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3 years ago

Describe the process you used to build a model

1 answer:

6 0

Step 1 : Get your supply list together

Step 2 : Pick what model you want to do

Step 3 : Ask for a partner

Step 4 : Complete the model and take your time.

Step 5 : Read the directions carefully

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Friction is generated when?

valkas [14]

Things are rubbed against each other

8 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

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

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

particle moves over three quarters of a circle of radius R what is the magnitude of its displacement

fomenos

Answer:

is equal to threee idont know

Explanation:

hindi ko alam ang explanation kwa kwek kwen hahahahhahahahaha

sanay akoy pag padendyahan niyo nay. hahahahhahahaha maawa kayo saakin mga aking mahal hahahhahahahahhaha hahah en then go char hahahhaa

6 0

3 years ago

A clear sheet of polaroid is placed on top of a similar sheet so that their polarizing axes make an angle of 30° with each other

xxMikexx [17]

Answer:

$\dfrac{I}{I_0}=\dfrac{3}{8}$

Explanation:

Given that

Angle ,θ = 30°

From Malus law,Intensity given as

$I=\dfrac{I_0}{2}cos^2\theta$

Io=Intensity of unpolarized light

I=Intensity of emerging light

Now by putting the value of angle

$I=\dfrac{I_0}{2}cos^2\theta$

$I=\dfrac{I_0}{2}cos^230^{\circ}$

We know that

$cos30^{\circ}=\dfrac{\sqrt{3}}{2}$

$I=\dfrac{I_0}{2}\times \dfrac{3}{4}$

$\dfrac{I}{I_0}=\dfrac{3}{8}$

Therefore ratio will be $\dfrac{3}{8}$

7 0

3 years ago

A person walks 20.0° north of east for 3.20 km. How far would she have to walk due north and due east to arrive at the same loca

aivan3 [116]

Answer:

1.09 km, 3 km

Explanation:

displacement, d = 3.20 km at 20° North of east

A vector quantity has two components one along the x axis and the other is along Y axis. the component along X axis is called the horizontal component and it is due east.

The component along y axis is called vertical component and it is due North.

Displacement due North, dy = 3.20 Sin 20° = 1.09 km

Displacement due east, dx = 3.20 Cos 20° = 3 km

8 0

3 years ago