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Minchanka [31]
4 years ago
14

How does the map scale help to interpret the map?

Engineering
2 answers:
krok68 [10]4 years ago
6 0

Map scale refers to the relationship (or ratio) between distance on a map and the corresponding distance on the ground. For example, on a 1:100000 scale map, 1cm on the map equals 1km on the ground.

blondinia [14]4 years ago
6 0

Answer:

c

Explanation:

You might be interested in
Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and
jonny [76]

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:    

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ =  0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

p = 8 bar = 800 kPa

Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ +  17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

3 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
2.11 Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface
alexandr402 [8]

Answer:

The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.

The cost associated with the heat loss through the windows for an 8-hour flight is:

For aerogel windows: $17.472 (most efficient)

For polycarbonate windows: $262.08

For soda-lime glass windows: $1,238.016 (least efficient)

Explanation:

To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:

\frac{dQ}{dt}=\dot Q =-kS\nabla \vec{T}

In this case:

\dot Q =k\frac{S}{L} \Delta T

If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).

To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).

To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):

Cost=C_{hc}\cdot \dot Q \cdot t \cdot n=1\frac{\$}{Kwh} \cdot \dot Q \cdot 8h \cdot 130

6 0
3 years ago
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approxim
diamong [38]

Answer:

Explanation:

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:

[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )

where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.

3 0
4 years ago
What do shoes have in the toe area?
Savatey [412]

Answer: it called a toe box is the part of a shoe that covers and protects the toes. Toe boxes come in a variety of shapes and styles depending on the type of shoe, but they should always be wide and long enough to accommodate the toes comfortably.

Explanation:

6 0
4 years ago
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