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3 years ago

8

The slope of a moment diagram is the load. a)-True b)-False

1 answer:

Alex17521 [72]3 years ago

8 0

Answer:

true.

Explanation:

but i am not 100% sure

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A vertical piston-cylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a c

Ann [662]

Answer:

Amount of air left in the cylinder=m $_{2}$ =0.357 Kg

The amount of heat transfer=Q=0

Explanation:

Given

Initial pressure=P1=300 KPa

Initial volume=V1=0.2 $m^{3}$

Initial temperature=T $_{1}$ =20 C

Final Volume= $V_{2}$ =0.1 $m^{3}$

Using gas equation

$m_{1}=((P_{1}*V_{1})/(R*T_{1}))$

m1==(300*0.2)/(.287*293)

m1=0.714 Kg

Similarly

m2=(P2*V2)/R*T2

m2=(300*0.1)/(0.287*293)

m2=0.357 Kg

Now calculate mass of air left,where me is the mass of air left.

me=m2-m1

me=0.715-0.357

mass of air left=me=0.357 Kg

To find heat transfer we need to apply energy balance equation.

$Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})$

Where me=m1-m2

And as the temperature remains constant,hence the enthalpy also remains constant.

h1=h2=he=h

Q=(me-(m1-m2))*h

me=m1-me

Thus heat transfer=Q=0

6 0

3 years ago

Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?

WITCHER [35]

Answer:

You can create high drag which allows a steeper angle without increasing your air speed on landing. you can reduce the length of landing role. Flaps are also used to increase the drag they are retracted when they are not needed. it is adviseable to down he flaps during the time of take off.

4 0

3 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

7 0

3 years ago

Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa a

Elena L [17]

Answer:

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

$S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1})$

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

$S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50})$

$S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888)$

$S_2 - S_1 = 0.5864 - 0.6968$

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

6 0

3 years ago

A 10 wt % aqueous solution of sodium chloride is fed to an evaporative crystallizer which operates at 80o C. The product is a sl

VikaD [51]

Answer: The feed rate is

17,020kg/he and the rate is 13,520kg/h

Check the attachment for step by step explanation

3 0

3 years ago