00:01
In this question, we need to find first the number of tubes or paths required to condense 2 .3 kilogram per seconds of steam.
00:10
So, performing an energy balance, we have this formula that is q is equals to mh, h, f, g is equals to c, c, c, t, c, point o, minus, t c point i for hot liquid at t is equals to 373 kelvin so we have our hfg which is equals to 2 ,257 kilojoules per kilogram our q is equal to 2 .3 kilogram per 2 ,257 kilojoules per kilogram so our q is equals to 5191 .1 kilowatts so our c is equals to mccpc so based on this for cold fluid at t is equals to 298 which is a constant temperature plus 288 kelvin over 2, this would be equal to 295 kelvin.
01:45
Now for cpc, so we have cpc which is equal to 4 .181 kilojoules per kilogram kelvin.
01:57
Our p is 998 kilogram per cubic meter.
02:02
Our k is 0 .606 watts per meter per kelvin we have our micrometer or micro which is 0 .959 times 10 raised the part negative 3 and seconds per second meter and our pr is equals to 6 .62 now we have the flow rate you need to calculate the flow rate per tube so to calculate the rate per tube we were going to use a second equation which is mc1 which is equals to pi d squared p um u m over 4 so this would be the second question or the second equation so the rate of your per tube you have pi times 0 .0 1 14 meter seconds times 998 kilogram per cubic meter times 3 .5 meter per seconds over 4.
03:19
So your flow rate per tube is equal to 0 .537 kilogram per seconds.
03:30
Now in this part we need to calculate the flow rate of the cold fluid so to calculate the flow rate so we have the flow rate of the cold fluid as mc is equals to n mc .1 so our mc is equals to 0 .5377 n kilogram per second our cc is equals to 0 .377 n kilogram per second times 4181 joules per kilogram kelvin and our cc is we have 2 ,248 .12 n w over k for the reynolds number so we have the raynodes number.
04:50
This is our third equation, which is raynodes number.
04:55
So this would be the raynodes number, which is dt.
05:01
So we have dt, which is equals to d -u -m over micro.
05:10
So this would be the third equation.
05:14
So raynodes number is equals to 998 kilogram per.
05:20
Cubic meter times 0 .014 meter times 3 .5 meter per seconds over 0 .959 times 10 raise power of negative 3 n seconds per square meter.
05:39
For the nassault number we have the nassault number so the nonsault number so we have this equation so based on this we have the given so we have h i times 0 .014 so 0 .14 meters over 0 .0606 m per kelvin and the nozzle number will be 0 .0 .032 3 .3 so this would be the nozzled number for the nozzled number from the d2's boiler correlation so we have nozzled number from the d2's boiler correlation.
06:52
So we have a nozzle number from the d2's boiler correlation so we have the nozzle number is equal to 0 .023 we have our r .e 0 .8 dt.
07:16
Pr one -third.
07:19
Nuzzle number is equals to 0 .023 times 50992 .7, raise the par of 0 .6 .8 times 6 .62 as our pr over 1 third.
07:37
So the nasal number from the detus boiler coloration will be 251.
07:45
Point nine seven two and according to the four and number five which is in number four we have 0 .0231 for the nuzzle number is equals to 251 .972 overall the heat transfer coefficient will be one over u a is equal to 1 over u subscript t, a subscript t is equal to 1 over u subscript o, a subscript o, which is equal to 1 over h t, a, a, t, plus l n, d, o over d t, over 2 pi k l, plus 1 ,000, and, 0 over d t, over 2 pi k l, plus 1, over h .o .a .o.
08:56
So this would be our six.
09:01
Accordingly we have from the six part, so from the six part we have one over ua is equals to one over hta plus one over hoa.
09:17
So you have u is equals to one over ht plus one over hoa.
09:18
So you have u is equals to one over h -t plus one over h -o.
09:25
Raised to the power of negative 1...