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2 years ago

Arnold has a contract to design a hospital. Which factors should he incorporate into the design of the building?

Engineering

2 answers:

goblinko [34]2 years ago

7 0

D best answer hope it helps

Guest

1 year ago

whats d then?

kykrilka [37]2 years ago

4 0

Chile idk I just need mine, sorry

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

2 years ago

Air whose density is 0.082 lbm/ft3 enters the duct of an air-conditioning system at a volume flow rate of 450 ft3/min. If the di

Akimi4 [234]

Answer:

$Given:\\V_{flow} =450 ft^3/min\\P=0.082lbm/ft^3\\A=16In$

where P is density of air and v flow is volume flow rate, so in order to find velocity of air and mass flow rate formula is given by :

$M_{flow} =V_{flow} *P\\\\M_{flow} =450*0.082=36.9ib/min$

so velocity is:

$v=V_{flow} /A=339.03m/min$

4 0

2 years ago

Read 2 more answers

The total floor area of a building, including below-grade space but excluding unenclosed areas, measured from the exterior of th

alex41 [277]

Answer:

Gross building area

Explanation:

The Gross building area refers to the entire area of a building covering all the floors. The measurement is expressed in square feet. The Gross building area also includes basements, penthouses, and mezzanines. It is calculated by estimating the exterior dimension of the building. Storage rooms, laundries, staircases are also a part of the gross building area.

6 0

2 years ago

A sample of normally consolidated clay was subjected to a CU triaxial compression test that was carried out until the specimen f

QveST [7]

Answer:

Check the explanation

Explanation:

Given

1) CU traixial compression test

2) Devatoric stress at failure = бd = 50 kN/ $m^2$

3) Confining pressure at failure = бd = 48 kN/ $m^2$

4) Pore pressure at failure = u = 18 kN/ $m^2$

5) Unconfined compression stress = q = 20 kN/ $m^2$

6) Undrained cohesion = q/2 = 20 kN/ $m^2$

To find:

1) Effective and total stress strength failure envelope

Kindly check the attached image below .

7 0

2 years ago

Why is it so dangerous to use a ground lift on a metal cased power tool

MaRussiya [10]

Answer:

Removal of the safety ground connection on equipment can expose users to an increased danger of electric shock and may contradict wiring regulations. ... If a fault develops in any line-operated equipment, cable shields and equipment enclosures may become energized, creating an electric shock hazard.

7 0

1 year ago