1. Làm thế nào để đảm bảo tính khả thi của văn bản hành chính ?

How long should the shafts remain in the furnace to achieve a desired centerline temperature of 800K? 2) Determine the temperatu

Phantasy [73]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0

3 years ago

Looking at the response vehicles (pictured above), explain two options you have in order to abide by the Move

SIZIF [17.4K]

Answer:

slow down
change lanes

Explanation:

The "Move Over law" varies by state, but generally requires you vacate the adjacent lane (the one you're currently traveling in), or slow down. Some states have specific speed requirements; others require only "safe and prudent" speed.

The sort of parked vehicles that require you to "move over" also vary by state. It would be "safe and prudent" to move over for <em>any</em> vehicle parked on the shoulder, especially if there are people or animals around those vehicles.

8 0

3 years ago

Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0

2 years ago

A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

strojnjashka [21]

Answer:

$\dot W_{out} = 399.47\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0$

The work done by the turbine is:

$\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}$

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

$P = 10\,MPa$

$T = 520\,^{\textdegree}C$

$h = 3425.9\,\frac{kJ}{kg}$

Outlet (Superheated Steam)

$P = 1\,MPa$

$T = 280\,^{\textdegree}C$

$h = 3008.2\,\frac{kJ}{kg}$

The work output is:

$\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW$

$\dot W_{out} = 399.47\,kW$

5 0

3 years ago

A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the

ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0

3 years ago