A resistor, an inductor, and a capacitor are connected in series to an ac source. What is the condition for resonance to occur?.

At a certain location, wind is blowing steadily at 7 m/s. Determine the mechanical energy of air per unit mass and the power gen

Kaylis [27]

Answer:

Explanation:

From the information given;

The velocity of the wind blow V = 7 m/s

The diameter of the blades (d) = 80 m

Percentage of the overall efficiency $\eta_{overall} = 30\%$

The density of air $\rho = 1.25 kg/km^3$

Then, we can use the concept of the kinetic energy of the wind blowing to estimate the mechanic energy of air per unit mass by using the formula:

$e_{mechanic} = \dfrac{mV^2}{2}$

here;

m = $\rho AV$

= $1.25 \times \dfrac{\pi}{4}(80)^2 \times 7$

= 43982.29 kg/s

∴

$W = e_{mechanic} = \dfrac{mV^2}{2}$

$= \dfrac{43982.29 \times 7^2}{2}$

$= 1077566.105 \ W$

$\mathbf{ =1077.566 \ kW}$

The actual electric power is:

$W_{electric} = \eta_{overall} \times W$

$W_{electric} = 0.3 \times 1077.566$

$\mathbf{W_{electric} =323.26 \ kW}$

8 0

2 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration

Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

4 0

3 years ago

Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a

ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s

Explanation:

Given:-

- The compressor exit conditions are given as follows:

Pressure ( Pe ) = 825 KPa

Temperature ( Te ) = 150°C

Velocity ( Ve ) = 10 m/s

Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

Pressure = Pi = 100 KPa

Temperature = Ti = 20.0

Velocity = Vi = 1.0 m/s

Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

Qe = Ae*Ve

Where,

Ae: The exit cross sectional area

Ae = π*de^2 / 4

Therefore,

Qe = Ve*π*de^2 / 4

Qe = 10*π*0.05^2 / 4

Qe = 0.01963 m^3 / s

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

Pe / ρe = R*Te

Where,

Te: The absolute temperature at exit

ρe: The density of air at exit

R: the specific gas constant for air = 0.287 KJ /kg.K

ρe = Pe / (R*Te)

ρe = 825 / (0.287*( 273 + 150 ) )

ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

m^ = ρe*Qe

= ( 6.79566 )*( 0.01963 )

= 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

Pi / ρi = R*Ti

Where,

Ti: The absolute temperature at inlet

ρi: The density of air at inlet

R: the specific gas constant for air = 0.287 KJ /kg.K

ρi = Pi / (R*Ti)

ρi = 100 / (0.287*( 273 + 20 ) )

ρi = 1.18918 kg/m^3

Using continuity expression:

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*1

Ai = 0.11217 m^2

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

Qi = Ai*Vi

Where,

Ai: The inlet cross sectional area

Qi = 0.11217*1

Qi = 0.11217 m^3 / s

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*Vi

Ai ( V ) = 0.11217 / Vi .... Eq 1

Inlet flow rate ( Qi ):

Qi = 0.11217 m^3 / s ... constant Eq 2

6 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago

A negative pressure respirator brings fresh air to you through a hose<br>A) True<br>B)False

madreJ [45]

Answer:FALSE

Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.

Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.

8 0

3 years ago

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