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3 years ago

Discuss the applications of numerical weather forecasting

1 answer:

3 0

Numerical weather prediction (NWP) uses mathematical models of the atmosphere and oceans to predict the weather based on current weather conditions. Though first attempted in the 1920s, it was not until the advent of computer simulation in the 1950s that numerical weather predictions produced realistic results. A number of global and regional forecast models are run in different countries worldwide, using current weather observations relayed from radiosondes, weather satellites and other observing systems as inputs.

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Sam decides to remove his oversized hooded jacket before he works near the pto driveline is it safe or unsafe

Salsk061 [2.6K]

Its safe because it isn't something with electricity

4 0

2 years ago

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula:

$\dot m = \frac{\eta \cdot \dot W}{g\cdot H}$ (1)

Where:

$\dot W$ - Pump power, in watts.

$\eta$ - Efficiency, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$H$ - Hydrostatic column, in meters.

If we know that $\eta = 0.72$ , $\dot W = 5000\,W$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the mass flow rate of water is:

$\dot m = 14.683\,\frac{kg}{s}$

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump ( $\Delta P$ ), in pascals, is determined by this equation:

$\Delta P = \rho\cdot g\cdot H$ (2)

Where $\rho$ is the density of water, in kilograms per cubic meter.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the pressure difference is:

$\Delta P = 245175\,Pa$

The pressure difference across the pump is 245.175 kilopascals.

4 0

3 years ago

Velocity components in an incompressible flow are: v = 3xy + x^2 y: w = 0. Determine the velocity component in the x-direction.

cupoosta [38]

Answer:

Velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

Explanation:

v=3xy+ $x^{2}$ y

We know that for incompressible flow

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

$\frac{\partial v}{\partial y}=3x+x^{2}$

So $\frac{\partial u}{\partial x}+3x+x^{2}=0$

$\frac{\partial u}{\partial x}= -3x-x^{2}$

By integrate with respect to x,we will find

$u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ +C

So the velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

3 0

3 years ago

Write a statement to print the data members of InventoryTag. End with newline. Ex: if itemID is 314 and quantityRemaining is 500

Advocard [28]

Answer:

#include <stdio.h>

typedef struct InventoryTag_struct {

int itemID;

int quantityRemaining;

} InventoryTag;

int main(void) {

InventoryTag redSweater;

redSweater.itemID = 314;

redSweater.quantityRemaining = 500;

/* Your solution goes here */

printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);

getchar();

return 0;

}

Explanation:

7 0

3 years ago

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W

cupoosta [38]

Answer:

It will not experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of KIc, Y, and the largest value of a in the material. This requires that we solve for σc from the following equation:

σc = KIc / (Y*√(π*a))

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 0

3 years ago

Discuss the applications of numerical weather forecasting​

Discuss the applications of numerical weather forecasting