There are two unknowns here and I am mot sure how to solve the problem. I know i

Question

There are two unknowns here and I am mot sure how to solve the problem. I know it is not difficult. Please help! I rate high!
A paper mill releases 1 m^3/s of treated effluent at 75 C, which is mixed with captured rain water at 13 C before it is released to a nearby river. Regulations dictate that the effluent discharge cannot be over 20 C. How many liters per day of captured rain water are necessary to allow for effluent discharge within the regulatory limits?
I am missing two volumes and I do not know how to solve! Thank You

Explanation / Answer

if you do not have charts or any other information the answer below is not valid and i am sorry for wasting your time.

i may be miss reading the problem but arent you given the volume of the treated effuent? isnt it just 1 m^3/s mulitplied by the number if seconds in a day (86400)
meaning the total volume of treated effluent is 86400 m^3

if this is true i believe you are still missing alot of information. First, we will have to assume that this is an open system meaning we will need to know our h values for both water and the effluent. you must also need to know the weight of the effulent and water (kg/m^3).

if you have charts for both liquids and can do the conversion from volume to mass of effulent then this is the answer if not this solution is really reeally reeeaaalllly wrong.

do an energy balance, we will neglect kinetic and potential energy meaning all we are left with is

m(effulent)* h (effulent) + m (water)*h(water) = m(combined)*h(combined)

for the m above they are in untils kg/s

with is you will know that m(combined) is just the other 2 m values combined, meaning that you have 2 equations and 2 unknows becuase your knowns are m(effulent), h(effulent),h(water), and h(combined) this leaves m(water) and m(combined). so there are 2 unknows and 2 equations the second one being m(combined) is the summation of them both.

solving you will find m(water) this m is in kg/s meaning you must multiply by 86400 and divide it by the number of cubic meters per kilogram of water which is 10^-3. this gives you m^3 of water per day

multiply this by 1000 and that is your number of liters.

if you dont have charts or this is a very poor question you might be doing it this way ( i dont think this is what the problem wants, but idk it doesnt make any sense to do it this way and it is physically wrong)

for this you are assuming that the h vaule is the same and doesnt change with temp. you will just solve the following equations

75 (C) * 1 (m^3/s) + 13 (C) * X(m^3/s) =20(C) * (X+1)(m^3/s)

so solve for X this will give m^3/s for the water multiply this by 1000 and 86400 to find liters.

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