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2 years ago

Hi plz delete this question i had to edit it cuz it was wrong question

Engineering

1 answer:

Rina8888 [55]2 years ago

3 0

Answer:

Explanation:ok

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Implement the function F(A, B, C, D) = Σm(1, 2, 3, 4, 6, 9, 12, 13, 15) as a minimal hazard-free SOP circuit. In your solution,

Marina86 [1]

Answer:

See attachment below

Explanation:

3 0

3 years ago

The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along

lora16 [44]

Answer:

Using the formula

V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2

Hence fluid speed at x axis =20x/(x^2+y^2)^1/2

While the fluid speed at y axis =20y/(x^2+y^2)^1/2

Now the angle at 1, 5

We substitute into the formula above

V= 20×5/(1+25)^1/2= 19.61

For x we have

V = 20× 1/(1+25)^1/2= 3.92

Angle = 19.61/3.92= 5.0degrees

Angel at 5, and 2

We substitute still

V = 20×5/(2+25)^1/2=19.24

At 2 we get

V= 20×2/(2+25)^1/2=7.69

Dividing we get 19.24/7.69=2.5degrees

At 1 and 0

V = 20/(1)^1/2=20

At 0, v =0

Angel at 2 and 0 = 20degrees

At 5 and 2

V = 100/(25+ 4)^1/2=18.56

At x = 2

40/(√29)=7.43

Angle =18.56/7.43 = 2.49degrees.

6 0

2 years ago

Read 2 more answers

Steam enters a turbine operating at steady state at 2 MPa, 323 °C with a velocity of 65 m/s. Saturated vapor exits at 0.1 MPa an

Lera25 [3.4K]

Answer:

$\dot Q_{out} = 13369.104\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0$

The rate of heat transfer between the turbine and its surroundings is:

$\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}$

The specific enthalpies at inlet and outlet are, respectively:

$h_{in} = 3076.41\,\frac{kJ}{kg}$

$h_{out} = 2675.0\,\frac{kJ}{kg}$

The required output is:

$\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW$ $\dot Q_{out} = 13369.104\,kW$