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3 years ago

What is pixel's intensity ?

Engineering

1 answer:

bonufazy [111]3 years ago

7 0

Answer:

Small points of colored light arranged in a grid, each is formed from three colored lights: red,green,and blue. (RGB), nothing is absorbed,nothing is reflected, just see pure colored lights.

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Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th

Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0

3 years ago

Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i

Brilliant_brown [7]

Answer:

$\dot Q_{in} = 372.239\,MW$

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

$w_{in} + h_{in}- h_{out} = 0$

$h_{out} = w_{in}+h_{in}$

$h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}$

$h_{out} = 203.81\,\frac{kJ}{kg}$

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

$h_{out} = 2685.4\,\frac{kJ}{kg}$

The rate of heat transfer in the boiler is:

$\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)$

$\dot Q_{in} = 372.239\,MW$

3 0

3 years ago

Read 2 more answers

Can someone explain it to me!

Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0

2 years ago

The insulation resistance of a motor operated by an electronic drive is to be tested using a megger. What precaution should you

EleoNora [17]

Use protective gear. Use insulated tools, Wear flame resistant clothing, safety glasses, and insulation gloves, Remove watches or other jewelry, Stand on an insulation mat. 03. Never connect the insulation tester to energized conductors or energized equipment and always follow the manufacturer's recommendations. When installing new electrical machinery or equipment, testing insulation resistance is important for two reasons. First, it ensures that the insulation is in adequate condition to begin operation. ... The test is accomplished by applying DC voltage through the de-energized circuit using an insulation tester. Insulation resistance should be approximately one megohm for each 1,000 volts of operating voltage, with a minimum value of one megohm. For example, a motor rated at 2,400 volts should have a minimum insulation resistance of 2.4 megohms.

4 0

3 years ago

What is pixel's intensity ?​

What is pixel's intensity ?