Steam is to be condensed on the shell side of a 1 shell pass...

Steam is to be condensed on the shell side of a 1-shell-pass and 8-tube-passes condenser, with 50 tubes in each pass at 90°F (hfg = 1043 Btu/lbm). Cooling water (cp = 1.0 Btu/lbm·°F) enters the tubes at 60°F and leaves at 73°F. The tubes are thin-walled and have a diameter of 3/4 in and a length of 5 ft per pass. If the overall heat transfer coefficient is 600 Btu/h·ft^2·°F, determine (a) the rate of heat transfer, (b) the rate of condensation of steam, and (c) the mass flow rate of cold water.

Transcript

00:02 The saturation temperature of stream let's say this is tha which is equals to 90 degree furnace and cooling water heat capacity cp is given which is 1 divided by alp meter degree fahrenheit so here we are given the inlet temperature which is represented as c tc i which is equal to the 60 degree finite and and the outlet temperature is given which is represented by tco is 73 degree finite.

00:37 Here we are given the diameter of tube which is equal to 3 divided by the 4.

00:42 L length of tube l is given which is 5 feet per pass.

00:48 And the number of passes np is given which is 8 and the number of tubes in 1 passes and t is given which is 50.

00:55 So overall heat transfer coefficient cu is given which is 600.

01:00 B t divided by h f t square degree celsius cp is given which is 18 t u lbm degree finite and hfg is given which is 1043 b t u per l bm so here we have to considering the parallel flow agile so here we are considering the parallel flow so according to this, according to this parallel flow hg we can make our diagram here let's say this is the diagram here this is t set this is cooling water the water the area between this two is t del t e this is t h i this is t h i this is t c not this is delta t i so from here let tm be the log mean temperature so here tm is equal to delta t i minus delta t divides by the ln of delta t i divides by the delta t so from here we can say that t of m is equal to t h i minus t c i minus t h i minus tc e divides fadda ln of t h i minus t c i divides by a t h d h dv divided by a t h e so after simplifying this term we can write it as that delta t m is equal to 30 minus 17 divides by ln of 30 divides by the 17 after solving this we get the value of t m as to 2 .887 degree final so this is the value of for the dm here now we have to find out the so we have to find out the total heat transfer at g which is equals to the a which is equal to the np multiplied by the np so here the value of nt is 0 .9817 and that of np's 8 .585 multiplied by the 50 so from here we get the value of a which comes out to be 392 .683 fit square...

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