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:
d2 = 4Qab/πUab…..(1)
d= (4×2.5/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(m3/s) |
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 |
|
S2a-B |
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:
Balancing the system Pressure drop:
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:
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 m3/s
Next, we calculate speed and power by considering the total pressure drops in the system:
?Pt = ρghL
= ρg [Frictional loss + Specific loss]
= ρg [ ]
Where:
ρ = density of air = 1.225 kg/m3
g = acceleration due to gravity = 9.81 m/s2
hL = head loss, m
u = velocity, m/s
de= 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.
This is done by considering each section pressure loss in the system:
(i) Pressure loss at exit of silencer 2:
Given,
A1 = π/4d2= 3.142/4 x 0.62= 0.2828m2
A2 = 3.142×0.52= 0.7855m2
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) =
= 1.225×4.12/2 x 0.135= 1.389Pa
(ii) Pressure loss in duct B to C:
Given K = 0.45
Pressure loss =
= 1.225×4.22/2 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)= 2.5m3/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
Effective length, l= 1m and Q= 0.3m3/s, u= 3.0m/s
P=1 x 0.35= 0.35Pa
Effective length, l= 1m and Q= 0.3m3/s, u= 3.0m/s
P=1 x 0.35= 0.35Pa
Effective length, l= 1m and Q= 0.3m3/s, u= 3.0m/s
P=1 x 0.35= 0.35Pa
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 m3/s & u = 4.1 m/s
P/L = 1.3 Pa/m (from chart1)
P = 1.3 x 0.35×3 = 1.365 Pa
Length: L = 11m, Q = 1.2 m3/s & u = 4.1 m/s
P/L = 1.3 Pa/m (from chart1)
P = 1.3 x 0.35x 11 = 5.005 Pa
Outlet 5:
Effective length, l= 1.5m and Q= 0.4m3/s,
P=1 x 0.35×1.5= 0.525Pa
Effective length, l= 1.5m and Q= 0.4m3/s,
P=1 x 0.35×1.5= 0.525Pa
Effective length, l= 1m and Q= 0.3m3/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+0.35×3+4.2+1.365+5.005+ (0.525×3)
= 31.567 Pa
To find the speed & power of the fan to meet the design conditions
We have Pt α Q2
Pt = K Q2
Where K is the system resistance
K = Pt / Q2 = 31.567/ (1.52) =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 |
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= 2.75m3/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 2.5m3/s in the system then the speed of fan is given by:
2.75/1000= 2.5/N2
N2 = 909.09rpm
Now for the speed of fan N2 = 909.09 the pressure drop across the fan is given by the relation:
75/10002= P2/909.092
P2= 75x (909.092/10002)= 61.9833
To find the fan power (P) we have the following relationship:
P= QP2/n= 2.75×61.9833/0.8 = 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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