Burnham BH Selection Procedure Steam Application, Given Design Conditions, Application Procedure

C.SELECTION PROCEDURE - STEAM APPLICATION

Example 1:

GIVEN DESIGN CONDITIONS:

Unit Heater Model: BH-63

Steam Pressure: 10 lb

Entering Air Temperature: 70°F

APPLICATION PROCEDURE:

1.	Capacity:	BtuA= BtuS x Heating Capacity Factor	￿	The standard capacity of an BH-63 is
		= 63,000 x 1.06 = 66,780 Btu/hr		63,000 Btu/hr from Table 3.
			￿	Steam heating capacity conversion factor
				for given circumstance is1.06 from Table 5.
2.	Condensate:	Condensate Rate = BtuA / Latent Heat of Steam	￿	The latent heat of steam at 10 lb.
		= 66,780 / 952.5 = 70.11 lb./hr		is 952.5 Btu/lb., from Table 8.
3.	Final Air	ATRA = (FATS-EATS) x Air Temp. Rise Factor	￿	The final air temperature of an BH-63 at
		= (12°F - 60°F) x 107 = 55.64°F		std. conditions is 112°F from Table 3.
	Temperature:	FATA = EATA + ATRA = 70°F + 55.64°F	￿	Air temperature rise conversion factor for
		= 125.64°F		given circumstance is 1.07 from Table 6.
			￿	The max. mounting height at std. condition
4. Max. Mounting		MMHA = MMHS x Correction Factor		is 14 feet, from Table 3.
	Height:	= 14 feet x 0.94 = 13.16 feet	￿	The mounting height correction factor is
				0.94 from Table 7.

Example 2:

GIVEN DESIGN CONDITIONS:

Vertical Air Delivery requested

Heating Load: 100,000 Btu/hr

Entering Air Temperature: 50°F

Steam Pressure: 2 lb

APPLICATION PROCEDURE:

1. Unit Selection:	BtuS = BtuA / Heating Capacity Factor	￿	From Table 3, an BV-95 model will meet the
	=100,000 / 1.07 = 93,458 Btu/hr		heating requirements with its rated capacity of
	(at standard conditions). Select BV-95.		95,000 Btu/hr at standard conditions.
2. Actual Heating	The capacity of an BV-95 at actual conditions	￿	Steam heating capacity conversion factor for
Capacity:	will be BtuA = BtuS x Heating Capacity Factor		given circumstance is 1.07 from Table 5.
	= 95,000 x 1.07 = 101,650 Btu/hr.

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