Heat engineering; a text book of applied thermodynamics for engineers and students in technical schools . warm substance flowing in the oppositedirection from the flow of the cool substance the arrangement iscalled counter flow (Fig. 27). A minus sign is required beforethe dtc as the temperature decreases with an increase of dF. If tic is the temperature of the cool substance at the end cor-responding to the entrance of the warm substance and t2c is thatof the cool substance at the point of exit of the warm substance,the following is true. Mhch(tlh - t2h) = T Mccc(tlc - t^) (14) HEAT TRANSMISS
Heat engineering; a text book of applied thermodynamics for engineers and students in technical schools . warm substance flowing in the oppositedirection from the flow of the cool substance the arrangement iscalled counter flow (Fig. 27). A minus sign is required beforethe dtc as the temperature decreases with an increase of dF. If tic is the temperature of the cool substance at the end cor-responding to the entrance of the warm substance and t2c is thatof the cool substance at the point of exit of the warm substance,the following is true. Mhch(tlh - t2h) = T Mccc(tlc - t^) (14) HEAT TRANSMISSION 83 The upper (minus) sign is for parallel flow and the lower(plus) sign is for counter flow in this equation. This of courseassumes that all the heat leaving the warm substance goes intothe cool substance and hence there is no radiation loss. The following notation may be used: hh ~ he = Ati hh — he = A^2 and from (13) it is seen that Mhch(tlh - txh) = + Mccc(tic - txc)± Mhch or txr. — Mccc ft txh) ~f he Atx = L * _ f _L MhCh / X MhCh + tXc — txh Jl n/r _ txh • 71/T _ t\h Mccc Mccc (15). Temperature of Warm -Counter current flow. Now Fig. 27 ^ = l1 ± J^J dAtxMhchMccc clL :.dt xh 1 ± Substituting this in (13) gives KAtxdF = T7— dAh 1 ± MhCh dF = 3600 McCcMhch dAt, K Mhch At, 3600K ~ McCc Mhch ± Mccc (16) (17) Aii 84 HEAT ENGINEERING Now H = i£(mean At)F = 3600Mhch(h - t2) A, 3600ilf^fti - h)meanA£ = ™ Substituting for F its value from (17) Mhch(h - h) meanA£ = Mhch 1 + _i *h Mhch 10g At2 (18) ~ McCc From (14) , Mhch _ tic — tic - McCc tlh tlh ? tic ? (tlh — t2c) .. Equation (18) reduces to mean At tihAh - AUtm — foh Ah — At* hh log, At,AU (18) (19) This is independent of the direction of flow. The form of theexpression is the same for parallel or counter current flow. It tl* \*h t*h yxc hc trc
Size: 2309px × 1083px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, bookpublishernewyo, bookyear1915
