University HVAC Systems: Numerical Solution and Design Analysis

This document presents a comprehensive numerical solution for an HVAC system design assignment. It begins by calculating the net volume flow rate and determining the total pressure and efficiency of the system. The solution then delves into detailed calculations of pressure drops across various components, including acoustic attenuators, inlets, gradual expansions, bends, elbows, fire dampers, and grill meshes. The calculations utilize the Darcy-Weisbach equation and K-factors from the CIBSE guide to determine head losses due to geometry and friction. The solution includes calculations for fan power required and the necessary fan speed. The assignment also addresses design considerations, such as the use of dampers for pressure regulation and the importance of vanes in bends and elbows to minimize pressure drops and turbulence. References to relevant resources are also included.

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Solutions to HVAC Assignment
Name of Student:
Name of University:
Author’s Note:

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Answer to Q1: Part a)
By the given data,
Net volume flow rate from ducts A, B, C and D = 0.5+1+0.5+1 = 3 m3/s
By table Q1, total pressure = 660 Pa and Efficiency = 85%
We know that, Total pressure = Static pressure + Dynamic Pressure ( or velocity pressure)
Face velocity = Volume flow rate/ Face area
Therefore, by duct dimensions, Face Area = 0.4 x 0.8 = 0.32 m2
Face velocity = 3/(0.32) = 9.375 m3/s
According to the tables given, drop in pressure due to acoustic attenuators A and B are, 39 Pa
and 47 Pa respectively. Loss in pressure head for A will be = 39/g and for Loss of head for
B = 47/g, where g is acceleration due to gravity. Therefore head loss for A and B are 3.25m
and 3.92m of air respectively.
We must now determine the loss of head at the different parts of the duct using the K or loss
coefficient determined from the index provided in the CIBSE guide. We then add them all
together to get net loss in head due to geometry. The general equation for calculating loss of
head is, Hloss = K (V2 /2g), where V is the face velocity.
Using the Darcy-Weisbach equation for calculating loss of head, we calculate the loss of head
due to geometry first and then add it with loss of head due to friction for simplicity.
Head lost at inlet HI = 0.5 Vi2/2g = 0.56 m
Head lost at gradual Hg = 0.04 vf2/2g = 0.179m
Head lost at acoustic attenuator A and B 3.25m and 3.92m respectively.

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Head lost at division Hd = 0.09 vf2/2g = 0.403m
Branch (3-6-7):
1. Head lost at right angled elbow with vanes Hev= 0.35 vf2/2g = 1.569 m
2. Head lost at fire damper HFD= 0.2 vf2/2g = 0.897m
3. Head lost at right angled elbow duct C = 3.7 vc2/2g = 5.814m
4. Head lost at exit C = vc2/2g = 1.57m
5. Head lost at exit D = vD2/2g = 1.992m
6. Head lost at grill mesh = 0.4 vf2/2g = 0.63m
7. Head lost at grill mesh = 0.4 vf2/2g = 1.79m
Branch (3-4-5):
1. Head lost at right angled sweep bend Hsb = 0.67 vf2/2g = 3.004m
2. Head lost at right angled sweep with 1 turning vane Htv = 0.23vf2/2g = 1.031 m
3. Head lost at fire damper Hfd= 0.2vf2/2g = 0.897m
4. Head lost at exit A = vA2/2g = 1.57 m
5. Head lost at exit B = vB2/2g = 1.992 m
6. Head lost at grill mesh = 0.4 vf2/2g = 0.63m
7. Head lost at grill mesh = 0.4 vf2/2g = 1.79m
Therefore, net loss of head due to friction can be calculated using the hydraulic diameter of
the rectangular duct and calculating the coefficient of friction from the Moody Chart.
Therefore net friction loss will occur by eq. (i) hf = (f L V2)/ 2gD , and different duct sections
will have different hydraulic diameters. 4 Sections of the system have different cross-sections
and therefore will have a different equivalent diameter and coefficient of friction, therefore
total head calculated by equation (i).

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hf-total = 0.0022+1.39+6.52+0.496 = 8.408 m
Therefore total static head loss Htot = 8.408 + 25.176 = 33.584m
Therefore total pressure lost = Htot g = 403.18 Pa , and dynamic pressure = 256.82 Pa.
Power x Efficiency = 3 x 256.82
or, P x 0.85= 770.46 W , or, P= 906.42 W ( Power Required By the Fan)
By performance laws,since ductwork requires flow rate of 6.55m3/s required fan speed will
be,
N1= Q2/Q1 x N1 = 6.55/3 x 1200rpm = 2620 rpm (Speed of fan)
Part b)
Dampers are used to regulate the pressure drop along the system and so in this system it is
preferable to add dampers in the ducts A and C to regulate pressure drop with blade angles
adjusted to 30° for appropriate damping, by the performance data provided by Solid
Air(2016).
Part c)
A good design includes vanes in bends and elbows to reduce unwanted turbulence that
encourage pressure drops. In the given design the sweep branch bend at (3) incurs excessive
pressure drop due to absence of vanes. The 90° elbow at for both A and C ducts are too sharp
and should have a more gradual slope towards the branching for lower loss factors.

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References
Solid Air.(2016). Ch 9. https://solid-air.com/_files/file/linked_en/catalogi/solid-air-en-
volume-control-dampers-chapter-9.pdf