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BSAppsMechworksheetRevB.pptx

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BSApps

Mechanical Worksheet

Rev B

Summary

Workshop	Question	Pg No	Notes
M1	1 to 5 Answers	4-8 9	U values and Heat loss
M2	6 to 11 Answers	10-16 17	Heat gains
M3	12 to 13 Answers	18-21 22	Radiator Selection
M4	14 to 16 Answers	23-25 26	Flow rates
M5	17 to 18	27-41	System Pressure Drop
		42	Concluding thoughts

At this point I will be providing answers only.

Full working out will be provided later in the year.

Formula

Equation 1

Where:

QM = Mechanical load required to maintain a steady internal temperature (note that if this is positive then heating is required, negative means cooling)

Qc = Conduction Heat Loss

QI = Infiltration Heat Loss

QV = Ventilation Heat Loss

QOCC = Occupancy

QL = Lighting

QE = Equipment

QS = Solar

For heat loss only the equation can be substituted with:

Air:

x =1.2kg/m3

Water:

= 4.19 kg/m3

Workshop M1

Q 1-5

U value and heat gains

M1 : Q1-5

Typical U-Block Wall Build Up

A typical detail is shown opposite, which can be simplified to the following table:

M1 : Q1-5

Typical U-Block Wall Build Up

Calculate the U value for this element

Assuming all insulation (mineral wool and kooltherm) is replaced with straw.

What is the U value with the same thickness?

Straw λ = 0.06W/m.K

When replacing the insulation with straw. How thick does the wall need to be to achieve the same thermal performance?

M1 : Q1-5

Calculate the overall heat loss (in Watts) using the information below

Also make note of the fabric coefficient [] and Infiltration coefficient (NV/3).

External air temperature = -3°C

Internal temperature = 20°C

Room Dimensions:

Length = 20.9m

Width = 9.7m

Height = 5.5m

Glazed area = 35.7m2

Infiltration rate = 0.2 ACH

No internal gains, assume unoccupied.

The space has a floor below and above it which are at the same temperature.

Assume the wall to the south is the same length as the wall to the North (I.e. 9.7m)

U values

Glazing: 1.6 W/m2K

Wall : 0.25 W/m2K

Calculations:

M1 : Q1-5

What would be the result if there was a roof above the office?

External air temperature = -3°C

Internal temperature = 20°C

Room Dimensions:

Length = 20.9m

Width = 9.7m

U value of roof equivalent to minimum Building Regulations Part L:

Make a note of the following figures for later:

Revised coefficient of Fabric Heat Loss, Cf = ΣUA

Answers

- Rtotal = 7.52 m2.K/W

- U value = 0.133 W/m2.K

- Rtotal = 4.08 m2.K/W

- U value = 0.25 W/m2.K

- Insulation thickness = 416mm

- Overall Buildup of wall = 595mm

- = 102.7 W/K

- NV/3 = 74.3 W/K

- Heat loss = 4,070 W

- Heat Loss = 5,236 W

- = 153.4 W/K

Notes

Always provide units!

You may have slightly different answers, which if within 2-3% is likely to be from rounding rather than your calculations….but do double check.

If you are struggling with the answer, then come and ask me

Workshop M2

Q6 – 11

Heat Gains

M2 : Q6-11

Calculate the occupancy, lighting and equipment heat gain within the meeting room:

Use the tables in the following page

Make a suitable assumption about the heat output of people (W/person) (table 6.3)

Work out the occupant density of each space and calculate the number of people (table 6.2)

Work out the total heat gains from occupancy (based on above numbers) and lighting/equipment based on table 6.2

At what external temperature does the occupancy alone within the space balance the thermal losses?

Keeping an operative temperature of 20oC

Ignore all other internal gains

What occupancy heat gain would you use if this was an activity studio?

(take a look back at the previous tables)

Complete the solar gain exercise on the following pages

M2 : Q6-11

Determine the peak solar gain for the space during July and August when external ambient temperatures are likely to be hotter from the tables* included on the following page using the solar gain equation included in the lecture notes.

Façade glazing areas:

Glazed area to façade A = 10.2m2

Glazed area to façade B = 18.6m2

Glazed area to façade C = 6.9m2

Façade glazing properties:

Effective G-value = 0.4

*We are using data for London as it is the closest location for these data sets which are only available for London/ Manchester/Edinburgh

M2 : Q6-11

At any point on a building:

Where:

Ag = Area of glazing

= Incident Solar Radiation

geff = Glazing G value

(proportion of heat transmitted)

These terms can be derived from literature and where multiple rooms exist then the load needs to be calculated for a specific time of day

M2 : Q6-11

Solar gains incident on the are building are provided by the following chart from CIBSE Guide A

M2 : Q6-11

10.2m2

6.9m2

18.6m2

10.2m2

6.9m2

18.6m2

Use the formula to work out the solar gains at each time of day then find the maximum

Answers

Q6 - 75 W/person

Q7 - 17 people

Q8 - Qocc = 1,267 W

- QE = 3,041 W

- QL = 2,027 W

Q9 Can you get to this formula:

Q10 This answer is subjective. What do you think?

Q11 QS

QS = 4,506 W (what month and time of day is this?)

Workshop M3

Q12 – 13

Radiator Selection

M3 : Q12-13

Based on the result from Question 5 select a suitable radiator for the space based on keeping the room to a minimum of 12°C overnight.

Make the following assumptions:

Heating circulation water temperatures – Flow = 75°C Return = 65°C

Internal temperature = 12°C

External temperature = -5 °C

Assume that you are going to use 4 equally sized radiators

Hint: Correct the radiator outputs using the table opposite

(ΔT = Mean Water Temperature – Room Temperature)

Radiator table on the next page

M3 : Q12-13

Calculate the steady state heat loss during the day and reselect a suitable radiator for the space from the information provided.

Make the following assumptions:

External air temperature = 0°C

Internal temperature = 20°C

Assume space is occupied with the heat gains you worked out within question 8.

The office has a daytime occupancy of 17, requiring 10 l/s/person

Assume that the space is ventilated with the minimum required

We will assume it’s a cold overcast day, so ignore any solar gains.

Steady State Heat losses are calculated using the following

When selecting the radiator use the following:

Heating circulation water temperatures – Flow = 75°C Return = 65°C

Internal temperature = 20°C

Assume that you are going to use 4 equally sized radiators

Assume space is occupied internal gains exist (assume the loads calculated previously in Question 8)

Hint: Correct the radiator outputs using the table opposite

(ΔT = Mean Water Temperature – Room Temperature)

Answers

Q12 - 969 W per radiator.

Selection of radiator is variable depending on the height and choice of radiator

For example P+ type with dimensions 600mmx600mm (don’t forget to inc the k factor)

What other options are there?

Q13 - 597 W per radiator.

For example K1 radiator from the radiator selection table which is 600mm high by 700mm wide

What other options are there?

What do these two results tell you?

Workshop M4

Q14-18

Water Flow Rates and Pressure Drop

M4 : Q14-16

AHU used to provide heating

Instead of naturally ventilating the space it is assumed that a small air handling unit will be used instead. This is shown below right and has a fan and a heating coil. What is the heat load in the heating coil?

Assume again that the external temperature = 0°C

Internal temperature is = 20°C

Occupancy as calculated earlier (17 people)

Air Flow Rate = Minimum Ventilation Rate

Hint: Use the following equation to establish the heat load required by the air

For air

=1.2kg/m3

=1.02kJ/kg.K

Heater

Fan

Filter

M4 : Q14-16

If the air handling unit was also to be used to maintain room temperature in the space.

Consider the following:

Part a)

What would be the required air supply temperature,

Maintain the same volume of air as calculated in Q14

Calculate the heat load balance using the heat gains from earlier

Formula:

Part b)

What is the new heat load in the heating coil?

Calculate the heating water flow rates required for the following:

Part a)

Radiators used at night. Calculate the flow per radiator and the combined flow of all 4 radiators in the room.

Use the load calculated from Q12

Part b)

The air handling unit coil used to heat the space during the day

Use the load calculated from Q15.

Answers

Q14a -

Q14b -

Q15 - 2,393W

Note: see any similarities with Q13 answer?

Q16a -

Q16b -

Q17 and 18 answers will be provided after the workshop.

M4 : Q17-18 : Pressure Drop

These questions are provided along with an excel spreadsheet on blackboard.

The questions include a worked through example for part of the work. Follow this through before trying to undertake the questions.

The next slide shows the overall schematic for the system.

This drawing is an extract from the heating layouts for the SU

It shows the heating pipework connecting a number of underfloor heating systems on the first floor.

M4 : Q17-18 : Layout

In this worked example we investigate the pipework connecting manifolds to the riser.

(in effect assume the riser is the plantroom)

The first step in developing a schematic is to label key nodes

These nodes are either junctions or system end points as shown

M4 : Q17-18 : Layout

M4 : Q17-18 : Schematic

The schematic should then set out the pipework connections clearly using the same node referencing as the layout.

The reason we select nodes as junctions is that between junctions the pipework size should always be consistent based on the flow rate in that section

M4 : Q17-18 : Index Run

The index run is the pipework run with the greatest overall pressure drop

This is the pressure that the pump driving the system needs to generate

M4 : Q17-18 : Index Run

Index Run Possibilities:

Option 1 = ΔP1-2 + ΔP2-3 + ΔP3-4

Option 2 = ΔP1-2 + ΔP2-3 + ΔP3-8

Option 1 = ΔP1-2 + ΔP2-5 + ΔP5-7

Option 1 = ΔP1-2 + ΔP2-5 + ΔP5-6

For a thorough analysis of pressure drop in a system it is necessary to calculate pressure drops in the section between each set of nodes.

All possible combinations are then examined to find the highest possible pressure drop.

For this exercise we are assuming all the UFH manifolds have the same flow and pressure drop through them …..……… so for this case the furthest one will be the highest pressure drop.

So manifold at node 4 has the highest pressure drop

M4 : Q17-18 : Index Run

Maximum demand on the manifold is 20kW

We now need to work out the mass flow rate

= ?

M4 : Q17-18 : Pipe Pressure Loss

Maximum demand on the manifold is 20kW

= 4.18

= 20oC

= 0.24 l/s

Note:

If you use 4.18 and a kW load you will immediately get l/s

Use tables or spreadsheet provided by CIBSE to size pipework.

Select pipe size based on following constraints:

Pipework should be as small as possible to reduce cost

Limit velocities 0.5-1.5m/s (in the most part to limit noise)

Limit pressure drop to below 300Pa/m

M4 : Q17-18 : Pipe Sizing

Taking the flow rate of 0.24kg/s.

Referring to the CIBSE table opposite it can be seen that to stay below a pressure drop of 300Pa/m we need to use a pipe size of 20mm diameter.

This results in 0.69m/s flow velocity of and 280Pa/m pressure drop.

Note it’s possible to be achieve higher accuracy through interpolation but rarely required.

0.24

280

0.69

These numbers can now be added to the call out label as shown.

M4 : Q17-18 : Bend Pressure Drop

Pressure loss factor derivation:

Velocity in this section = 0.69m/s

Pipe size = 20mm

Assume bend

Therefore ζ = 0.92

We can then calculate the pressure drop:

219Pa

Note there are three bends in this section so overall an allowance of 3x219Pa is required (657Pa)

M4 : Q17-18 : Bend Pressure Drop

Note the flow arrangement

For flow pipe we have Diverging flow:

For flow pipe we have:

Diverging flow

Pipe size = 32mm

Converged velocity = 0.83m/s

Converged flow = 0.48kg/s

Branch flow = 0.24kg/s

Flow ratio = 0.24/0.48 = 0.5

Therefore ζ = 0.63

M4 : Q17-18 : Branch Pressure Drop

The pressure drop due to a branch is derived from the following equation:

217Pa

Complete the schematic (note letters denote the reference to the crosshairs)

Work out the pressure loss through the system from the riser

Fill in the red blocks in the table opposite

Note: tables etc are available on blackboard

Concluding Thoughts

In your future designs be careful not to over design heating systems, consider careful what loads you use, how you apply diversity and what factors of safety you add.

Modern insulation standards mean that the heat losses are low and we need often to think how to prevent overheating and free cooling

In hydronic systems think carefully about system temperatures and the balance of how your heat emitters work and maximising efficiency in your heat source

Ensure that systems are variable flow and operate efficiently at part load