5.4. Additional exercises

Exercise 5.7

For De Bilt we know the following climatologies:

Temperature	10.1 \(^\circ\)C
Precipitation	900 mm/year
Relative humidity	82 %
Wind speed	3.4 m/s
Incoming solar radiation	115 W/m\(^2\)

You might need the parameters below:

Albedo	0.24
Specific heat of air	1004 J/kg/K
Density of air	1.205 kg/m\(^3\)
Density of water	1000 kg/m\(^3\)
Latent heat of vaporization	2.45 \(\cdot\) 10\(^6\) J/kg
Psychrometer constant	0.066 kPa/\(^\circ\)C
Stefan Bolzmann constant	4.90 \(\cdot\) 10\(^{-3}\) J/d/m\(^2\)/K\(^4\)

Answer the following questions based on these climatologies and parameters.

a) Compute the annual reference evaporation using Makkink.

b) The annual mean Penman evaporation is 38 W/m\(^2\). Using the Penman evaporation and the Budyko relationship, what is the discharge in [mm/year]?

c) What is the annual combined runoff and evaporation?

d) If the annual mean Penman evaporation is 38 W/m\(^2\), what is then the net mean upwelling longwave radation in W/m\(^2\)?

Answer Exercise 5.7a

\[ E_{Makkink} = C_e \frac{s}{s+\gamma} \cdot \frac{R_c}{\rho \lambda_w} \]

\[ C_e = 0.65 \]

\[ \gamma = 0.066 \text{ kPa/$^\circ$C} \]

\[ R_c = 115 \text{ W/m$^2$} = 115 \cdot 365 \cdot 86400 = 3626640000 \text{ J/m$^2$} \]

\[ \rho = 1000 \text{ kg/m$^3$} \]

\[ \lambda_w = 2.45 \cdot 10^6 \text{ J/kg} \]

\[ e_s = 0.61 \text{ exp}({\frac{19.9 \cdot 10.1}{273+10.1}}) = 1.2407 \text{ kPa} \]

\[ s = \frac{5430}{(273+10.1)^2} \cdot 1.2407 = 0.08406 \text{ kPa/$^\circ$C} \]

\[ E_{makkink} = 0.65 \cdot \frac{0.08406}{0.08406+0.066} \cdot \frac{3626640000}{1000 \cdot 2.45 \cdot 10^6} = 0.539 \text{ m/year} = 539 \text{ mm/year} \]

Answer Exercise 5.7b

\[ \overline{E} = \overline{P}(1 - \text{ exp}(- \frac{\overline{E_{penman}}}{P})) \]

\[ \overline{P} = 900 \text{ mm/year} \]

\[ \overline{E_{penman}} = 38 \text{ W/m$^2$} = \frac{38 \text{ J/(s m$^2$)}}{2.45 \cdot 10^6 \text{ (J/m$^3$)}\cdot 10^3 \text{ (kg/m$^3$)}} = 489.1298 \text{ mm/year} \]

\[ \overline{E} = 900 \cdot (1 - \text{ exp}(- \frac{489.1298}{900})) = 377.35 \text{ mm/year} \]

\[ Q = \overline{P} - \overline{E} = 900 - 377.35 \ \ (mm/year) \]

Answer Exercise 5.7c

\[ \frac{dS}{dt} = P - E - Q = 0 \]

\[ \overline{P} = \overline{E} + \overline{Q} \]

\[ \overline{P} = 539 + 377 = 916 \text{ mm/year} \]

Answer Exercise 5.7d

\[ \rho \lambda E = \frac{s \cdot R_n + c_p \cdot \rho_a \frac{e_s(T) - e_a(T)}{r_a}}{s +\gamma} \]

\[ T = 10.1 \text{ $^\circ$C} \]

\[ h = 82 \ \ [\%] \]

\[ u = 3.4 \text{ m/s} \]

\[ e_s(10.1) = 1.24 \text{ kPa} \]

\[ e_a = 0.82 \cdot 1.24 = 1.02 \text{ kPa} \]

\[ s = 0.084 \text{ kPa} \]

\[ r_a = 1.21 \cdot 10^{-3} \text{ d/m} = 104.88 \text{ s/m} \]

\[ \rho \lambda E_{penman} = 38 \text{ W/m$^2$} \]

\[ 38 = \frac{0.084 \cdot R_n + 1004 \cdot 1.205 \frac{1.24 - 1.02}{104.8}}{0.084 + 0.066} \]

\[ R_n = 37.2 \text{ W/m$^2$} \]

\[ R_n = (1 - \alpha)R_c + R_b \]

\[ 37.2 = (1 - 0.24) \cdot 115 + R_b \]

\[ R_{l,n} = 50.2 \text{ W/m$^2$} \]

Exercise 5.8

You want to calculate the evaporation using the Penman equation based on the data from a weather station in Delft for May 7th. For the Penman equation, you need temperature, wind speed, net radiation, and humidity. The following is given:

Variable	Equivalent value	Unit
albedo	0.24
Specific heat capacity of air	1004	J/kg/K
Air density	1.205	kg/m\(^3\)
Latent heat of vaporization	2.45*10\(^6\)	J/kg
Psychrometric constant	0.066	kPa/\(^\circ\)C
Stefan-Boltzmann constant	4.90*10\(^{-3}\)	J/d/m\(^2\)/K\(^4\)

Unfortunately, the temperature measurement in Delft did not go well. A colleague advises you to use Makkink instead of Penman.

a) Why is this not a solution? (multiple answers possible)

1. Makkink is not applicable in the Netherlands

2. The constant 0.65 is temperature-dependent

3. The slope of the vapor pressure curve is temperature-dependent

4. Makkink only provides potential evaporation

5. Longwave radiation is temperature-dependent

You still want to use the Penman equation and decide to use meteorological data from nearby stations in Rotterdam. The temperature there was 19.4° Celsius.

The other data from your weather station in Delft are reliable:

• Wind speed: 1.8 m/s

• Relative humidity: 78%

• Incoming shortwave radiation: 17 MJ/m2/day

b) Calculate the actual vapor pressure in [kPa].

c) Calculate the slope of the vapor pressure curve in [kPa/°C].

d) Calculate the aerodynamic resistance in [s/m].

e) Calculate the net radiation, assuming that the net outgoing longwave radiation is 4 MJ/m2/day. Provide your answer in [MJ/m2/day].

f) How much is the net radiation expressed in [W/m2]?

g) Calculate the evaporation using the Penman equation in [mm/d].

Answer Exercise 5.8a

The right answer is: ”3. The slope of the vapor pressure curve is temperature-dependent”.

Answer Exercise 5.8b

Using eq. (5.8) gives:

\[ e_s(T=19.4) = 0.61 \exp \bigg(\frac{19.9*19.4}{273+19.4}\bigg) = 2.28 \ \ [kPa] \]

For the actual vapor pressure (\(e_a (T=19.4)\)) we need to use eq. (5.9). With the given \(h = 78 \% \) and solving for the unknown \(e_a(T=19.4)\) gives:

\[ e_a (T=19.4) = (e_s(T=19.4) * h) / 100 = 2.28 * 78/100 = 1,783314726 \ \ [kPa] \]

Answer Exercise 5.8c

Using eq. (5.11) and having solved for \(e_s(T=19.4)\) from the previous answer, the following applies:

\[ s =\frac{\textrm{d}e_s}{\textrm{d}T} = \frac{5430*e_s(T=19.4)}{(273+T)^2} = \frac{5430*2.28}{(273+19.4)^2} = 0,145 \ \ [kPa/°C] \]

Answer Exercise 5.8d

Using eq. (5.7) and with the given wind speed of \(u = 1.8 \ \ [m/s^2]\), we get:

\[ r_a = \frac{245}{(0.54*1.8+0.5)} = 166.44 \ \ [s/m] \]

Answer Exercise 5.8e

Using eq. (5.14) with the given albedo \(r = 0.24\), the net radiation (\(R_n\)) is:

\[ R_n = (1-r)R_{s,in}-R_{l,n} = (1-0.24)*17 - 4 = 8.92 \ \ [ MJ/m^2/day] \]

Answer Exercise 5.8f

We use the following converstion ratio between watt (\(W\)) and joules per second (\(J/s\)):

\[ 1 W = 1 J/s \]

Then because we have our answer in \([MJ/m^2/day] = [MJ/day /m^2]\) we need to convert it to \(J/s/m^2\) as follows:

\[ 1 \ \ [MJ/day] = \frac{10^6}{24*60*60} \ \ [J/s] = \frac{10^6}{24*60*60} \ \ [W] \]

Thus net radiation in [\(W\)]:

\[ R_n = 8,92 * \frac{10^6}{24*60*60} = 103.24 \ \ [ W/m^2 ] \]

Answer Exercise 5.8g

Using eq. (5.23) and solving it with the given and calculated values gives:

\[ E_{pm} = 7.63*10^6 \ \ [J/day/m^2 ] = 3.11 \ \ [kg/d/m^2 ] = 3.11*10^{-3} \ \ [m/d ] \]

Thus final answer is equal to:

\[ E_{pm} = 3.11 \ \ [mm/d ] \]

Exercise 5.9

Via the website of the KNMI, the following data is downloaded from weather station De Bilt for a beautiful late summer day:

Variable	Value	Unit
\(T \)	20.9	Celsius
\(h \)	78	%
\(u\)	1.8	m/s
\(R_{s,in}\)	17.33	MJ/day/m\(^2\)

Furthermore, the following data applies:

Variable	Value	Unit
\(c_p\)	1004	J/kg/K
\(ρ_a\)	1.205	kg/m\(^3\)
\(λ\)	\(2.45*10^{10}\)	J/kg
\(γ\)	0.066	kPa/°C
\(r_c\)	70	s/m
\(ρ_w\)	1000	kg/m\(^3\)
\(albedo\)	0.24
\(σ\)	\(4.9*10^{-3}\)	J/d/m\(^2\)/K\(^4\)

To estimate evaporation, KNMI calculates the Makkink evaporation because this equation uses global radiation, which is easy to measure. This is in contrast to, for example, the Penman-Monteith equation, which requires net radiation.

a) Calculate the Makkink reference evaporation for grass in mm/day. Also, provide the saturation vapor pressure (kPa) and vapor pressure slope (kPa/°C).

b) Assuming that the Makkink reference evaporation for grass is equal to the Penman-Monteith reference evaporation, what is the net radiation in MJ/day/m\(^2\)? Also, provide the actual vapor pressure (kPa) and aerodynamic resistance (s/m).

c) How much is the net radiation expressed in W/m\(^2\)?

d) What is the net outgoing longwave radiation in W/m\(^2\)? (+ indicates net longwave radiation from surface to atmosphere; - indicates net longwave radiation from atmosphere to surface).

e) Assuming we can neglect advection and warming of the Earth’s surface, what is the convective heat transport to the atmosphere (also known as ‘sensible heat’) in W/m\(^2\)?

Answer Exercise 5.9a

Using equations (5.8), (5.11) and (5.26) respectively we get the following final answers:

\[ e_s = 2.51 \ \ [kPa] \]

\[ s = 0.16 \ \ [kPa/°C] \]

\[ E_{mak} = 3.24 \ \ [mm/d] \]

Answer Exercise 5.9b

We first need to calculate \(e_a\) using eq. (5.9), then we calculate \(r_a\) using eq. (5.7). After that we solve for \(R_n\) from eq. (5.24), the final answers should yield the following values (pay attention to units):

\[ e_a = 1.96 \ \ [kPa] \]

\[ r_a = \frac{245}{(0.54*1.8+0.5)} = 166.44 \ \ [m/s] \]

\[ R_{n} = 10.5 \ \ [MJ/dag/m^2] \]

Answer Exercise 5.9c

\[ R_n = 10.5 * \frac{10^6}{24*60*60} = 121.53 \ \ [ W/m^2 ] \]

Thus, final answer:

\[ R_{n} = 121.53 \ \ [W/m^2] \]

Answer Exercise 5.9d

Albedo \(r\) is given, \(R_{s,in}\) is also given (be careful of units!) and we calculated the net radiation \(R_n\). We use eq. (5.14) to solve for \(R_{l,n}\) in [\(W/m^2\)] to get:

\[ R_n =(1-r)R_{s,in}-R_{l,n} \]

Rearranging to \(R_{l,n}\) gives:

\[ R_{l,n} =(1-r)R_{s,in}-R_n = (1-0.24)*17.33* (\frac{10^6}{24*60*60}) - 121.53 = +30.91 [W/m^2] \]

Answer Exercise 5.9e

\[ H = 29.2 \ \ [Wm^2] \]

Exercise 5.10

In Cabauw, the KNMI (Royal Netherlands Meteorological Institute) accurately measures the weather. Cabauw is located in a polder, and the measurement field consists of well-watered grass with an albedo of 0.23. On a particular day, the following meteorological variables are measured:

Variable	Value	Unit
Incoming shortwave radiation	25 \(\cdot\) 10\(^6\)	J/d/m\(^2\)
Net outgoing longwave radiation	4 \(\cdot\) 10\(^6\)	J/d/m\(^2\)
Temperature	20	degrees Celsius
Wind speed	1	m/s
Relative humidity	60	%

Furthermore, the following data applies:

Variable	Value	Unit
\(c_p\)	1004	J/kg/K
\(ρ_a\)	1.205	kg/m\(^3\)
\(λ\)	\(2.45*10^{10}\)	J/kg
\(\gamma\)	0.066	kPa/°C
\(ρ_w\)	1000	kg/m\(^3\)
\(albedo\)	0.23
\(σ\)	\(4.9*10^{-3}\)	J/d/m\(^2\)/K\(^4\)

To estimate evaporation, KNMI calculates the Makkink evaporation because this equation uses global radiation, which is easy to measure. This is in contrast to, for example, the Penman-Monteith equation, which requires net radiation.

a) Calculate the saturation vapour pressure in KPa.

b) Calculate the slope of the saturation vapor pressure curve in kPa/°C.

c) Calculate the reference evaporation using Makkink in mm/day.

d) Calculate the convective heat transport to the atmosphere in J/d/m\(^2\).

Answer Exercise 5.10a

Using equations (5.8), we get the following final answers:

\[ e_s(T=20) = 0.61 \exp \bigg(\frac{19.9*20}{273+20}\bigg) = 2.37 \ \ [kPa] \]

Answer Exercise 5.10b

Using equations (5.11), we get the following final answers:

\[ s = \frac{5430}{(273+20)^2} \cdot 2.37 = 0.15 \text{ [kPa/$^\circ$C]} \]

Answer Exercise 5.10c

Using equations (5.26), we get the following final answers:

\[ E_{makkink} = 0.65 \cdot \frac{0.15}{0.15+0.066} \cdot \frac{25000000}{1000 \cdot 2.45 \cdot 10^6} = 4.61 \text{ mm/d} \]

Answer Exercise 5.10d

Using equations (5.26), we get the following final answers:

\[ R{n} = 1.53\cdot10^7 \ \ [J/d/m^2] \]

\[ H = 3.96\cdot10^6 \ \ [J/d/m^2] \]

Exercise 5.11

On a well-irrigated field in Ghana, maize is cultivated. The following data is known about the meteorological conditions and the crop:

Variable	Value	Unit
Incoming shortwave radiation	25 \(\cdot\) 10\(^6\)	J/d/m\(^2\)
Net radiation	170	W/m\(^2\)
Air Temperature	25	degrees Celsius
Wind speed	3	m/s
Relative humidity	50	%
\(albedo\)	30	%

Furthermore, the following data applies:

Variable	Value	Unit
\(c_p\)	1004	J/kg/K
\(ρ_a\)	1.205	kg/m\(^3\)
\(λ\)	\(2.45*10^{10}\)	J/kg
\(\gamma\)	0.066	kPa/°C
\(ρ_w\)	1000	kg/m\(^3\)
\(albedo\)	0.23
\(σ\)	\(4.9*10^{-3}\)	J/d/m\(^2\)/K\(^4\)

To estimate evaporation, KNMI calculates the Makkink evaporation because this equation uses global radiation, which is easy to measure. This is in contrast to, for example, the Penman-Monteith equation, which requires net radiation.

a) Calculate the aerodynamic resistance in s/m.

b) Calculate the saturation vapor pressure in kPa.

c) Calculate the actual saturation vapor pressure in kPa.

d) Calculate the slope of the vapor pressure curve in kPa/C.

e) Calculate the evaporation using the Penman-Monteith equation in W/m\(^2\) and in mm/day.

Due to poor maintenance, the irrigation system no longer operates optimally, and water is no longer supplied to the maize field. This causes stress to the crop, and the crop resistance increases to 150 s/m.

g) What is the absolute difference in convective heat transport between the well-functioning and poorly functioning irrigation systems in W/m\(^2\)?

Answer Exercise 5.11a

Using equations (5.7), we get the following answer:

\[ r_a = \frac{245}{(0.54 \cdot 3+0.5)} = 115.57 \ \ [s/m] \]

Answer Exercise 5.11b

Using equations (5.8), we get the following final answers:

\[ e_s(T=20) = 0.61 \exp \bigg(\frac{19.9*25}{273+25}\bigg) = 3.24 \ \ [kPa] \]

Answer Exercise 5.11c

Using equations (5.9):

\[ h = \frac{e_a(T)}{e_s(T)} \times 100\% \]

And solving for \(e_a(T)\) gives:

\[ e_a(T) = \frac{h \cdot e_s(T)}{100\%} = \frac{50\% \cdot 3.24}{100\%} = 1.62 \ \ [kPa] \]

Answer Exercise 5.11d

Using equations (5.11), we get the following answer:

\[ s = \frac{5430}{(273+25)^2} \cdot 3.24 = 0.198 \text{ [kPa/$^\circ$C]} \]

Answer Exercise 5.11e

Using equations (5.24), we get the following answer:

\[ E_{p}= 157.62 \ \ [W/m^2] = 5.56 \ \ [mm/dag] \]

Answer Exercise 5.11g

Using the energy balance principle:

\[ R_{net} = H + E \]

We get the following answers: $\( H_{well watered} = 12.38 \ \ [W/m^2] \)$

\[ E_{pm stress} = 144.75 \ \ [W/m^2] \]

\[ H_{stress} = 25.25 \ \ [W/m^2] \]

\[ \Delta H = 25.25 - 12.38 = 12.87 \ \ [W/m^2] \]

from jupyterquiz import display_quiz

Exercise 5.12

Choose the correct statement.