Additional exercises

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
\[ \rho \lambda E = \frac{s \cdot R_n + c_p \cdot \rho_a \frac{e_s(T) - e_a(T)}{r_a}}{s + \lambda} \]
\[ T = 10.1 \text{ $^\circ$C} \]
\[ h = 82 \ \ [\%] \]
\[ u = 3.4 \text{ m/s} \]
\[ e_s(10.1) = 1.24 \text{ kPa} \]
\[ e_a = 0.82 \cdot 1.24 = 1.02 \text{ kPa} \]
\[ s = 0.084 \text{ kPa} \]
\[ r_a = 1.21 \cdot 10^{-3} \text{ d/m} = 104.88 \text{ s/m} \]
\[ \rho \lambda E_{penman} = 38 \text{ W/m$^2$} \]
\[ 38 = \frac{0.084 \cdot R_n + 1004 \cdot 1.205 \frac{1.24 - 1.02}{104.8}}{0.084 + 0.066} \]
\[ R_n = 37.2 \text{ W/m$^2$} \]
\[ R_n = (1 - \alpha)R_c + R_b \]
\[ 37.2 = (1 - 0.24) \cdot 115 + R_b \]
\[ R_{l,n} = 50.2 \text{ W/m$^2$} \]

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/m2

Furthermore, the following data applies:

Variable

Value

Unit

\(c_p\)

1004

J/kg/K

\(ρ_a\)

1.205

kg/m3

\(λ\)

\(2.45*10^{10}\)

J/kg

\(γ\)

0.066

kPa/°C

\(r_c\)

70

s/m

\(ρ_w\)

1000

kg/m3

\(albedo\)

0.24

\(σ\)

\(4.9*10^{-3}\)

J/d/m2/K4

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/m2? Also, provide the actual vapor pressure (kPa) and aerodynamic resistance (s/m).

c) How much is the net radiation expressed in W/m2?

d) What is the net outgoing longwave radiation in W/m2? (+ 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/m2?

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):

\[ e_a = 1.96 \ \ [kPa] \]
\[ r_a = \frac{245}{(0.54*1.8+0.5)} = 166.44 \ \ [m/s] \]
\[ R_{net} = 10.5 \ \ [MJ/dag/m^2] \]
Answer Exercise 5.9c
\[ R_n = 10.5 * \frac{10^6}{24*60*60} = 121.53 \ \ [ W/m^2 ] \]

Thus, final answer:

\[ R_{net} = 121.53 \ \ [W/m^2] \]
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
\[ H = 29.2 \ \ [Wm^2] \]