<p>ANSWER:</p>
<p><span class="math-container">$h={\Delta}t/{\Delta}l$</span></p>
<p><span class="math-container">$h=$</span> height of LCL (unit depends on unit height of lapse rates)</p>
<p><span class="math-container">${\Delta}t=T-T_d$</span></p>
<p><span class="math-container">${\Delta}l=l_t-l_d$</span></p>
<p><span class="math-container">$T$</span> is surface temp (deg)</p>
<p><span class="math-container">$T_d$</span> is surface dewpoint temp (deg)</p>
<p><span class="math-container">${\Delta}l=$</span> difference between lapse rates (deg)</p>
<p><span class="math-container">$l_t=$</span> dry adiabatic lapse rate of temperature (deg/unit height)</p>
<p><span class="math-container">$l_d=$</span> lapse rate of the saturation mixing ratio corresponding to the surface dewpoint (deg/unit height)</p>

CC BY-SA 4.0
2020-12-27T11:50:09.270Z
Welcome on the ES SE! Your post needs to answer the question (on the top), I can not see, how it does.
peterh
CC BY-SA 4.0
2020-12-28T06:23:46.040Z
@peterh-ReinstateMonica looks valid to me... he does say his formula calculates LCL. Not sure I've seen the formula using lapse rates before, but in a way it looks the most elegant\understandable, since the usual hand method uses the intersection of the two lapse rate lines... and so boiling it down to the distance to cover (ΔT) being found by rate of decrease per height * the height [the old D=R*t idea], and solving for h makes sense. Ted it wouldn't hurt to try to give a little more explanation of reasoning for why this is the answer (or at least the source of the equation), but looks valid.
JeopardyTempest
CC BY-SA 4.0
2020-12-28T10:05:55.083Z
@JeopardyTempest Ok, thanks! I would happily fix his Latex, but I am not sure, where to start.
peterh
CC BY-SA 4.0
2020-12-27T06:02:53.357Z
2020-12-28T17:25:52.917Z
2020-12-28T17:25:52.917Z
Ted MacKechnie
<p>Well, I think if it starts at T=30, and the dew point is T=15, then once it has dropped 15 degrees clouds will start to form.  </p>

<p>The DALR is about 10 degrees/km.  SO, first, I think you need to solve for the altitude at which T will have dropped from 30 to 15.</p>

CC BY-SA 3.0
2018-03-29T04:12:27.400Z
2018-03-29T04:12:27.400Z
eSurfsnake
<p>I have a parcel of air at 30°C with a dew point temp of 15°C.  It goes up and then back down a mountain 4km in height. Assuming the normal and dry adiabatic lapse rates apply, and an ELR=7.5°C/km, with the environment also at 30°C.</p>

<p>From Wikipedia it says that to find cloud base I:</p>

<p>(1). Start at the initial temperature (T) and pressure of the air parcel and follow the dry adiabatic lapse rate line upward (provided that the RH in the air parcel is less than 100%, otherwise it is already at or above LCL).</p>

<p>So then the initial temp is 30*C not sure what it means to follow the dry adiabatic lapse rate line upwards.</p>

<p>I know that lapse rate is the rate at which Earths atmosphere increases and decreases in conguence with the altitude. So the temperature increases when the altitude decreases and the temperature decreases when the altitude increases. So a dry unsaturated lapse rate is one where the parcel lacks water content, and the saturated lapse rate is one where the parcel is moist and contains water content. I understand that when they warm or cool both types move in up and down in direction.</p>

<p>(2). From the initial dew point temperature (Td) of the parcel at its starting pressure, follow the line for the constant equilibrium mixing ratio (or "saturation mixing ratio") upward.</p>

<p>Not sure what this means</p>

<p>(3). The intersection of these two lines is the LCL.</p>

<p>Not sure what this is asking me to do.</p>

<p>Things I have been asked to provide include:</p>

<blockquote>
  <p>a) the elevation at which clouds will occur (the lifting condensation
  level);</p>
  
  <p>b) the temperature of the parcel at the top of the mountain;</p>
  
  <p>c) the temperature the parcel will have when it descends to the bottom
  on the far side of the mountain;</p>
  
  <p>d) the portions of its journey where it follows the wet and dry rates</p>
  
  <p>e) explain why the parcel is warmer/cooler/same temperature on the
  back side of the mountain as compared to when it started.</p>
</blockquote>

CC BY-SA 3.0
2018-03-09T11:26:23.003Z
A hint: you can use for example an emagram to solve this.
Communisty
<atmosphere><homework>
How do you calculate the elevation at which clouds will form (lifting condensation level)?
yre
# How do you calculate the elevation at which clouds will form (lifting condensation level)?

(Asked by: username_0 on 2018-03-07)

I have a parcel of air at 30°C with a dew point temp of 15°C. It goes up and then back down a mountain 4km in height. Assuming the normal and dry adiabatic lapse rates apply, and an ELR=7.5°C/km, with the environment also at 30°C.

From Wikipedia it says that to find cloud base I:

(1). Start at the initial temperature (T) and pressure of the air parcel and follow the dry adiabatic lapse rate line upward (provided that the RH in the air parcel is less than 100%, otherwise it is already at or above LCL).

So then the initial temp is 30\*C not sure what it means to follow the dry adiabatic lapse rate line upwards.

I know that lapse rate is the rate at which Earths atmosphere increases and decreases in conguence with the altitude. So the temperature increases when the altitude decreases and the temperature decreases when the altitude increases. So a dry unsaturated lapse rate is one where the parcel lacks water content, and the saturated lapse rate is one where the parcel is moist and contains water content. I understand that when they warm or cool both types move in up and down in direction.

(2). From the initial dew point temperature (Td) of the parcel at its starting pressure, follow the line for the constant equilibrium mixing ratio (or "saturation mixing ratio") upward.

Not sure what this means

(3). The intersection of these two lines is the LCL.

Not sure what this is asking me to do.

Things I have been asked to provide include:

> a) the elevation at which clouds will occur (the lifting condensation
> level);
>
> b) the temperature of the parcel at the top of the mountain;
>
> c) the temperature the parcel will have when it descends to the bottom
> on the far side of the mountain;
>
> d) the portions of its journey where it follows the wet and dry rates
>
> e) explain why the parcel is warmer/cooler/same temperature on the
> back side of the mountain as compared to when it started.

**Tags:** atmosphere, homework

> **Comments:**
> **username_1**: A hint: you can use for example an emagram to solve this.  

---

## 2 answers

### Answer by username_2 on 2020-12-27. Score: 2

ANSWER:

$h={\Delta}t/{\Delta}l$

$h=$ height of LCL (unit depends on unit height of lapse rates)

${\Delta}t=T-T\_d$

${\Delta}l=l\_t-l\_d$

$T$ is surface temp (deg)

$T\_d$ is surface dewpoint temp (deg)

${\Delta}l=$ difference between lapse rates (deg)

$l\_t=$ dry adiabatic lapse rate of temperature (deg/unit height)

$l\_d=$ lapse rate of the saturation mixing ratio corresponding to the surface dewpoint (deg/unit height)

> **Comments:**
> **username_3**: Welcome on the ES SE! Your post needs to answer the question (on the top), I can not see, how it does.  
> **username_4**: @username_3-ReinstateMonica looks valid to me... he does say his formula calculates LCL. Not sure I've seen the formula using lapse rates before, but in a way it looks the most elegant\understandable, since the usual hand method uses the intersection of the two lapse rate lines... and so boiling it down to the distance to cover (ΔT) being found by rate of decrease per height * the height [the old D=R*t idea], and solving for h makes sense. Ted it wouldn't hurt to try to give a little more explanation of reasoning for why this is the answer (or at least the source of the equation), but looks valid.  
> **username_3**: @username_4 Ok, thanks! I would happily fix his Latex, but I am not sure, where to start.  

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### Answer by username_5 on 2018-03-29. Score: 1

Well, I think if it starts at T=30, and the dew point is T=15, then once it has dropped 15 degrees clouds will start to form.

The DALR is about 10 degrees/km. SO, first, I think you need to solve for the altitude at which T will have dropped from 30 to 15.


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