it is assumed that the flow of mass through the system is kept constant (Kroos and Potter, 2014). Another assumption is the balance between the energy into the system and the energy out of the system i.e. they are equal. The energies involved in the system are; kinetic energy, heat and work, potential energy, flow and internal energy.

According to Kroos and Potter (2014), in Fluid Dynamics, the Steady Flow Energy Equation is simply the mathematical equation that states that for a steady flow the mass flow rate into a control volume is equal to the mass flow rate out of the control volume. Since mass flow rate vector can be expressed as product of density r, volume flow rate vector V and area vector A perpendicular to the flow, SFEE gives the product of these integrated over the control surface to be equal to zero (Kroos and Potter, 2014).

In Thermodynamics, SFEE equation mathematically states that for a control volume in steady state the incoming energy is equal to outgoing energy (Kroos and Potter, 2014). By definition, heat (QQ) is always the incoming energy, work (WW) is done by the fluid is outgoing energy and difference in enthalpy (HH) and in kinetic energy (K=m(v2)/2K=m(v2)/2) which is outgoing, then

Q=W+H+K

The purpose of this experiment was to prove the steady flow energy equation using a system of air flow through a heated duct. The steady flow energy equation is shown below;

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Where:

u = internal energy (J)

v = volume (m3)

P = pressure (N/m2)

C = velocity (m/s)

Z = height above a datum (m)

W = work (J)

g = acceleration due to gravity (m/s2)

Q = heat flow (J)

P .v = displacement or flow energy.

C2 / 2 = kinetic energy.

g. Z = potential energy.

However, other important formulas play a part in finding the values of the above equation parameters, which are;

Air mass flow rate,[kg/s]

Where,[n/m2]

rw=1000 kg/m3Air inlet area,[m2]

Air exit area,[m2]

Velocity of air at inlet,[m/s]

Velocity of air at outlet,[m/s]

Power of fan wf = vf af[w]

Power of heater wh = vh ah[W]

Equipment provided

Measuring tape

Digital vernier calliper

Measurement procedures

The system was set as shown in the figure below and measurements were done in a precise manner repeatedly to obtain the best results. The system was given some time to stabilize before any measurements were taken or recorded.

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Using a thermometer, the temperature of the air at inlet t1 was measured and recorded in degrees Celsius (oC).

The air temperature at outlet t2 was also measured and recorded in degrees Celsius (oC)

The pressure difference across the orifice was also determined and recorded in cm water gauge.

The height of air inlet above datum line was also observed and the value recorded in the table in meters as Z2.

The height of the outlet above datum line was also observed and recorded as Z2 in meters in the records table.

The diameter of the circumferential air inlet was also determined through measurement using a Vanier calliper and recorded as d1 in meters on the table.

The width of the circumferential air inlet was also measured using a Vanier calliper and this value recorded on the table as W1.

The diameter of air outlet pipe was measured using a Vanier calliper and the value recorded as d2 in meters on the table.

Heater voltage was also determined by use of a voltmeter and recorded on the table in volts as VH.

Heater current then was measured using a voltmeter in current mode, and the reading was recorded on the table in amperes as AH.

Results and Discussion

Measurements (raw data)

Examination of air flow in a heated duct Test Observations

Air Temperature at inlet t1 [C] 19.6oC

Air Temperature at outlet t2 [C] 60.7 oC

Pressure difference across orifice Ho [cm water gauge] 0.048m

Height of air inlet above datum line Z1 [m] 0.60m

Height of air outlet above datum line Z2 [m] 1.485m

Diameter of circumferential air inlet d1 [m] 0.1302m

Width of circumferential air inlet w1[m] 0.0107m

Diameter of air outlet pipe d2 [m] 0.033m

Heater Voltage VH [volts] 145V

Heater Current AH [amperes] 3.0A

Fan Motor voltage VF [volts] 460V

Fan motor current AF [amperes] 2.05A

Analysis and results

Conservation of Energy (First Law)

dE = dQ-dW

where

rate of energy transfer to the system as heat

rate of work done by the system

For steady-state (VW, S & B: 6.3)

so

(units J/s)

or

Neglecting potential and chemical energy (PE and CE)

Where c is the speed of the fluid, and c2/2 is the kinetic energy of the fluid per unit mass relative to some coordinate system.

If we divide through by the mass flow and set the inlet of the control volume as station 1, and the outlet as station 2, then

It is also more convenient to divide the work into two terms: 1) the flow work done by the system which is p2v2-p1v1, and 2) any additional work which we will term external work or shaft work, ws. Then we have

or

This is the steady flow energy equation.

For an ideal gas dh = cp dT so

To elaborate and simplify our calculations, we will observe the system and make calculations according to the types of energies involved. According to our system, we have kinetic energy developed by the moving air, the potential energy, and the heat energy.

The system is doing work hence the convention of the work done by the system will be positive while Q (heat transfer) will be to the environment hence the direction of Q will be negative.

From the above equations and neglecting the w on the left-hand side of the equation because there is no additional work.

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Q =

T2 =60.7

T1 = 19.6

C1 = 53.56

C2 = 216.85

Cp = 1.005

Q = 1.005x60.7+216.8522-1.005x19.6+53.5622Q = 32889.68 J

Q = -32.9 Kj

Discussion

Our calculations have proved that there is a relationship between the work done by the system and the change in energy as well as the energy transferred from the system to the environment through the boundary. This supports the steady flow energy equation as we have shown above. The calculations also included the different forms of energies involved in this experiment and their vital roles played by them in the steady flow energy equation during our calculations. These energies are outlined below, and their importance explained;

According to Huilgo (2015), enthalpy is best beneficial for unraveling flow work from exterior work (as might be produced by a shaft crossing the control volume boundary for instance). In the figure shown below. Heat is added, a compressor is doing work on the system, the flow entering the system does work on the system (work = -p1V1), and work is done by the system through pushing out the flow (work = +p2V2) (Huilgo, 2015). Huilgo, also stated that the first law relates the change in energy between states 1 and 2 to the difference between the heat added and the work done by the system. Frequently, however, we are interested only in the work that crosses the system boundary, not the volumetric or flow work. In this case, it is most convenient to work with enthalpy.

This also leads to a direct physical interpretation for enthalpy. In an open flow system, enthalpy is the amount of energy that is transferred across a system boundary by a moving flow (Huilgo, 2015). This energy is composed of two parts: the internal energy of the fluid (u) and the flow work (pv) associated with pushing the mass of fluid across the system boundary.

Conclusion

It is evident from our analysis through this experiment that the steady flow energy equation is correct and that energy cannot be created nor destroyed. The transfer of energy from the system to the environment happens through convention because the system is a solid from the outside boundary and the environment is the air. The direction of the flow of this energy out of the system is negative. This experiment was conducted in an environment whose temperatures were lower that the temperatures of the system as on the boundary of the system hence the transfer. There are also different types of energy in this experiment hence the derivation of the overall formula to account for all the fundamental energies in this system and their contribution to the energy transfer.

References

Huilgo, R. R. (2015). Fluid mechanics of viscoelasticity. Heidelberg [Germany] : Springer.

Kroos, K. & Potter, M. C. (2014). Thermodynamics for engineers. Independence: Nelson Engineering.

Lienhard, J. H. (2011). A heat transfer textbook. Mineola, N.Y.: Dover Publications.