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3 years ago

What type of degree is needed to be an architect?

2 answers:

3 0

Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license.

Vadim26 [7]3 years ago

3 0

Answer:

A professional Bachelor's or Master's degree in arcgitecture.

Explanation:

Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license. The median annual salary for an architect in May 2015 was $74,110.

MARK AS BRAINLIEST!

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125

Gekata [30.6K]

Answer:

a) $\dot m_{3} = 135\,\frac{lbm}{s}$ , b) $h_{3}=168.965\,\frac{BTU}{lbm}$ , c) $T = 200.829\,^{\textdegree}F$

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

$\dot m_{1} + \dot m_{2} - \dot m_{3} = 0$

The mass flow rate exiting the tank is:

$\dot m_{3} = \dot m_{1} + \dot m_{2}$

$\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}$

$\dot m_{3} = 135\,\frac{lbm}{s}$

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

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

$h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}$

Properties of water are obtained from tables:

$h_{1}=180.16\,\frac{BTU}{lbm}$

$h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)$

$h_{2}=29.032\,\frac{BTU}{lbm}$

The specific enthalpy at outlet is:

$h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }$

$h_{3}=168.965\,\frac{BTU}{lbm}$

c) After a quick interpolation from data availables on water tables, the final temperature is:

$T = 200.829\,^{\textdegree}F$

8 0

3 years ago

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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago

The Canadair CL-215T amphibious aircraft is specially designed to fight fires. It is the only production aircraft that can scoop

ioda

Answer:

Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪

Explanation:

check attached files for explanation

6 0

3 years ago

You don't have to notify employees that a lockout/tagout is about to begin?

Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

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3 years ago

Which of the following lists the steps of a process in the correct order?

Vladimir79 [104]

Answer: C. Problem identification, solution design, implementation, evaluation.

Explanation:

4 0

3 years ago

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