"Mechanical Systems For the palestinian ECONOMIC POLICY RESEARCH INSTITUTE"   

 Supervisor :
                              Eng. Ramez Al Khaldi

Prepared by :
  Bahaa Yousef Malhis
  Nodar Hisham Sabbah
  Ameer Ghazi Malhis
  Ala’a Abd An-Naser Diab




 Presentation out line:

- Building Description. 
- Heating load calculation.
- Cooling load calculation.
- Duct Design.
- Plumbing System.
- Fire Fighting System.
- Equipment Selection.


Building Description
 Building Location :
            Country: Palestine.
            City: Ramallah
              Elevation: 840 m above sea level Latitude: 32˚.
            Building face sits at the south orientation.
            The wind speed in Ramallah is above 5 m/s.

 Building  Details :


The Palestinian Economic Policy research institute  consists of 6 floors, 5 up  ground, one below  ground. 


Overall Heat Transfer Coefficient, U overall
The U overall is given by:


In our project the method was used as following:
    


Where:
      U: The overall heat transfer coefficient [W/ m².˚C].
      Ri: Inside film temperature [m².˚C/W].
      Ro: Outside film temperature [m².˚C/W].
      K1, 2, …, n : Thermal conductivity of the material [W/m.˚C].
      X1, 2, …, n : Thickness of each element of the wall construction [m].









Results of overall heat transfer coefficient for each element
	Type	Overall Heat Transfer Coefficient U(W/m2.K)
	Outside wall	0.74
	Inside wall	2.6
	Ceiling	0.88
	Floor	0.88
	Glass	3.5
	Wood door	3.5



Heat Load Calculations
Heat loss by conduction and convection heat transfer through any surface is given by:

Where:
     Q = heat transfer through walls, roof, glass, etc.
     A = surface area.
     U = overall heat transfer coefficient.
     ΔT = Difference in outside temperature and inside temperature.

Select inside & outside design condition  (from Palestinian code) :




	Parameters	Winter	Parameters	Winter
	Tin	22	Win	8.3
	Tout	4.7	Wout	4.1
	Φin	50%	Tun	30.65
	Φout	62%		



General Procedures for Calculating Total Heat Load: 
Select inside design condition (Temperature, relative humidity).
Select outside design condition (Temperature, relative humidity).
Select unconditioned temperature (Tun).
Find over all heat transfer coefficient Uo for wall, ceiling, floor, door, windows, below grade.
Find area of wall, ceiling, floor, door, windows, below grade.
Find Qs conduction.
Find Vinf, V vent.
Find Qs, QL vent, inf.
Find Q total and Q boiler.


Equations and Results for heating load calculations :



Heating load summary :





	Floor	Q Total(KW)
	Ground Floor	83.8
	First Floor	48.1
	Second Floor	13.385
	Third Floor	32.74
	Roof	15.942



Cooling Load Calculations :

Select inside & outside design condition  (from Palestinian code) :


	Parameters	summer	Parameters	summer
	Tin	22	Win	9.6
	Tout	30	Wout	16.1
	Φin	50%	Tun	27.3
	Φout	57%		




     In calculations of cooling load, orientation is an important basic factor during calculation, the general equation in cooling calculations is:
                   
               

    For transmitted through glass:


   For convection through glass:
 










Cooling Load Results:
	Floor	Qtotal(KW)	Qtotal(TON)
	Ground floor	89.3	25.5
	First floor	47.7	13.6
	Second floor	45.9	13.1
	Third floor	47.4	13.5
	Roof floor
	35.9	10.25



Duct Design :    
   Design procedures
Number of grills and diffusers are calculated and distributed uniformly.
The total sensible heat of floor is calculated.
The Vcirculation of floor is calculated.
The main branch duct velocity is 5 m/s.
The pressure drop (∆P/L)  method is achieved for duct design (by using ∆P/L(0.6 Pa/m)).
The main diameter is calculated ,at the same (∆P/L).
The height and width of the rectangular ducts are determined from software program .



Equations For Design duct :
 




The equal pressure drop method for sizing in second floor room (12,13,14) :









	FCU
(No.)	section	Vcirc
(l/s)	Velocity(m/s)	∆P/L
(Pa/m)	H
(cm)	W
(cm)
	4	AB	424.95	5	0.6	30	37.5
		BC	271.25	3.6	0.6	30	25
		CD	174	3.6	0.6	30	200






Plumbing System:
Potable water:
Find number of unit for fixtures from tables , the results  as shown in table: 
	Floor	Hot F.U	Cold F.U
	Hot
 size(in)	Cold
size(in)
	  Basment 	3.75	6.75	1	1.25
	Ground floor	10	40	1.25	1.5
	First floor	7.5	22.5	1.25	1.5
	Second floor	7.5	22.5	1.25	1.5
	Third Floor	7.5	22.5	1.25	1.5
	Roof	12.75	23.75	1.25	1.5
	Sum	49	138	2	2.5



Potable water piping inside floor
For example 2nd  floor:






Riser:
cold water:
	Size(in)A	Demand (l/s)	Fixture	Section
	2.5	3.24	138	a-b
	2.5	2.91	114.25	b-c
	2	2.57	91.75	c-d
	2	2.148	69.25	d-e
	1.5	1.45	46.75	e-f
	1.25	1.052	6.75	f-6



Hot water:
	Size(in)	Demand(l/s)	Fixture	Section
	2	1.81	49	a-b
	2	1.78	45.25	b-c
	1.5	1.58	35.25	c-d
	1.5	1.4	27.75	d-e
	1.5	1.25	20.25	e-f
	1.25	1.034	12.75	f-6






Drainage System:
Two separated stacks one for soil and the other for waste. Each stack 4 in.
(standard 4 in for WC,2 in for other fixtuers,4 in floor drain).





Fire Fighting System
We used in fire system class 3 as standpipe system (cabinet and landing valves).

Class III Systems: A Class III standpipe system shall provide 11⁄2 in. (40 mm) hose stations to supply water for use by trained personnel and 21⁄2 in. (65 mm) hose connections to supply a larger volume of water for use by fire departments and those trained in handling heavy fire streams.
One riser so 500 GPM for 40 min operating the tank size = 80m^3.
Pressure for cabinet = 65 Psi and 100 Psi for landing valve.





 Equepments Selction
Boiler:
The heating load for building is 193.9 Kw, and load for boiler selected =1.1*193.9=213.3 Kw, since we have one boiler, Because that we choose from mansour catalog of steam boiler is MS-10 which has a capacity is 470 KW.






Chiller:
For the building the total cooling load is equal 272.23kw, so the minimum capacity for the chiller should be equal to:
272.23x1.1=300 kw, we select chiller that fits the needs, in tons units=85.7 tons
Flow rate= 85.7*2.4=210.48 GPM and 50 HZ.
From the figure below we select the chiller of model W P S  a  135-2D




Fan coil units:
Sample 2nd floor:
 Fan04
1000 Cfm
DC P 10 H/C  3





Circulation pumps for chiller
S series S55 with  258.688 Pa/m and 6.5 L/s flow rate 
Hot water pump  demand = 28.5 GPM  pressure= 6Psi
Potable water pumps:
From last chapter(plumping) we calculate the total demand  cold water of the building which is equal=51.17 GPM     pressure=6.7 Psi
 
 
Fire water pumps:
According to number of risers which  equal 1 
Now:
Flow rate=500 GPM 
Pressure= 128.4 PSI
Jocky pump = 128.4+10=138 .4 Psi



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