One of the most critical applications faced by plumbing engineers is selecting a water heater for a facility where there is a combination eyewash and emergency shower.  These fixtures are required by the ANSI Z358.1 code to deliver a specific amount of water flow for a specific period of time.  

These are the critical variables that need to be determined: How many emergency events will need to be served at a time? What water temperature will be delivered from the shower? What is the expected design cold water temperature? What is the expected design hot water temperature in the water heater? Will the system be fed by a tepid water system, or will the water be delivered by the domestic hot and cold water system and one mixing valve per fixture? What is the recovery efficiency of the water heater? What is the usable storage capacity factor for the hot water storage tank?

The basic requirement here is to deliver 23 gpm of 80°F (87 L/m of 26.7°C) water to the emergency fixture for 15 minutes.  Initially we need to determine how many gallons are required for this event.  (note 80°F (26.7°C) is a sample temperature; ANSI Z358.1 requires tepid water to be delivered between 60°F and 100°F (15.6°C and 37.8°C).)

Volume Required = 23 gpm × 15 minutes = 345 gallons of 80°F water

[= 87 L/m x 15 minutes = 1,305 liters of 26.7°C ]

There are three methods of supplying this amount of tepid water to serve this critical load: Provide sufficient recovery capacity to deliver all of this water “instantaneously;” Provide sufficient capacity from storage accounting for the usable storage capacity factor; and Provide a combination of recovery capacity and storage.

The governing equations here are:

Equation 1 (volume is all from recovery)

V_E =  r_E × t ×  hour/(60 minutes)

Equation 2 (volume is all from storage)

V_E =  M × S_tE

Equation 3 (volume is from a combination
of recovery and storage)

V_E = ( x ) × r_E × t ×  hour/(60 minutes) + ( 1 – x ) × M × S_tE

where

V_E = Available hot water, gallons (liters)

x = Proportion of hot water from recovery, 

dimensionless

r_E = Recovery rate for the water heater to deliver 100% capacity by recovery as required for the emergency
fixture, gph (L/h)

t = Event duration, minutes

(1-x) = Proportion of hot water from storage,
dimensionless

M = Usable storage capacity factor, dimensionless (same as used for the storage water heater or its storage tank.

StE = Storage capacity to deliver 100% capacity from storage as required for the emergency fixture, gallons (liters) 

For this application, let’s look at an example. We have a small industrial facility that includes a couple of rest rooms with hand sinks, a break room sink, three showers, several service sinks, and two emergency fixtures.  By Method 1: Average Hourly Demand outlined in Table 6-1 in Chapter 6 of the 2014 American Society of Plumbing Engineers (ASPE) Volume 2 Plumbing Engineering Design Handbook (PEDH), before considering the emergency fixture, we have a need for 147.7 gph (558.5 L/h) of recovery and 183 gallons (693 liters) of storage (assumes M = 0.75 usable storage capacity factor).  In this example, we are using 45°F (7.2°C) for the cold water design temperature, 80°F (26.7°C) for the tepid water temperature, and 140°F (60°C) for the hot water temperature in the storage tank.

Using the formulas discussed in the water mixing section in Chapter 6 of the ASPE Volume 2 PEDH, this 345 gallons [1,305 liter] of water for the emergency fixture can be broken down into the following mixture:

Cold water delivered: 217.9 gallons at 45°F [824 L at 7.2°C]

Hot water delivered:    127.1 gallons at 140°F [481 L at 60°C] - This is the critical water flow of consideration.

For determining the “instantaneous” capacity for One Event, remember this is a 15-minute event and not a 60-minute event.  

OPTION 1 if 100% recovery from instantaneous water heating:

Equation 4

r_E = V_E × 1/t × (60 min)/hr

where

r_E = Recovery Rate for the Emergency Fixture, gph (L/h)

V_E = Emergency Volume, Gallons (liters) – this is a calculated value, 

= 127.1 gallons (481 liters) for this example.

t = Event time, minutes – normally 15-minutes for emergency fixtures

Using Equation 4

r_E = 127.1 × 1/(15 min) × (60 min)/hr = 508.4 gph 

[ r_E = 481 × 1/(15 min) × (60 min)/hr × m^³/(1,000 L) = 1.924 m³/h ]

It should be noted here that the recovery rate for the emergency fixture is nearly 3.5 times the recovery rate for ALL of the normal plumbing fixtures in the building and as a result will have a sizable impact on the water heater sizing exercise.

Substituting the result of Equation 4 into Equation 6-2 of Chapter 6 of the ASPE Volume 2 PEDH

q = 508.4 gph × ( (1 Btu)/(lb-°F) × (8.33 lb)/gal  × ( 140 – 45°F ) ) = 402,000 Btuh

[ q = 1.924 m³/h × ( (4.188 kJ)/(kg-°K) × (999.6 kg)/m^³   × ( 60 – 7.2°K ) ) = 425,000 kJ/h ]

To attain this recovery, if electric were used:

Using Equation 6-4 of Chapter 6 of the ASPE Volume 2 PEDH

q =  (508.4 gph × ( 140 – 45°F ) )/410   = 118 kW

[ q =  (1,924 L/h × ( 60 – 7.2°C ) )/860   = 118 kW ]

If natural gas, propane, LPG, LNG, or fuel oil are used to heat the water, the following equation is used to determine the actual input fuel rate required to produce the recovery rate required:

Equation 5

Qin = q / n

where

Qin = Time rate of energy input, British thermal units per hour (Btuh) (kilojoules per hour)

n = Water heater efficiency, dimensionless

q = Time rate of heat transfer - output, British thermal units per hour (Btuh) (kilojoules per hour)

Using Equation 5, if natural gas were used with a thermal efficiency of 90%:

Qin = 402,000 Btuh / 90% = 447,000 Btuh

[ Qin = 425,000 kJ/h / 90% = 472,200 kJ/h ]

OPTION 2 if 100% of capacity from storage:

Equation 6

S_tE = V_E / M

where

S_tE = Storage capacity to deliver 100% capacity from storage as required for the emergency fixture, gallons (liters)

V_E = Hot water required for the Event, gallons (liters)

M = Usable storage capacity factor (dimensionless), same as used for the water heater.

As was seen, the amount of recovery required to heat 100% of the water for the emergency fixture is much, much larger than the building domestic water recovery requirement. If the amount of energy were 100% stored, the 127.1 gallons (481 liters) required for our example would need to be added to the storage capacity for the building’s water heater system. You would need to include the usable storage capacity factor for this application. 

Using Equation 6:

S_tE = 127.1 gallons / 0.75 = 169.5 gallons

[ S_tE = 481 liters / 0.75 = 642 liters ]

OPTION 3 if combination of recovery capacity and storage:

In this example we varied the amount of required capacity that is provided by recovery and the amount of required capacity that is provided by storage.  “x” for recovery and “( 1 – x )” for storage. We then varied “x” until we had a combination of water heater storage capacities and recovery rates until the combination was optimized to match both the storage capacity and the heat input capacity of a standard water heater package furnished by one of the water heater manufacturers. For this example, we will use equation 3.

For instance, we determined that using 318 gallons (1,204 liters) of storage and two gas water heaters with 110,000 Btuh (116,000 kJ/h) input at 90% efficiency provided optimal use of the water heater selections in our catalog.  

The building’s domestic hot water requirement was 147.7 gph (558.5 L/h) which equates to 130,000 Btuh (137,100 kJ/h) input with a 90% efficiency.  

So the capacity available for the emergency fixture is (2) x 110,000 – 130,000 = 90,000 Btuh ( (2) x 116,000 – 137,100 = 94,900 kJ/h ). 90,000 Btuh (94,900 kJ/h) represents 20.1%   of the required emergency fixture capacity (447,000 Btuh (472,200 kJ/h)).  Therefore, 79.9% of the capacity must come from storage. 79.9% of 169.5 gallons (642 liters) is 135 gallons (511 liters). The original storage requirement for the building’s domestic hot water system was 183 gallons (693 liters).  

So total required storage for the building and the emergency fixture is 318 gallons (1,204 liters). The plumbing engineer can modify the percentage going to storage (“1 - x”) and percentage going to recovery (“x”) until the size of his water heater and storage is optimized.

Once the plumbing engineer is assured there will be adequate tepid water available to supply the emergency event for 15-minutes, the final step is to assure that tepid water will be delivered at the required temperature. The emergency fixture(s) need to be provided individually or as a group with an ASSE 1071 thermostatic mixing valve. The ASSE 1071 valve is different from other thermostatic mixing valves in that it will provide a cold water bypass in the event that hot water becomes unavailable. The valve also provides positive shut off in the event of cold water pressure failure.

The final step is to insure there is adequate water pressure to deliver the tepid water to the fixture outlet. The emergency fixture standards call for the fixtures to deliver the rated flow at 30 psig (207 kPa) inlet pressure. This does not include the pressure drop encountered by the ASSE 1071 thermostatic mixing valve. Typical thermostatic mixing valve pressure drops range from 5 psig (34 kPa) for large valves to 25 psig (172 kPa) for a more modestly sized valve. Therefore, the required delivery pressure for cold and hot water at the fixture needs to be between at least 35 to 55 psig (241 to 379 kPa).    The plumbing engineer also needs to determine the pressure drop for the flow from the “street” to the fixture to finalize this design.

Keep in mind that if you size the system for 70°F [21.1°C] delivery temperature and set the discharge temperature of the ASSE 1071 mixing valve for 80°F [26.7°C], you have the potential to run out of hot water after maybe 10 minutes at which time only cold water will be delivered.

Paul Baker, PE, CPD, GPD is a senior mechanical engineer at Jacobs Engineering Group, Inc., in Fort Worth, Texas. Baker graduated with a BSME in Mechanical Engineering from Lehigh University. He has more than 39 years of practical mechanical engineering experience. He has been an ASPE member since 2006. Baker is a registered professional engineer in nine states. He would like to thank Brodie Heflin, P.E., of Jacobs for the engineering content of this article. Email him at paul.baker@jacobs.com.