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Sunny_sXe [5.5K]
3 years ago
11

Can you solve this question​

Engineering
2 answers:
Alecsey [184]3 years ago
6 0

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

PIT_PIT [208]3 years ago
4 0
This should help it’s very helpful it helps with everything

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If you know that the change in entropy of a system where heat was added is 12 J/K, and that the temperature of the system is 250
SVEN [57.7K]

Solution:

Given:

Change in entropy of the system, ΔS = 12J/K

Temperature of the system, T_{o} = 250K

Now, we know that the change in entropy of a system is given by the formula:

ΔS = \frac{\Delta Q}{T_{o}}

Amount of heat added, ΔQ = \Delta S\times T_{o}

ΔQ = 3000J

3 0
3 years ago
The air standard efficiency ofan Otto cycle compared to diesel cycle for thie given compression ratio is: (a) same (b) less (c)
Svetach [21]

Answer:

Correct sentence: (c) more.

Explanation:

Otto cycle:

An ideal Otto cycle models the behavior of an explosion engine. This cycle consists of six steps, as indicated in the figure. Prove that the performance of this cycle is given by the expression

η = 1 - (1/r^(ρ-1))

where r = VA / VB is the compression ratio equal to the ratio between the volume at the beginning of the compression cycle and at the end of it. ρ=1.4

Diesel cycle:

The ideal Diesel cycle is distinguished from the ideal Otto in the combustion phase, which in the Otto cycle is assumed at a constant volume and in the Diesel at constant pressure. Therefore the performance is different.

If we write the performance of a Diesel cycle in the form

:

η = 1 - (1/r^(ρ-1)) × ( (r^ρ)-1)/(ρ^(r-1)) )

we see that the efficiency of a Diesel cycle differs from that of an Otto cycle by the factor in parentheses. This factor is always greater than the unit, therefore, for the same compression reasons r

                   diesel performance is less than otto performance

\ eta_ \ mathrm {Diesel} <\ eta_ \ mathrm {Otto} \,

5 0
3 years ago
Which conditions are required for nuclear fusion to begin
Thepotemich [5.8K]
Fusion processes require fuel and a confined environment with sufficient temp, pressure, and confinement time to create a plasma in which fusion can occur.
6 0
3 years ago
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
Two piezometers have been placed along the direction of flow in a confined aquifer that is 30.0 m thick. The piezometers are 280
Dafna11 [192]

Answer:

time = 224 days

Explanation:

given data

thick = 30 m

piezometers =  280 m

head between  two = 1.4 m

aquifer hydraulic conductivity = 50 m

porosity =  20%

solution

we get here average pole velocity that is get by using Darcy law that is

Va = \frac{k}{\eta } \times \frac{\Delta h}{L}   ................1

here Va is average pole velocity and k is hydraulic conductivity and \eta is porosity  

here v is = k \times  \frac{dh}{dl}   ...........2

v = 50 × \frac{1.4}{280}

v = 0.25 m/day

and here average linear velocity Va will be

Va = \frac{v}{\eta }  

Va = \frac{0.25}{0.2}  

Va = 1.25 m/day  

travel time for water will be

time = \frac{280}{1.25}  

time = 224 days

8 0
3 years ago
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