A Sample Dryer Project

In this our last post on impingement dryer design we illustrate a simple set of dryer design calculations.

Given:

Tweb entry=60°F

Tweb exit=135°F

T_{Air}=220°F, T_{Web}=120°F or the wet web/coating temperature in the dryer

Production rate for the web=100 lbs/hr

Web width = 2 ft

Web Speed = 50 ft/min=3000 ft/hr

Web Specific Heat= 0.4 btu/lbm·°F

Production rate for the coating solids=100 lb/hr

Coating Solids Specific Heat= 0.35 Btu/lbm·°F

Production rate for the coating liquid=50 lb/hr

Coating liquid is toluene with specific heat of 0.41 Btu/lbm·°F, latent heat of 156 Btu/lb and a boiling temperature of 230°F

Allowable Heat Transfer Coefficient = 15 Btu/(hr·°F·ft^{2})

We calculate the product load:

Product Load_{substrate}=100 lb/hr x 0.4 Btu/(lbm∙℉) x(135℉-60℉)=3,000 Btu/hr

Product Load_{solids}=100 lb/hr x0.35 Btu/(lbm∙℉) x(135℉-60℉)=2,625 Btu/hr

Product Load_{liquid}=100 lb/hr x (156+0.41 Btu/(lbm∙℉) x(230-60))=22,570 Btu/hr

Total Product Load = 28,195 Btu/hr

We then calculate the required dryer area by dividing the product load by the product of the heat transfer coefficient and the difference in temperature between the impinging air and the web:

Required dryer area=28,195 Btu/hr÷[15 Btu/(hr∙℉∙ft^{2} ) x(220℉(T_{Air})-120℉(T_{Web}))]=18.8 ft^{2}

With a 2 ft web width the dryer length is 18.8 ft^{2}/2ft=9.4 ft