What should be a concern as a weldment becomes larger as more parts are added?

A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i

kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

= 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

= 25050 ft

using super elevation formula

$e + f = \dfrac{v^2}{rg}$

$0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}$

$r = \dfrac{95.33^2}{32.2\times 0.18}$

r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

$L = \dfrac{\piR\Delta}{180}$

$L = \dfrac{\pi\times 1574\times 38}{180}$

L = 1044 ft

Tangent of the curve calculation

$T = R tan(\dfrac{\Delta}{2})$

$T = 1574 tan(\dfrac{38}{2})$

T = 542 ft

The station PC and PT are

PC = PI - T

PC = 25050 - 542

= 24508 ft

= 245 + 8 ft

PT = PC + L

= 24508 + 1044

=25552

= 255 + 52 ft

the middle ordinate calculation

$MO = R(1-cos\dfrac{\Delta}{2})$

$MO = 1574\times (1-cos\dfrac{38}{2})$

MO = 85.75 ft

degree of the curvature

$D = \dfrac{5729.578}{R}$

$D = \dfrac{5729.578}{1574}$

D = 3.64°

8 0

4 years ago

Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha

Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 0

4 years ago

A First Stage in a turbine receives steam at 10 MPa, 800 C with an exit pressure of 800 KPa. Assume the stage is adiabatic and r

9966 [12]

Answer

given,

P₁ = 10 MPa T₁ = 800

P₂ = 800 KPa T₂ = ?

Using formula

$\dfrac{T_2}{T_1} = (\dfrac{P_2}{P_1})^{\dfrac{n-1}{n}}$

For steam n = 1.33

$\dfrac{T_2}{800+273} = (\dfrac{800\times 10^{3}}{10\times 10^{6}})^{\dfrac{1.33-1}{1.33}}$

T₂ = 573.368 K

T₂ = 573.368 - 273 = 300.368 °C

$W = \dfrac{P_1V_1-P_2V_2}{n-1}$

= $\dfrac{mR(T_1-T_2)}{n-1}$

= $\dfrac{1\times 0.287\times (800 - 300.368)}{1.33-1}$

W = 434.53 kJ/kg

7 0

3 years ago

I have five brainliest why is it only showing 2?

svp [43]

Answer: Either your computer is malfunctioning or it is glitched and still thinks you only have 2. Also good job

Explanation: Leave a brainliest it helps

4 0

3 years ago

Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the cycle is modified with regeneration that

Elza [17]

Answer:

The correct answer is OPTION A (the turbine work output will decrease.)

Explanation:

In open feedwater heaters, there is a mixing of cold feedwater and steam which raise the temperature of feed water causing the fluid average temperature to rise and thereby increasing the thermodynamic efficiency of the cycle.

In a simple ideal Rankine cycle, heating of the water is converted into work done by the cycle in potential and kinetic energy forms. If there is a fixed boiler flow rate, and the simple ideal Rankine cycle is modified with regeneration, turbine work output, and heat supplied, will decrease.

3 0

3 years ago

​What should be a concern as a weldment becomes larger as more parts are added?

What should be a concern as a weldment becomes larger as more parts are added?