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2 years ago

15

A ____ is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does no

t exceed about 30 psig. Group of answer choices

1 answer:

Alika [10]2 years ago

3 0

Answer:

A relief valve is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does not exceed about 30 psig.

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A 3 m aluminum pole is kept at a residential site for construction

Aliun [14]

Answer:

I don't know sorry

Explanation:

5 0

3 years ago

The transfer function of a typical tape-drive system is given by

maw [93]

Answer:

the range of K can be said to be : -3.59 < K< 0.35

Explanation:

The transfer function of a typical tape-drive system is given by;

$KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}$

calculating the characteristics equation; we have:

1 + KG(s) = 0

$1+ \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0$

${s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0$

$s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0$

$s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0$

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as follows:

$S^5$ 1 5.1 2+ K

$S^4$ 1.9 6.2 4K

$S^3$ 1.83 $\dfrac{1.9 (2+K)-4K}{1.9}$ 0

$S^2$ $\dfrac{11.34-1.9(X)}{1.83}$ 4K 0

S $\dfrac{XY-7.32 \ K}{Y}$ 0 0

$\dfrac{1.9 (2+K)-4K}{1.9} = X$

$\dfrac{11.34-1.9(X)}{1.83}= Y$

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34 - 1.9(X) > 0

$11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0$

$11.34 - (3.8 - 2.1K)>0$

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

$(\dfrac{3.8+1.9K-4K}{1.9})[11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]$

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be : -3.59 < K< 0.35

4 0

3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Ilya [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

8 0

3 years ago

An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40

Helen [10]

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

6 0

3 years ago

thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power

IrinaK [193]

Answer:

(a) the percent thermal efficiency is 27.94%

(b) the temperature of the cooling water exiting the condenser is 31.118°C

Explanation:

3 0

3 years ago