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3. a large hotel with over 1000 rooms has a distributed...

# Question: a large hotel with over 1000 rooms has a distributed...

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A large hotel with over 1,000 rooms has a distributed air conditioning system with a centralized chilled water cooling design as shown in Figure 1.

Figure 1 – Chilled water system supplying to guest rooms

The chilled water coils are located in the guest room to cool down air in the room with a fan coil arrangement as shown in Figure 2.

Figure 2 – Guest room air handling unit

Thermo interactions between the room’s internal air volume, conduction from outside, heat generated within the room and heat taken out by the chilled water system can be represented in Figure 3.

Figure 3 – Schematic of thermo interactions in guest room

The heat from outside conducts into the room through the double layer room wall.  The room size is standardized to be 6 m by 8 m (internal dimension) with a wall height of 2.5 m.  The wall is made of hollow concrete panels with panel thickness of 100 mm (0.1 m) and a gap of 500 mm (0.5 m) in between two panels.  The ceiling is made with the same type of double layer panels.  Note that no heat is conducted through the ground.

With air circulation, the heat taken out of the room will reduce the temperature of the air in the room.  The rate of temperature dropping is proportional to the net heat flow:

${H}_{r}={C}_{p}{M}_{r}\left(-\frac{d{T}_{r}}{dt}\right)$

where

• Mr = Mass of air in the room (kg)
• Cp = Specific heat of air at constant pressure = 1,000 J/kg/oC
• Note that density of air at atmospheric pressure = 1.204 kg/m3

The chilled water removes heat by circulating cold water through a cooling coil which has air circulating around it as shown in Figure 4.

Figure 4 – Schematic of chilled water cooling coil and circulating air

The chilled water is controlled by a water valve that opens and closes instantly.  The amount of water flowing into the room’s chilled water system is a pre-set constant Q (m3/s).

The amount of heat absorbed by the chilled water is then proportional to the amount of chilled water through the cooling coil as well as temperature difference between the room and the coil.  The heat absorbed by the chilled water is given by:

${H}_{w}=Q\rho {C}_{w}\mathrm{\Delta }{T}_{w}$

where

• Q = Actual volume flow rate of water through the cooling coil (m3/s)
• r = Density of water = 1,000 kg/m3
• Cw = Specific heat of water = 4,184 (J/kg/K)
• $\mathrm{\Delta }{T}_{w}={T}_{c}-{T}_{in}$
• Tin = Temperature of water from chiller (oC)
• Tc = Temperature of water in the cooling coil (oC)

Due to the time required to conduct heat into the cooling cool water, the chilled water temperature in the cooling coil and hence flowing out of the room is governed by the total amount of heat at the time of valve opening:

$\frac{d{T}_{c}}{dt}{V}_{t}\rho {C}_{w}=Q\rho {C}_{w}\left({T}_{in}-{T}_{c}\right)$

where

• Vt = volume of water in the cooling coil (m3)

The heat generated in the room can be represented by three sources:

• Two guests in the room at 100 W per person.
• Water kettle boiling water at 1000 W, average 5 minutes per hour
• Lights in the room total 200 W, average half time use over the duration of analysis

The chilled water is pumped into the cooling coil system and hence requires power from the boost pump.  The power of boost pump to boost chilled water is proportional to the total amount of water through the cooling coil system.

The duration of analysis is 1.5 hours.

Operating conditions of the system are:

• Initial room air temperature = 30 oC
• Initial air temperature in panel gap = 30 oC
• Initial water temperature in chilled water cooling coil = 30 oC
• Outside temperature = 30 oC
• Chilled water temperature flowing into the room’s cooling coil Tin = 8 oC
• Desirable room air temperature to be cooled down = 25 oC

Ideally, the system design should:

• Consume low pump power – indicated by the total amount of chilled water used
• Does not exceed an average temperature variation of 1 oC during the duration of analysis

Which system design parameter set would you select?

Formulae applicable to this question:

The double layer wall provides further insulation to outside heat.  Figure 5 shows the arrangement with annotated symbols.

Figure 5 – Double layer wall heat flow and temperature variations

Heat conducted through panel walls to the room is given by:

${H}_{o}={c}_{r}{S}_{r}\frac{{T}_{g}-{T}_{r}}{{\delta }_{r}}$

where

• Sr = surface area of room walls and ceiling (m2)
• ${\delta }_{r}$ = wall thickness (m)
• cr = thermal conductivity of wall material = 1.6 W/m oC

Behaviour of air in the room is governed by the perfect gas law:

${P}_{0}=\frac{{M}_{r}}{V}R{T}_{r}$

where

• P0 = Pressure of air in room = 1 atmospheric pressure = 101,325 N/m2
• Tr = Temperature of air in room (K)
• Mr = Mass of air in room (kg)
• R = Air gas constant = 286 (J/kg/K)
• V = Volume of air in room (m3)

Heat conducted from outside of room through the outside panel is given by:

${H}_{e}={c}_{r}{S}_{o}\frac{{T}_{o}-{T}_{g}}{{\delta }_{r}}$

where

So = surface area of room walls and ceiling (m2)

Behaviour of air in the gap is governed by the perfect gas law:

${P}_{0}=\frac{{M}_{g}}{{V}_{g}}R{T}_{g}$

where

• P0 = Pressure of air in room = 1 atmospheric pressure = 101,325 N/m2
• Tg = Temperature of air in room (K)
• Mg = Mass of air in gap (kg)
• R = Air gas constant = 286 (J/kg/K)
• Vg = Volume of air in room (m3)

Hence, balancing heat gained in the gap air volume:

$\frac{d{T}_{g}}{dt}=\frac{{H}_{e}-{H}_{o}}{{C}_{p}{M}_{g}}$

Heat removed by conducting through the cooling coil:

${H}_{w}={c}_{r}{S}_{t}\frac{{T}_{r}-{T}_{c}}{{\delta }_{t}}$

where

• St = surface area of cooling coil exposed to circulating air (m2)
• ${\delta }_{t}$ = wall thickness of cooling coil (m)
• ct = thermal conductivity of cooling coil brass material = 110 W/m oC

The surface area of cooling cool is given by:

${S}_{t}=\pi {l}_{t}\left({d}_{t}+2{\delta }_{t}\right)$

where

• dt = internal diameter of cooling coil tube (m)
• lt = length of cooling coil exposed in room (m)

${\delta }_{r}$${\delta }_{r}$${\delta }_{r}$ = room wall panel thickness = 0.1 m

dg = gap between wall panels = 0.5 m

${\delta }_{t}$${\delta }_{t}$${\delta }_{t}$ = cooling coil brass tube wall thickness = 0.0015 m

dt = internal diameter of cooling coil tube = 0.05 m

lt = length of cooling coil tube = 2.0 m

Q = Maximum chilled water flow rate = 0.1 litre/min

${\delta }_{r}$${\delta }_{r}$ = room wall panel thickness = 0.1 m

dg = gap between wall panels = 0.5 m

${\delta }_{t}$${\delta }_{t}$ = cooling coil brass tube wall thickness = 0.0015 m

dt = internal diameter of cooling coil tube = 0.02 m

lt = length of cooling coil tube = 1.0 m

Q = Maximum chilled water flow rate = 0.1 litre/min

${\delta }_{r}$${\delta }_{r}$${\delta }_{r}$ = room wall panel thickness = 0.1 m

dg = gap between wall panels = 0.5 m

${\delta }_{t}$${\delta }_{t}$${\delta }_{t}$ = cooling coil brass tube wall thickness = 0.0015 m

dt = internal diameter of cooling coil tube = 0.004 m

lt = length of cooling coil tube = 0.3 m

Q = Maximum chilled water flow rate = 0.0005 litre/min

${\delta }_{r}$${\delta }_{r}$${\delta }_{r}$ = room wall panel thickness = 0.1 m

dg = gap between wall panels = 0.5 m

${\delta }_{t}$${\delta }_{t}$${\delta }_{t}$ = cooling coil brass tube wall thickness = 0.0015 m

dt = internal diameter of cooling coil tube = 0.004 m

lt = length of cooling coil tube = 0.2 m

Q = Maximum chilled water flow rate = 0.00045 litre/min

${\delta }_{r}$${\delta }_{r}$ = room wall panel thickness = 0.1 m

dg = gap between wall panels = 0.5 m

${\delta }_{t}$${\delta }_{t}$ = cooling coil brass tube wall thickness = 0.0015 m

dt = internal diameter of cooling coil tube = 0.004 m

lt = length of cooling coil tube = 1.0 m

Q = Maximum chilled water flow rate = 0.1 litre/min

${\delta }_{r}$${\delta }_{r}$ = room wall panel thickness = 0.1 m

dg = gap between wall panels = 0.5 m

${\delta }_{t}$${\delta }_{t}$ = cooling coil brass tube wall thickness = 0.0015 m

dt = internal diameter of cooling coil tube = 0.02 m

lt = length of cooling coil tube = 2.0 m

Q = Maximum chilled water flow rate = 0.1 litre/min