University of Nottingham
Dept of Architecture and Built Environment
ABEE2033: Architectural Engineering Design 3
Autumn 2024/25 - Exam Solution
Student ID: [INSERT ID HERE]
Question 1
(a) Ventilation Strategy Selection
When debating between retaining a mechanical system or converting to a natural ventilation
strategy for the deep-plan lecture theatre in Edinburgh, the CIBSE ventilation strategy selection
process highlights several critical constraints that make natural ventilation unsuitable.
1. Heat Gains: The internal heat gains are stated as 100 W/m2 . Natural ventilation is generally limited to cooling loads of approximately 30–40 W/m2 . A load of 100 W/m2 is significantly beyond the capacity of air buoyancy and wind pressure to remove heat without
causing excessive drafts or requiring impractical opening sizes.
2. Building Geometry: The space is a ”deep-plan” theatre enclosed on three sides. Singlesided natural ventilation is effective only up to a depth of approximately 2.5×ceiling height,
and cross-ventilation up to 5 × ceiling height. A deep-plan structure often exceeds these
limits, leading to poor air distribution and stagnant zones.
3. External Environment: The theatre is located by a ”busy road.” Natural ventilation requires openings (windows/vents), which would allow the ingress of traffic noise and pollution (particulates/NOx). This conflicts with acoustic requirements for a lecture theatre and
Indoor Air Quality (IAQ) standards.
4. Control Requirements: The brief specifies that operative temperature and relative humidity (50%) must be ”tightly/closely controlled.” Natural ventilation is inherently variable
and weather-dependent; it cannot actively dehumidify air to maintain 50% RH, nor can it
guarantee tight temperature tolerances.
Conclusion: Based on the CIBSE selection process, a mechanical ventilation or air conditioning system is the only viable strategy to meet the thermal, acoustic, and humidity control
requirements given the high heat loads and external constraints.
(b) Thermal Comfort Model
Given the requirement for tight control in a mechanically conditioned environment, **Fanger’s
Heat Balance Model** (predicted Mean Vote - PMV) is the appropriate model. The Adaptive
Thermal Comfort model is generally reserved for naturally ventilated or free-running buildings.
Thermo-regulatory Process: The human body maintains a core temperature of approx.
37◦ C.
- Vasodilation: Blood vessels near the skin expand to increase blood flow and heat loss
when hot.
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- Vasoconstriction: Blood vessels contract to reduce blood flow and conserve heat when
cold.
- Sweating: Secretion of moisture for evaporative cooling.
- Shivering: Involuntary muscle activity to generate metabolic heat.
Parameters and Heat Transfer Processes:
1. Physical Parameter 1: Metabolic Rate (M ) - Internal heat generation.
2. Physical Parameter 2: Clothing Insulation (Icl ) - Resistance to heat loss (Conduction/Convection/Radiation).
3. Environmental Parameter 1: Air Temperature (Ta ) - Convection (heat transfer between
skin/clothing and surrounding air).
4. Environmental Parameter 2: Mean Radiant Temperature (Tr ) - Radiation (heat exchange
with surrounding surfaces).
5. Environmental Parameter 3: Air Velocity (v) - Convection (accelerates heat removal) and
Evaporation (sweat efficiency).
6. Environmental Parameter 4: Relative Humidity (RH) - Evaporation (limits the rate of
latent heat loss via sweating/respiration).
(c) Compressive Refrigeration Cycle
The vapour compression refrigeration cycle moves heat from a low-temperature reservoir (the
building interior) to a high-temperature reservoir (outside) using work input.
The Cycle Components:
1. Evaporator: Low-pressure liquid refrigerant absorbs heat from the indoor air (or chilled
water loop), boiling into a low-pressure gas. (Heat Absorption).
2. Compressor: The gas is compressed, significantly increasing its pressure and temperature. (Work Input).
3. Condenser: The hot, high-pressure gas rejects heat to the outside air (or cooling tower
water), condensing back into a high-pressure liquid. (Heat Rejection).
4. Expansion Valve: The high-pressure liquid passes through a restriction, dropping its pressure and temperature rapidly before entering the evaporator to repeat the cycle.
Performance Metric: The performance is quantified by the **Coefficient of Performance
(COP)**.
Cooling Output (kW)
COP =
Electrical Power Input (kW)
A higher COP indicates a more efficient system (e.g., a COP of 3 means 3 kW of cooling is provided for every 1 kW of electricity consumed).
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(d) Post Occupancy Evaluation (POE) Brief
Objective: To evaluate the actual performance of the lecture theatre refit against the design
intent and CIBSE benchmarks, focusing on occupant satisfaction and system efficacy.
Focus Areas:
1. Thermal Comfort: Is the ”tight control” actually delivering comfort?
2. Acoustics: Is the mechanical system noise intrusive? Does the facade block road noise
effectively?
3. Air Quality: Is the air perceived as fresh?
Research Question: ”Do the operative temperature and humidity levels in the lecture theatre result in a user satisfaction rate of > 80%, consistent with CIBSE Category I/II comfort criteria?”
Method/Metric: We will utilize a quantitative **Occupant Satisfaction Survey** (using a 7point Likert scale) combined with physical monitoring.
- Metric: Percentage of occupants voting between -1 and +1 on the PMV scale (or ”Satisfied”
on the survey).
- Physical Data: Simultaneous logging of Ta , RH, and CO2 to correlate physical conditions
with subjective responses.
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Question 2
(a) Comfort Cooling vs. Air Conditioning
Difference:
- Comfort Cooling: Primarily controls air temperature (sensible cooling). It generally cannot actively control humidity (though some incidental dehumidification occurs).
- Air Conditioning: Offers full control over the atmospheric environment, including temperature, humidity (both humidification and dehumidification), air velocity, and cleanliness.
System Identification (Figure Q2a): Referring to the description of Figure Q2a (”Added at
30°C Ratio for Water” and the process lines on the chart), the system involves specific moisture
changes. If the process shows a vertical line or a specific moisture addition/removal to hit a
target point (like 50% RH), it is an **Air Conditioning** system. The requirement to supply
10 L/s of outdoor air mixed with recirculation and condition it implies full treatment. The chart
likely shows mixing, cooling/dehumidifying (moving left and down), and potentially reheating
or humidifying. Given the context of ”office block” and typical charts showing mixing + cooling
coil lines, it is an Air Conditioning system because it actively manages the psychrometric state
(Enthalpy/Moisture) rather than just dropping the Dry Bulb temperature.
(b) Load Estimation
Note: As values from the chart must be read visually, standard calculation methods are presented
below with derived estimates.
Assumptions:
- Occupancy: 15 people.
- Outdoor Air Supply: 10 L/s/person = 150 L/s = 0.15 m3 /s.
- ρ = 1.2 kg/m3 .
- Mass flow rate of outdoor air ṁ = 0.15 × 1.2 = 0.18 kg/s.
Method 1: Enthalpy Difference (Total Load) Using enthalpy values (h) from the chart at the
Room condition (R) and Supply condition (S).
Qtotal = ṁ(hR − hS )
Separate into sensible and latent using the Sensible Heat Ratio (SHR) from the protractor.
Method 2: Mass Flow Equations (Specific Loads)
1. Sensible Cooling Load (Qs ):
Qs = ṁ · cp · (Troom − Tsupply )
where cp ≈ 1.00 kJ/kg K (or 1.02).
2. Latent Load (Ql ):
Ql = ṁ · hf g · (groom − gsupply )
where hf g = 2450 kJ/kg and g is moisture content in kg/kg.
3. Heating Load: If the cycle involves reheat (e.g., winter or humidity control), Qheat = ṁcp (Tsupply −
Tof f _coil ).
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(c) Pressure Loss and Regain
Data:
- Ptotal,1 = 200 Pa.
- Velocities: v1−2 = 10 m/s, v3−4 = 4.1 m/s.
- Friction losses: ∆Pf,1−2 = 25 Pa, ∆Pf,3−4 = 5 Pa.
- Expansion Loss Factor K = 0.59.
Calculations: 1. Dynamic Pressures (Pv ):
Pv = 0.5 × ρ × v 2
Pv1,2 = 0.5 × 1.2 × 102 = 60 Pa
Pv3,4 = 0.5 × 1.2 × 4.12 ≈ 0.6 × 16.81 = 10.09 ≈ 10.1 Pa
2. Expansion Loss (Point 2 to 3): Standard fitting loss equation: ∆Ploss = K · Pv,inlet
Loss = 0.59 × 60 = 35.4 Pa
3. Pressure Progression:
- Point 1: Pt = 200, Pv = 60. Ps = 200 − 60 = 140 Pa.
- Point 2: Friction loss 25 Pa. Pt = 200 − 25 = 175 Pa. Pv = 60. Ps = 175 − 60 = 115 Pa.
- Point 3 (After Expansion): Pt = Pt2 − Loss = 175 − 35.4 = 139.6 Pa. Pv = 10.1 Pa. Ps =
139.6 − 10.1 = 129.5 Pa.
- Point 4: Friction loss 5 Pa. Pt = 139.6 − 5 = 134.6 Pa. Pv = 10.1. Ps = 134.6 − 10.1 = 124.5 Pa.
4. Static Regain (3-4? Usually Regain is across expansion 2-3): The question asks for
regain between 3 and 4? No, friction occurs 3-4. Regain occurs across the expansion (2 to 3).
Correction based on question phrasing: ”calculate the static-pressure regain between points 3 and
4”. This is physically unusual as 3-4 is a duct section with friction (5 Pa loss), so static pressure
drops. However, assuming the question implies the regain achieved by the expansion (realized
at 3), or strictly asking for ∆Ps between 3 and 4. Ps4 − Ps3 = 124.5 − 129.5 = −5 Pa. If referring
to the expansion (2 to 3): Static Regain = Ps3 − Ps2 = 129.5 − 115 = 14.5 Pa. (We will list pressures
as calculated).
Table Q2c: System Air Pressures (Pa)
Pressure Point
Total
Static
1
200
140
2
175
115
3
140
130
4
135
125
Table 1: *
Values rounded to nearest integer or 2 s.f. consistent with exam instructions.
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(d) Mechanical Filters
Two Purposes:
1. Protect Equipment: Prevent dust accumulation on heating/cooling coils, fans, and ducts
which reduces efficiency and lifespan.
2. Protect Occupants: Remove harmful particulates (PM2.5, PM10), pollen, and pathogens
to ensure healthy Indoor Air Quality (IAQ).
Mechanisms:
- Straining: Particle is larger than the gap between fibers.
- Impaction: Heavy particles carried by inertia hit fibers as air streamlines bend.
- Interception: Particle passes within one radius of a fiber and adheres.
- Diffusion: Very small particles (Brownian motion) erratically hit fibers.
Efficiency vs Diameter: The efficiency curve is typically ”U” shaped (or V-shaped) with a
minimum efficiency at the Most Penetrating Particle Size (MPPS), typically around 0.1–0.3µm.
- Large particles (> 1µm): High efficiency (Impaction/Straining).
- Small particles (< 0.1µm): High efficiency (Diffusion).
- Medium particles: Lower efficiency.
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Question 3
(a) IAQ Airflow
Data: Floor 600 m2 , Volume 6000 m3 , Density 2 m2 /person.
- Occupancy: N = 600/2 = 300 people.
- Recommended Rate: CIBSE Guide A typically recommends 10 L/s/person for excellent IAQ
(or a minimum of 8 L/s).
- Total Airflow (Q):
Q = 300 × 10 L/s = 3000 L/s = 3.0 m3 /s
- Air Change Rate (ACH):
ACH =
- Turnover Time:
t=
Q × 3600
3.0 × 3600
=
= 1.8 ACH
V
6000
1
1
=
hours ≈ 33.3 minutes
ACH
1.8
Comment on Summer Dissipation: For summer cooling, the primary driver is heat removal
rather than pollutant dilution. The required air change rate would likely need to be significantly
higher (e.g., 5–10 ACH or more) to flush out solar and internal heat gains effectively.
(b) Radon Ventilation Rate
Data:
- Threshold (Climit ) = 200 Bq/m3 .
- Ambient (Co ) = 0 Bq/m3 .
- Source Area (A) = Floor Area = 600 m2 (Granite floor is the source).
- Emission Rate (E) = 0.02 Bq/s/Pa/m2 .
- Pressure Diff (∆P ) = 2 Pa.
Calculation: 1. Total Emission (S):
S = E × ∆P × A = 0.02 × 2 × 600 = 24 Bq/s
2. Steady State Balance Equation:
Climit = Co +
S
24
=⇒ 200 = 0 +
Q
Q
3. Minimum Ventilation Rate (Q):
Q=
24
= 0.12 m3 /s
200
Benchmark: This required rate (0.12 m3 /s) is significantly lower than the IAQ requirement
(3.0 m3 /s). Therefore, the ventilation design driven by occupant needs will essentially eliminate
the radon risk by dilution.
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Surface
Area (m2 )
Temp (◦ C)
A×T
Floor
Ceiling
Front
Rear
Left
Right
600
600
60
60
100
100
15
20
21
15
19
18
9000
12000
1260
900
1900
1800
Total
1520
26860
(c) Radiant and Operative Temperatures
(i) Mean Radiant Temperature (Tr ) Tr =
P
(Ai ·Ti )
P
Ai
26860
≈ 17.67◦ C
1520
(ii) Lowest Air Temperature (Ta ) Assumptions:
Tr =
- Target Operative Temperature (Tc ) for comfort in a church = 21◦ C (Standard assumption
for sedentary occupancy).
- Air velocity v = 0.2 m/s.
- For v < 0.2 m/s, Tc ≈ 12 Ta + 12 Tr . At 0.2 m/s, this approximation remains valid for estimation.
Calculation:
21 = 0.5Ta + 0.5(17.67)
42 = Ta + 17.67
Ta = 42 − 17.67 = 24.33◦ C
The air temperature must be raised to approximately **24◦ C** to compensate for the cold
stone surfaces.
(d) Estimating Heating System Size
To estimate the size (kW capacity) of the heating system, follow these steps:
1. Define Design Conditions: Establish the desired internal temperature (Tint ) and the winter external design temperature (Text ) for Cornwall (e.g., −2◦ C or −4◦ C).
2. Calculate Fabric Heat Loss (Qf ): Determine the U-values for all building elements (walls,
roof, floor, windows).
X
Qf =
(U × A × ∆T )
3. Calculate Ventilation Heat Loss (Qv ): Based on the air infiltration rate and mechanical
ventilation rate (N air changes per hour).
Qv = 0.33 × N × V × ∆T
4. Apply Margins:
- Intermittency Factor: Since the church is heated intermittently, add a significant
margin (e.g., 20–50%) to allow for rapid pre-heating of the high thermal mass (stone).
- Height Factor: Adjust for stratification in the high-ceiling space.
5. Total Plant Size: Sum the losses and margins to select the boiler capacity.
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