In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50 m3/s. The pressure at the inlet is to be 315612 Pa. If the nozzle is lying in a horizontal plane. Jet A is 10 cm in diameter, jet B is 12 cm in diameter, and the pipe (1) is 30 cm in diameter. The x-component of force (Rx) acting through the flange bolts is required to hold the nozzle in place is:

Answers

Answer 1

Solution :

Given data :

p = 315612 Pa

[tex]$V_1=7.07 \ m/sec$[/tex]

At exit of B,

p = [tex]$P_{atm}$[/tex]

[tex]$V_B = 26.1 \ m/sec$[/tex]

At exit of A,

[tex]p=P_{atm}[/tex]

[tex]$V_{A} = 26.1 \ m/s$[/tex]

We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.

From figure,

[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]

[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]

[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]

Substitute all the values,

[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]

[tex]$F_x = -18.2733 \ kN$[/tex]

Therefore, the force required to hold the nozzle in its place along horizontal direction.

[tex]$F_x = -18.2733 \ kN$[/tex]

In The Figure Below, This Double Nozzle Discharges Water (at 10C, Density= 1000 Kg/m3) Into The Atmosphere

Related Questions

To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 297(106)ft2. Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.

Required:
What is the area measurement, 293 (106) ft^2, in SI units?

Answers

This question is incomplete, the complete question is;

To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.  

Required:

What is the area measurement, 293 × 10⁶ ft², in SI units?

293 × 10⁶ ft² = ?km²

Answer:

the area measurement is  27.221 km²

Explanation:

Given the data in the question;

What is the area measurement, 293 × 10⁶ ft², in SI units

we are to the result of the measured area from ft² to km²

we know that;

1 meter = 3.2808 ft

1 km = 1000 m

1 ft = (1 / 3.2808)m

1 m = ( 1/1000 ) km

since our measured are is 293 × 10⁶ ft²

hence

A = 293 × 10⁶ × [ (1 / 3.2808)m ]²

A = 27221252.74 m²

A = 27221252.74 × [ ( 1/1000 ) km ]²

A = 27.221 km²

Therefore, the area measurement is  27.221 km²


8. The operation of a TXV is controlled by the
O A. thermostatic spring.
O B. temperature bulb.
O C. external pressure of the evaporator.
O D. modulating valve.

Answers

Answer should be C hopefully

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor​

Answers

78950W the answer

Explanation:

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

convert 25 inches / min to mm/hour​

Answers

Answer:

25 mins into hours = 0.416667 hours

25 inches as mm = 635

Explanation:

If the two 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mmmm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the temperature at which the average normal stress in either the magnesium or the steel first becomes 12.0 MPaMPa .

Answers

Answer: hello some data related to your question is missing attached below is the missing data

answer:

T2 = 265°C

Explanation:

First step : calculate sum of vertical forces

∑ y = 0

Fmg - 2(0.5 Fst ) = 0

∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )

          = 1884.96 N

Also determine the Compatibility equation in order to determine the change in Temperature

ΔT = 250°C

therefore Temperature at which average normal stress becomes 12.0 MPa

ΔT = T2 - T1

250°C = T2 - 15°C

T2 = 250 + 15 = 265°C

attached below is the detailed solution

What must you do when you reach a steady yellow traffic light?

Answers

Answer:

When you come up on a steady yellow traffic light you should always yield to cross traffic if you can yield safely. The flashing yellow light is there to inform drivers to be careful and to slow down.

Explanation:

hope it helped!

Hi guys I hope you have a good weekend too long and we love the guys and piper love and miss y’al

thiết kế ic 555 và code để ic hoạt động

Answers

Answer:

here you go.

screenshot 2 should give you some basic idea

When a voltage (v=353 sin (251t+30) is applied to two elements impedance a current (i =7.07 cos 251 t) is passing. Find the nature and the value of the elements and the circuit power ​

Answers

Answer:

A.C. voltage, V= V0 sin ωt As,t = πω = 12.2πω = 12T, therefore, first half cycle (T/2). Hence, average value of AC voltage, Eav = 2V0π.

In a certain pressing operation, the metallic powder fed into the open die has a packing factor of 0.5. The pressing operation reduces the powders to 70% of their starting volume. In the subsequent sintering operation, shrinkage amounts to 10% on a volume basis. Given that these are the only factors that affect the structure of the finished part, determine its final porosity.

Answers

Answer:

0.2063

Explanation:

Given data:

packing factor = 0.5

percentage of reduction of powders = 70%

Calculate the final porosity

after sintering Bulk specific volume = 0.9 * 0.7 = 0.63

assuming true specific volume = 1

packing factor = 0.5 , bulk specific volume = 2

packing factor after pressing and sintering

= 1 / ( 2 * 0.63 ) = 0.7937

hence : porosity = 1 - packing factor

                            = 1 - 0.7937 = 0.2063

A stream of ethylene glycol vapor at its normal boiling point and 1atm flowing at a rate of 175 kg/min is to be condensed at constant pressure. The product stream from the condenser is liquid g lycol at the condensation temperature.
a. Calculate the rate at which heat must be transferred from the condenser (kW).
b. If heat were transferred at a lower rate than that calculated in part (A), what would the state of the product stream be? (Dedu ce as much as you can about the phase and the temperature of the stream.)
c. If heat were transferred at a higher rate than that calculated in part (A), what could you deduce about the state of the product stream?

Answers

Answer: hello attached below is the question properly written

a) 2670 Kw

b) product will be made up of vapor and liquid

c) Product will be a super cooled liquid

Explanation:

mass Flow rate ( m ) = 175 kg/min

pressure = 1 atm

molecular weight of ethylene glycol ( mw ) = 62.07 g/mol

enthalpy of vaporization ( ΔHv ) = 56.9 KJ/mol

Using values from the table 8.1 related to the question

a) Determine the rate at which heat must be transferred from condenser

Using values from the table 8.1 related to the question

ΔH = 2670 Kw

b) If heat is transferred  at a lower temperature the product will be made up of vapor and liquid

c) If heat was transferred at a higher temperature the product will be a super cooled liquid

nơi nào có điện tích thì xung quanh điện tích đó có :

Answers

Explanation:

sory sorry sorry sorrysorrysorry

Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has a power rating of 2250 W. To dry one typical load of clothes the dryer will run for approximately 45 minutes. In Ontario, the cost of electricity is $0.11/kWh. Calculate the costs to run the dryer for your family for one year.

Answers

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

Các đặc điểm chính của đường dây dài siêu cao áp .

Answers

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

An astronaut orbits the earth in a space capsule whose height above the earth is equal to the earth's radius. How does the mass of the astronaut in the capsule compare to her mass on the earth?

Answers

Answer:

The weight will be [tex]\frac{1}{4}[/tex] of its weight on earth surface. A further explanation is provided below.

Explanation:

According to the question,

h = R

The value of gravitational acceleration at height equivalent to radius of earth R.

⇒ [tex]g=\frac{g_0}{(1+\frac{h}{R} )^2}[/tex]

or,

⇒ [tex]g=\frac{g_0}{(1+\frac{R}{R} )^2} =\frac{g_0}{4}[/tex]

here,

[tex]g_0[/tex] = gravitational acceleration earth's surface

then,

⇒ [tex]mg=\frac{mg_0}{4}[/tex]

Thus, the above is the appropriate solution.

In a series circuit, there ______ for current to flow.
a.
are at least two paths

b.
are multiple paths

c.
is only 1 path

d.
are very small paths

Answers

Answer:

Option C = Only 1 path for current to flow.

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after traveling a distance d.

a. What is the magnitude of the electric field?
b. What is the direction of the electric field?

1. in the direction of the electron's motion
2. opposite to the direction of the electron's motion
3. perpendicular to the direction of the electron's motion

Answers

Answer:

The answer is below

Explanation:

a) The work done is equal to the loss in kinetic energy (KE)

Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy

Final KE = [tex]\frac{1}{2}mv_f^2[/tex]

But the final velocity is 0 (at rest). Hence:

Final KE = [tex]\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0[/tex]

ΔKE = 0 - K = -K

W = ΔKE = -K

Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)

W = qEd

but q = -e, hence:

W = -e * E * d

Using:

W = ΔKE

-e * E * d = -K

E= K / (e * d)

b) The electric field is in the direction of the electrons motion

An assembly line has 3 fail safe sensors and one emergency shutdown switch.The line should keep moving unless any of the following conditions arise:
(1) If the emergency switch is pressed
(2) If the senor1 and sensor2 are activated at the same time.
(3) If sensor 2 and sensor3 are activated at the same time.
(4) If all the sensors are activated at the same time
Suppose a combinational circuit for above case is to be implemented only with NAND Gates. How many minimum number of 2 input NAND gates are required.

Answers

Answer:

1 NAND gate

Explanation:

The minimum number of 2 input NAND gates that can be used to implement the combinational circuit = 1

The only true combinations conditions that can produce a false result ( i.e. condition/result different from the expected result as stated in the question )

Sensor 2 activated + Emergency switch pressed = False ( Line will keep moving )

For a sixth-order Butterworth high pass filter with cutoff frequency 3 rad/s, compute the following:

a. The locations of the poles.
b. The transfer function H(s).
c. The corresponding LCCDE description.

Answers

Solution :

Given :

A six order Butterworth high pass filter.

∴ n = 6, [tex]w_c=1 \ rad/s[/tex]

a). The location at poles :

    [tex]$s^6-(w_c)^6=0$[/tex]

   [tex]$s^6=(w_c)^6=1^6$[/tex]

  ∴ [tex]$s^6 = 1$[/tex]

Therefore, it has 6 repeated poles at s = 1.

b). The transfer function H(S) :

    Transfer function H(S) [tex]$=\frac{1}{1+j\left(\frac{w_c}{s}\right)^6}$[/tex]

                                         [tex]$=\frac{1}{1-\left(\frac{w_c}{s}\right)^6}$[/tex]

  ∴    H(S) [tex]$=\frac{s^6}{s^6-(w_c)^6}=\frac{s^6}{s^6-1}$[/tex]

   H(S) [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]

c). The corresponding LCCDE description :

  [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]

   [tex]$Y(s)(s^6-1) = s^6 \times (s)$[/tex]

   [tex]$Y(s)s^6-y(s).1 = s^6 \times (s)$[/tex]

By taking inverse Laplace transformation on BS

   [tex]$L^{-1}[Y(s)s^6-Y(s)1]=L^{-1}[s^6 \times (s)]$[/tex]

   [tex]$\frac{d^6y(t)}{dt^6}-y(t)=\frac{d^6 \times (t)}{dt^6}$[/tex]

  Hence solved.

What is the differences between total revenue and total costs? Make

sure that your answer will cover all aspects related with two

mentioned concepts. With any supported simple example about

petroleum industrial?

Answers

Answer:

The basic difference between Total cost and total revenue is that the total cost includes the total expenditure incurred on the production of a commodity whereas total revenue refers to the money received from selling that commodity.

Explanation:

Draw the logic circuit for each of the following. For each gate, determine if it generates either EVEN or ODD parity bit and finds the output for the given input data: (Remember: A XOR generates EVEN parity bit. A XNOR generates Odd Parity bit, whatever how many inputs they have.)
Data Inputs Which kind parity bit can it generate?
4-input XOR, input data-1001 Even Parity Bit ODD Parity Bit
5-input XOR, input data-10010
6-input XOR, input data-101001
7-input XOR, input data 1011011

Answers

Answer:

a) 4-input XOR, input data-1001  = 0 Even parity Bit

b)  5-input XOR, input data-10010 = 0 Even parity Bit  

c) 6-input XOR, input data-101001 = 1 Even parity Bit

d) 7-input XOR, input data 1011011 = 1 Even parity Bit

Explanation:

a) 4-input XOR, input data-1001  ;  generates 0 Even parity Bit

b)  5-input XOR, input data-10010 ; generates 0 Even parity Bit  

c) 6-input XOR, input data-101001 ; generates 1 Even parity Bit

d) 7-input XOR, input data 1011011 ; generates 1 Even parity Bit

Attached below is the Logic circuits of the data inputs

In 2009 an explosive eruption covered the island of Hunga Ha'apai in black volcanic ash. What type of succession is this?

Answers

Answer:

The type of succession is:

Primary succession

Explanation:

This is a type of succession that occurs after a volcanic eruption or earthquake; it involves the breakdown of rocks by lichens to create new, nutrient rich soils.

Primary succession is one of the two types of succession we have. It begins on rock formations, such as volcanoes or mountains, or in a place with no organisms or soil.

find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters and feet​

Answers

Answer:

The volume for this is 29.7

Explanation:

Trust me on this I'm an expert

R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic energy to a pressure of 165 kPa. Find the exit temperature and the ratio of the exit pipe diameter to that of the inlet pipe (Dex/Din) so that the velocity stays constant.

Answers

Solution :

For R-134a, we are given :

[tex]$T_i = 25^\circ C$[/tex]

[tex]$P_i=750 \ kPa$[/tex]

[tex]$P_e=165 \ kPa$[/tex]

Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :

[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]

We also know that the gas is throttled and there is no change in the kinetic energy.

So, [tex]$v_e=v_i$[/tex]

Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,

[tex]$h_i=h_e$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.

[tex]$T_e=-15^\circ C$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :

[tex]$h_f = 180.19 \ kJ/kg$[/tex]

[tex]$h_{fg} = 209 \ kJ/kg$[/tex]

Calculating the exit flow quality factor,

[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]

    [tex]$=\frac{234.59-180.19}{209}$[/tex]

   = 0.26

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :

[tex]$v_f = 0.00746 \ m^3/kg$[/tex]

[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]

Calculating the exit specific volume :

[tex]$v_e=v_f+x_e(v_{fg})$[/tex]

   = 0.000746 + 0.26 (0.11932)

   = 0.0318 [tex]m^3/kg[/tex]

The mass flow is equal to :

[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]

  [tex]$=A_e . \frac{v}{v_e}$[/tex]

So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]

Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]

[tex]$\frac{D_e}{D_i}=6.19$[/tex]

     

You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210 centipoise [cP]. What is the kinematic viscosity of Fluid A, in units of stokes [St]

Answers

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

Calculate the kinematic viscosity of Fluid A

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

3. According to the drag equation the velocity of an object moving through a fluid can be modeled by the equation -- -ky- where k is a constant.
(a) Find the general solution to this equation.
(b) An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. What will the velocity be after ten seconds?​

Answers

A find the ganeral solution to this equation

An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.

What is velocity?

Velocity is defined as the speed at which an object's position changes in relation to time and a frame of reference.  Speed is the rate at which an object travels along a path over time, whereas velocity is the speed and direction of an item's motion. In other words, speed is a scalar value, but velocity is a vector.

As given

Initial velocity = 40 m / sec

Velocity after 2 seconds = 30 m / sec

So the velocity after 10 seconds will be = 0 m / sec

That is the object will stop moving.

Thus. an object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.

To learn more about velocity, refer to the link below:

https://brainly.com/question/28738284

#SPJ2

Your question is incomplete, but probably your complete question was

Please label the following statements as either True (T) or False (F).
(a) The true stress is higher than the engineering stress for a sample under tension.
(b) Creep test is carried out with a dynamic stress under elevated temperature.

Answers

Answer:

a. True

b. False

Explanation:

a. Since true stress, σ' = σ(1 + ε) where σ = engineering stress and ε = engineering strain.

Also under tension ε > 0, so, (1 + ε) > 1

Since (1 + ε) = σ'/σ > 1, ⇒ σ' > σ

So, the true stress is greater than the engineering stress.

So, the statement is true

b. Creep is a time-dependent deformation under certain applied load.

Creep occurs at high temperatures under different types of stress.

But, Creep test is carried out at constant high temperature and constant stress.

This statement is false.

Just because I seen someone else ask but they didn't have enough information.

If a filesystem has a block size of 4096 bytes, this means that a file comprised of only one byte will still use 4096 bytes of storage. A file made up of 4097 bytes will use 4096*2=8192 bytes of storage. Knowing this, can you fill in the gaps in the calculate_storage function below, which calculates the total number of bytes needed to store a file of a given size?

Answers

Answer:

Following are the program to the given question:

def calculate_storage(filesize):#definging a method calculate_storage that takes filesize as a parameter  

 block_size = 4096#definging block_size that holds value

 full_blocks = filesize//block_size#definging full_blocks that divides the value and hold integer part

 partial_block_remainder = filesize%block_size#definging partial_block_remainder that holds remainder value

 if partial_block_remainder > 0:#definging if that compare the value

   return block_size*full_blocks+block_size#return value

 return block_size*full_blocks#return value

print(calculate_storage(1))    # calling method by passing value

print(calculate_storage(4096)) # calling method by passing value

print(calculate_storage(4097)) # calling method by passing value

Output:

4096

4096

8192

Explanation:

In this code, a method "calculate_storage" is declared that holds a value "filesize" in its parameters, inside the method "block_size" is declared that holds an integer value, and defines "full_blocks and partial_block_remainder" variable that holds the quotient and remainder value and use it to check its value and return its calculated value. Outside the method, three print method is declared that calls the method and prints its return value.

good housekeeping can increase production in a work place is it true or false​

Answers

False, Good housekeeping eliminates accident and fire hazards. It also maintains safe, healthy work conditions; saves time, money, materials, space, and effort; improves productivity and quality; boosts morale; and reflects an image of a well-run, successful organization.

Hope it helps you❤️

The pressure gage on a 2.5-m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank (mass in kg) if the temperature is 28°C and the atmospheric pressure is 97 kPa.

Answers

Answer:

[tex]n=5.36kg[/tex]

Explanation:

From the question we are told that:

Volume [tex]V=2.5m^3[/tex]

Pressure[tex]\rho=500Kpa[/tex]

Temperature [tex]T=28^o[/tex]

Atmospheric pressure [tex]\rho_{atm} =97 kPa.[/tex]

Generally the equation for an Ideal gas is mathematically given by

 [tex]PV=nRT[/tex]

Therefore

 [tex]n=\frac{500*2.5}{8.314*28}[/tex]

 [tex]n=5.36kg[/tex]

The view factor for radiation emitted by surface 1 to surface 2 was calculated to be 0.4. The working area of surface 1 is 0.01 m2, the working area of surface 2 is 0.04 m2. What is the view factor for radiation emitted by surface 2 to surface 1?

Answers

Answer:

The view factor for radiation emitted by surface 2 to surface 1 is 0.1

Explanation:

Given

[tex]F_{12} = 0.4[/tex]

[tex]A_1 = 0.01m^2[/tex]

[tex]A_2 = 0.04m^2[/tex]

Required

Determine [tex]F_{21}[/tex]

To do this, we make use of the following equivalent ratio

[tex]A_1 * F_{12} = A_2 * F_{21}[/tex]

Make [tex]F_{21[/tex] the subject

[tex]F_{21} = \frac{A_1 * F_{12}}{ A_2}[/tex]

Substitute values into the equation

[tex]F_{21} = \frac{0.01m^2 * 0.4}{0.04m^2}[/tex]

[tex]F_{21} = \frac{0.01 * 0.4}{0.04}[/tex]

[tex]F_{21} = \frac{0.004}{0.04}[/tex]

[tex]F_{21} = 0.1[/tex]

Other Questions
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