Mechanical Academy - Mechanical Team



The University of Jordan – Faculty of Engineering and Technology

Mechanical Engineering Department

ME 0904467: Design of Sanitary Systems

Instructor: Dr. Jamil Al Asfar- Fall, 2010

Example 2: November 1st , 2010

Q1) Size the cold water supply system (flow rate and pipe sizing) of the plumbing system shown below for an apartment in the floor plan given in the figure using the Simplified Sizing Method. The apartment is located at the last level of a four-storey building in which every apartment is supplied individually and indirectly from the mains. Use copper piping.

Table 4.3 Water supply fixture units for sanitary fixtures

| Load values assigned |

|Water supply |

|fixture unit |

|Fixture |Occupancy |Type of supply control |Cold |Hot |total |

|Water closet |Public |Flush valve |10 | |10 |

|Water closet |Public |Flush tank |5 | |5 |

|Urinal |Public |24.5 mm flush valve |10 | |10 |

|Urinal |Public |19 mm flush valve |5 | |5 |

|Urinal |Public |Flush tank |3 | |3 |

|Lavatory |Public |Faucet |1.5 |1.5 |2 |

|Bathtub |Public |Faucet |3 |3 |4 |

|Showerhead |Public |Mixing valve |3 |3 |4 |

|Service sink |Offices, etc |Faucet |2.25 |2.25 |3 |

|Kitchen sink |Hotel, restaurants |Faucet |3 |3 |4 |

|Drinking fountain |Offices, etc |9.52 mm valve |0.25 | |0.25 |

|Water closet |Private |Flush valve |6 | |6 |

|Water closet |Private |Flush tank |3 | |3 |

|Lavatory |Private |Faucet |0.75 |0.75 |1 |

|Bathtub |Private |Faucet |1.5 |1.5 |2 |

|Shower stall |Private |Mixing valve |1.5 |1.5 |2 |

|Kitchen sink |Private |Faucet |1.5 |1.5 |2 |

|Laundry trays |Private |Faucet |2.25 |2.25 |3 |

|Combination fixture |Private |Faucet |2.25 |2.25 |3 |

|Dishwashing machine |Private |automatic | |1 |1 |

|Laundry machine 3.6 kg |Private |automatic |1.5 |1.5 |2 |

|Laundry machine 3.6 kg |Public or general |automatic |2.25 |2.25 |3 |

|Laundry machine 7.3 kg |Public or general |automatic |3 |33 |4 |

Table 4.4 Minimum size of fixture supply pipes

|Fixture |Size mm |

|Water closet, Flush tank |9.5 |

|Water closet, Flush valve |25.4 |

|Urinal, flush valve |25.4 |

|Urinal, flush valve |19.0 |

|Urinal Flush tank |12.7 |

|Lavatory |9.5 |

|Bathtub |12.7 |

|Showerhead |12.7 |

|Service sink |12.7 |

|Kitchen sink domestic |12.7 |

|Kitchen sink commercial |19.0 |

|Drinking fountain |9.5 |

|Dishwashing machine |12.7 |

|Hose bib |12.7 |

|Wall hydrant |12.7 |

[pic]

Solution: The first step in sizing the piping system is to lay out the piping network and draw a three-dimensional view with lengths assigned as shown in Fig. 4.xxx. The cold water distribution pipe is laid within the walls at 0.4 m height above the terrazzo and underneath it in the floor at 0.05 m depth (see Fig. 4.x. and 4.xx). Note that the height of different appliances was taken into account. For example, the height of the kitchen sink is 0.9 m above the terrazzo (0.5 +0.4m). This is easily seen from summing up the height of line A-a which is 0.5 m and the height above the terrazzo to point A which is 0.4 m. The objective of the design is to size the piping sections to satisfy the flow requirements. The "Simplified Sizing Method" is based as follows:

Step 1: Figure 4.xxx shows the sizes selected. In this step we have sized the supply pipes to water outlets according to Table 3, p. 30 of the code. This is indicated as the third entry in the rectangle (see insert designated Key). The fixture units and the flow rate are also indicated therein.

Step 2: All other parts of the system are sized according to velocity limitations using Fig. 4, p. 41 of the code (for copper pipes). The size is indicated the box. Although we have indicated there the load in fixture units as the sum of fixture units of the fixtures following the piping section in question it must be noted that the flow rate (load in [pic]/s) does not correspond to the fixture units indicated. For example, the branch line (B-C) to the bathroom has 7 fixture units which are translated into 0.74[pic]/s (see Table x.x.) Instead, we used 0.55 [pic]/s which is the sum of the bathtub and the basin flow rates. This is based on the assumption that these two fixtures may be used simultaneously. The 20 mm diameter pipe selected corresponds to a velocity limit of 1.75 m/s (Fig. 4, p. 41) which is less than 3 m/s recommended by the code (p. 31).

Note that all the diameters selected in this step correspond to velocities less than 2.5 m/s.

[pic]

Fig. x.x. Floor plan of apartment showing water supply piping.

[pic]

Figure 4.6. Three-diemnsional layout of cold water system showing piping line segments (all dimensions in m)

[pic]

Figure 4.7. Pipe sizing diagram

Step 3: Adequacy Check: Now that all pipe sections have been sized as shown in Fig. 4.7 we proceed to check the adequacy of the design. The first step is to select the Basic Design Circuit (BDC) (or the Critical Run). In our case, it is clear that it is line (OABCDEF) but we shall pretend that we don't know and try to find it in a systematic way as follows:

a) Label the different sections by letters. For example, the pipe section to the kitchen sink is OAa and the pipe section to the lavatory is OABb1 and so on.

b) Select the BDC candidates. We select 3 candidates:

i) OAa

ii) OABb2

iii) OABCDEF

c) Find the frictional and minor losses for each candidate line as shown in the following Tables:

Table (i) for OAa

|Pipe Section |Pipe Size|Actual Length|Equiv. length$$ of |Total |Flow Rate |Friction & Minor |Total Losses |

| | | |fittings & valves |Length |[pic]/s |Losses(Pa/m) |Pa |

| | | | | | |(Fig. 4, p.41) | |

|OA |25 |4.7 |2.35 |7.05 |1.15 |2000 |14100 |

|Aa |15 |0.7$ |0.35 |1.05 |0.25 |1400 |1470 |

| Total: |

|15570 |

$we added 0.2 m for the unmarked lengths. We will do the same in the rest of the tables.

$$ we used 50% of the physical length

Table (ii) for OABb2

|Pipe Section |Pipe Size |Actual Length|Equiv. length of |Total |Flow Rate |Friction & Minor |Total Losses|

| |(mm) |(m) |fittings & valves |Length |([pic]/s) |Losses(Pa/m) |(Pa) |

| | | | | | |(Fig. 4, p.41) | |

|OA |25 |4.7 |2.35 |7.05 |1.15 |2000 |14100 |

|AB |25 |3.25 |1.625 |4.875 |0.9 |1250 |6094 |

|B-b1* |15 |2.15 |1.075 |3.225 |0.35 |2250 |7256 |

|b1*-b2 |15 |1.2 |0.6 |1.8 |0.15 |500 |900 |

| Total |

|28350 |

Table (ii) for OABb2

|Pipe Section |Pipe Size |Actual |Equiv. length of |Total |Flow Rate |Friction & Minor |Total Losses |

| |(mm) |Length |fittings & valves |Length |([pic]/s) |Losses(Pa/m) |(Pa) |

| | |(m) | | | |(Fig. 4, p.41) | |

|OA |25 |4.7 |2.35 |7.05 |1.15 |2000 |14100 |

|AB |25 |3.25 |1.625 |4.875 |0.9 |1250 |6094 |

|B-C |20 |4.45 |2.225 |6.675 |0.55 |1750 |11681 |

|CDEF |20 |4.65 |2.325 |6.975 |0.4 |1000 |6975 |

| Total 38850|

Obviously, the third piping run (OABCDEF) has the highest pressure loss. Also, it has the minimum hydrostatic head available since it has the highest outlet. Therefore, this line is the Basic Design Circuit or the Critical Run.

d) Find the residual pressure Pra available to spend on the losses calculated in (c). If it is equal or greater than the losses in (c) then the design is satisfactory. If it turns out to be less then a pump must be used.

Pra = Ptotal – Poutlet - Pstatic

= [pic]-0.2x105-[pic]

= [pic]-0.2x105-[pic]

= -2342 Pa

Clearly, a pump is needed.

e) Find the pump size:

The pressure as calculated in (c) Pr is given by:

Pr = Ptotal +Ppump-Poutlet-Pstatic

Solving for pump pressure:

Ppump = Pr + Poutlet + Pstatic - Ptotal

Ppump = 38850-(-2342) = 41192 Pa

The pump head in meters: hpump = [pic]

and the discharge is = 1.15 [pic]/s . The pump should be installed at the tank outlet.

Q2) A cold water cistern 4500 [pic] capacity located on the roof of a multi-storey building is supplied from the city main as shown. The pressure in the city main is maintained at 360 kPa for just one hour/day. The vertical height from the city main to the ball valve is 24 m and the horizontal length from the city main to the riser is 5.5 m. It is required to size the supply pipe as follows:

a) Neglect the minor losses and use Thomas Box formula

b) Use the result in (a) as a first estimate and evaluate the minor losses due to valves, fittings, elbows, ..etc. from Table 6 (Code, p. 40). All four valves are used for system maintenance purposes (i.e., gate valves). You may need to iterate.

c) Use the result from (b) as a first estimate to calculate Le and size the piping system for the following materials:

1) copper tubing (Fig. 4, p.41, Code)

2) steel pipe (Fig. 5, p. 42, Code)

3) PVC pipe (Fig. 6, p. 43 Code)

Observe velocity limitations.

[pic]

Solution: We should note here that Thomas Box formula is an approximate one, and accurate calculations should make use of Darcy's formula coupled with moody chart for the friction factor or Colebrooke’s formula.

Solution of Part (a):

1. The flow rate,[pic], the available residual head, hr, and the length L, are found as follows:

(i) [pic]

(ii) [pic] = hCM –hM – hfix – hst

where hCM is the city main head

hM is the head loss due to meter

hfix is the head required at the fixture outlet (normally 0.2 bar)

hst is the static head of the fixture (height)

[pic]

[pic] = 4.54 m

This is the head available to spend on friction through straight sections of the pipe and minor losses.

(iii) The total length is the sum of actual total length, Lactual, and the equivalent total length, Le,t which is taken to be zero as per the problem statement.

.

L = Lactual + Le,t

L = (5.5. + 24.0 + 0.5) + 0 = 30 m

3. Solving for the diameter from Thomas Box formula:

[pic]=30.4 mm=31 mm

Recall that pipes are available in standard sizes. Therefore, as a rule we choose the next available size which is 32 mm.

Solution of Part (b): The diameter found in part (a) is 32 mm. This enables us to evaluate the minor losses as (see Table 6, p. 40 of Code) in the following Table.

|Component |Le @ 32 mm Diameter |Quantity |Total Equivalent Length per component (m) |

|Gate valve |0.2 |4 |0.8 |

|Ball valve |6.0 |1 |6.0 |

|90o elbow |1.2 |2 |2.4 |

|Stop cock |5.0 |1 |5.0 |

|Coupling |0.4 |4 |1.6 |

| Total 15.8 m |

Table 4.1. Hot water piping heat loss in W/m for different piping materials

and insulation conditions

[pic]

Q3) Design a hot and cold water system for a building having the following properties:

1- Residential and consists of four floors and a basement.

2- Height of each floor is 3m

3- Water storage is an underground cistern. Its lowest point is 5 m below ground level.

4- Each floor has 3 bathrooms (each has 1 WC, 1 Lavatory, and 1 Bathtub,) a kitchen with one sink and a dishwasher, and a laundry room with washing machine and a sink.

5- The basement has 1 WC, 1 lavatory and one shower tray.

6- There is one hose bib for the garden.

The following information is also available.

1- Use copper pipes

2- Assume indirect system

3- No hot water circulation

4- Pressure drop across the water heater is 10 kPa

5- Assume number of building habitants is 4x6 = 24 people.

The riser diagram can be as follows:

[pic]

SOLUTION:

A- pipe sizing:

| |Supply control |WSFU |Flow rate |Pipe D (mm) |

| | |Hot |Cold |Hot |Cold |Hot |Cold |

|Sink, S |Faucet |1.5 |1.5 |0.25 |0.25 |19 |19 |

|Washing Machine WM |Private |1.5 |1.5 |0.25 |0.25 |19 |19 |

|Lavatory L |Faucet |0.75 |0.75 |0.14 |0.14 |12.7 |12.7 |

|Bath B |Faucet |1.5 |1.5 |0.25 |0.25 |19 |19 |

|Water Closet WC |Flush Tank | |3.0 |- |0.41 |- |25.4 |

|Dish Washer DW |Automatic |1.0 |- |0.19 |- |12.7 |- |

|Shower Tray ST |Mixing valve |1.5 |1.5 |0.25 |0.25 |19 |19 |

Starting from the end point of each branch and using table 4. to find the WSFU and corresponding flow rate and pipe diameter mark the resulting data as follows:

{WSFU}

(Flow rate in [pic])

[pipe diameter ]

use blue color for cold water lines and red color for hot water lines.

[pic]

B- Pump selection:

Flow resistance in each section of the piping network is estimated using the proper table (table xxxx). The resulting information is tabulated as show in the table below.

|Section |Size (mm) |Length, (m) |Friction |Total Equivalent |Flow rate, |Resistance, kPa/m |Total friction, |

| | | |equivalent |Length (m) |([pic]) | |kPa |

| | | |length, Le | | | | |

| | | |(m) | | | | |

|A-B |63.8 |7.0 |3.5 |10.5 |3.5 |0.225 |2.363 |

|C-D |50.8 |4.0 |2.0 |6.0 |1.83 |0.2 |1.2 |

|D-F |50.8 |6.0 |3.0 |9.0 |1.63 |0.15 |1.35 |

|E-F |38.1 |6.0 |3.0 |9.0 |1.22 |0.350 |3.150 |

|F-G |31.8 |5.0 |2.5 |7.5 |0.92 |0.450 |3.375 |

|G-H |31.8 |3.0 |1.5 |4.5 |0.78 |0.375 |1.688 |

|H-I |25.4 |3.0 |1.5 |4.5 |0.59 |0.550 |2.475 |

|I-J |25.4 |3.0 |1.5 |4.5 |0.37 |0.250 |1.125 |

|J-K |12.7 |1.0 |0.5 |1.5 |0.19 |1.150 |1.725 |

= 18.45 kPa

The total flow rate resulted from part a is (3.5 [pic]for cold and hot lines + 0.32 [pic]for the hose bib) = 3.82 [pic]

The pressure head that the pump has to overcome is 18.45 kPa resistance due to friction + 10.0 kPa pressure drop within the water heater + 105 kPa pressure drop in the dish washer + 16.0 m * 9.81 for the static head = 290.41 kPa

Therefore the pump should have the following characteristics

Head = 290 kPa head = 29 meter head

Flow rate = 3.8 [pic]

C- Water heater sizing

Estimating the number of inhabitants of the building as follows:

Number of apartments = 4

Number of people per apartment = 6

Total number of people = 24 + 1 guard = 25

Using table xxxx to estimate the power needed per person = 1.2 kW

Therefore the total power needed for the water heater is 25[pic]1.2 = 30 kW

D- Cold Water storage sizing

Depending on the water supply to the building (could be different from different countries), the number of days that the storage should be adequate for can be determined. Assuming that water storage should be adequate for five days, then;

Using the estimated water consumption by an individual in that country (assume to be 100 liters per day)

Then the storage capacity is : 25 people[pic]5 days[pic]100 liters = 12,500 liters = 12 cubic meters

Q4) The attached Figure shows a schematic of a three storey building having one apartment (dwelling) on each floor. They have a common cold and hot water supply systems. The boiler and hot water storage-type heater (indirect heating) are located in the basement as shown. The flow rate of hot water to each apartment is 1.15 [pic].

• Size the riser (line 13-12) using a velocity limitation of 2.4 m/s.

• Select the type of hot water circulation system

• Size the return line

The lengths designated in this problem are: 13-a1 = 2.5 m

a1-a2 = 3.5 m

a2-12 = 3.5 m

11-12 = 0.5 m

10-13 = 0.5 m

Assume minor losses to constitute 20% of the frictional losses.

Solution:

a) The riser is sized based on the flow rates and velocity limitation given, i.e., 2.4 m/s. The hot water riser consists of sections 13-a1, a1-a2, and a2-12. The corresponding flow rates and diameters are:

|Section |L (m) |Flow rate ([pic]) |Diameter (mm) |Remarks |

|13-a1 |2.5 |1.15*3=3.45 |50 |Steel |

|a1-a2 |3.5 |1.15*2=2.3 |40 | |

|a2-12 |3.5 |1.15*1=1.15 |25 | |

b) The type of hot water circulation system (i.e., forced or natural) depends on whether the "thermal head" is sufficient to sustain the required circulation rate in the given circulation piping. The selection is made as follows:

1. The total height of the circulation loop is:

[pic]

= 2.5 + 3.5 + 3.5 + 0.5 = 10 m

2. The developed head is:

h = 0.01*H = 0.1 m

[pic]

3. To determine whether the available head is sufficient to sustain the required circulation rate we proceed as follows:

3.1. The heat loss rate of the circulation piping is found from Table 5.1. The diameter of return line is assumed:

a) Heat loss through the HW supply is (assume bare steel piping).

|Section |L (m) |Flow rate (l/s) |Diameter (mm) |Remarks |

|13-a1 |2.5 |1.15*3=3.45 |50 |Steel |

|a1-a2 |3.5 |1.15*2=2.3 |40 | |

|a2-12 |3.5 |1.15*1=1.15 |25 | |

3.2. Heat loss through return line. Assume d=0.5[pic]*50 =25mm

Return line length:

[pic]

Total heat loss = 643 + 510 = 1153 W.

3. The circulation rate required:

[pic]

3.4. Frictional head loss

[pic]

= 0.0006600m

[pic] So, natural circulation will do.

Another reference for indirect systems

The indirect hot-water system

The basic layout of the combined central heating and indirect hot-water service system is shown in Fig. 6.5. The cylinder is insulated with 75mmfibre glass and should have a thermostat attached to its surface at the level of the primary return. Water is stored at 65◦C, and when fully charged the thermostat closes the motorized valve on the primary return. This ‘off’ signal may also be linked into the pump and boiler control scheme to complete the shut-down when the central heating controls are satisfied. Hot-water pipes are insulated with a minimum of 25 mm of insulation, as are tanks exposed to frost. The primary system feed and expansion tank has a minimum capacity of 50 l, and the cold-water storage tank has a capacity of at least 230 l. Hot-water storage requirements at 65◦C are as follows: office, 5 l/person; dwelling, 30 l/person; hotel and sports pavilion, 35 l/person (CIBSE, 1986).

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