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# Exercises

Exercises. Problem-1. The ozone (O 3 ) concentration in urban areas with photochemical smog problem reaches a value of 0.1 ppm over a 1-hr period. Determine by what percentage this level exceeds the ambient air quality standard of 160 μ g/m 3 if the temperature is 25°C. . Solution. Step 1

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## Exercises

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1. Exercises

2. Problem-1 • The ozone (O3) concentration in urban areas with photochemical smog problem reaches a value of 0.1 ppm over a 1-hr period. Determine by what percentage this level exceeds the ambient air quality standard of 160 μg/m3if the temperature is 25°C.

3. Solution Step 1 [ Note: molecular weight in g/mol]

4. Step 2:

5. Problem-2 • Find the height above sea level where the pressure is 90 kpa for an atmosphere pressure of 101.3 kpa and an air density of 1.23 kg/m3.

6. Solution Step1: • dP = - (Rho) g dZ     static equilibrium Step 2: Integrate from 0 to z: (101.3-90) kPa = 1.23 kg/m3* 9.81 m/s2* (z-0) • z = 0.9365 kPa*m3/(kg*m/ s2)        [note- N = kg*m/ s2,  N/m2 = Pa] z = 936.5 m above sea level

7. Problem-3 • Determine the potential temperature at 1100 ft if the pressure and temperature are 980 mbar and 690 F respectively.

8. Solution θ = T (P0/P)R/Cp θ = T (P0/P)0.288 Where, P0 = 1000 mbar Given: T = 690 F = 20.560C = 20.56 + 273 = 293.56 0K P = 980 mbar θ =  293.56 * (1000/980)0.288 = 295.27 0K

9. Problem-4 • During a field program carried by a utility in Ohio, the atmospheric lapse rate on July 29, 2009 was found constant up to 800 m. At ground level the pressure (Po) is 1020 mbar and the temperature To is 15oC. A radio sound  measurement indicates that at some elevation z the pressure and temperature are 975 mbar and 11.5oC respectively. Determine the atmospheric temperature gradient dT/dz in degree Kelvin per meter.

10. Solution EQUATIONS • dp = -(Rho)gdz      static equilibrium • p= (Rho)RT           ideal gas law • dT/dz = b        lapse rate definition • T –T0 = bz       for constant lapse rate, initial z = 0   dT=b dz UNITS • J = N*m • N = kg*m/s2 • T0 = 15 + 273 = 288 K          use absolute temperatures • T = 11.5 + 273 = 284.5 K

11. Solution Step 1: Calculate ideal gas law constant for air in appropriate units: R(air) = Ru/MWair =    8.315 kJ/(kgmol*K)/(28.96 kg/kgmol)           =   0.28712 *103 (kg*m/s2*m/(kgmol*K))/(kg/kgmol)                     =   287.12 m2/s2/K

12. Step2: Divide the static equilibrium equation by the ideal gas law equation and substitute dT/b for dz from the lapse rate definition. dp/dz = -(p/RT)*g dp/p = -(g/R)*dz/T or, dp/p = -(g/Rb)*dT/T Lapse rate b = -(g/R)*ln(284.5/288)/ln(975/1020) = -{(9.81 m/s2/(287.12 m2/s2/K))}*{(-.012227)/(-.045120) = -0.926 K/100 m atmospheric temperature gradient

13. Problem-5 • Calculate the equilibrium-incremental increase in surface temperature from 1850 to 2050 due to carbon dioxide. The concentrations for CO2 are as follows: • 1850: 280 ppm • 2050: 562 ppm (estimated)

14. Climate Change Calculations • CO2 Emission Calculation: CO2 = (Pi)*(Foi)*Ci • P: Amount of fuel • FO : Fraction oxidized • C : Average carbon content • i: Particular fuel type • Change in radiative forcing: For CO2 ΔFCO2 = K3ln [(co2)/(co2)0] • K3 : 6.3 (W/m2)

15. Cont… • Surface temperature change: ΔTs = ΔF * λ • ΔTs: Surface temperature • ΔF: Radiative forcing • λ: Climate sensitivity factor (0.57 in 0C/W/m2)

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