a flow of 2 kgs completely dry air at t1 100 kpa is cooled down to 10pc by spraying liquid water

Step 1: Calculate the exit moist air humidity ratio using the formula ( omega = frac{0.622 cdot

A flow of 2 kgs completely dry air at t1 100 kpa is cooled...

A flow of 2 kg/s of completely dry air at T1 = 100 kPa is cooled down to 10°C by spraying liquid water at 108°C, 100 kPa into it, so it becomes saturated moist air at 10°C. The process is steady-state with no external heat transfer or work. Find the exit moist air humidity ratio and the flow rate of liquid water. Find also the dry air inlet temperature T1.

Transcript

00:01 We are given here flow.

00:02 So here we have a flow that is dm by d t which is equal we can wait 2 kilogram per second that completely dry at a d1 that is 100kcal that is the pressure that pum which is equal to 100 kilo pascal and that cool down to a 10pc.

00:22 Now here in this question we want to find here in this question first way we want to find this question first we want to find this pressure find that work we want to find so let's solve this question so how to find it we are we know that p which is equal to p which is equal to p which is equal we can wait 1 .2 to 76 kilo pascal and here we know that at tens less is we are given here that hfg which is equal we can write 2477 .7 .7 kilojoule per kilogram and now here we're using formula of w2 which is equal we can write here 0 .622 here we can write here 0 .622.

01:11 What level we can write here pv divided by we can write here p a minus pv now we just substitute value into the formula so here you can write w2 which is equal to 6 0 .6 2 1 .6 to whatever we can write here value of pv which is equal you can 1 .2 to 76 divided by we can write here value of p a which is 100 minus 2.

01:33 We can write here that if pv which is equal to 1 .2 -76 now we just simply find we can write here value of work which is equal to answer here is 0 .0703 so this is the value of that work let's follow our second question so suppose let's follow our second question in the conservation of mass so we can write m as we can write m l i q which is equal we can write m a plus 1 plus m a multiple we can write a 1 plus w 2 now we can write m that is a i like you which is equal we can write m a multiple by w2 now is a substitute value into the somatres where you can write w multiple we can write here 0 .0 7 7 3 now we just simply find we can write here answer is so answer here is 0 .01506 that is is in the kilogram per second and that are all in the rate of mass this is the we are given here that is here changing the mass for unit time that is the data here we are given here dm by d t now that's all our thoughts are part of this question so here we can write that conservation of mass so here we can write here to m a that is h1 plus we can write m i like u whats however h f which is equal we can write here m a mother we can write h a 2 plus we can write w 2 here we can write hg 2 here we can write here h a 1 minus each equal to we can write cpa multiple by t 1 minus t 2 which is equal we can write w 2 multiple by h g2 minus we can write w2 multiplied 2 multiple我有 2 now here we just substitute of value into this formula and then we just simplify and we can write here answer is so here we can write that w to hgf now we just substitute of value into this formula so here you can write 0 .0 0703 multibra we can write here 2477 .75 now we just simplify and we can write here answer is answer here is 9 .15 kilojoule for kilo now let's solve the here last question...

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