For a solid propellant rocket motor with a sea level thrust of 207,000 lbf, determine: (a) the (constant) propellant mass flow rate and the specific impulse I at sea level, (b) the altitude for optimum nozzle expansion as well as the thrust and specific impulse at this optimum condition, and (c) at vacuum conditions. The initial total mass of the rocket motor is 50,000 Ibm and its propellant mass fraction is 0.90. The residual propellant (called silvers, combustion stops when the pressure falls below the deflagration limit) amounts to 3% of the burnt propellant. The burn time is 50 seconds, the nozzle throat area (4) is 164.2 in2 and its area ratio (A/A) is 10. The chamber pressure (pi) is 780 psia and the pressure ratio (pi/p2) across the nozzle is taken as 90.0. Neglect any start/stop transients and use the altitude data attached.
a) m = 873 lbm/sec, 237 sec.
b) 13.900 ft, F-216.900 lof. 1,-248.5 sec.
c) F = 23 1,000 lbf Isp = 295 sec. 2 0.

Answer:

Trash rack

Open channel

Fore bay

Pen stock

Inlet valve

Turbine

Tailrace

Generator

Power house

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

Answer:

h = 6.35 W/m².k

Explanation:

In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:

Rate of Heat Transfer Through Convection = Energy Stored in Plate

- h A (Ts - T₀) = m C dT/dt

where,

h = convection heat transfer coefficient = ?

A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²

Ts = Surface Temperature of hot thin plate = 225⁰C

T₀ = Ambient Temperature = 25°C

m = mass of plate = 3.75 kg

C = Specific Heat = 2770 J/kg. k

dT/dt = rate of change in plate temperature = - 0.022 K/s

Therefore,

- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)

h = (- 228.525 W)/(- 36 m².k)

h = 6.35 W/m².k

Answer:

c. 96.676 sq. Inches and 0.671 sq. Feet

Explanation:

From the list of the given option, we are told to chose the correct area in sq. inches that correspond to sq. Feet.

If we recall from the knowledge of our conversion  table that,

1 sq feet = 144 sq inches

Then, let's confirm if the option were true.

a.  966.76 sq. Inches and 8.056 sq. Feet

Assuming

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 1 is wrong

b. 96.676 sq. Inches and 8.056 sq. Feet

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 2 is wrong/

c. 96.676 sq. Inches and 0.671 sq. Feet

if 1 sq feet = 0.671

in sq inches, we have ( 0.671 × 144 ) sq inches

=  96.624 sq. Inches which is closely equal to 96.676 sq. Inches

Therefore, this is the correct answer as it proves that 96.676 sq. Inches = 0.671 sq. Feet

Answer: true

Explanation:

Answer: Central Processing Unit

Answer:

[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]

Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.

Explanation:

For equation to be dimensionally homogeneous both side of the equation must have same dimensions.

For given Equation:

F= Force, μ= viscosity, D = Diameter, V = velocity, ρ= Density

Dimensions:

[tex]F=\frac{ML}{T^2}[/tex]

[tex]\mu=\frac{M}{LT}[/tex]

[tex]D=L\\\\V=\frac{L}{T}\\ \\\rho=\frac{M}{L^3}[/tex]

Constants= 1

Now According to equation:

[tex]\frac{ML}{T^2}=[\frac{M}{LT}][L] [\frac{L}{T}] + [\frac{M}{L^3}][\frac{L^2}{T^2}][L^2][/tex]

Simplifying above equation, we will get:

[tex]\frac{ML}{T^2}=2*\frac{ML}{T^2}[/tex]

Ignore "2" as it is constant with no dimensions. Now:

[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]

Hence it is proved that Stokes-Oseen formula  is  dimensionally homogenous.

Answer:

645 lb/ft

Explanation:

The height of the wall = 15 ft

For the 2 x 4 in. studs:

The minimum dead load for a 2 x 4 in. studs (from table 1-3) = 4 psf = 4 lb/ft²

Therefore for a 15 ft tall wall, the dead load for a 2 x 4 in. studs = 15 ft × 4 lb/ft² = 60 lb/ft

For the 4 in. clay brick:

The minimum dead load for a 4 in. clay brick (from table 1-3) = 39 psf = 39 lb/ft²

Therefore for a 15 ft tall wall, the dead load for a 4 in. clay brick = 15 ft × 39 lb/ft² = 585 lb/ft

Therefore the average load = load for the 4 in. clay brick + load for the 2 x 4 in. studs = 585 lb/ft + 60 lb/ft = 645 lb/ft

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

              b) 5.398 kw

              c) 2.14

              d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

The quality that will exist at the inlet of the refrigerant's evaporator would be:

a). [tex]0.4801[/tex]

The load of the refrigeration would be          

b). [tex]5.398 kW[/tex]

The refrigerator's COP would be:

c). [tex]2.14[/tex]

The maximum refrigeration load would be as follows:          

d) [tex]12.72[/tex]

a). The determination of the quality of the refrigerant at the inlet of the evaporator would be:

[tex]h4 = hf_{4} + x_{4}(hfx_{4})[/tex]  

As given,

[tex]hf_{4}[/tex] [tex]= 3.841[/tex]

[tex]hf_{2,4}[/tex][tex]= 223.95[/tex]

[tex]h_{4} = 111.37[/tex]

Now,

solving for [tex]x_{4}[/tex] [tex]= (111.37 - 3.841)/223.95[/tex]

[tex]= 0.4801[/tex]

b). Refrigeration load

[tex]Q_{l}[/tex] [tex]= m(h_{1} - h_{4})[/tex]

[tex]= 0.0455(230.01-111.37)[/tex]

[tex]= 5.398 kW[/tex]

c). COP of the refrigerator

[tex]Q_{l}[/tex]/[tex]m(h_{2} - h_{1}) - Q_{m}[/tex]

by putting the values, we get

∵ COP [tex]= 2.14[/tex]

d). Theoretical maximum refrigeration load

[tex](Q_{l})[/tex]max [tex]= COPr.rev[/tex] × [tex]{m(h_{2} - h_{1}) - Q_{in}[/tex]

by putting the values, we get

∵ [tex]Q_{L}max = 12.72[/tex]

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Answer:

Explanation:

Given that:

From process 1 → 2

[tex]P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3[/tex]

[tex]PV^{1.5} = \ constant[/tex]

[tex]\gamma = 1.5[/tex]

Process 2 → 3

The volume is constant i.e [tex]V_2 =V_3 = 4m^3[/tex]

[tex]P_3 = 10 \ bar[/tex]

Process 3 → 1

P = constant  i.e the compression from state 1

Now, to start with 1 → 2

[tex]P_1V_1^{1.5} = P_2V_2^{1.5}[/tex]

[tex]P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}[/tex]

[tex]P_2 = 10 \times (\dfrac{1}{4})^{1.5}[/tex]

[tex]P_2 =1.25[/tex]

The work-done for the process  1 → 2 through adiabatic expansion is:

[tex]W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1][/tex]

We know that 1 bar = [tex]10^5 \ N/m^2[/tex]

∴

[tex]W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1][/tex]

[tex]W =1000000 \ J[/tex]

[tex]W_{1 \to 2} = 1000 kJ[/tex]

For process 2 → 3

Since V is constant

Thus:

W = PΔV = 0

[tex]W_{2 \to 3} = 0[/tex]

For process 3 → 1

W = PΔV

[tex]W _{3 \to 1} = P_3(V_1-V_3)[/tex]

[tex]W _{3 \to 1} = 10 \times 10^5 (1-4)[/tex]

[tex]W _{3 \to 1} = 10 \times 10^5 (-3)[/tex]

[tex]W _{3 \to 1} = -3 \times 10^6 \ J[/tex]

[tex]W _{3 \to 1} = -3000 \ kJ[/tex]

The net work-done now  for the entire system is :

[tex]W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }[/tex]

[tex]W_{net} = (1000 + 0 + (-3000)) \ kJ[/tex]

[tex]W_{net} =-2000 \ kJ[/tex]

The sketch of the processes on p -V coordinates can be found in the image attached below.

A) The work done for each process are :

B) The net work for the cycle = -2000 kJ

Given Data :

For process (1 -2)      For process ( 2 - 3 )       process ( 3 - 1 )

P₁ = 10 bar                   P₃ = 10 bar                  constant pressure compression

V₁ = 1 m³                  constant volume heating

V₂ = 4 m³

PV[tex]^{1.5}[/tex] = constant

A) Determine work done for each process

Calculate work done for process (1 - 2)

W₁ ₋ ₂ = [tex]\frac{P_{1}V_{1} - P_{2}V_{2} }{n -1 }[/tex] * 100

         = [ ( 10*1 ) - ( 1.25 * 4 ) ] / 1.5 - 1

         = [ 10 - 5 ] / 0.5

         = 10 * 100 = 1000 kJ

Calculate work done for process ( 2-3 )

given that there is constant volume heating

W₂₋₃ = 0 kJ

Calculate work done for process ( 3-1)

W₃₋₁ = P ( Δ V )     given that p = constant

       = 10 * 100 ( -3 )

       = - 3000 kJ

B) The net work for the cycle

W₁ ₋ ₂  +  W₂₋₃  + W₃₋₁

= 1000 kJ  + 0 kJ  +  - 3000 kJ

= - 2000 kJ

Hence we can conclude that the ) The work done for each process are :

and The net work for the cycle = -2000 kJ

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Attached below is the P-V sketch of the process

Answer: IMA = 12.042

Explanation:

Given that;

Height from the ground h is 3 ft

Slope of ramp s is 1:12

Horizontal length of the ramp x will be 3 × 12 = 36 ft

Now to get the IMA ( ideal mechanical advantage,)

IMA = length of the ramp over / height of the ramp

IMA = (√( 36² + 3²)) / 3

IMA = (√ 1305 ) / 3

IMA = 36.124 / 3

IMA = 12.042

Answer:

The answer would be ISP address

Explanation:

An Internet Protocol address is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication. An IP address serves two main functions host or network interface identification and location addressing which can help you a lot.

The type of address the computer use to find something on network is ISP address.

ISP is simply Internet service provider. They are companies which provides internet services to people. They charge based on the quantity of their service used at a particular time. ISP companies also provide softwares for its users.

An Internet Protocol address refers to a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication.

Therefore, the correct answer is option D

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Answer:

599.7 m

approximately 600 m

Explanation:

initial speed of the rocket = 0

net acceleration upwards = 7 m/s^2

the engine cuts out 10 sec after take off

maximum height reached = ?

we neglect air resistance

To get the velocity of the rocket at the point where the engine cuts off, we use the equation

v = u + at

where

v is the velocity at this point where the engine stops = ?

u is the initial velocity of the rocket from rest = 0 m/s

a is the net acceleration upwards = 7 m/s^2

t is the time the engine runs = 10 s

substituting, we have

v = 0 + (7 x 10)

v = 70 m/s

to get the distance from the ground to this point, we use the equation

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as

where

v is the final velocity at the the height where the engine is cut out = 70 m/s

u is the initial speed at the ground = 0 m/s

a is the net acceleration on the rocket = 7 m/s^2

s is the distance from the ground to this point

substituting, we have

[tex]70^{2}[/tex] = [tex]0^{2}[/tex] + 2(7 x s)

4900 = 14s

s = 4900/14 = 350 m

After this point when the engine cuts out, the rocket experiences an acceleration proportional to the acceleration due to gravity 9.81 m/s^2 downwards, and slows down gradually before coming to a stop at the maximum height.

To get this height, we use the equation

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] - 2gs   (the negative sign is due to the downward direction of the acceleration g)

where

v is the final velocity at the maximum height = 0 m/s (it comes to a stop)

u is the speed at the instance that the engine is cut out = 70 m/s

g is the acceleration due to gravity = 9.81 m/s^2

s is the distance from this point to the maximum height

substituting values, we have

[tex]0^{2}[/tex] = [tex]70^{2}[/tex] - 2(9.81 x s)

0 = 4900 - 19.62s

4900 = 19.62s

s = 4900/19.62 = 249.7 m

The maximum height that will be reached = 350 m + 249.7 m = 599.7 m

approximately 600 m

Explanation:

1. Weld only in authorized areas. Make sure the area is dry, chemical free, and well ventilated.

2. Inspect the equipment before starting to use it.

3. Keep other people away, unless they are authorized to be there and are wearing the appropriate personal protective equipment.

Answer:

Set high and clear expectations for quality work

Don't attempt to cram all your studying into one session

Explanation:

Participate in Class

eat a well balanced diet

Find perfect place to Study.

HavE a NIce dAY ;}

Answer:

Portable appliance testing (PAT) is the term used to describe the examination of electrical appliances and equipment to ensure they are safe to use. Most electrical safety defects can be found by visual examination but some types of defect can only be found by testing.

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Answer:

(a) the maximum tensile stress is 68.2 psi

(b) the maximum normal strain is 2.35 x 10⁻⁶

Explanation:

Given;

modulus of elasticity, E = 29,000 ksi = 29 x 10⁶ psi

(a) the maximum tensile stress

[tex]\tau = \frac{f}{A}[/tex]

f is the maximum force suspended = 20 x 60 = 1200 lb

A is the area of W10 x 60 = 17.6 in²

[tex]\tau = \frac{1200}{17.6} \\\\\tau = 68.2 \ psi[/tex]

(b) the maximum normal strain.

According to Hook's law stress is directional to strain

τ = Eε

[tex]\epsilon = \frac{\tau}{E}\\\\\epsilon = \frac{68.2}{29*10^{6}}\\\\\epsilon = 2.35*10^{-6}[/tex]

Answer:

V = 20.6 m/s

Explanation:

Given that the temperature of the neon molecules = 356 Kelvin

Pressure = 0.9 bar

The mass number of Neon = 21.

Using the formula below

1/2 m v^2 = (3kT)/2

Where

T = temperature = 356 k

K = Bolzmann constant

= 1.38 × 10^-23jk^-1

NA = 6.02214076×10²³ mol⁻¹

Where NA = Avogadro number

Substitute all the parameters Into the formula

1/2 × 21/NA × v^2 = (3 × k × 356)/2

1.744×10^-23V^2 = 7.3692 × 10^-21

Make V^2 the subject of formula

V^2 = (7.37×10^-21)/1.744×10^-23

V^2 = 422.55

V = sqrt( 422.55)

V = 20.55

Therefore, the average (root mean square) velocity (m/s) of Neon molecules at 356 Kelvins and 0.9 bars is 20.55 m/s

Answer:

b. satellite surveying

c. aerial surveying

d. optical tooling

e. marketing surveying

f. control surveying

h. statistical surveying

i. telephone surveying

k. alignment surveying

l. mine surveying

m. solar surveying

Explanation:

A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.

Some type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying

A Survey means an examination of a features, subject or material under consideration.

Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

However, some other type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying.

Read more about surveying

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Answer:

A) 48

B) Pitch diameters : pinion = 42.164 mm, Gear = 101.19 mm

C) standard addendum : pinion = 46.3804, Gear = 105.406

    standard dedendum : pinion = 37.265 mm, Gear = 96.312 mm

D) 71.672 mm

E) 94.989 N , 101.0858 N,  34.573 N

Explanation:

Given Data :

∅ = 20⁰ , Tp = 20 ( tooth pinion ),

diameter pitch = 12, Np = 1776 rpm ,

Ng = 740 rpm,

attached below is the detailed solution of the given problems

Answer:

i)The results are consistent at ; Rows ( 1,2,4,6 ), which simply means that it is fairly consistent and makes sense

ii) Row 5 shows a gain of -99 which is ≈ -100 and there is a possibility of experimental error of (1%)

ii ) The omitted values are : row 3 = 0.99, Row 7 = 5.049

Explanation:

To calculate/determine the missing/omitted values we have to apply this formula:

Vo = ( V1 - V2 ) G

Vo = output = 1

G = gain = - 100

V1 =

we find  V1 along line 3 ( row 3 )

(V1 - V2) G = 1

( V1 - 1 ) * -100 = 1

hence V1 = 0.99

we find V2 along row 7

( V1 - V2 ) G = 1

( 5.10 - V2 ) * -100 = 1

hence V2 = 5.049

This question is incomplete, the complete question is;

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.

The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.

Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Explanation:

Given that;

T = 300k

B = 3.56×10¹⁴ (cm⁻³ K^-3/2)

Eg = 1.42 eV

we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K

so to find the ni for gallium arsenide;

ni = B×T^(3/2) e^ ( -Eg/2kT)

we substitute

ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))

ni =  (3.56×10¹⁴)(5196.1524)e^-27.4651

ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)

ni = 2.1837 × 10⁶ cm⁻³

Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Answer:

Fit test

Explanation:

it's on the automotive technology sight : )

Answer: In a clean plastic or metal container with a tightly sealed lid

Answer:

some connecting rods have spit holes

Some connecting rods have crankshaft to help lubricate the cylinder wall or piston pin.

A crankshaft is a mechanical component that converts reciprocating motion in a piston engine into rotational motion.

The crankshaft is a rotating shaft with one or more crankpins that drive the pistons via the connecting rods.

Reduced fuel economy is one issue you'll face if your car's crankshaft isn't working properly. The reason for this is that your vehicle's fuel injectors will not efficiently direct gas to the engine.

The connecting rods' big ends are lubricated with oil that travels up an oilway drilled through the crankshaft.

From the main bearings to the big end bearings, oil is pushed. It will lubricate the cylinder walls in some engines by passing through a passage drilled in the connecting rod.

Thus, the answer is crankshaft.

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Answer:

Explanation:

In mechanics of materials, the bending stress of a beam in bending can be determined by the equation

Answer:

Technician B is correct.

Explanation:

Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate.

The statement made by Technician A saying that an open in one of the parallel branches could be the cause has no ground because an opening in any branches has not to do with the issue since there is no disconnection The statement made by Technician B saying that a shorted (or stuck closed) brake lamp switch wired in series could be the cause is correct.