Results 1 to 5 of 5

Math Help - Could someone double check this for me?

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    4

    Could someone double check this for me?

    Could someone double check my math on this?

    I have a countertop of 6.97 square meters made of a stone that emits 30.20 Bq of Radon per day and I need to figure out how much the stone adds to the Radon level in the home in pico Curies per liter of air.

    30.20 Bq x 27.02 (Bq to pCi conversion factor) = 816.004 pCi x 6.97 sq meters of countertop = 5,687.54 pCi of Radon emitted per day.

    12' x 15' kitchen with 8' ceilings would be 12 x 15 x 8 = 1,440 cubic feet -351 cubic feet for cabinets and appliances= 1089 cubic feet of air present.

    28.3 liters per cubic foot so 1089 cubic feet x 28.3 = 30,818.7 liters of air present in the room.

    5,687.54 pCi of Radon mixed into 30,818.7 liters of air = 0.1845 pCi/liter increase.

    Thanks for taking the time.
    Al
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by solid surface View Post
    Could someone double check my math on this?

    I have a countertop of 6.97 square meters made of a stone that emits 30.20 Bq of Radon per day and I need to figure out how much the stone adds to the Radon level in the home in pico Curies per liter of air.

    30.20 Bq x 27.02 (Bq to pCi conversion factor) = 816.004 pCi x 6.97 sq meters of countertop = 5,687.54 pCi of Radon emitted per day.

    12' x 15' kitchen with 8' ceilings would be 12 x 15 x 8 = 1,440 cubic feet -351 cubic feet for cabinets and appliances= 1089 cubic feet of air present.

    28.3 liters per cubic foot so 1089 cubic feet x 28.3 = 30,818.7 liters of air present in the room.

    5,687.54 pCi of Radon mixed into 30,818.7 liters of air = 0.1845 pCi/liter increase.

    Thanks for taking the time.
    Al
    First you will need to know the other Radon sources in the kitchen,
    also the level of Radon in the external air, and finaly the air replacement
    rate for the kitchen. Then you will have to set up the Radon balance
    equation.

    You might get away with the Radon level without the countertop, the
    countertop emission rate and the air exchange rate.

    But either way there is not enough information given to solve the problem.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View Post
    First you will need to know the other Radon sources in the kitchen,
    also the level of Radon in the external air, and finaly the air replacement
    rate for the kitchen. Then you will have to set up the Radon balance
    equation.

    You might get away with the Radon level without the countertop, the
    countertop emission rate and the air exchange rate.

    But either way there is not enough information given to solve the problem.

    RonL
    We can do a sample calculation assuming a nominal air exchange rate of
    0.35 \mbox{ /hr}. Then the equilibrium conditions give

    Outflow of Radon=Inflow of Radon

     <br />
0.35 \times C_e=S<br />

    where C_e is the equilibrium Radon load, and S in the hourly input rate from
    the source. So

     <br />
C_e=S/0.35<br />

    Here you have a source rate of \approx 474 \mbox{ pCi/hr}, so:

     <br />
C_e \approx 1354 \mbox{ pCi}<br />

    which you can now divide by the room volume to get a concentration of

    \approx 0.044 \mbox{ pCi/l}

    (this has ignored the Radon load from other sources and in the ambient air
    which may not be negligable)

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2007
    Posts
    4
    Quote Originally Posted by CaptainBlack View Post
    We can do a sample calculation assuming a nominal air exchange rate of
    0.35 \mbox{ /hr}. Then the equilibrium conditions give

    Outflow of Radon=Inflow of Radon

     <br />
0.35 \times C_e=S<br />

    where C_e is the equilibrium Radon load, and S in the hourly input rate from
    the source. So

     <br />
C_e=S/0.35<br />

    Here you have a source rate of \approx 474 \mbox{ pCi/hr}, so:

     <br />
C_e \approx 1354 \mbox{ pCi}<br />

    which you can now divide by the room volume to get a concentration of

    \approx 0.044 \mbox{ pCi/l}

    (this has ignored the Radon load from other sources and in the ambient air
    which may not be negligable)

    RonL

    Thanks, CaptainBlack.

    I see that without considering ventilation and other sources of Radon it would be impossible to figure the Radon present in the room, but what I am looking for is the added Radon. How much is given off exclusively by the granite into the volume of air in the kitchen without any air exchange.

    I'm a little confused where the 474 pCi came from. The daily total is around 5,600 or so and I am assuming it would need to be divided by 24 to get hourly rate of 233 or so. Am I doing something wrong with my math?

    Can I assume that the basic method of figuring the Radon emitted is accurate, not considering air exchange or other Radon imputs?

    Thanks for taking the time to help out. It is appreciated.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by solid surface View Post
    Thanks, CaptainBlack.

    I see that without considering ventilation and other sources of Radon it would be impossible to figure the Radon present in the room, but what I am looking for is the added Radon. How much is given off exclusively by the granite into the volume of air in the kitchen without any air exchange.

    I'm a little confused where the 474 pCi came from. The daily total is around 5,600 or so and I am assuming it would need to be divided by 24 to get hourly rate of 233 or so. Am I doing something wrong with my math?

    Can I assume that the basic method of figuring the Radon emitted is accurate, not considering air exchange or other Radon imputs?

    Thanks for taking the time to help out. It is appreciated.
    You are right I divided by 12 rather than 24, so all the numbers in my
    calculation can be halved so that the rate is 237 pCi/hr.

    The calculation will work if you are only interested in the added Radon.

    (You will need to check that 0.35 air changes per hour is the applicable
    rate for this particular kitchen)

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. can someone double check this for me?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 18th 2010, 01:55 PM
  2. Double Check
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 15th 2009, 05:11 PM
  3. double check
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 21st 2008, 10:03 AM
  4. can someone please just double check
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 1st 2008, 09:54 PM
  5. Double check please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 4th 2006, 07:09 PM

Search Tags


/mathhelpforum @mathhelpforum