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liberstina [14]
3 years ago
15

A client wants Mila’s team to maximize the energy efficiency of their new office building. The client is interested in acquiring

a large plot of land for the new building in a suburban area north of a major city. From the city, the main access to this area is by a couple of major roads that have heavy traffic at almost all times of day. The client’s current building is easier to access and large enough for projected staffing needs, but it is also old, ugly, and high in maintenance costs. How should Mila’s team best approach this project while adhering to principles of green building?
Agree to build on the land the client is interested in acquiring and concentrate on maximizing the efficiency advantages of the new building.

Persuade the client to tear down the existing building instead and build a new high-efficiency building from the ground up in the same spot.

Persuade the client to renovate their existing building both aesthetically and in terms of efficiency and utility of function.

Find a different location for the new building in a more advantageous spot both in terms of transportation access but also in taking advantage of daylighting.
Engineering
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

Persuade the client to renovate their existing building both aesthetically and in terms of efficiency and utility of function.

Explanation:

Just would not make sense to tear the building down just to rebuild it when it can be restored.

You might be interested in
A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of ins
Stella [2.4K]

Answer:

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

Explanation:

Generally, thermal resistance for conduction heat transfer in a sphere.

R_{cond} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi K}}  

Where R_{cond} is the thermal resistance for conduction, K is the thermal conductivity of the material, r_{i} is the inner radius of the sphere, and r_{o} is the outer radius of the sphere.

The surface area of sphere, A_{s} is given by

A_{s}=4\pi {r^2}

For aluminum sphere, the thermal resistance for conductive heat transfer is given by

Calculate the thermal resistance for conductive heat transfer through the aluminum sphere.

R_{cond,s{\rm{ - 1}}} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}}

Where K_{Al} is aluminum’s thermal conductivity at T_{s}

Thermal resistance for conductive heat transfer through the insulation.

R_{cond,1{\rm{ - 2}}} = \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}}

Thermal resistance for convection is given by

R_{conv} = \frac{1}{{hA}}

Where h is convective heat transfer coefficient, R_{conv} is thermal resistance for convection and A is the cross-sectional area normal to the direction of flow of heat energy

Thermal resistance for convective heat transfer in-between the outer surface of the insulation and the ambient air.

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h{A_s}}}

Where h represents convective heat transfer coefficient at the outer surface of the insulation. Since A_{s} is already defined, substituting it into the above formula yields

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h\left( {4\pi {r^2}} \right)}}

To obtain radial distance of the outer surface of the insulation from the center of the sphere.

r = r_{o} + t where t is thickness of insulation

r=0.21+0.15=0.36m

Total thermal resistance

R_{eq} = {R_{cond,s{\rm{ - 1}}}} + {R_{cond,1{\rm{ - 2}}}} +{R_{conv,2{\rm{ - }}\infty }}

Where R_{eq} is total thermal resistance

R_{eq} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}

Consider the thermal conductivity of aluminum at temperature T_{s} as 234W/m.K

Rate of heat transfer for the given process

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{{R_{eq}}}}

Where \dot Q_{s - \infty }} is the steady state heat transfer rate in-between the inner surface of the sphere and the ambient air.

Substituting \left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right) for R_{eq} we obtain

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{\left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right)}}

\begin{array}{l}\\80{\rm{ W}} = \frac{{250{\rm{ }}^\circ {\rm{C}} - 20{\rm{ }}^\circ {\rm{C}}}}{{\left( {\frac{{\left( {\frac{1}{{0.18{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.21{\rm{ m}}}}} \right)}}{{4\pi \left( {234{\rm{ W/m}} \cdot {\rm{K}}} \right)}} + \frac{1}{{30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}\left( {4\pi {{\left( {0.36{\rm{ m}}} \right)}^2}} \right)}}\frac{{\left( {\frac{1}{{0.21{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.36{\rm{ m}}}}} \right)}}{{4\pi {K_{ins}}}} + } \right)}}\\\\80{\rm{ W}}\left( {{\rm{0}}{\rm{0.020737 K/W}} + \frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}}} \right) = 230{\rm{ K}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = \frac{{230{\rm{ K}}}}{{80{\rm{ W}}}} - {\rm{0}}{\rm{0.020737 K/W}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = {\rm{2}}{\rm{.854263 K/W}}\\\end{array}

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

7 0
3 years ago
) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is app
Zarrin [17]

Answer:

F=710KHZ

Explanation:

From the question we are told that:

Frequency selectivity ratio R=10

AM range 750 kHz to 2600 kHz

Therefore Bandwidth is

 B=2100-680

 B=1420KHZ

Generally the equation for The intermediate frequency is mathematically given by

Intermediate frequency=\frac{Bandwidth}{2}

 F=\frac{B}{2}

 F=\frac{1420}{2}

 F=710KHZ

8 0
3 years ago
Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in
Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time =   ∑  (\frac{Clock cyles}{Clock rate})

Clock cycles can be determined using following formula

Clock cycles = (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

Clock cycles = ( 50 x 10^{6} x 1) + (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

    = \frac{512(10^{6}) }{2(10^{9}) }

 Initial execution Time =  256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

(\frac{Clockcycles}{2} )=   (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

CPI_{FP improved} x No. FP instructions  =  (\frac{Clockcycles}{2} ) -[ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)]

CPI_{FP improved} x 50 x 10^{6}  = ( \frac{512(10)^{6} }{2} ) - [ (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)]

CPI_{FP improved} x 50 x 10^{6}  =  - 206 x 10^{6}

CPI_{FP improved}  = - 206 x 10^{6} / 50 x 10^{6}

CPI_{FP improved} = - 4.12 < 0

3 0
3 years ago
The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. Ifh=3ft, determin
KATRIN_1 [288]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
Adding new equipment or processes may require changes to the PPE requirements for
Yuki888 [10]
I think it’s is false I’m not that sure
5 0
2 years ago
Read 2 more answers
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