Heat engineering; a text book of applied thermodynamics for engineers and students in technical schools . v 2Fig. 1.—Adiabatic on the pV plane. 6 HEAT ENGINEERING idly and the height of the curve is practically zero after a shortdistance to the right. If u does not change on the curve (the isodynamic) the heatadded is equal to the work done. Jq = I pdv (12) I pdv = 0, or dv = 0, and v = constant. Jq = ub — ua (13) If there is no this case GRAPHICAL REPRESENTATION OF HEAT ON PLANE In Fig. 2 let ab be any path. Draw from a and b two adia-batics to infinity and from a draw the isodyn
Heat engineering; a text book of applied thermodynamics for engineers and students in technical schools . v 2Fig. 1.—Adiabatic on the pV plane. 6 HEAT ENGINEERING idly and the height of the curve is practically zero after a shortdistance to the right. If u does not change on the curve (the isodynamic) the heatadded is equal to the work done. Jq = I pdv (12) I pdv = 0, or dv = 0, and v = constant. Jq = ub — ua (13) If there is no this case GRAPHICAL REPRESENTATION OF HEAT ON PLANE In Fig. 2 let ab be any path. Draw from a and b two adia-batics to infinity and from a draw the isodynamic until it strikesthe adiabatic from b at 12 v Fig. 2.—Graphical representation of heat added. Now la oo = ua 2b co = ub lab2 = 1 pdv 26 oo + la&2 = la&co f = ub + 1 pdv lab oo — la oo = oo ab oo r = Ub + 1 pdv - t/ a ub — ua-\- p t/ a J? FUNDAMENTAL THERMODYNAMICS 7 This gives the important relation that the area on the pvplane between any path and the two adiabatics from the ex-tremities of the path to infinity is equal to the heat added on thepath. This area is infinite in extent and consequently cannot berepresented graphically. To make it finite and definite, the iso-dynamic ac was drawn. Now ua = uc. Hence 3c °° = la °° and lab co — 3c oo = Jq = labcS or the area beneath any path added to that beneath an adiabaticfrom the second point of that path to the intersection of theadiabatic and the isodynamic from the first point is equal to theheat added on the path. It will be seen that lab2and f = I pdv 26c3 = Ub — Uc = Ub — ua This latter statement should be clear since uc = ua. SCALE OF TEMPERATURE
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