<p>I'm trying to calculate the saturated moist adiabatic lapse rate with the following formula:</p>
<p><span class="math-container">$L_m = \frac{y_d}{1+\frac{L_v}{c_p}\frac{d_q(s)}{dT}}$</span></p>
<p>with:</p>
<ul>
<li><span class="math-container">$y_d = 9.8 (K\ km^{-1})\\\\$</span>,</li>
<li><span class="math-container">$L_v = 2.5*10^6 (J\ kg^{-1})\\\\$</span>,</li>
<li><span class="math-container">$c_p = 1004 (J\ kg^{-1} K^{-1})\\\\$</span>,</li>
<li><span class="math-container">$T$</span> = temperature <span class="math-container">$(K)\\\\$</span>,</li>
<li><span class="math-container">$q(s)$</span> = saturation specific humidity <span class="math-container">$(kg/kg)\\\\$</span></li>
</ul>
<p>Imagine air at a pressure level of 760 hPa that has a temperature of 4 degrees Celsius. If I fill in the formula:</p>
<p>First calculate the saturation specific humidity for 2 temperatures:</p>
<p><span class="math-container">$qs_1$</span> for <span class="math-container">$T_1$</span> (3°C) = 0.622*((6.11*exp(19.85 - 5423/(273.15+3))*100))/(76000) * 1000
= 6.18 g/kg = 0.006180 kg/kg</p>
<p><span class="math-container">$qs_2$</span> for <span class="math-container">$T_2$</span> (5°C) = 0.622*((6.11*exp(19.85 - 5423/(273.15+5))*100))/(76000) * 1000
= 7.12 g/kg = 0.007120 kg/kg</p>
<p>Fill in the formula:</p>
<p><span class="math-container">$L_m = \frac{0.0098}{1+\frac{2.5*10^6}{1004}\frac{0.007120-0.006180}{5-3}} = 8.34480 \ K\ km^{-1}$</span></p>
<p>This seems rather high. Am I doing something wrong?</p>

CC BY-SA 4.0
2022-11-25T14:58:23.707Z
@gansub according to this calculator (http://kishou.u-gakugei.ac.jp/tools/calc/gammaw.html) the answer should be 5.44 K km-1. I don't know how to get to that result. I'd prefer to be able to understand and calculate it myself rather then using pre-existing packages.
user28035
CC BY-SA 4.0
2022-11-25T21:02:56.040Z
@gansub thank you for your answers! I found that equation in Wallace and Hobbs (2006) chapter 3: atmospheric thermodynamics, p. 107 (the equation with the blue square to the right of it, see https://gibbs.science/teaching/efd/handouts/wallace_hobbs_ch3.pdf)
user28035
<meteorology><atmosphere><thermodynamics>
How to calculate the moist adiabatic lapse rate?
user28035
# How to calculate the moist adiabatic lapse rate?

(Asked by: username_0 on 2022-11-25)

I'm trying to calculate the saturated moist adiabatic lapse rate with the following formula:

$L\_m = \frac{y\_d}{1+\frac{L\_v}{c\_p}\frac{d\_q(s)}{dT}}$

with:

* $y\_d = 9.8 (K\ km^{-1})\\\\$,
* $L\_v = 2.5\*10^6 (J\ kg^{-1})\\\\$,
* $c\_p = 1004 (J\ kg^{-1} K^{-1})\\\\$,
* $T$ = temperature $(K)\\\\$,
* $q(s)$ = saturation specific humidity $(kg/kg)\\\\$

Imagine air at a pressure level of 760 hPa that has a temperature of 4 degrees Celsius. If I fill in the formula:

First calculate the saturation specific humidity for 2 temperatures:

$qs\_1$ for $T\_1$ (3°C) = 0.622\*((6.11\*exp(19.85 - 5423/(273.15+3))\*100))/(76000) \* 1000
= 6.18 g/kg = 0.006180 kg/kg

$qs\_2$ for $T\_2$ (5°C) = 0.622\*((6.11\*exp(19.85 - 5423/(273.15+5))\*100))/(76000) \* 1000
= 7.12 g/kg = 0.007120 kg/kg

Fill in the formula:

$L\_m = \frac{0.0098}{1+\frac{2.5\*10^6}{1004}\frac{0.007120-0.006180}{5-3}} = 8.34480 \ K\ km^{-1}$

This seems rather high. Am I doing something wrong?

**Tags:** meteorology, atmosphere, thermodynamics

> **Comments:**
> **username_0**: @gansub according to this calculator (http://kishou.u-gakugei.ac.jp/tools/calc/gammaw.html) the answer should be 5.44 K km-1. I don't know how to get to that result. I'd prefer to be able to understand and calculate it myself rather then using pre-existing packages.  
> **username_0**: @gansub thank you for your answers! I found that equation in Wallace and Hobbs (2006) chapter 3: atmospheric thermodynamics, p. 107 (the equation with the blue square to the right of it, see https://gibbs.science/teaching/efd/handouts/wallace_hobbs_ch3.pdf)  

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## 0 answers

