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dezoksy [38]
3 years ago
6

Which statement describes a possible limitation on a experimental design? A. Collecting samples to analyze is expensive B. The e

xperiment has a very simple procedure C. The subject of the experiment has been studied by other scientist's D. There is a large number of constants
Engineering
2 answers:
AveGali [126]3 years ago
8 0

Answer:

collecting samples to analyze is very expensive is correct^

Explanation:

just took the quiz, thanks btw

exis [7]3 years ago
7 0

Answer:

A. Collecting samples to analyze is expensive

Explanation:

Experimental research designs may be referred to as an analytical approach which is based on establishing a scientific hypothesis through by subjecting observations or subjects to certain treatment carried out in a controlled environment. In a typical experimental research design, we have two groups which are the constant group or observation and the variable group which are the actual experimetal group which are subjected to treatment and whose behavior are compared to that of the constant group. From the options given, one actual limitation of the experimetal research design is the associated cost particularly those which require longer time to study and analyse. Other limitations include ; difficulty in measuring human response and skepticism over applicability of findings in a natural environment.

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Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
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