15. A electric utility wants to build a new coal fired power...

15. A electric utility wants to build a new coal fired power plant in an urban setting. The regulatory agency has determined that in order for the plant to be permitted it must limit its emissions to prevent the ground level SO$_2$ concentration from exceeding 0.015 g/m$^3$ at the site fence line 0.5 km directly downwind (y = 0) from the stack. What is the maximum SO$_2$ discharge rate (g/s) the plant can emit? Use the following: Effective stack height = 50 m Wind speed = 1 m/s Stability Class C, using Martin's equation: $s_y = 104.0X^{0.894}$ and $s_z = 61.0X^{0.911}$ where the dispersion coefficient is in meters and the X distance is in km $x = \left(\frac{Q}{\pi s_y s_z u}\right) \exp\left(-\frac{1}{2}\left(\frac{y}{s_y}\right)^2\right) \exp\left(-\frac{1}{2}\left(\frac{H}{s_z}\right)^2\right)$

Transcript

00:01 Hello students welcome here in this question.

00:03 We have a muldron cold burning pub plant operates the steam at t is equal to 540 so this is the t max value and t minimum is at 30 degrees celsius and we have here the heat input is given one charles and other things that are given we need to find out the maximum heat that can be added to the room we need to find out here so here we have to find out the maximum work so w max divided by max q1 is equal to 1 minus the temperature it is 273 plus 273 so this is a t minimum divided by t max.

01:05 So 1 minus 303 divided by 800 and so this is approximately 13.

01:16 810 .13 is a value.

01:19 So this one is equal to 0 .6273.

01:28 So as it is 1.

01:30 The value is equal to eta.

01:34 So that is efficiency.

01:39 Efficiency we can say.

01:41 So as we know the efficiency, it is in watt second.

01:47 So it is joust.

01:48 Jaws can be written as watt second.

01:51 And in kilowatt second, it is 0 .6273 multiplied by 10 power minus 3, kilowatts multiplied by 1 divided by 3 ,600 r inverse we need to say so it is equal to 1 .742 multiplied by 10 power minus 7 kilowatts per hour now kilowatts hours we can say not per hour it is in seconds so we just converted this one into hours so this is ours now efficient efficiency max is equal to efficiency max maximum.

02:40 So is equal to 1 minus t minimum divided by max t max.

02:46 So that is equal to w divided by eta from this one is equal to q, you can write.

02:56 So q is equal to, q1 is equal to.

03:01 So this is 500 multiplied by 10 power 6 divided by 0 .673 then q is equal to 797 .0668 multiplied by 10 power this is 668 10 power 6 jous per second so then we can write it is 797 .0668 mega jous per second this one is now amount of coal amount of coal burn m is equal to q divided by caliphic value value so that is equal to q is 797 .000668 mj per second divided by this one is 297 is a weight of the weight a caligraphic value of aid so then we will have 27 .485 cases per second and now we will convert this one 27 .485 so this value is converted into hours as it is in seconds so multiplied by 3 ,600 then you will get here 98...

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