engineering report heat exhanger b(1)

ENGINEERING REPORT

SUBJECT TO THE REPORTThe report shall be limited to the Design, Installation, Operation and Maintenance of Hot Water System Using Heat Pump to Include Detailed Calculation of Piping, Storage Tank Design, Pumps and Insulation of five (5) floors of Maxims Hotel guest room and the Recovery of Cold Air for Machine Room Ventilation.

The heat pump system will incorporate also the interfacing of the system to Building Management System for start/stop of air-conditioning of elevator machine room, exhaust fan of electrical room and air-handling unit of guest room for fresh air supply. The said Building Management System will play a big role in energy conservation or recovery of the system. The report will also describe the advantages of using heat pump due to its recovery system and environmental compliance. GENERAL DESCRIPTION OF WORKThe project involves the Design, Installation, Operation and Maintenance of Hot Water System Using Heat Pump to Include Detailed Calculation of Piping, Storage Tank Design, Pump & Insulation and the Recovery of Cold Air for Machine Room Ventilation for Maxims Hotel located in Newport Blvd., Newport City, Pasay City. Hot water system mainly consist of heat pump with built-in primary circulating pump, hot water storage tank, secondary circulating pump, piping distribution and make up water. The focus of the design will be the supply of hot water in hotel guest rooms were numerous capacity of water required. The hotel used heat pump with built-in primary circulating pump located at the roof deck of Maxims Hotel. Each heat pump consists of four (4) units of compressors. Heat pump is an energy collecting and transferring system mainly consists of compressor, condenser, expansion valve and evaporator.Hot water produced by heat pump will convey in copper pipe to the guest rooms by means of pressurized hot water tank. They use rubber for the insulation of copper pipes and rock wool for the storage tank with aluminum cladding. Copper pipes are use for header, sub header and riser and polypropylene pipes inside the guest room.The system is advantageous and efficient due to its recovery in terms of operation. It is also environment friendly compare to boiler system which has flue gas discharge and chemical blow down. Since the required hot water temperature of guest room is not too high, the applicant suggests to use heat pump due to its simplicity of design & maintenance. Operators are ready since the system is the same as air-conditioning system, no need to train technical person for heating equipment like boiler. To gain the benefits of energy conservation or recovery of discharge air of heat pump, it can also be used as ventilation of elevator machine room, ventilation of electrical room and utilized for the fresh air of hotel guest room. The sequence of the said energy recovery is when the heat pump is in operation, the discharge air of the heat pump will be utilized as ventilation of elevator machine room, ventilation of electrical room and fresh air of hotel guest room by means of ducting. The interfacing of Building Management System to Hot Water System will synchronized the start/stop of controller of air-conditioning and exhaust fan and the heat pump. When the heat pump turns on, the air-conditioning and exhaust fan turns off. And when the heat pump meets the desired temperature, the heat pump will automatically turn off and the air-conditioning of elevator machine room, exhaust fan of electrical room and air handling units turns on. DESIGN PARAMETERS AND CONSIDERATIONSDesign RequirementsThe hot water system was composed of the following requirements; Design Total Heat Capacity Heat Capacity of hot water at the Guest Room Detailed Calculation of Pipe Sizing to Guest Room Required Capacity and Thickness of Hot Water Tank Calculation of Heat loss of Tank and Pipe Total flow rate and total dynamic head of secondary circulating pump Area of Heating Coil Duct Sizing of Recovery Cold Air

The design reference is set in the code of practice and regulation for most of the countries and to the standard of the following;a. Maximum daily demand (100 or more units) 15 gal/unit, Table 7 Hot Water Demand and Various Types of Buildings (see page 18 of Service Water Heating of ASHRAE HVAC Application).b. Maximum hourly demand (100 or more units) 4 gal/unit, Table 7 Hot Water Demand and Various Types of Buildings (see page 18 of Service Water Heating of ASHRAE HVAC Application).c. 68F Design cold water temperatured. 77F - Design ambient air temperaturee. 118.4F - Hot water supply of Guest Room and tank design temperature.f. 140F - Heat Pump Design Temperatureg. 0.5 to 8 ft/s - Hot Water Velocity (see page 5 of Air Heating Coil of ASHRAE HVAC Systems and Equipment).h. Copper pipe nominal diameter of 67mm, 52mm 35mm with wall thickness of 2.4mm, 2.1mm and 1.6mm respectively Type L domestic water. (http://www.engineeringtoolbox.com)i. 1 to 2 hrs - Capacity of storage tank (see page 19 of ASHRAE HVAC Application).j. 20% - Expansion and sludge and reserve of tank (see page 8 of Medium and High Temperature Water Heating of ASHRAE HVAC Systems and Equipment).k. 125 to 150 psig - Design pressure rating for below 350F (see page 8 of Medium and High Temperature Water Heating of ASHRAE HVAC Systems and Equipment).l. 2 Btu / hr-ft2-F - Outside surface coefficient free convection (Thermodynamics by Virgil Mourning Faires)m. 312 Btu-in / hr-ft2-F - Film Coefficient for steel (Thermodynamics by Virgil Mourning Faires)n. 0.28 Btu-in / hr-ft2-F - Film Coefficient for rockwool(Thermodynamics by Virgil Mourning Faires)o. 1,400 Btu-in / hr-ft2-F - Film Coefficient for Aluminum(Thermodynamics by Virgil Mourning Faires)p. 1.65 Btu / hr-ft2-F - surface coefficient inside building wall(Thermodynamics by Virgil Mourning Faires)

q. 1,000 Btu / hr-ft2-F - inside surface coefficient saturated Water flowing (Thermodynamics by Virgil Mourning Faires)r. 2,616 Btu-in / hr-ft2-F Film Coefficient of Copper(Thermodynamics by Virgil Mourning Faires)s. 0.27 Btu-in / hr-ft2-F Film Coefficient of rubber insulation (Insulac Brand)t. 3 Btu-in / hr-ft2-F Film Coefficient of Polyethylene (http://www.engineeringtoolbox.com)u. 0.69 Btu-in / hr-ft2-F Film Coefficient for polypropylene (http://www.engineeringtoolbox.com)v. 60-80 Btu / hr-ft2-F Overall heat transmission of copper(http://www.engineeringtoolbox.com)w. 1 Gal/min for 1- inches and 1- inches diameter per riser of circulating pump (see page 7 Water Heating of ASHRAE HVAC Application).

Hot Water Pressure At 45 psi, variable speed drive pump is required to maintain the operating pressure of Hot Water System. The pressure of hot water system must be the same as cold water line.

Hot Water TemperatureHot Water Circulating Pump and installation of proper insulation is required to maintain the temperature of hotel guest room. By doing this, we can save power for heat lost of the system and meet client requirements.

Hot Water Pipes Because of their inherent resistance to corrosion and easy installation, copper pipe are often used in water supply installation for hotel.

System ControlDue to its recovery, the control will interface different system. It will help the activation or sequence required during operation by means of Building Management System.

Duct Layout and DamperThe duct layout should follow required thickness and duct support to lessen the noise level due to its location at roof deck. This is to avoid transmission of sounds to guest room. Proper insulation is needed for ducting used as recovery system for the supply of fresh air at the guest room.

Electric MotorHigh Efficiency electric motor, Totally Enclosed Fan Cooled (TEFC), IEC standard, IP 55 and class F insulation and NEMA 3R controller for hot water circulating pump.DESIGN CALCULATIONSDesign Total Heat CapacityTo determine the total required heat capacity of guest room, add the heat load of water consumption of guest room, heat lost of tank and heat lost of pipe by the following equation;

QH=QL + QLT + QLP

Where:QH=Required Total Heat Capacity (Btu/hr)QL= Heat Load of Water Consumption of Guest Room (Btu/hr)QLT= Heat Lost of Tank (Btu/hr)QLP= Heat Lost of Pipe (Btu/hr)

2.3.1.1 Heat Load of Water Consumption of Guest Room

QL=Mhw x Cpw ( Thw Tcw )

Where:Mhw=Mass of Hot Water Required in the Guest Room (lb/hr)Cpw=Specific Heat of Water (1 Btu/lbs F)Thw=Hot Water Temperature Supply in the Guest Room (F)Tcw=Cold Water Temperature (F)V=Volumetric Flow Rate (ft3/hr)=Density of Water (62.4 lb/ft3)

To determine the total heat capacity required for the guest room, get the maximum load demand required in the guest room using 90% occupancy of 172 guest room. The design average room occupancy is 2 person per unit (see Appendices for Architectural Layout Annex 1A-1D).

V=No. of unit x daily maximum demand x Occupancy % x no. of personV=172 units x 15 gals/day-unit x 90% x 2 personV=4,644 gals/day or 17,577.54 Liters/day

Say;18,000 Liters/day (Daily Consumption)

On ASHRAE standard we can determine the percentage of peak load by getting the maximum hourly demand. The maximum hourly demand is 4 gals/unit.

V=No. of unit x hourly maximum demand x Occupancy % x no. of personV=172 units x 4 gals/day-unit x 90% x 2 personV=1,238.4 gals/hr or 4,687.34 Liters/hr

The computed percentages of peak load demand is approximately 26% based on maximum hourly demand over the maximum daily demand. The percentage acquire will be use to simulate the average usage of Hot Water System of Maxims Hotel per hour.

Table 1.1 Hot Water ProfileTime of dayHot Water Usage (Liter)% Usage

1:0000%

2:0000%

3:0000%

4:0000%

5:0000%

6:003602%

7:009005%

8:004,68026%

9:001,4408%

10:003602%

11:003602%

12:001801%

13:0000%

14:0000%

15:001801%

16:003602%

17:003602%

18:007204%

19:001,0806%

20:004,68026%

21:001,0806%

22:007204%

23:005403%

24:0000%

18,000100%

Based on Table 1.1 we can get the average hourly demand of Hot Water System approximately 1,125 Liters/hr or 39.70 ft3/hr;

In mass flow rate;

Mhw= V x Mhw=39.70 ft3/hr x 62.4 lb/ft3Mhw=2,477.28 lb/hr

To determine the heat load water consumption of guest room, substitute the mass flow rate in the equation;QL=Mhw x Cpw ( Thw Tcw )QL=2,477.28 lb/hr x 1 Btu / lbs F x (118.4F - 68F)QL=124,854.91 Btu/hr

Detailed Calculation of Pipe Sizing to Guest RoomIn Item 2.3.1.1 the maximum hourly demand is 4,687.34 Liters/hr or 165.40 ft3/hr. We can now compute the size of pipe header of Hot Water System.

A=V / v

Where:V=Volumetric Flow Rate (165.40 ft3/hr or 0.046 ft3/s)A=Area of Pipe (in2) v =Velocity of Hot Water (1.8 ft/s)d=Diameter of Pipe (inches)=Pi, (3.1416)

For hot water header pipe @ peak load capacity of 165.40 ft3/hr or 0.046 ft3/s.

In Area;A=V / vA=0.046 ft3/s / 1.8 ft/s A=0.026 ft2 or 3.744 in2In diameter;

A= (d2) / 4d= sqrt (3.744 in2 x 4) / d=2.2 inches

Say;2.5 inches diameter for pipe header

For sub header volumetric flow rate @ 50% capacity of 82.7 ft3/hr;V=Volumetric Flow rate (82.7 ft3/hr or 0.023 ft3/s)

In Area;

A=V / vA=0.023 ft3/s / 1.8 ft/s A=0.013 ft2 or 1.9 in2

In diameter;

A= (d2) / 4d=sqrt (1.9 in2 x 4) / d= 1.6 in

Say; 2 inches diameter for pipe sub-header

For riser volumetric flow rate @ 25% capacity of 41.35 ft3/hr;V=Volumetric Flow rate (41.35 ft3/hr or 0.011 ft3/s)

In Area;

A=V / vA=0.011 ft3/s / 1.8 ft/s A=0.0061 ft2 or 0.9 in2

In diameter;

A= (d2) / 4d=sqrt (0.9 in2 x 4) / d=1.1 inches

Say;1- inches diameter for riser

The pipe sizes are 2.5 inches, 2 inches and 1- inches for header, sub header and riser, respectively (See Appendices for schematic diagram of hot water piping in Annex 2A-2B).

Required Capacity and Thickness of Hot Water TankBased on Table 1.1, Hot Water Profile, the storage capacity of tank that can be used is 900 Liters and 4,680 Liters in time of 7am and 8am. This is based on ASHRAE that can be used 1 to 2 hours storage capacity peak demand of the day.We can get the capacity of Tank by the equation of;

VT=V1 + V2 + V3

Where:VT=Total Volume CapacityV1=Peak Capacity of Hot WaterV2=Expansion CapacityV3=Sludge and Reserve

V2 + V3 = 20% (V1)

VT=V1 + V2 + V3 VT=V1 + 20% (V1)VT=(900 Liter + 4,680 Liter) + 20% (900 Liter + 4,680 Liter)V1=5,580 Liters + 1,116 LitersV1=6,696 Liters

Say; 2 x 3,348 L or 2 x 118.14 ft3, total volume of tank

To find the dimension of the tank determine first the volume of ellipsoid and the volume of cylinder (see Appendices Annex 3A-3B);

VT=VC + VEWhere:VT=Total Volume CapacityVC=Volume of CylinderVE=Volume of EllipsoidhT=Overall Height of TankdT=Diameter of Tank

Given the diameter of tank at 4.6 ft with ellipsoid radius vertical distance of 0.98 ft, where major axis and minor axis are the same. (see Appendices Annex 3A-3B).

VE=4/3 x x a x b x c

Where:a=Radius of Major Axisb=Radius of Minor Axisc=Radius of Vertical DistanceVE=4/3 x x a x b x c VE=4/3 x x 2.3 x 2.3 x 0.98VE=21.72 ft3

For cylinder;

VC=VT - VEVC=118.14 21.72VC=96.42 ft3For the height of the cylinder;

VC= x r2 x hh=96.42 / ( x 2.32) h=5.8 ft

Adding the height of cylinder and ellipsoid;

hT=c + hhT=(2 x 0.98 ft) + 5.8 fthT=7.8 ft

Say; 4.6 ft diameter x 7.8 ft height

And the wall thickness of tank using 125 psi operating pressure and 4.6 ft or 55.2 inches diameter;

tm =(P x r x FS) / TS x E

Where:tm =wall thickness (inches)P=maximum internal pressure in 125 psir=radius of tank (inches)TS=Tensile Strength (55,029 psi)FS=5 for new constructionE=Efficiency 90%tm =(P x r x FS) / TS x Etm =(125 psi x 27.6 in x 5) / 55,029 psi x 0.90tm =17,250 / 49,526.10tm =0.348 in

Say; 3/8 inches standard commercial thickness

Calculation of Heat loss of Tank and PipeWe can compute the heat lost of tank by following equation QLT= QLT1 + QLT2QLT1=(Tin x Tout) / (1/ho) + (1/hi) + A/KQLT2=AT x (Tin x Tout) / (1/ho) + (1/hi) + L/K

Where: QLT=Total Loss of Tank (Btu/hr)QLT1=Heat Loss of Cylindrical Tank Surface Area (Btu/hr)QLT2=Heat Loss of Ellipsoid Cover of Tank Surface Area (Btu/hr)AT=Area of Tank (in2)L=Length of Pipe or height of Tank (ft)Tin=Design Temperature Inside the Tank (F)Tout=Design Ambient Temperature (F)ho=Outside Surface Coefficient (Btu/hr- ft2-F)hi=Inside Surface Coefficient (Btu/hr- ft2-F)

K=Film Coefficient (Btu-in / hr- ft2-F)R= Resistance (hr-F / Btu)

For cylindrical tank heat lost;

QLT1=(Tin x Tout) / (1/ho) + (1/hi) + A/K orQLT1=(Tin x Tout) / R

To determine the heat lost, get first the summation of resistance;For R1, given the height of cylindrical tank 5.8 ft and diameter of tank 4.6 ft with thickness of 3/8 inches, we can get the internal diameter of tank in 4.57 ft;

For inside resistance of tank;

R1=1 / Ai x hiR1=1 / ( x 4.57 ft x 5.8 ft x 1,000 Btu/hr- ft2-F)R1=1 / 83,271.25 hr-F / BtuR1=1.2 x 10-5 hr-F / Btu

For resistance of tank given the inside and outside diameter of tank and the film coefficient;

R2=ln (Do / Di) / (2 x x L x K)R2=ln (4.6 ft /4.57 ft) / (2 x x 5.8 ft x 312/12 Btu-in / hr- ft2-F)R2=0.0065 / 947.51R2=6.86 x 10-6 hr-F / Btu

For resistance of insulation given the thickness of insulation 0.17 ft and the film coefficient;

R3=ln (Do / Di) / (2 x x L x K)R3=ln (4.77/4.6) / (2 x x 5.8 ft x 0.28/12 Btu-in/hr-ft2-F)R3=0.036 / 0.85R3=4.24 x 10-2 hr-F / Btu

For resistance of cladding given the thickness of cladding 0.0066 ft and the film coefficient;

R4=ln (Do / Di) / (2 x x L x K)R4=ln (4.7766/4.77) / (2 x x 5.8 ft x 1,400/12 Btu-in/hr-ft2-F)R4=0.0014 / 4,251.63R4= 3.29 x 10-7 hr-F / Btu

For outside resistance of tank natural convection;

R5=1 / Ao x hoR5=1/ ( x 4.7766 ft x 5.8 ft x 2 Btu/hr- ft2-F)

R5=1 / 174.07 hr-F / BtuR5=5.74 x 10-3 hr-F / Btu

Summation of resistance as follows;

R=R1 + R2 + R3 + R4 + R5R=1.2 x 10-5 + 6.86 x 10-6 + 4.24 x 10-2 + 3.29 x 10-7 + 5.74 x 10-3 R=0.048 hr-F / Btu

QLT1=(Tin x Tout) / RQLS1=(118.4 F 77 F) / 0.048 hr-F / BtuQLS1=862.5 Btu/hr

Compute the surface area of top and bottom of tank with the formula of ellipsoidal;

SA=4 x ((ap bp + apcp+ bpcp)/3)1/p

Where:a=Radius of Major Axis (2.3 ft)b=Radius of Minor Axis (2.3 ft)c=Radius of Vertical Distance (0.98 ft)P=1.6075 SA=4 x ((ap bp + apcp+ bpcp)/3)1/pSA=4 x ((2.31.6075 x 2.31.6075 + 2.31.6075 x 0.981.6075 + 2.31.6075 x 0.981.6075) / 3)1/1.6075SA=4 x ((3.81 x 3.81) + (3.81 x 0.97) + (3.81 x 0.97)/3) 0.622 SA=4 x ((14.52) + (3.69) + (3.69)/3) 0.622 SA=4 x x (21.9/3) 0.622SA=43.27 ft2

Substitute the thickness of tank, thickness of insulation and thickness of cladding;

L/K=0.0313 ft / (312/12) Btu-in/hr- ft2-F + 0.17 ft / (0.28/12) Btu-in/hr-ft2-F + 0.0066 ft / (1,400/12) Btu-in/hr-ft2-F) L/K=0.0012 Btu / hr- ft2-F + 7.29 Btu / hr- ft2-F + 0.000057 Btu / hr- ft2-F L/K=7.29 hr- ft2-F / Btu

Given the surface area of top and bottom of tank and the summation of L/K, substitute the values in the equation;

QLS2=AT x (Tin x Tout) / (1/ho) + (1/hi) + L/KQLS2=43.27 ft2 (118.4 F 77F) / (1/1,000 Btu /hr-ft2-F) + 1/2 Btu/hr- ft2-F + 7.29 hr- ft2-F / Btu QLS2=1,791.38 / (0.001 + 0.5 + 7.29)

QLS2=1,791.38 / 7.791QLS2=229.93 Btu/hr

For total heat lost of tank;

QLT=QLS1 + QLS2 QLT=862.5 Btu/hr + 229.93 Btu/hr Heat lost of TankQLT=1,092.43 Btu/hr

Since the selected number of hot water tank is two (2) the total heat lost of tank has a total of 2,184.86 Btu/hr.

Total heat lost of pipe sizes 2.5 inches, 2 inches, 1- inches, inches supply to guest room and 2 inches return pipe to hot water tank (see Appendices for figure reference annex 4).

Based on schematic diagram and architectural layout, we can compute the length of pipe .

QLP=(Tin x Tout) / (1/ho) + (1/hi) + A/K

Where:Tin=Temperature Inside the Pipe (F)Tout=Average Ambient Temperature (F)ho=Outside Surface Coefficient (Btu / hr-ft2-F)

hi=Inside Surface Coefficient (Btu / hr-ft2-F)K=Film Coefficient (Btu-in / hr-ft2-F)

Total heat lost of pipe size 2.5 inches;For inside resistance of pipe size 2.5 inches;Given the length 44 ft and the nominal diameter of pipe 2.64 inches or 0.22 ft with thickness of 0.094 inches, we can get the internal diameter of tank in 2.55 inches or 0.2125 ft;

R1=1 / Ai x hiR1=1/ ( x 0.2125 ft x 44 ft x 1,000 Btu/hr- ft2-F)R1=1 / 29,373.96 hr-F / BtuR1=3.4 x 10-5 hr-F / Btu

For resistance of copper pipe 2.64 inches given the inside and outside diameter of pipe and the film coefficient;

R2=ln (Do / Di) / (2 x x L x K)R2=ln (0.22/0.2125) / (2 x x 44 ft x 2,616/12 Btu-in / hr- ft2-F)R2=0.035 / 60,268.45R2=5.81 x 10-7 hr-F / Btu

For resistance of insulation given the thickness of insulation 1 in or 0.125 ft and the film coefficient;R3=ln (Do / Di) / (2 x x L x K)R3=ln (0.345/0.22) / (2 x x 44 ft x 0.27/12 Btu- in/hr-ft2-F)R3=0.4499 / 6.22R3=7.23 x 10-2 hr-F / Btu

For resistance of polyethylene tape given the thickness of 0.0066 ft and the film coefficient;

R4=ln (Do / Di) / (2 x x L x K)R4=ln (0.352/0.345) / (2 x x 44 ft x 3/12 Btu-in/hr-ft2-F)R4=0.02 / 69.12R4=2.89 x 10-4 hr-F / Btu

For outside resistance of pipe;

R5=1 / Ao x hoR5=1/ ( x 0.352 ft x 44 ft x 2 Btu/hr- ft2-F)R5=1 / 97.31 hr-F / BtuR5=1.03 x 10-2 hr-F / Btu

R=R1 + R2 + R3 + R4 + R5R=3.4 x 10-5 + 5.81 x 10-7 + 7.23 x 10-2 + 2.89 x 10-4 + 1.03 x 10-2 R=0.08 hr-F / Btu

QLP=(118.4 77) / 0.08QLP=517.5 Btu/hr Heat Lost @ 2.5 inches

2.3.1.4.2.2 Total heat lost of pipe size 2 inches;For inside resistance of pipe size 2 inches;Given the length 334 ft and the nominal diameter of pipe 2.13 in or 0.18 ft with thickness of 0.08 inches, we can get the internal diameter of tank in 2.05 in or 0.17 ft;



R2=ln (Do / Di) / (2 x x L x K)R2=ln (0.18/0.17) / (2 x x 334 ft x 2,616/12 Btu-in / hr- ft2-F)

R2=0.057 / 457,492.36R2=1.25 x 10-7 hr-F / Btu

For resistance of insulation given the thickness of insulation 1 in or 0.08 ft and the film coefficient;

R3=ln (Do / Di) / (2 x x L x K)R3=ln (0.26/0.18) / (2 x x 334 ft x 0.27/12 Btu- in/hr-ft2-F)R3=0.37 / 47.22R3=7.84 x 10-3 hr-F / Btu


R4=ln (Do / Di) / (2 x x L x K)R4=ln (0.267/0.26) / (2 x x 334 ft x 3/12 Btu-in/hr-ft2-F)R4=0.027 / 524.65R4=5.15 x 10-5 hr-F / Btu


R5=1 / Ao x hoR5=1/ ( x 0.267 ft x 334 ft x 2 Btu/hr- ft2-F)

R5=1 / 560.32 hr-F / BtuR5=1.78 x 10-3 hr-F / Btu

R=R1 + R2 + R3 + R4 + R5R=5.61 x 10-6 + 1.25 x 10-7 + 7.84 x 10-3 + 5.15 x 10-5 + 1.78 x 10-3 R=9.7 x 10-3 hr-F / Btu

QLP=(118.4 77) / 9.7 x 10-3QLP=4,268.04 Btu/hr Heat Lost @ 2.0 inches

Total heat lost of pipe size 1- inches;For inside resistance of pipe size 1- inches;Given the length 1,300 ft and the nominal diameter of pipe 1.38 in or 0.155 ft with thickness of 0.0059 inches, we can get the internal diameter of tank in 1.374 in or 0.115 ft;

R1=1 / Ai x hiR1=1/ ( x 0.115 ft x 1,300 ft x 1,000 Btu/hr- ft2-F)R1=1 / 469,669.2 hr-F / BtuR1=2.13 x 10-6 hr-F / Btu


R2=ln (Do / Di) / (2 x x L x K)R2=ln (0.155/0.115) / (2 x x 1,300 ft x 2,616/12 Btu-in / hr- ft2-F)R2=0.298 / 1,780,658.88R2=1.67 x 10-7 hr-F / Btu


R3=ln (Do / Di) / (2 x x L x K)R3=ln (0.235/0.155) / (2 x x 1,300 ft x 0.27/12 Btu- in/hr-ft2-F)R3=0.42 / 183.78R3=2.29 x 10-3 hr-F / Btu


R4=ln (Do / Di) / (2 x x L x K)R4=ln (0.242/0.235) / (2 x x 1,300 ft x 3/12 Btu-in/hr-ft2-F)R4=0.029 / 2,042.04R4=1.42 x 10-5 hr-F / Btu


R5=1 / Ao x hoR5=1/ ( x 0.242 ft x 1,300 ft x 2 Btu/hr- ft2-F)R5=1 / 1,976.69 hr-F / BtuR5=5.06 x 10-4 hr-F / Btu

R=R1 + R2 + R3 + R4 + R5R=2.13 x 10-6 + 1.67 x 10-7 + 2.29 x 10-3 + 1.42 x 10-5+ 5.06 x 10-4R=2.81 x 10-3 hr-F / Btu

QLP=(118.4 77) / 2.81 x 10-3QLP=14,733.10 Btu/hr Heat Lost @ 1- inches

Total heat lost of pipe size 3/4 inches;For inside resistance of pipe size 3/4 inches;Given the length 4,500 ft and the nominal diameter of pipe 0.87 in or 0.072 ft with thickness of 0.078 inches, we can get the internal diameter of tank in 0.786 in or 0.066 ft;

R1=1 / Ai x hiR1=1/ ( x 0.066 ft x 4,500 ft x 1,000 Btu/hr- ft2-F)R1=1 / 933,055.20 hr-F / BtuR1=1.07 x 10-6 hr-F / Btu

For resistance of polypropylene pipe 0.87 inches given the inside and outside diameter of pipe and the film coefficient;

R2=ln (Do / Di) / (2 x x L x K)R2=ln (0.072/0.066) / (2 x x 4,500 ft x 0.69/12 Btu-in / hr- ft2-F)R2=0.09 / 1,625.78R2=5.54 x 10-5 hr-F / Btu


R3=ln (Do / Di) / (2 x x L x K)R3=ln (0.152/0.072) / (2 x x 4,500 ft x 0.27/12 Btu- in/hr-ft2-F)R3=0.75 / 636.17R3=1.18 x 10-3 hr-F / Btu


R4=1 / Ao x hoR4=1/ ( x 0.152 ft x 4,500 ft x 1.65 Btu/hr- ft2-F)R4=1 / 3,545.61 hr-F / BtuR4=2.82 x 10-4 hr-F / BtuR=R1 + R2 + R3 + R4R=1.07 x 10-6 + 5.54 x 10-5 + 1.18 x 10-3 + 2.82 x 10-4 R=1.52 x 10-3 hr-F / Btu

QLP=(118.4 77) / 1.52 x 10-3QLP=27,236.84 Btu/hr Heat Lost @ 3/4 inches

2.3.1.4.2.5 Total heat lost of pipe size 2 inches;For inside resistance of pipe size 2.0 inches (Return Pipe);Given the length 400 ft and the nominal diameter of pipe 2.13 in or 0.18 ft with thickness of 0.08 inches, we can get the internal diameter of tank in 2.05 in or 0.17 ft;



R2=ln (Do / Di) / (2 x L x K)R2=ln (0.18/0.17) / (2 x x 400 ft x 2,616/12 Btu-in / hr- ft2-F)R2=0.057 / 547,895.04R2=1.04 x 10-7 hr-F / BtuFor resistance of insulation given the thickness of insulation 1 in or 0.08 ft and the film coefficient;

R3=ln (Do / Di) / (2 x L x K)R3=ln (0.26/0.18) / (2 x x 400 ft x 0.27/12 Btu- in/hr-ft2-F)R3=0.37 / 56.55R3=6.54 x 10-3 hr-F / Btu


R4=ln (Do / Di) / (2 x L x K)R4=ln (0.267/0.26) / (2 x x 400 ft x 3/12 Btu-in/hr-ft2-F)R4=0.027 / 628.32R4=4.23 x 10-5 hr-F / Btu


R5=1 / Ao x hoR5=1/ ( x 0.267 ft x 400 ft x 2 Btu/hr- ft2-F)R5=1 / 671.05 hr-F / BtuR5=1.49 x 10-3 hr-F / Btu

R=R1 + R2 + R3 + R4 + R5R=4.68 x 10-6 + 1.04 x 10-7 + 6.54 x 10-3 + 4.23 x 10-5 + 1.49 x 10-3 R=8.08 x 10-3 hr-F / Btu

QLP=(118.4 77) / 8.08 x 10-3QLP=5,123.76 Btu/hr Heat Lost @ 2.0 inches

The total heat lost of pipe is;

[email protected] + QLP@2 + QLP@1-1/4 + QLP@3/4 + QLP@2(RETURN)QLP=517.5 + 4,268.04 + 14,733.10 + 27,236.84 + 5,123.76QLP=51,879.24 Btu/hr Total Heat lost of pipe

Total Heat load of the guest room;

QH=QL + QLT + QLP QH=124,854.91 + 2,184.86 + 51,879.24QH=178,919 Btu/hr

Say;180,000 Btu/hr, Total Flow Rate and Total Dynamic Head of Secondary Circulating PumpTotal Volumetric Flow RateTo get the total volumetric flow rate of secondary pump, 1 gal/min per each riser for 1- inches or 1- inches pipe size are given in ASHRAE 1995 HVAC Application page 7. Based on architectural layout and pipe chase, they have 23 risers;

VS=No. of Riser x 1 gal/minVS=23 x 1 gal/minVS=23 gal/min or 3.07 ft3/min

Total Dynamic HeadFor total dynamic head of secondary pump get the friction lost of pipe and valves, required pressure of the fixture. The fixture that has maximum pressure is shower, with 12 psig based on Table 8 Proper flow and pressure required during flow for differential fixtures ASHRAE Fundamentals.

Hf=f LV2 / d2g Darcy equation

Where:Hf=Friction Head Loss (inches)d=Diameter of Pipe (inches)g=Gravitational Force (32.2 ft/s)

V=Velocity of Pipe (ft/s)f=Friction Coefficient

All data are given except for the value of friction coefficient but we can now determine the value by the equation;

f=0.00181 + 0.0011787 (z / dvS)0.355

Where: z = fluid viscosity, (0.60 centipoises 120F Table 14-8 Viscosities centipoises page 606 Power Plant Engineering by Frederick Morse).S = Specific Gravity of WaterV=1.8 ft/s or 0.55 m/s

Friction Head of 2.5 inches diameterValue of friction at 0.065 m internal diameter, friction coefficient is;

f=0.00181 + 0.0011787 (z / dvS)0.355f=0.00181 + 0.0011787 (0.60 centipoises / 0.065 m x 0.55 m/s x 1)0.355f=0.005

Compute the friction head given the effective total length of supply hot water pipe to guest room pipe (see Appendices Annex 6).

Hf=f LV2 / d2gHf=0.005 x 64.3 ft (1.8 ft/s)2 / 0.21 ft x 2 x 32.2 ft/s2Hf=1.04 / 13.52Hf=0.08 ft

Friction Head of 2.0 inches diameterValue of friction at 0.052 m internal diameter, friction coefficient is;

f=0.00181 + 0.0011787 (z / dvS)0.355f=0.00181 + 0.0011787 (0.60 centipoises / 0.052 m x 0.55 m/s x 1)0.355f=0.0053

Compute the friction head of supply hot water pipe to guest room, given the effective total length of pipe (see Appendices Annex 6).

Hf=f LV2 / d2g Hf=0.0053 x 458.1 ft (1.8 ft/s)2 / 0.17 ft x 2 x 32.2 ft/sHf=7.87 / 10.95Hf=0.72 ft

Friction Head of 1- inches diameterValue of friction at 0.035 m internal diameter friction coefficient is;

f=0.00181 + 0.0011787 (z / dvS)0.355f=0.00181 + 0.0011787 (0.60 centipoises /0.035 m x 0.55 m/s x1)0.355f=0.0059

Compute the friction head of supply hot water pipe to guest room given the effective total length of pipe (see Appendices Annex 6).

Hf=f LV2 / d2g Hf=0.0059 x 2,481.9 ft (1.8 ft/s)2 / 0.115 ft x 2 x 32.2 ft/sHf=47.44 / 7.406Hf=6.41 ft

Friction Head of 2.0 inches diameterUsing the value of friction for 0.052 m internal diameter, we can get the friction head of return hot water pipe to guest room given the effective total length of pipe (see Appendices Annex 6).

Hf=f LV2 / d2g Hf=0.0053 x 2,039 ft (1.8 ft/s)2 / 0.17 ft x 2 x 32.2 ft/sHf=35.01 / 10.95Hf=3.20 ft

Summation of friction head;

Hf=0.08 ft + 0.72 ft +6.41 ft + 3.2 ftHf=9.67 ftTo get the total dynamic head, given the summation of friction head add the pressure required of fixture;

TDH=Hf + fixture pressure required (12 psig or 27.72 ft)TDH=9.67 ft + 27.72 ftTDH=37.39 ft

Say; 40 ft total dynamic head

The flow rate of secondary circulating pump and total dynamic head are 3.07 ft3/min and 40 ft respectively.

EQUIPMENT SELECTION AND COMPUTATIONSelection of Hot Water TankReference to 2.3.1.3 the total storage required for two hours is 5,580 Liters. Commercial selection has 2,550 Liters storage capacity. The researcher suggests to use 2 units of 2,550 Liters. Each tank has 6m2 heating coil. The capacity of the selected tank will be shown in Table 1.2, simulation of equipment capacity.

Heat Pump Capacity Based on total heat lost of hot water supply from guest room, add the heat lost of pipe and hot water tank in volumetric flow rate. The total heat capacity of heat lost is 54,064.10 Btu/hr, 30% of design capacity of 180,000 Btu/hr. By simulating reference on Table 1.1 Hot water profile, 70% of volumetric flow rate is 18,000 Liters/hr based on 124,854.91 Btu/hr. The 30% heat lost in volumetric flow rate is 7,714 Liters/hr. To get the average hourly basis divide the 7,714 Liters/hr in 24 hours. The average volumetric flow rate is 321.4 Liters/hr.The selected Heat pump capacity model 268CW2-134C @ 77F with the hot water production 2,108 liters/hr, the hot water production will be added to the hot water tank recovery during operation. To make sure the required capacity of guest room will meet the requirements on the selected equipment (see Table 1.2).

Table 1.2 Hot Water Profile with Hot Water ProductionTime of dayUsage Liter/hr% UsageLosses Liter/hrTank Capacity LiterHot Water Production Liter/hrExcess Cap,Liter

1:0000%321.4510004,557

2:0000%321.4510004,236

3:0000%321.4510003,914

4:0000%321.4510003,593

5:0000%321.451002,1082,108

6:003602%321.451002,1083,535

7:009005%321.451002,1084,421

8:004,68026%321.451002,1081,528

9:001,4408%321.451002,1081,874

10:003602%321.451002,1083,301

11:003602%321.451002,1084,728

12:001801%321.451001,0545,280

13:0000%321.451005275,486

14:0000%321.4510005,164

15:001801%321.451005275,190

16:003602%321.451005275,036

17:003602%321.451005274,881

18:007204%321.451001,5815,421

19:001,0806%321.451001,0545,073

20:004,68026%321.451002,1082,180

21:001,0806%321.451002,1082,887

22:007204%321.451002,1083,953

23:005403%321.451002,1085,200

24:0000%321.4510004,878

18,000100%28,985

In Table 1.2, we can say that the selected capacity is enough to the requirements of the hotel. At peak load usage of hot water of guest room, the hot water production of heat pump using 4 compressors running has almost 10% excess capacity of peak load. Below is graph of the simulation.

FIGURE 1.4

Figure 1.4 shows the line graph of hot water requirements of guest room in blue line, the green line is the hot water production of selected capacity of heat pump and red line is the excess capacity in normal operation of heat pump a day.

Say; 2 units of heat pump 1 operational, 1 stand by

Area of Heating CoilTo counter check the selected capacity 2,550 Liters at 6m2 or 64.5 ft2 model IT 2500, get the area of copper pipe inside the indirect calorifier with the selected Heat Pump model. The heat output of the Heat Pump is 300,967 Btu/hr with compressor capacity 4 x 9hp reference on Quantum Heat Pump. The total capacity of discharge air of heat is 21,172 ft3/min at 1.6 inches total static pressure.

Given the 300,967 Btu/hr, we can compute the area of coil inside the tank by the equation;

QC=U x A x LMTDQC=300,967 Btu/hr

Where:QC=Heat Required by the Cooling CoilU=Overall Coefficient of Heat Transfer of Copper PipeLMTD=Logarithmic Mean Temperature DifferenceA=Area of Heat TransferTo determine the Logarithmic Mean Temperature Difference, see appendices for the graph to get the temperature difference of cross flow heat transfer (see Appendices Annex 7).

LMTD=(max min) / ln (max / min)LMTD=(57.6 21.6) / ln (57.6 / 21.6)LMTD=36 / 0.98LMTD=36.73

QC=U x A x LMTDA=300,967 Btu/hr / 70 Btu/hr-ft2-F x 36.73A=117.06 ft2 or 10.88 m2

Since we have two tanks we can the divide the computed area into 2. The area of copper of each tank is 5.44 m2.

Say; 6 m2, the commercial size tank meets the requirement of heat pump production.

Therefore, the heat output of the equipment 300,967 Btu/hr with compressor capacity 4 x 9hp references on Quantum Heat Pump and the area of copper of indirect calorifier meets the requirement of Hot Water System.Calculate the Power (hp) Requirements of Secondary Circulating Pump;The volumetric flow rate and total dynamic of secondary circulating pump is 3.07 ft3/min and 40 ft (refer to item 2.3.2). To determine the horsepower required of hot water secondary circulating pump by the equation;

HP= VP x x H

Where: VP=Volumetric Flow Rate (ft3/min)HP=Horsepower Required=Density of Water (62 lb/ft3) H=Total Head (ft)

HP= VP x x HHP= (3.07 ft3/min x 62.4 lb/ft3 x 40 ft) / (33,000 ft-lb/min-hp x 0.33)HP= 7,662.72 / 10,890HP=0.70

Say; 2 units of hot water circulating pump 1 operational, 1 stand by (See Appendices Annex 8 Equipment layout).

RECOVERY OF COLD AIRIn every design recovery system is very important especially in 24 hours operation. One of the advantages of the designer is by using heat pump. The discharge air of heat pump can be utilized as ventilation of elevator machine room and electrical room. Existing capacity of window type air-conditioning of elevator system is 2 x 2.5hp, 1 x 0.7hp exhaust fan for electrical room and 7.5hp air handling unit each.Since the Maxims Hotel are equip with Building Management System, the additional controls will be put in the hot water system, air conditioning of elevator machine room, exhaust fan of electrical room and air handling unit to be integrated in Building Management System. This will be used for start/stop of the system during the operation to gain the recovery system.Additional ducting should be also included. This is to convey the air discharge of heat pump. The sizes will be determined by the equation;

VAIR=A x v

Where:VAIR=Volumetric Flow Rate of Air (ft3/s)v=Velocity of Duct (ft/s)A=Area (ft2)

Given the total capacity of discharge air 36,000 m3/hr or 21,172.50 ft3/min at selected capacity of heat pump. The dimension of duct header at 2,500 ft/min velocity of duct;

A=VAIR/ vA=21,172.50 ft3/min / 2,500 ft/minA=8.47 ft2

Say for 2 ft height of duct;

A=Width x HeightW=8.47 ft2 / 2 ftW=4.24 ft

Say; Dimension of duct is 4.24 ft x 2 ft (Duct Header)

The required air flow for two units air handling unit for fresh air supply of guest room is 2 x 8,000 ft3/min. The size of duct is;

A=VAIR/ vA=16,000 ft3/min / 2,500 ft/minA=6.4 ft2Say for 2 ft height of duct;A=Width x HeightW=6.4 ft2 / 2 ftW=3.2 ft

Say; Dimension of duct is 3.2 ft x 2 ftWith the total air discharged of heat pump, 21,172.50 ft3/min less the air requirement of air handling unit of 16,000 ft3/min is 5,172.5 ft3/min. The excess capacity will run thru the elevator machine room and discharge to electrical room by means of relief damper.

For elevator machine room;

A=VAIR/ vA=5,172.5 ft3/min / 2,500 ft/minA=2 ft2


A=Width x HeightW=2 ft2 / 1 ftW=2 ft

Say; Dimension of duct is 2 ft x 1 ft

For size of diffuser for machine room at 800 ft/min;

A=VAIR/ vA=5,172.5 ft3/min / 800 ft/minA=6.47 ft2


A=Width x HeightW=6.47 ft2 / 2 ftW=3.23 ft

Say; Dimension of louver is 3.23 ft x 2 ftMATERIALS AND EQUIPMENT SPECIFICATIONAir Source Heat Pumps The heat pump water heater shall be source equipment, factory assembled, charged, and tested. The heat pump have the capability of producing not less than 140 F water, with heating capacity as indicated on Summary of Technical Data. Heat Pumps shall be tested in compliance with health and safety standards for heating Equipments that come into contact with hot and cold potable water.Heat pump unit should consist of compressor, condensers, evaporator coil, fans, water circulating pump, pipings, controls and should be factory assembled, charged, and tested. The heat pump shall contain the following components, and features.a. Cabinet should be Type 304L stainless steel. Support, channels and beams should also be constructed type 304L stainless steel. Compartment should have large access doors for servicing. Compressor and condenser should be located in separate compartment from fan and evaporator for in-operation serving. Base section under evaporator coil should have stainless steel drip pan for condensate. Cabinet should be designed for outdoor operation.

b. Fans The fan arrangement should be draw-through design. Fan should be centrifugal, direct driven at no more than 1075 RPM (60hz). Fan diameter should not be greater than 9 (229mm). Construction should be epoxy coated steel. The motor should be mounted to blower scroll with vibration isolating rubber grommets, with stainless steel fasteners. It should also be marine duty corrosion protected.

c. Evaporator Coils should be constructed with aluminum waffle plate fins mechanically bonded to seamless copper tubing. All tube joints to be brazed with silver rod. Coil should have corrosion inhibiting coated fins and type 304 stainless steel coil casings and drain pan.

d. Refrigerant should be R-134a. This refrigerant is readily available.

e. Expansion Valves should be specifically designed for heat pump use with adjustment superheat feature.

f. Scroll type, suitable for high temperature operation suitable for R-134a refrigerant.

g. Compressor Control: Compressor controls / accessories must include the following:1. Crack case Heater2. Crankcase Heater Relay3. High pressure safety switch4. Low Pressure Safety Switch

h. Indicator lights for:1. power on2. Normal Stop 3. High pressure Fail4. Low pressure Fail

j. Condenser Coaxial vented double wall type. Suitable for high temperature operation for potable water.1. Filter-Driers: Sweat connection type2. Site glass: Moisture indicating typek. Unit should be factory wired to allow a maximum of twelve compressor short cycling and allow time for suction and discharge pressure to equalize permitting the compressor to start in an unloaded condition.l. Water circulating pumps should be in-line all bronze stainless steel body centrifugal type able to deliver rated flow against the external head.

m. Controls The heat pump unit shall be factory wired for fully automatic operation. Safeties shall include and fan motor thermal overload protection, manual reset pressure stats, anti-cycling compressor relays, plus standard items recommended by the equipment manufacturer.

Hot Water Storage Tank with Heat Exchangera.Combination of Hot Water Storage and Heat Exchanger System, constructed with clear separation between domestic water and storage tank in accordance with health and safety standard for heating equipment. Hot water storage tank should be provided with coil type heat exchanger using hot water from the air source heat pump as the primary heating source.

b.Hot water storage tank should have a minimum storage capacity of 2,550 Liters vertical mounting, installed at the roof level of the hotel floor.

c.It should be insulated using Rockwool. External insulations should be provided with gage 24 aluminum cladding suitable for outdoor installation.

d.Heat exchanger should be designed for primary hot water temperature of 149 F and maximum domestic hot supply temperature of 140 F for legionella Bacteria flushing and average or normal domestic hot water supply of 113 F for shower and bathtub usage for guest room. Domestic hot water surface area of heat exchanger is 64 ft2 or 6 m2.

e.Tank should be designed for a minimum working pressure of 125 psig.

HOT WATER PIPINGCopper type L hard drawn in accordance with ASTM B-88.1 - 94. Minimum working pressure psig.1. Fittings:Wrought Copper2. Joints:Solder joints for sizes up to 50mm (2.0 inches)and mechanical grooved coupling suitable for copper tubes or pipes for the risers both up feed and down feed pipes. Polypropylene Pipes should be temperature service of 75 -C and nominal pressure 300 psig.

VALVE PRESSURE RATINGUnless otherwise indicated, use valves suitable for 150 psig, working pressure. For circulating pump discharge valves pressure should be suitable for 300 psig working pressure.line all bronze stainless steel body centrifugal type able to deliver rated flow againts e for suction and discharge press

Basic OperationThe Hot Water System using heat pump is the primary hot water supply of the hotel guest room. The system is 24 hours in operation depend on the load requirements of the guest room. The heat pump has four (4) compressors simultaneously running depend on the temperature setting to make sure the hot water temperature will meet the desired temperature requirements.At peak load, it will run in four (4) compressors to sustain the hot water storage and storage recovery. It means the energy recovery will maximize also. The discharge air will be supply to elevator machine room, electrical room and serve as the fresh air of the guest room.When the systems run in three (3) compressors, the capacity of discharge is not enough to maintain the required air requirements. One (1) air handling unit will turn off and switch to normal mode of air conditioning system (Chilled Water System) by means of Building Management System. On the other hand, one (1) air handling unit using recovery system, will give the exact quantity of air requirement and the excess will go to elevator machine room. Since the elevator machine room has no return, the owner will put relief damper to electrical room. The discharge air of the electrical room will goes out to louver door.The system operation on two (2) compressors running is the same as three (3) compressors the discharge air of heat pump will be utilized in one (1) air handling unit supply of fresh air of guest room, ventilation of elevator machine room.In low demand load of hot water of guest room the systems run in one (1) compressor, the rooms that can be utilized are the elevator machine room and the electrical room. The two (2) air handling units will run in normal mode system the Chilled Water System.And when the Heat Pumps System turns off, it means the required temperature setting meet the normal air-conditioning are running; chilled water system in air handling unit, window type air-conditioning in elevator machine room and typical exhaust fan in electrical room.The start/stop of the system will run throughout the Building Management System by different setting and by means of dry contact. The Building Management System plays a big role in the energy recovery system. Proper settings and control of the parameter are required in the system.For volume damper, the actuators are interlock with the system. The volume damper of heat pump is normally open when the heat pump is in operation and normally closed when the heat pump turns off. Same as in air handling unit the motorized damper is open when the ducting of the recovery system were connected and the fresh air of air handling unit in take is closed. When the recovery air is not enough, the air handling unit itself will run thru chilled water system. The motorized damper where the recovery duct is closed and the fresh air intake will be open.In elevator machine room, all excess air of heat pump will be supply to elevator machine room. Since the elevator machine room has no return or any opening the relief damper of the electrical room will be open. This is to maintain the air pressure of the elevator room.Since the electrical room has louver door, the air relief by the damper from elevator room will goes out to surrounding by means of the said louver door.

ECONOMIC ASPECTSAs Mechanical Engineers, it is our responsibility to design the most efficient and economical systems for our client, which needs Engineering Knowledge, Skills and Research to check what are the latest developments or technology in the field of work. Because even we select and spend so much investment to the updated and efficient equipment but if we install it in an inefficient system, the result is just the same. That will spend so much in operating cost and maintenance cost during the life span of the equipment.Since the report is limited to the Design, Installation, Operation and Maintenance of Hot Water System Using Heat Pump including Detailed Calculation of Piping, Storage Tank Design, Pump & Insulation and the Recovery of Cold Air for Machine Room Ventilation of Maxims Hotel in Newport Blvd, Newport City. The applicant at the beginning of the report, emphasizing that the Heat Pump is more advantageous than the other systems unlike electric water heater and boiler system due to its recovery. And by installing the ductwork to the machine room it will help a lot to reduce the operating cost plus the knowledge of the applicant in the importance of system efficient.Therefore the economic analysis of this report will focus on the following two (2) major items:Operating cost analysis diverting discharge air of heat pump to elevator machine room, electrical room and supply as fresh air of guest room.Investment Analysis of Heat Pump System over the Heat Pump using Recovery System.The initial investment of Heat Pump System without recovery is slightly lower than the Heat Pump System with energy recovery. But with the study, the additional cost for the energy recovery can be recouped in few years only. The investment for ducting and other accessories are the items sometimes missed by the owners that will lead to additional cost. Table 1.3 shows the comparative initial investment of the said two (2) systems. Table 1.3 Cost Comparison of EquipmentItem No.DescriptionHeat Pump SystemHeat Pump System w/ Recovery

1Heat Pump, 1 operational 1 Back up Including Primary Pump and Installation5,700,0005,700,000

2Controller relay of Air Conditioning System and Exhaust Fan15,00040,000

3Secondary Circulating Pump including Controller25,00025,000

4Wires and Contacts of BMS20,000200,000

5Ducting Works including support and miscellaneous worknone1,200,000

Total CostP5,760,000.00P7,165,000.00

The amount of Calorifier and piping system to guest room will be Php12,500,000.00 which include Contractors Profit (Hot Water System only, fixtures and cold water line are not included).Ducting works include hangers and support, grilles, filter and insulation of duct for fresh air system. Additional wires and relay to the controller will synchronize the start/stop of the Building Management System.

Annual Operating Analysis using Heat Pump Without Air Recovery System.Annual operating cost using Heat Pump without recovery is the operation where the discharge air of the heat pump is disposed on the atmosphere. The tabulation below is the summary of electrical consumption without recovery system;

Table 1.4 Heat Pump with No RecoveryTime of dayHEAT PUMP WITH NO RECOVERY

Comp, kWFan KW PCP KWSCP KWWAC, KWEF, KWAHU, KWTotal KW

1:0000.0000.553.750.511.216.0

2:0000.0000.553.750.511.216.0

3:0000.0000.553.750.511.216.0

4:0000.0000.553.750.511.216.0

5:0026.81.300.70.553.750.511.244.8

6:0026.81.300.70.553.750.511.244.8

7:0026.81.300.70.553.750.511.244.8

8:0026.81.300.70.553.750.511.244.8

9:0026.81.300.70.553.750.511.244.8

10:0026.81.300.70.553.750.511.244.8

11:0026.81.300.70.553.750.511.244.8

12:0013.40.660.70.553.750.511.230.8

13:006.70.330.70.553.750.511.223.7

14:0000.000.00.553.750.511.216.0

15:006.70.330.70.553.750.511.223.7

16:006.70.330.70.553.750.511.223.7

17:006.70.330.70.553.750.511.223.7

18:0020.10.990.70.553.750.511.237.8

19:0013.40.660.70.553.750.511.230.8

20:0026.81.300.70.553.750.511.244.8

21:0026.81.300.70.553.750.511.244.8

22:0026.81.300.70.553.750.511.244.8

23:0026.81.300.70.553.750.511.244.8

24:0000.000.00.553.751.511.217.0

Total KW/Day:783.9

With the total capacity of 783.9 KW-Day as reference for the annual operating cost;

Total KW of Heat Pump operating in 24 hrs:783.9 KW/DayAnnual KW of Heat Pump @ 365 days:286,123.50 KW/year

Say, cost of electricity is P 6.24 / KW-hr annual operating cost is;285,904.5 KW-year x P6.24 / KW-hr = Php1,785,410.64

The value of KW/hr is based on primary metering cost of 13.4KV.Annual Operating Cost = Php1,785,410.64 (Without Recovery)

Table 1.5 Building Management SequenceTime of dayHEAT PUMPAHU1 8,000 CFMAHU2 8,000 CFMAC 550 CFMEF 300 CFMEXCESS CFM OF HEAT PUMP

Comp kWFan KWFan CFM

1:0000.000stopstopstopstop0




5:0026.81.3021,172startstartstartstart4,332







12:0013.40.6610,586startstopstartstart1,736

13:006.70.335,293stopstopstartstart4,443












Based on tabulation of building management sequence, the equipments running is tabulated in power consumption to get the power savings during operation. You will notice the recovery is based on the number of compressor running.

Table 1.6 Heat Pump with Recovery SystemTime of dayHEAT PUMP WITH RECOVERY

Comp kWFan KW PCP KWSCP KWWAC kWEF, kWAHU1 kWAHU2kWTotal kW

1:0000.0000.553.750.55.65.616.0

2:0000.0000.553.750.55.65.616.0

3:0000.0000.553.750.55.65.616.0

4:0000.0000.553.750.55.65.616.0

5:0026.81.300.70.550.000.00.00.029.4

6:0026.81.300.70.550.000.00.00.029.4

7:0026.81.300.70.550.000.00.00.029.4

8:0026.81.300.70.550.000.00.00.029.4

9:0026.81.300.70.550.000.00.00.029.4

10:0026.81.300.70.550.000.00.00.029.4

11:0026.81.300.70.550.000.00.00.029.4

12:0013.40.660.70.550.000.00.05.620.9

13:006.70.330.70.550.000.05.65.619.5

14:0000.000.00.553.750.55.65.616.0

15:006.70.330.70.550.000.05.65.619.5

16:006.70.330.70.550.000.05.65.619.5

17:006.70.330.70.550.000.05.65.619.5

18:0020.10.990.70.550.000.00.05.627.9

19:0013.40.660.70.550.000.00.05.620.9

20:0026.81.300.70.550.000.00.00.029.3

21:0026.81.300.70.550.000.00.00.029.3

22:0026.81.300.70.550.000.00.00.029.3

23:0026.81.300.70.550.000.00.00.029.3

24:0000.000.00.553.750.55.65.616.0

Total KW/Day:566.7

With the total capacity of 566.7 KW/Day as reference for the annual operating cost;

Total KW of Heat Pump operating in 24 hrs:566.17 KW/DayAnnual KW of Heat Pump @ 365 days:206,845.5 KW/year

Say, cost of electricity is P 6.24 / KW-hr annual operating cost is;

206,845.5 KW-year x P6.24 / KW-hr = Php1,290,715.92

Annual Operating Cost = Php1,289,349.36 (With Recovery)

FIGURE 1.5Figure 1.5 shows the comparison of power consumption of heat pump system using recovery system in red hidden line is lower than the blue line heat pump without recovery.

OBSERVATION, COMMENTS AND RECOMMENDATIONObservationsThe Hot Water System using heat pump of Maxims Hotel in Newport Blvd, Newport City Pasay City is one of the effective, efficient and economical design in the country. The flexibility of design and recovery assured a healthy and safe environmental.

The Mechanical Engineer and the contractor is not having a hard time during the testing and commissioning of the equipment especially the validation and verification of the capacity. The simplicity of the design and start up procedure make advantage to the owner to start ahead of schedule.

Comments and RecommendationsInstalling the heat pump is an excellent and wise decision of the owner and Mechanical Engineer by the small investment in the installation and space saver especially in Newport City, a prime lot city. The owner will enjoy the great benefits of it in terms of operating cost and the assurance of safe environment.Since the discharge air temperature is approximately 25 C or 77F, diverting and installation of duct to machine room, electrical room and fresh air of hotel guest room will gain additional benefits. Installation of relay or sensor to the controller of air-conditioning of elevator machine room and exhaust fan using Building Management System, this is to maintain the required temperature of the room. Once it meet the required temperature the heat pump will automatically turn off. The air-conditioning of the machine room and exhaust fan of electrical room will turn on. Vice-versa.

No matter how high is the efficiency of our equipment and install it in an inefficient system the result is inefficient, the premium that we pay for the high efficiency equipment is wasted. Thats why it is highly recommended that the design engineer should properly study the basic engineering knowledge in his/her field of work.

CONCLUSIONHeat pump system is one of the neglected areas in the project, owner does not want to invest additional cost for recovery system so much in hot water system, but if we realize the impact of energy recovery in operation, we can say that it is very efficient.Not because of the development in the hot water system which is the heat pump recovery, the project owner can now realize that even we spend additional cost in the operation requirements an additional saving and advantages in the operation and space will gain.It is important to consider the heat pump system to comply or to minimize the environmental concern especially the waste discharge by the other hot water system.It is also equally important that the system should be properly operated and maintained by a qualified and competent Mechanical Engineer, in that way we are sure that the life span expected of the system will achieve. And its function will be throughout its useful life.REFERENCESThe Philippine Mechanical CodeThe PSME Code and Standards CommitteeRoom 401, 4/F Don Lorenzo Bldg. 889 P. Paredes St.Sampaloc, Manila

ASHRAE Handbook HVAC ApplicationsAmerican Society of Heating, Refrigeration and Air-Conditioning Engineer, Inc.1791 Tullie Circle, N.E., Atlanta, GA 30329

ASHRAE Handbook HVAC SystemsAmerican Society of Heating, Refrigeration and Air-Conditioning Engineer, Inc.1791 Tullie Circle, N.E., Atlanta, GA 30329

ASHRAE Handbook FundamentalsAmerican Society of Heating, Refrigeration and Air-Conditioning Engineer, Inc.1791 Tullie Circle, N.E., Atlanta, GA 30329

ASHRAE Equipments and SystemsAmerican Society of Heating, Refrigeration and Air-Conditioning Engineer, Inc.1791 Tullie Circle, N.E., Atlanta, GA 30329

The Theory and Practice of StationaryElectric Generating PlantFrederick Morse

ThermodynamicsVirgil Morning FairesClifford Max Simmang

Corporate FinanceBreadley / Myers / Allen

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