HW1

ATSC 201 Fall 2014
Assignment 1 Answer Key
Note: All equation numbers are those given in the 3rd edition (final version) of Stull's textbook
Chapter 1: A1d, A3d, A5d, A6d, A9d, A14d, A15d
A1d)
Find wind the direction (degrees) and speed (m/s), given the (U,V)
components: (6, 6) m/s
Given:
U=
V=
6 m/s
6 m/s
Find:
M, alpha
Using eq. 1.1:
M = (U^2 + V^2)^1/2
Using eq. 1.2a:
alpha = 90deg - (360deg/C)*arctan(V/U) + 180deg
M=
alpha =
8.49 m/s
225.00 degrees
Winds are coming from 225 degrees (south-west) at 8.49 m/s.
Checks
Discussion
A3d)
Units ok. Physics ok.
The wind is coming exactly out of the direction of the
bisector of south and west, which makes sense given
equal U and V components.
Convert the following UTC times to local times in your own time zone:
0920
Given:
Find:
9:20 UTC
local time
ST = UTC - alpha
DT = UTC - beta
equations found in table 1.1
Our Zone is "U", so alpha = 8 and beta = 7.
Currently, local time is PDT.
DT =
2:20 PDT
for partial marks: ST = 1:20 PST
Checks
Discussion
A5d)
Units ok.
Vancouver is 8 hours behind UTC in in DST.
Find the pressure in kPa at the following heights above sea level,
assuming an average T = 250K:
25 km above sea level
Given:
avg T =
z=
Find:
P (kPa)
250 K
25 km
Using eq. 1.9a:
P = Po*e^(-(a/T)*z)
where a=0.0342 K/m, Po = 101.325 kPa
Before using eq 1.9a, the height above the ground must be converted to meters
z=
25000 m
P=
3.31459 kPa
Checks
Discussion
A6d)
Units ok. Physics Ok.
At 25 km above sea level the pressure is 3% that
of sea level.
Use the definition of pressure as a force per unit area, and consider a column
of air that is above a horizontal area of 1 square meter. What is the mass of air
in that column:
d) above a height where the pressure is 85 kPa?
Given:
Pbottom =
Ptop =
A=
Find:
Δm =
85 kPa
0 kPa
1 m^2
?
kg
Δm = (A / g) * (Pbottom - Ptop)
(eq.1.11)
Convert Pbottom and Ptop to Pa:
Pbottom =
85000 Pa
Ptop =
0 Pa
(note that 1 Pa = 1 kg m^-1 s^-2)
So,
Δm = (1 m^2) / (9.8 m s^-2) * (70000 kg m^-1 s^-2 - 50000 kg m^-1 s^-2 )
m=
8664.6 kg
Checks
Discussion
A9d)
(eq. 1.11)
Units ok. Physics ok.
This is just over 100 times the mass of a human body
for only a small fraction of the atmosphere!
Convert the following temperatures:
d) 15 degs C = ? Degs F
Given:
T=
Find:
T=
15 degs C
?
deg F
First convert temperature to degrees C
Using eq. 1.6a:
T (in deg F) = ((9/5)*T(deg C)) + 32
T(deg C)=
59.00 degs F
So,
T=
59.00 degs F
Checks
Discussion
A14d)
Units ok. Physics ok.
What is the geometric height and geopotential , given the geopotential height?
d) 11 km
Given:
H=
Find:
z=
phi =
11 km (geopotential height)
?
?
km (geometric height)
m^2/s^2 (geopotential)
To find geometric height, use eq. 1.14b:
z = R0*H/(R0 - H)
where
R0 =
6356.766 km
So,
z=
11.01906783 km
To find geopotential, use eq. 1.15:
phi = | g | * H
where
|g| =
9.8
H=
11000.00
So,
phi =
107800
(note: use H in metres)
m/s^2
m
m^2/s^2
Checks
Discussion
A15d)
Units ok. Physics ok.
The difference between the geometric height and
geopotential height is 19 m at 11 km.
What is the standard atmospheric temperature, pressure, and density at
the following geopotential heights?
d) 8 km
Given:
H=
Find:
T=
P=
rho =
8 km
?
?
?
degs C
kPa
kg/m^3
To find the standard atmospheric temperature, use eq.1.16 for H <= 11km:
Put H in km H=
8.00 km
T = 288.15 K - (6.5 K/km)*H
So,
T=
236.15 K
T=
-37 degs C
To find the standard atmospheric pressure, use eq. 1.17 for H <= 11km:
P = (101.325 kPa)*(288.15K/T)^(-5.255877)
So,
P=
35.59980514 kPa
To find the standard atmospheric density, use the ideal gas law for dry air (eq.1.18)
rho = P / (Rd * T)
where
Rd =
0.287053 kPa K^-1 m^3 kg^-1
So,
rho =
0.525167184 kg/m^3
Checks
Discussion
Units Ok, Physics ok.
The standard atmospheric pressure, temperature
and density have varied a lot from the base state at
the surface.

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