1. A variable speed fan is used to supply the air for the ventilation system. The characteristics for the fan, in terms of total pressure, for a speed of 1000 r.p.m.,

(i) Calculate the speed at which the fan should operate in order to deliver the required volume flow rates.

(ii) Calculate the power required by the fan.

2. List any assumptions made during your calculations. Is there any action you would take to improve the design of this system?

## Calculating Required Fan Speed and Power

Firstly, we use the continuity equation which is based on mass conservation to determine the flow rates in each ducting:

The flow rate Q= Uab xAab

But since it is a circular one, the duct size (diameter) is obtained using the formula:

d^{2 }= 4Qab/πUab…..(1)

d= (4x2.5/πx5)^{0.5}= 0.7979 approximately d= 0.8m

Then we can now use the chart in Appendix from the course book, we read off the velocity and duct size by considering the flow rate and pressure drop.

But firstly we will establish the common pressure drop in the entire system that will be used as a basis for finding the missing parameters at every designated point within the system:

And this is found as: P/l = 0.35

And then we can turn back to the individual ducts to size them using the mentioned chart hence table 1 provides the approximated values for velocity and duct size.

Table 1: Duct Sizing

DUCT |
FLOW RATE Q(m |
VELOCITY U(m/s) |
Duct size (m) |

H-7 |
0.4 |
3.2 |
0.4 |

H-6 |
0.4 |
3.2 |
0.4 |

B-H |
0.4+0.4+0.4=1.2 |
4.1 |
0.6 |

E-3 |
0.3 |
3.1 |
0.35 |

E-4 |
0.4 |
3.2 |
0.4 |

D-E |
0.4+0.3=0.7 |
3.8 |
0.5 |

C-D |
0.3+0.7=1.0 |
4.0 |
0.58 |

C-1 |
0.3 |
3.0 |
0.35 |

B-C |
0.3+1.0=1.3 |
4.2 |
0.6 |

S |
1.3+1.2=2.5 |
5.0 |
0.8 |

F2-S1(a) |
2.5 |
5.0 |
0.8 |

F1-S1(b) |
2.5 |
5.0 |
0.8 |

A-S1(a) |
2.5 |
5.0 |
0.8 |

We assume the following for the above case:

- The maximum velocity in main duct AB is fixed at 5m/s
- The duct sizes of F2-S1, F1-S1, A-S1, S2(a)-B are diametrically equal
- The pressure drops in the silencers and fan are balanced by the frictional factor hence in the calculation, pressure drops for the silencers have not been considered
- To improve performance and ensure that the system operates optimally, we will need to include dampers to provide effective control
- There are no flow losses in the silencer and fan

Balancing the system Pressure drop:

Considering duct B-H:

The equivalent length, Le= 3+2+2+2+15+1.5=25.5m

Pb-h= Le(P/Le)=25.5 x 0.35= 8.925Pa

Now, considering duct B-E

Equivalent length Le= 2+1.5+2+1.5+2+1.5+2= 12.5

Pb-e= 12.5 x 0.35= 4.375Pa

It is clear that B-H has higher losses than B-E

Hence a damper is added to equalize the pressures between duct branches B-H and B-E

(Damper) Pb-e= Pb-h –Pb-e = 8.925- 4.375= 4.55 Pa

In this question, we consider a variable speed fan which is used to supply the air for the ventilation system. The characteristics for the fan, in terms of total pressure, for a speed of 1000 r.p.m., are given in Table 1 below.

Table 2: Fan characteristics at 1000rpm |
||

Flow rate |
Total Fan Pressure |
Fan Efficiency (%) |

Q |
Pt |
n |

0.5 |
110 |
40 |

1.0 |
115 |
60 |

1.5 |
115 |
75 |

2.0 |
107 |
82 |

2.5 |
95 |
82 |

3.0 |
75 |
75 |

3.5 |
55 |
62 |

4.0 |
35 |
40 |

Hence the following are determined from the given data:

- Calculating the speed at which the fan should operate in order to deliver the required volume flow rates.

The system fan supplies air to six outlet ducts as shown in the illustrated schematic. The total volume flow rate in the given system is computed using continuity equation hence:

Q = 0.4+0.4+0.4+0.7+0.3+0.3 = 2.5 m^{3}/s

Next, we calculate speed and power by considering the total pressure drops in the system:

?P_{t }= ρgh_{L}

_{ }= ρg [Frictional loss + Specific loss]

= ρg [ ]

Where:

ρ = density of air = 1.225 kg/m^{3}

g = acceleration due to gravity = 9.81 m/s^{2}

h_{L} = head loss, m

u = velocity, m/s

d_{e}= equivalent diameter

K = loss coefficient

We use chart 1 from the appendix course book to determine the pressure loss due to friction is determined from chart1 and pressure loss due to specific losses is calculated by finding relevant K value from CIBSE Guide tables and substituting in the above formula.

## Assumptions Made During Calculations

This is done by considering each section pressure loss in the system:

- Pressure loss from point (B) to E-4:

(i) Pressure loss at exit of silencer 2:

Given,

A1 = π/4d^{2}= 3.142/4 x 0.6^{2}= 0.2828m^{2}

A2 = 3.142x0.5^{2}= 0.7855m^{2}

A1/A2= 0.2828/0.7855 = 0.3600

θ1 = 45°, θ2 = 45°

θ = θ1 + θ2 = 90°

From CIBSE Guide C Table C4.80 (SMACNA):

For A2/A1 = 0.36 & θ = 180°; Loss Coefficient K = 0.135

Pressure loss from point at the entrance of duct (B) =

= 1.225x4.1^{2}/2 x 0.135= 1.389Pa

(ii) Pressure loss in duct B to C:

Given K = 0.45

Pressure loss =

= 1.225x4.2^{2}/2 x 0.45= 4.862Pa

(iii) Pressure loss in Silencer 2:

We check its velocity given the average pressure drop, u=3.0m/s

Pressure loss = (2+3+4+6+8+12)/6= 5.833

(iv) Pressure loss in Silencer 1:

We assume the same pressure drop as silencer 2 hence:

Pressure loss= 5.833

(v) Pressure loss in Fan:

Length: L = 1+ 1.5 = 2.5 m

For (Q)= 2.5m^{3}/s & u = 5.0 m/s;

Pressure loss due to friction is given by:

P/L = 1.3 Pa/m (from chart1)

P = 1.3 x 0.35= 0.455Pa

- Pressure loss in duct outlets:

Effective length, l= 1m and Q= 0.3m^{3}/s, u= 3.0m/s

P=1 x 0.35= 0.35Pa

Effective length, l= 1m and Q= 0.3m^{3}/s, u= 3.0m/s

P=1 x 0.35= 0.35Pa

Effective length, l= 1m and Q= 0.3m^{3}/s, u= 3.0m/s

P=1 x 0.35= 0.35Pa

- Pressure loss in Duct branch B-H:
- Pressure loss in the straight duct B to the corner:

Effective length, le = 3+2+2+2+1.5+1.5= 12m

For velocity u = 4.1 m/s;

P = 0.35x 12= 4.2Pa (from given data)

(ii) Pressure loss at the corner

Length: L = 3m, Q = 1.2 m^{3}/s & u = 4.1 m/s

P/L = 1.3 Pa/m (from chart1)

P = 1.3 x 0.35x3 = 1.365 Pa

- Pressure loss at the Junction to 5:

Length: L = 11m, Q = 1.2 m^{3}/s & u = 4.1 m/s

P/L = 1.3 Pa/m (from chart1)

P = 1.3 x 0.35x 11 = 5.005 Pa

- Pressure loss at the outlets:

Outlet 5:

Effective length, l= 1.5m and Q= 0.4m^{3}/s,

P=1 x 0.35x1.5= 0.525Pa

Effective length, l= 1.5m and Q= 0.4m^{3}/s,

P=1 x 0.35x1.5= 0.525Pa

Effective length, l= 1m and Q= 0.3m^{3}/s, u= 3.0m/s

P=1 x 0.35= 0.35Pa

Total Pressure loss in the system = 1.389+4.862+5.833+5.833+ 0.455+0.35x3+4.2+1.365+5.005+ (0.525x3)

= 31.567 Pa

To find the speed & power of the fan to meet the design conditions

We have P_{t} α Q^{2}

P_{t} = K Q^{2}

Where K is the system resistance

K = P_{t} / Q^{2} = 31.567/ (1.5^{2}) =14.0298

Table 3: Fan characteristics at 1000rpm |
|||

Flow rate Q |
Total Fan Pressure |
Fan Efficiency (%) |
System Pressure |

0 |
0 |
0 |
0 |

0.5 |
110 |
40 |
3.50745 |

1.0 |
115 |
60 |
14.0298 |

1.5 |
115 |
75 |
21.0447 |

2.0 |
107 |
82 |
28.0596 |

2.5 |
95 |
82 |
35.0745 |

3.0 |
75 |
75 |
42.0894 |

3.5 |
55 |
62 |
49.1043 |

4.0 |
35 |
40 |
56.1192 |

## Potential Improvements to the Design

By plotting system characteristics on the same graph as fan characteristics and we use it to find the operating point in which we look at the intersections hence Q= 2.75m^{3}/s (Kirby, 2018)

And the pressure loss?P = 15Pa

And efficiency η = 80% ; speed of fan, N = 1000 rpm

For the fan to deliver a volume flow rate of 2.5m^{3}/s in the system then the speed of fan is given by:

2.75/1000= 2.5/N2

N_{2 }= 909.09rpm

- Calculating the power required by the fan.

Now for the speed of fan N_{2 }= 909.09 the pressure drop across the fan is given by the relation:

75/1000^{2}= P2/909.09^{2}

P_{2}= 75x (909.09^{2}/1000^{2})= 61.9833

To find the fan power (P) we have the following relationship:

P= QP2/n= 2.75x61.9833/0.8 = 213.0678 W

P= 0.2130678kW

Assumptions made during calculations.

Frictional factor in the system was assumed to be balanced by the silencer and fan pressure drops

The fan system was assumed to be operating at a constant temperature

Losses due to leakage were never considered in the calculation

Action taken to improve the design of the system:

Minimizing losses by ensuring a more effective matching between system characteristic and fan characteristics

Deployment of energy efficiency strategies such as renewable energy use like solar PV in place of generators for powering

Installing a variable ventilation system such that during demand period, the performance is self-excited to peak and declines in less demanding situations within the day; one way is to have an automatic air inlet system that self-regulates the air inflow

A properly instituted preventive maintenance program need to be established such that all the components within the system are regularly inspected for reliability hence otherwise referred to as reliability-centered maintenance.

A huge chunk of costs normally comes from running of the plant; it is crucial that continuous energy audits be performed to ascertain the energy cost vis-à-vis performance in the building

Although the design of the ducts has been done, the design needs to be optimized so that transmission losses are minimized. For Example, we can check the number of bends and let them be brought down to a minimum number that optimizes the ventilation performance

Furthermore, the HVAC system engineer or technician will have to monitor and assess the performance of the system in and around all the designated rooms; in case, the performance is poor, he/she must undertake the system redesign. Common cause for this, according to Winter (2009), is often the presence of unsealed ducts which then attracts losses along the distribution channel.

References

Kirby ,R. (2018). Building Services Engineering Programme. Revised : Dr C König. Department of Mechanical Engineering: Brunel University London.

Winter, S. (2009). ImprovingVentilation in Existing or New Buildings with Central Roof Exhaust. National Center for Healthy Housing.

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