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2 years ago

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A 600-ha farmland receives annual rainfall of 2500 mm. There is a river flowing through the farmland with an inflow rate of 5 m3

/s and outflow rate of 4 m3/s. The annual water storage is the farmland increases by 2.5 x 106 m3. Based on the hydrologic budget equation, determine the annual evapotranspiration amount in mm. (1 ha

1 answer:

shepuryov [24]2 years ago

3 0

Answer:

E = 7333.33 mm

Explanation:

The annual evapotranspiration (E) amount can be calculated using the water budget equation:

$P*A + Q_{in}*\Delta t = E*A + \Delta S + Q_{out}*\Delta t$ (1)

<u>Where</u>:

<em>P: is the precipitation = 2500 mm, </em>

<em>Q(in): is the water flow into the river of the farmland = 5 m³/s, </em>

<em>ΔS: is the change in water storage = 2.5x10⁶ m³, </em>

<em>Q(out): is the water flow out of the river of the farmland = 4 m³/s.</em>

<em>Δt: is the time interval = 1 year = 3.15x10⁷ s </em>

<em>A: is the surface area of the farmland = 6.0x10⁶ m² </em>

Solving equation (1) for ET we have:

$E = \frac{P*A + Q_{in}*\Delta t - \Delta S - Q_{out}*\Delta t}{A}$

$E = \frac{2.5 m \cdot 6.0 \cdot 10^{6} m^{2} + 5 m^{3}/s \cdot 3.15 \cdot 10^{7} s - 2.5 \cdot 10^{6} m^{3} - 4 m^{3}/s \cdot 3.15 \cdot 10^{7} s}{6.0\cdot 10^{6} m^{2}}$

$E = 7333.33 mm$

Therefore, the annual evapotranspiration amount is 7333.33 mm.

I hope it helps you!

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Answer:

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$a_z = -5.70m/s^2$

$a_q = 0m/s^2$

Explanation:

We are given:

$Speed v_o = 28 m/s$

$Acceleration a_o= 7 m/s^2$

We first need to find the radial position r of washer in x-y plane.

Therefore

$r = \sqrt{300^2 + 400^2}$

r = 500 mm

To find length along direction OA we have:

$L = \sqrt{500^2 + 700^2}L = 860 mm$

Therefore, the radial and vertical components of velocity will be given as:

$V_r = V_o*cos(Q)$

$V_z = V_o*sin(Q)$

Where Q is the angle between OA and vector r.

Therefore,

$V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}$

$V_r = -16.28 m/s$

• $V_z = 28 * \frac{700}{860} = -22.8$

• $V_q = 0 m/s$

The radial and vertical components of acceleration will be:

$a_r = a_o*cos(Q)$

$a_z = a_o*sin(Q)$

Therefore we have:

• $a_r = 7* \frac{500}{860} = -4.07m/s^2$

• $a_z = 7 * \frac{700}{860} = -5.70 m/s^2$

• $a_q = 0 m/s^2$

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An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs

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Answer:

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the initial velocity $V_{1}$ =100 km/h-->100*10^3/3600=27.78 m/s

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<u>To find:</u>

the change in kinetic energy ΔKE

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<u>assumption:</u>

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ΔKE=5*m*( $V_{F}$ ^2- $V_{1}$ ^2)

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Answer:

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Explanation:

Given that

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a)

We know that volume V given as

$V=\pi r^2 h$

$V=\pi \times 1^2\times 3$

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b)

Mass = Density x volume

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1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

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Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

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$v=\dfrac{1}{40}\ ft^3/lb$

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d)

Specific weight(w) is the product of density and the gravity(g).

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Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

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The total amount of daily heat transfer is 1382.38 M w.

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<u>Explanation:</u>

Given data,

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$h_{0}$ = 250 w/ $m^{2}$ k

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Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

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$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

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$T_{3}$ - 10 = 7.96

$T_{3}$ = 17.96 ° C.

4 0

2 years ago