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riadik2000 [5.3K]
3 years ago
12

How is a scale model different from other types of models?

Engineering
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:

a scale model each size is a certain amount smaller

Explanation:

You might be interested in
A 500 turn coil is wound on an iron core. When a 120Vrms 60Hz voltage is applied to the coil, the current is 1A rms. Neglect the
Tju [1.3M]

Answer:

R = 7.854 x 10⁵ anpert turns / Wb

Relative permeability = 405.3

Explanation:

Detailed explanation is given in the attached document.

4 0
4 years ago
Read 2 more answers
Steam enters an adiabatic nozzle at l MPa, 260 C, 30 m/s and exits at 0.3 MPa and 160 'C. Calculate the velocity at the exit.
maxonik [38]

Answer:

v₂ = 35.57 m/s

Explanation:

Given :

Inlet steam pressure, P₁ = 1 MPa

Inlet steam temperature, T₁ = 260°C

Inlet velocity of steam, V₁ = 30 m/s

Outlet steam pressure, P₂ = 0.3 MPa

Outlet steam temperature, T₂ = 160°C

Now,

From steam table at pressure 1 Mpa and temperature 260°C, enthalpy, h₁ = 2964.8 kJ/kg

From steam table at pressure 0.3 Mpa and temperature 160°C, enthalpy, h₂ = 2782.14 kJ/kg

Therefore, for an open system from 1st law of thermodynamics, we get

Energy in = Energy out

E₁ = E₂

\left ( h_{1}+\frac{v_{1}^{2}}{2} \right ) = \left ( h_{2}+\frac{v_{2}^{2}}{2} \right )

v_{2}^{2}= 2\left [ h_{1}-h_{2}+\frac{v_{1}^{2}}{2} \right ]

v_{2}^{2}= 2\left [ 2964.8-2782.14+\frac{30^{2}}{2} \right ]

v_{2}^{2}=2 X 632.66

v₂ = 35.57 m/s

Therefore, outlet velocity, v₂ = 35.57 m/s

7 0
3 years ago
NO reacts with Br2 in the gas phase according to the following chemical equation: 2NO(g) +Br2(g)2NOBr(g) It is observed that, wh
klemol [59]

Answer:

a) rate=r=k[NO]^{2} [Br_{2}]^{1}

b) k=\frac{1}{s*M^{2}}

Explanation:

First of all you need to indicate the reaction order of each reactant ( NO and Br_{2}):

1.  Br_{2}

Note that if Br_{2} concentration ([Br_{2} ]) is reduced to 1/3 of its initial value, the rate of the reaction is also reduced to 1/3 of its initial value, it means:

[Br_{2} ]=1/3 then r=1/3

As the change in the rate of the reaction is equal to the change of the initial concentration of  Br_{2}, you could concluded that the reaction is first order with respect to  Br_{2}

2. NO

Now, note that if NO concentration ([NO]) is multiplied by 3.69, the rate of the reaction increases by a factor of 13.6. In this case, to know the ratio could be advisable divide the rate of the reaction (13.6) over the factor whereby was multiplied the concentration (3.69), as follows:

\frac{13.6}{3.69}=3.69

As the result is the same factor 3.69 you could concluded that the change of the rate of reaction is proportional to the square of the concentration of A:

r=[NO]^{2} =3.68^{2} =13.6

It means that the reaction is second order with respect to NO

3. Rate Expression

Remember that the rate expression of the reactions depend on the concentration of each reactant and its order. In this case we have 2 reactants: NO and Br_{2}, then we have a rate law depending of  2 concentrations, as follows:

<h2>rate=r=k[NO]^{2} [Br_{2}]^{1}</h2>

Note that the expression is the result of the concentration of each reactant raised to its reaction order (previously determined)

<em>Note: I hope that you do not mix up the use of the rates of reaction of each reactant, that is experimentally determined, with the stoichiometric coefficient, are different.</em>

4. Rate constant units (k)

Assuming concentration is expressed as \frac{mol}{L}=M and time is in second, to find the units of k we need to solve an equation with units and with supporting of the rate equation previously obtained, as follows:

r=k[NO]^{2} [Br_{2}]^{1}

Where:

[r]=[\frac{M}{s}]

[[NO]]=[M]

[ [Br_{2}]]=[M]

Then:

\frac{M}{s}=kM^{2} M^{1}

\frac{M}{s}=kM^{3}

\frac{M}{s*M^{3}}=k

The units of the rate constant k are:

k=\frac{1}{s*M^{2}}

8 0
4 years ago
Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at
garri49 [273]

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

4 0
3 years ago
Read 2 more answers
A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -30C by rejecting
chubhunter [2.5K]

Answer:

a) x = 0.4795

b) QL = 5.85 KW

c) COP = 2.33

d) QL_max = 12.72 KW

Explanation:

Solution:-

- Assuming the steady state flow conditions for both fluids R-134a and water.

- The thermodynamic properties remain constant for respective independent intensive properties.

- We will first evaluate the state properties of the R-134a and water.

- Compressor Inlet, ( Saturated Vapor ) - Ideal R-134a vapor cycle

              P1 = 60 KPa, Tsat = -36.5°C  

              T1 = -34°C , h1 = hg = 230.03 KJ/kg

              Qin = 450 W - surrounding heat  

- Condenser Inlet, ( Super-heated R-134a vapor ):

              P2 = 1.2 MPa , Tsat = 46.32°C  

              T2 = 65°C   , h2 = 295.16 KJ/kg

- Condenser Outlet, ( Saturation R-134a point ):

             P3 = P2 = 1.2 MPa , Tsat = 46.32°C

             T3 = 42°C   , h3 = hf = 111.23 KJ/kg

- R-134a is throttled to the pressure of P4 = compressor pressure = P1 = 60 KPa by an "isenthalpic - constant enthalpy pressure reduction" expansion valve.

- Inlet of Evaporator - ( liquid-vapor state )

             P4 = P1 = 60 KPa, hf = 3.9 KJ/kg , hfg = 223.9 KJ/kg

             h4 = h3 = 111.23 KJ/kg

- The quality ( x ) of the liquid-vapor R-134a at evaporator inlet can be determined:

             x4 = ( h4 - hf ) / hfg

             x4 = ( 111.23 - 3.9 ) / 223.9

             x4 = 0.4795      Answer ( a )        

- Water stream at a flow rate flow ( mw ) = 0.25 kg/s is used to take away heat from the R-134a.

- Condenser Inlet, ( Saturated liquid water ):

             Ti = 18°C , h = hf = 75.47 KJ/kg  

- Condenser Outlet, ( Saturated liquid water ):

             To = 26°C , h = hf = 108.94 KJ/kg

- Since the heat of R-134a was exchanged with water in the condenser. The amount of heat added to water (Qh) is equal to amount of heat lost from refrigerant R-134a.

- Apply thermodynamic balance on the R-134a refrigerant in the condenser:

             Qh = flow (mr) * [ h2 - h3 ]

Where,

flow ( mr ) : The flow rate of R-134a gas in the refrigeration cycle

             flow ( mr ) = Qh / [ h2 - h3 ]

             flow ( mr ) = 8.3675 / [ 295.16 - 111.23 ]

             flow ( mr ) = 0.0455 kg/s

- The cooling load of the refrigeration cycle ( QL ) is determined from energy balance of the cycle net work input ( Compressor work input ) - "Win" and the amount of heat lost from R-134a in condenser ( Qh ).

- Apply the thermodynamic balance for the compressor:

           

            Win = flow ( mr )*[ h2 - h1 ] - Qin

            Win = 0.0455*[ 295.16 - 230.03] KW - 0.45 KW

            Win = 2.513 KW

- The cooling load ( QL ) for the refrigeration cycle can now be calculated. Apply thermodynamic balance for the refrigeration cycle:

            QL = Qh - Win

            QL = 8.3675 - 2.513

            QL= 5.85 KW  .... Refrigeration Load, Answer ( b )

- The COP of the refrigeration cycle is calculated as the ratio of useful work and total work input required:

           

             COP = QL / Win

             COP = 5.85 / 2.513

             COP =  2.33      Answer ( c )            

- For a compressor to be working at 100% efficiency or ideal then the maximum COP for the refrigeration cycle would be:

           

             COP_max = [ TL ] / [ Th - TL ]

Where,

            TL : The absolute temperature of heat sink, refrigerated space

            TH : The absolute temperature of heat source, water inlet

                 

            COP_max = [ -30+273 ] / [ (18+273) - (-30+273) ]          

            COP_max = 5.063

- The theoretical ideal refrigeration load ( QL max ) would be:

     

           COP_max = QL_max / Win

           QL_max = Win*COP_max

           QL_max = 2.513*5.063

           QL_max = 12.72 KW     Answer ( d )

5 0
4 years ago
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