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3 years ago

Calculate the theoretical efficiency for an Otto cycle engine with γ=1.40 and compression ration r=9.50.

Engineering

1 answer:

DENIUS [597]3 years ago

7 0

Answer:

The theoretical efficiency of the Otto engine is 59.4 percent.

Explanation:

The theoretical efficiency for Otto engine ( $\eta$ ), dimensionless, is obtained by this formula after supossing that fluid is an ideal gas:

$\eta = 1-\frac{1}{r^{\gamma -1}}$

Where:

$r$ - Compression ratio, dimensionless.

$\gamma$ - Heat capacity ratio, dimensionless.

Given that $r = 9.50$ and $\gamma = 1.40$ , the theoretical efficiency of the Otto engine is:

$\eta = 1 - \frac{1}{9.50^{1.40-1}}$

$\eta = 0.594$

The theoretical efficiency of the Otto engine is 59.4 percent.

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The condition of irrotationality for a two-dimensional flow is satisfied when rotation w everywhere is (less than — equal to — m

Deffense [45]

Answer:

Zero.

Explanation:

Lets take velocity V is given as in 2 dimensional flow

V=u i +v j

V=f(x,y)

u=f(x,y)

v=f(x,y)

Rotational ability ω given as

$\omega =\dfrac{1}{2}\left (\dfrac{\partial v}{\partial x}-\dfrac{\partial u} {\partial x}\right)$

ω=0 ,then flow will be irrotational.

ω ≠0 ,then flow will be rotational.

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3 years ago

The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin

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Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.

The surface of the cross-section of the stapes was determined:

A_ab= 10 mm^2

A-cd= 15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab $\leq$ σ_allow

σ_cd = F_cd/A_cd $\leq$ σ_allow

In the next step we will determine the static size: Picture b).

We apply the conditions of equilibrium:

∑F_x=0

∑F_y=0

∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

==> F_ab = w*k*N

Now we determine the load w

Sector AB:

σ_ab = F_ab/A_ab $\leq$ σ_allow=300 KPa

= w/10*10^-6 $\leq$ σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

Sector CD:

σ_cd = F_cd/A_cd $\leq$ σ_allow=300 KPa

= 2*w/15*10^-6 $\leq$ σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

==> w=2.25 * 10^-3 kN/m

The solution is:

w=2.25 N/m

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3 years ago

Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it

juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

gcd(31,415,14,142) =gcd(14,142,3,131)

=gcd=(3,131, 1,618)

=gcd(1,618, 1,513)

=gcd(1,513, 105)

=gcd(105, 43)

=gcd(43, 19)

=gcd(19, 5)

=gcd(5, 4)

=gcd(4, 1)

=gcd(1, 0)

STEP 2

b) The number of comparison of given input with the algorithm based on checking consecutive integers and Euclid's algorithm is

The number of division using Euclid's algorithm =10 from part (a)

The consecutive integer checking algorithm:

The number of iterations =14,142 and 1 or 2 division of iteration.

14,142 ∠= number of division∠ = 2*14,142

Euclid's algorithm is faster by at least 14,142/10 =1400 times

At most 2*14,142/10 =2800 times.

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3 years ago

What are some tasks commonly performed in Environmental Service Systems jobs? Check all that apply.

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Answer: a,c,d,f

Explanation:

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4 years ago

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Airida [17]

Answer: they are connected in series.

Explanation:

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4 years ago