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# NATS 1780 A 6.0 Y (Summer XXXXXXXXXXAssignment 3 Due: No later than 11:00 pm EDT on August 12, 2020 via Moodle (Late Penalty: 25% per day - including weekends; 100% deduction after solutions are...

Answered Same Day Aug 07, 2021

## Solution

Sivaranjan answered on Aug 09 2021
1. For stations above Mean Sea Level (MSL), the pressure at the point of observation (pOBS ) can be co
ected to its MSL equivalent pressure (pMSL ) through use of the Hydrostatic Equation namely:
(Eqn 1)
In this equation, zOBS is the height of the observation point relative to MSL, ρ is the density of air at the point of observation and g is the acceleration due to gravity.
a. Obtain a screenshot of EMOS similar to Figure 1, and include it in your submission. For the same date and time, obtain a pressure measurement from a weather service for a station close to EMOS; also include this screenshot in your submission.
Fig 1(a) Screenshot of EMOS
Fig. 1(b) Weather report at same date and time
. Based on your EMOS screenshot (Question 1(a)), state the height (in m above MSL) of the EMOS meteorological observation station.
Ans: As per the screen shot at Fig. 1 (a), the altitude is 196 m above sea level.
c. Assuming that ρ is 1.2041 kg/m3 , g is 9.81 m/s2 , and pOBS is your EMOS station air pressure in your figure (Question 1(a)), calculate pMSL from the equation above (i.e., Eqn 1). [Recall: 1 hPa ≡ 100 Pa and 1 Pa = 1 kg / (m s2 ). Note that 1.2041 kg/m3 is the value for the density of air at STP - namely at 20 °C and 1013.25 hPa.]
Ans:
Given,
pOBS = 994.0 hPa = 99400 Pa
ρ = 1.2041 kg m-3
g = 9.81 ms-2
Since, hPa is the SI unit of pressure,
We have,
d. Compare and contrast the result of your calculation above (Question 1(c)) with the EMOS-reported value for MSL air pressure in your EMOS screenshot (Question 1(a)).
Ans:
Calculated value of pMSL = 101715.195 Pa
Given value of pMSL = 1017.2 hPa = 101720 Pa
Difference = 101720.000 - 101715.195 = 4.805 Pa
e. Assuming that the reason for this difference is entirely due to density, calculate the density at EMOS using the Ideal Gas Law , namely:
(Eqn 2)
Note that pOBS is the same EMOS station air pressure in the above figure, TOBS the EMOS air temperature from the above figure converted to K, and finally that R is a constant with the value 287.058 J/(kg K). [Note that the Joule, a unit of energy, is expressible as a (N m).]
Ans:
Given,
pOBS = 994.0 hPa = 99400 Pa
TOBS = 21.8 OC = 21.8 + 273.15 K = 294.95 K
R = 287.058 J kg-1 K -1
From Eqn 2:
f. Re-calculate pMSL from Eq’n 1 above using the ‘improved’ value for density determined in Question 1(e).
Ans:
Given,
pOBS = 994.0 hPa = 99400 Pa
ρ = 1.174 kg m-3
g = 9.81 ms-2
Since, hPa is the SI unit of pressure,
We have,
g. Compare and contrast the result of your calculation above (Question 1(f)) with the EMOS-reported value for MSL air pressure in your EMOS screenshot (Question 1(a)).
Ans:
Calculated value of pMSL = 101657.32 Pa
Given value of pMSL = 1017.2 hPa =...
SOLUTION.PDF