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lara [203]
3 years ago
15

The north-south streets of a grid have block lengths of 250 m and the east-west streets have block lengths of 200 m. Desired spe

eds of progression are 45 km/h north-south and 35km/h east-west. Determine the appropriate alternate systems, cycle, and actual speeds of progression for two-way

Engineering
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer:

See attached pictures.

Explanation:

See attached pictures for detailed explanation.

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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

               

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3 years ago
An organization sets its standards for quality according to the best product it can produce.
Marianna [84]
I believe it’s True, but please correct me if I’m wrong!
6 0
3 years ago
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An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa
maxonik [38]

Answer:

\dot m_{2} = 0.199\,\frac{kg}{s}

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

1 + y = z

Then, the expression is simplified afterwards:

\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0

\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

h = 2994.3\,\frac{kJ}{kg}

State 2 (Saturated liquid)

h = 1008.3\,\frac{kJ}{kg}

State 3 (Liquid-Vapor mixture)

h = 2444.22\,\frac{kJ}{kg}

The ratio of the stream at state 2 to the stream at state 1 is:

y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}

y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }

y = 0.397

The mass flow rate of the saturated liquid is:

\dot m_{2} = y\cdot \dot m_{1}

\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )

\dot m_{2} = 0.199\,\frac{kg}{s}

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3 years ago
Please help<br> describe the impact that a toy robot has had or could have on its intended audience
GuDViN [60]
Depending on the age the toy is made for it could teach younger children things such as letters and numbers and for a older kid it could teach them how different things are put in the robot to help it work
3 0
3 years ago
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10POINTS
deff fn [24]

Answer:

Chinese

Explanation:

I hope this helps

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