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Alenkasestr [34]
3 years ago
10

State 3 advantages and 3 disadvantages of unit rate contract​

Engineering
1 answer:
Zinaida [17]3 years ago
7 0
6 would be the answer
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
⚠️I mark BRIANLIST ⚠️The same engineering teams are able to design and develop the different subsystems for an airplane.
Gekata [30.6K]

Answer:

False

Explanation:

BRIANLIST?

5 0
2 years ago
Read 2 more answers
A heat pump with refrigerant-134a as the working uid is used to keep a space at 25C by absorbing heat from geothermal water that
Snezhnost [94]

Answer:

A) 0.03382 kg/s

B) 7.0372 Kw

C) 4.3982

D) 0.7396 kw

Explanation:

Given data:

Evaporator at 60 C

Space temperature = 25 C

power consumed by compressor = 1.6 kw

T1( evaporator temperature ) = 12°C

attached below is the detailed solution

8 0
3 years ago
A fair die is thrown, What is the probability gained if you are told that 4 will
Semmy [17]

Answer:

1/6

Explanation:

A dice has 6 sides, the probability of 4 appearing is 1/6.

7 0
2 years ago
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is enc
Aleks [24]

Answer:

a. 4\mu m

b. 1 m

Explanation:

According to the question, the data is as follows

The Density of water at 20 degrees celcius is 1000 kg/m^3

Viscosity is 0.001kg/m/.s

Velocity V = 25 cm/s

V = 0.25 m/s

Now

a. The creeping motion is

As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

= 4E - 06^m

= 4\mu m

b. Now the sphere diameter is

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

250,000 = (1,000 × 0.25 × d) ÷ 0.0001

d = (250,000 × 0.001) ÷ (1,000 × 0.25)

= 1 m

6 0
3 years ago
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