Write down about the water source selection criteria

A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated pow

Anna [14]

Answer:

Rated power = 1345.66 W/m²

Mechanical power developed = 3169035.1875 W

Explanation:

Wind speed, V = 13 m/s

Coefficient of performance of turbine, $C_p$ = 0.3

Rotor diameter, d = 100 m

or

Radius = 50 m

Air density, ρ = 1.225 kg/m³

Now,

Rated power = $\frac{1}{2}\rho V^3$

or

Rated power = $\frac{1}{2}\times1.225\times13^3$

or

Rated power = 1345.66 W/m²

b) Mechanical power developed = $\frac{1}{2}\rho AV^3C_p$

Here, A is the area of the rotor

or

A = π × 50²

thus,

Mechanical power developed = $\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3$

or

Mechanical power developed = 3169035.1875 W

8 0

3 years ago

A panel crimper cuts metal True or False

densk [106]

Answer:

True (They can cut through most metal panels, and probably heavier)

Explanation:

Hope this helps you as much as intended.

3 0

3 years ago

Explain what the engineering team should advise in the following scenario.

Evgesh-ka [11]

Answer: its c

Explanation:

7 0

3 years ago

What was the purpose of the vasa ship

goldfiish [28.3K]

The main purpose was for power. The vessel has come to symbolize Sweden's Great Power Period, when the nation became a major European power and controlled much of the Baltic.

8 0

3 years ago

A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0

3 years ago

Write down about the water source selection criteria​

Write down about the water source selection criteria