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3 years ago

determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004

Engineering

1 answer:

Afina-wow [57]3 years ago

6 0

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

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Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______

Alina [70]

Answer:

Multiplying impulse response by t ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

5 0

3 years ago

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are

Vika [28.1K]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle $W_{net$ = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ $^{Y-1$ = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = $($ T₂ / T₁ $)^{\frac{1}{Y-1}$

so we substitute

⇒ V₁ / V₂ = $($ 973 K / 303 K $)^{\frac{1}{1.4-1}$

= $($ 3.21122 $)^{\frac{1}{0.4}$

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c) the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂ = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

8 0

3 years ago

A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show

aleksandrvk [35]

Answer:

$GM$

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter $d=2m$

Length $l=2.5m$

Weight $W=22kN$

Specific weight of sea water $\mu= 10.25kN/m^3$

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

$W=(pwg)Vd$

Where

$V_d=\pi/4(d)^2y$

Therefore

$22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m$

Therefore

Center of Bouyance B

$B=\frac{y}{2}=0.26m\\\\B=0.75$

Center of Gravity

$G=\frac{I.B}{2}=2.6m$

Generally the equation for\BM is mathematically given by

$BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\$

Therefore

$BG=2.6-0.476\\\\BG=0.64m$

Therefore

$GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\$

Therefore

$GM$

So the bouy does not float with its axis vertical

4 0

3 years ago

An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+

nexus9112 [7]

Answer:

the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

Explanation:

From the question; the equations of the velocities profile in the system are:

$u = u_{max}(Ay^2+By+C)$ ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0 ----- (a)

At y = h; u =0 -----(b)

At y = $\frac{h}{2}$ ; u = $u_{max}$ ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

$u_{max}$ = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0$

Replacing the boundary condition (b) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah \ \ \ \ \ --- (d)$

Replacing the boundary condition (c) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 = \frac{Ah^2}{4} + \frac{h}{2}(-Ah) \\ \\ 1= \frac{Ah^2}{4} - \frac{Ah^2}{2} \\ \\ 1 = \frac{Ah^2 - Ah^2}{4} \\ \\ A = -\frac{4}{h^2}$

replacing $A = -\frac{4}{h^2}$ for A in (d); we get:

$B = - ( -\frac{4}{h^2})h$ $B = \frac{4}{h}$

replacing the values of A, B and C into the velocity profile expression; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)$

To determine the volume flow rate; we have:

$Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)$

Replacing $u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u$

$\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})$

$\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h$

Thus; the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

$\frac{Q}{b}= Vh$

Also; $\frac{Q}{b} = \frac{2}{3} u_{max} h$

Then;

$\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}$

Thus; the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

5 0

4 years ago

Evaluate the performance of the proposed heat pump for three locations Using R134a. Discuss the effect of outdoor temperature on

Phoenix [80]

Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).

1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42

.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43

-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82

A paired t-test is the standard procedure for testing this null hypothesis.

We use a paired t-test because each worker was measured twice, once for Paired t-test for

each workplace, so the observations on the two workplaces are dependent. paired data

Fast workers are probably fast for both workplaces, and slow workers are

slow for both. Thus what we do is compute the difference (standard − er-

gonomic) for each worker, and test the null hypothesis that the average of

these differences is zero using a one sample t-test on the differences.

Table 2.2 gives the differences between standard and ergonomic times.

Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-

ences in the sample. We assume that these differences are independent sam-

ples from a normal distribution with mean µ and variance σ

, both unknown.

Our null hypothesis is that the mean µ equals prespecified value µ0 = 0

(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the

workers to be faster in the ergonomic workplace.

The formula for a one sample t-test is

t =

¯d − µ0

s/√

where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test

sample size, and s is the sample standard deviation (of the differences)

s =

vuut

n − 1

i=1

(di − ¯d )

2 .

If our null hypothesis is correct and our assumptions are true, then the t-

statistic follows a t-distribution with n − 1 degrees of freedom.

The p-value for a test is the probability, assuming that the null hypothesis

is true, of observing a test statistic as extreme or more extreme than the one The p-value

we did observe. “Extreme” means away from the the null hypothesis towards

the alternative hypothesis. Our alternative here is that the true average is

larger than the null hypothesis value, so larger values of the test statistic are

extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of

freedom from the observed t-value to the right. (If the alternative had been

µ < µ0, then the p-value is the area under the curve to the left of our test

Explanation: The curve represents the sum total of the evaluation

4 0

4 years ago