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3 years ago

Where does Henrietta think William might be?

Engineering

2 answers:

lora16 [44]3 years ago

6 0

Answer:

B.) CEMETERY

Explanation:

Have a nice day!

umka2103 [35]3 years ago

5 0

Answer:

The cemetery

Explanation:

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Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 1

Elis [28]

Answer:

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W" $_{cv$ / [ (h₁ - h₂ ) - $\frac{V_2^2}{2}$ ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW × $\frac{1\frac{k^3}{s} }{1kW}$ ] / [ (932.93 - 503.02 )k³/kg - $\frac{100^2\frac{m^2}{s^2} }{2}$ | $\frac{ln}{kg\frac{m}{s^2} }$ || $\frac{1kJ}{10^3N-m}$ | ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[ $(\frac{8.314}{28.97}\frac{k^3}{kg.K})$ × $500 K$ × $(7.54 kg/s)$ ] / [(1 bar)(100 m/s )]} | $\frac{1 bar}{10N/m^2}$ || $10^3N.m/1k^3$

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²

8 0

3 years ago

The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of

weeeeeb [17]

Answer:

$P_{pump}= 68600Pa$

Explanation:

We understand that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

In this way,

$P_{tank} + P_{pump} = P_{nozzle}$

For the pressure we also know that it is given by

$P=\rho gh$

$\rho gh_1 + P_{pump} = \rho gh_2$

So,

$P_{pump} = \rho g(h2 - h1)$

$P_{pump}= 1000*9.8(27-20)$

$P_{pump}= 68600Pa$

7 0

3 years ago

What is the difference between white and yellow line?

Savatey [412]

Answer:

The question doesnt make any sense. Like what are the lines used for.

Explanation:

5 0

3 years ago

Read 2 more answers

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

Aliun [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.

5 0

3 years ago

What were some of the challenges to safety resulting from such radical airframe designs as highly swept wings, high wing loading

Andrew [12]

How to take off and land, stopping considerations (stopping distance), control system capability over a large speed range and flutter, and structural integrity for the wing platform and speed range.

8 0

3 years ago