700

4) Calculate surface temperature ts:

5) Calculate tm = (700+129)/2 = 415°F. The insulation thermal conductivity at 415°F is 0.49, which is close enough to the assumed value (see Appendix 15.1). Then, QF = 52 Btu/hr ft2.

6) Determine the outside area of insulated pipe. From Table 15.2, pipe radius = rl = 2.25-in., then, outside insulated area (ft2):

= 2n (rl+tk)(length)/(12 in./ft) = 2n (2.25 in+3.5 in.)(100 ft)/(12 in./ft) = 301 ft2

7) Calculate heat losses with insulation:

STEP 4. Determine heat losses savings Qjsavings.

= (435,600-15,652 Btu/hr)(4,160 hr/yr) (1 MMBtu/106 Btu)

STEP 5. Determine fuel cost savings. Assuming 1 MCF = 1 MMBtu:

Fuel savings

= (QTsavings)(conVersion factor)/

(combustion efficiency) = (1,747 MMBtu/yr)(1 MCF/MMBtu)/(0.8) = 2,184 MCF/yr

Then,

Fuel cost savings = (fuel savings) (fuel cost) = (2,184 MCF/yr) ($3/MCF) = $6,552/yr

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