Match each context to the type of the law that is most suitable for it.

Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s and e

ZanzabumX [31]

Answer:

exit temperature 285 K

Explanation:

given data

temperature T1 = 270 K

velocity = 180 m/s

exit velocity = 48.4 m/s

solution

we know here diffuser is insulated so here heat energy is negleted

so we write here energy balance equation that is

0 = m (h1-h2) + m × $(\frac{v1^2}{2}-\frac{v2^2}{2})$ .....................1

so it will be

$h1 + \frac{v1^2}{2} = h2 + \frac{v2^2}{2}$ .....................2

put here value by using ideal gas table

and here for temperature 270K

h1 = 270.11 kJ/kg

$270.11 + \frac{180^2\times \frac{1}{1000}}{2} = h2 + \frac{48.4^2\times \frac{1}{1000}}{2}$

solve it we get

h2 = 285.14 kJ/kg

so by the ideal gas table we get

T2 = 285 K

4 0

2 years ago

The loneliest people are to kindest

valkas [14]

Answer:

The most damaged people are the wisest is a fact

Explanation:

5 0

2 years ago

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What would happen to a plane if the weight force becomes greater than the lift force?

mrs_skeptik [129]

Answer:

When the lift is greater than the weight, the aircraft gains altitude. ... Drag must be overcome for the aircraft to move, and movement is essential to obtain lift. To overcome drag and move the aircraft forward, another force is essential. This force is thrust.

Explanation:

Hope this helps!

8 0

2 years ago

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A series AC circuit contains a resistor, an inductor of 250 mH, a capacitor of 4.40 µF, and a source with ΔVmax = 240 V operatin

slega [8]

Answer:

Explanation:

Inductance = 250 mH = 250 / 1000 = 0.25 H

capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)

ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A

a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms

b) capacitive reactance = $\frac{1}{2\pi fC}$ = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

c) impedance = $\frac{Vmax}{Imax}$ = 240 / 0.11 = 2181.82 ohms

7 0

3 years ago

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A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t

Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0

3 years ago