Saturated water vapor at 100c condenses in a 1 shell and 2...

Saturated water vapor at 100°C condenses in a 1-shell and 2-tube heat exchanger with a surface area of 0.5 m^2 and an overall heat transfer coefficient of 2000 W/m^2·K. Cold water (cpc = 4179 J/kg·K) flowing at 0.5 kg/s enters the tube side at 15°C. Determine (a) the heat transfer effectiveness, (b) the outlet temperature of the cold water, and (c) the heat transfer rate for the heat exchanger.

Transcript

00:01 This question we are given here, saturated paper at 100 celsius condensed.

00:04 So that is a condensate 100 celsius and that value we are given surface area, that is 0 .5 meters square.

00:14 Over a heat transfer coefficient, so we are given a coefficient of heat transfer temperature that is 2 ,000 watt per meter square for kelvin and the cold water, the specific heat we are given here and 0 .5 kilogram per second dienter's the tube at 15 celsius so temperature we are given here that is 15 celsius so here that's all our first part of this question so first understand that we have given a mass flow here we are given here mass flow that is a 0 .5 here we are given a that is a mass flow mw prime which is equal to we are given here 0 .5 kilogram per second 0 .5 kilogram per second and we are given a temperature that is a 13 that is a tc temperature here and this the temperature is t h in temperature and here the area is 0 .5 meter per second square and this is a value of u here this is a value of the over a heat transfer coefficient now let's sell our first of part of this equation so here you can wait in the first of part of this question we want to find heat transfer effectiveness so here we want to find heat transfer effectiveness before we finding heat transfer effectiveness let's find out heat capacity so here you can notice cc which is equal to am multiple we can write here cpc now we just substitute of value into the formula so here we can write 0 .5 multiple we can wait here 4179 now we just simplify we can write here answer is so answer is here 2089 .5 that is a watt for kilogram what for kelvin now let's find out the maximum heat transfer rate so here we can make maximum heat transfer rate which is equal to we know the formula is c minimum mulliverr we can write here t h in minus we can write t c in now we just substitute value into this formula so here we can write that is a 2089 .5 multiple we can wait here that is a hundred minus we can write 13 now you simplify and we can write here answer is 181 .786 kilowatt so this is the value of q max heat transfer maximum heat transfer now let's find out n -du.

02:33 So n -tu which is equal to we know the formula is here ua divided where we can write here c minimum which is equal which is simply which substitute value into this formula so here we can weigh 2 ,000, only but we can write here 0 .7 here we can write area we are given here.

02:48 Area is given here 0 .5 we can write here c minimum it is 2089 .5 now we just simply find we can write your value of n2 which is equal to answer here is 0 .67 we just simply find we can write here answer is 0 .67 now let's solve our first subpart so in the first subpart we want to find the effective nays so that value e we want to we want to find so we know the formula of q that is a heat transfer which is equal to we know the effective nase multiple i can wait here max here we can write as a q max now here we can write effective nis which is equal we can write 1 minus e xp by nta...

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