<p>Instead of just giving you the answer to a textbook problem (which would make SE a version of Chegg that future students would use to cheat on homework), I'll guide you in getting an answer.</p>
<ol>
<li>Which variable does <span class="math-container">$\overline{u'w'}$</span> affect most?</li>
<li>For that variable, identify the surface layer.</li>
<li>Imagine that you are on the line. If you go straight up (that is, <span class="math-container">$w'&gt;0$</span>) what would the difference be between the <span class="math-container">$M$</span> traced out and the graphed <span class="math-container">$M$</span>? What is the sign of <span class="math-container">$w' \times u'$</span>?</li>
<li>Do the same as number 3, except start at the top and go down (<span class="math-container">$w'&lt;0$</span>).</li>
</ol>

CC BY-SA 4.0
2020-10-22T05:00:46.090Z
Hi! Thanks for your answer. I’m not sure if I got it right. I posted and edit above with some questions. Thanks for your help!
meteo_96
CC BY-SA 4.0
2020-10-22T14:34:53.373Z
... I don't think you did it right. Try thinking of step 3 and step 4 as two different steps- start on the line for both.
BarocliniCplusplus
CC BY-SA 4.0
2020-10-23T02:32:00.753Z
When you said what the sign of w’ x u’ is, did u mean a cross product? Or something like, for step 3, since w>0 going up is (+) and u in this area would be (-), hence w’u’ < 0. Similarly, for step 4 going down, w’ < 0 and u’ > 0 giving w’u’ <0. So the average or net $\overline{w’u’}$ <0, meaning net momentum flux in this layer is downward? Thanks.
meteo_96
CC BY-SA 4.0
2020-10-23T03:30:16.853Z
No, I meant multiplication. You don't have to take an average. If you start on the line and go up. What is the difference between that point and the line? Consider page 52 of Stull.
BarocliniCplusplus
CC BY-SA 4.0
2020-10-25T08:20:29.010Z
Okay. So going up (w’ >0) , the difference is negative (u < 0), and the sign of w’u’ < 0 [step 3]. Doing the same thing going down (w’ < 0), the difference between the traced line and the surface layer wind is positive (u > 0), making w’u’ < 0 [step 4]. Now, since both steps 3 and 4 returned negative results for w’u’, is this enough to say that flux is towards the ground?
meteo_96
CC BY-SA 4.0
2020-10-25T21:59:01.037Z
No, look at the order that I said for the difference. So I mean $M_{line} - M_{parcel}$. To move the parcel up ($w'>0$), the parcel needs to move faster ($u'>0$).
BarocliniCplusplus
CC BY-SA 4.0
2020-10-21T17:47:05.853Z
2020-10-21T17:47:05.853Z
BarocliniCplusplus
<p>Given the question below, how would the kinematic momentum flux be at the surface layer?</p>
<p>For letter a, why would it be that way (positive, negative, or 0)?</p>
<p><a href="https://i.sstatic.net/gM8Qb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gM8Qb.jpg" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/OvUiP.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OvUiP.jpg" alt="enter image description here" /></a></p>
<p>Based on the answer below:</p>
<p>I’m not sure if I’ve done it correctly, but this is how I answered it based in your comment below.</p>
<ol>
<li>First, I assumed that <span class="math-container">$\overline{u’w’}$</span> = <span class="math-container">$\overline{w’u’}$</span>. So is this something like the mixing of horizontal momentum into areas of lower wind speed (as is the case in the surface layer)?</li>
</ol>
<p><a href="https://i.sstatic.net/612z0.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/612z0.jpg" alt="enter image description here" /></a></p>

<meteorology><planetary-boundary-layer>
Horizontal momentum flux in the surface layer
meteo_96
# Horizontal momentum flux in the surface layer

(Asked by: username_0 on 2020-10-19)

Given the question below, how would the kinematic momentum flux be at the surface layer?

For letter a, why would it be that way (positive, negative, or 0)?

[![enter image description here](https://i.sstatic.net/gM8Qb.jpg)](https://i.sstatic.net/gM8Qb.jpg)
[![enter image description here](https://i.sstatic.net/OvUiP.jpg)](https://i.sstatic.net/OvUiP.jpg)

Based on the answer below:

I’m not sure if I’ve done it correctly, but this is how I answered it based in your comment below.

1. First, I assumed that $\overline{u’w’}$ = $\overline{w’u’}$. So is this something like the mixing of horizontal momentum into areas of lower wind speed (as is the case in the surface layer)?

[![enter image description here](https://i.sstatic.net/612z0.jpg)](https://i.sstatic.net/612z0.jpg)

**Tags:** meteorology, planetary-boundary-layer


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## 1 answers

### Answer by username_1 on 2020-10-21. Score: 3 | Accepted answer

Instead of just giving you the answer to a textbook problem (which would make SE a version of Chegg that future students would use to cheat on homework), I'll guide you in getting an answer.

1. Which variable does $\overline{u'w'}$ affect most?
2. For that variable, identify the surface layer.
3. Imagine that you are on the line. If you go straight up (that is, $w'>0$) what would the difference be between the $M$ traced out and the graphed $M$? What is the sign of $w' \times u'$?
4. Do the same as number 3, except start at the top and go down ($w'<0$).

> **Comments:**
> **username_0**: Hi! Thanks for your answer. I’m not sure if I got it right. I posted and edit above with some questions. Thanks for your help!  
> **username_1**: ... I don't think you did it right. Try thinking of step 3 and step 4 as two different steps- start on the line for both.  
> **username_0**: When you said what the sign of w’ x u’ is, did u mean a cross product? Or something like, for step 3, since w>0 going up is (+) and u in this area would be (-), hence w’u’ < 0. Similarly, for step 4 going down, w’ < 0 and u’ > 0 giving w’u’ <0. So the average or net $\overline{w’u’}$ <0, meaning net momentum flux in this layer is downward? Thanks.  
> **username_1**: No, I meant multiplication. You don't have to take an average. If you start on the line and go up. What is the difference between that point and the line? Consider page 52 of Stull.  
> **username_0**: Okay. So going up (w’ >0) , the difference is negative (u < 0), and the sign of w’u’ < 0 [step 3]. Doing the same thing going down (w’ < 0), the difference between the traced line and the surface layer wind is positive (u > 0), making w’u’ < 0 [step 4]. Now, since both steps 3 and 4 returned negative results for w’u’, is this enough to say that flux is towards the ground?  
> **username_1**: No, look at the order that I said for the difference. So I mean $M_{line} - M_{parcel}$. To move the parcel up ($w'>0$), the parcel needs to move faster ($u'>0$).  

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