Answer:
COP(heat pump) = 2.66
COP(Theoretical maximum) = 14.65
Explanation:
Given:
Q(h) = 200 KW
W = 75 KW
Temperature (T1) = 293 K
Temperature (T2) = 273 K
Find:
COP(heat pump)
COP(Theoretical maximum)
Computation:
COP(heat pump) = Q(h) / W
COP(heat pump) = 200 / 75
COP(heat pump) = 2.66
COP(Theoretical maximum) = T1 / (T1 - T2)
COP(Theoretical maximum) = 293 / (293 - 273)
COP(Theoretical maximum) = 293 / 20
COP(Theoretical maximum) = 14.65
Answer:
3) the pressure drop across high MERV filters is significant.
Explanation:
MERV (Minimum-Efficiency Reporting Value) is used to measure the efficiency of filter to remove particles. A filter of high MERV can filter smaller particles but this causes an increase in reduced air flow that is an increase in pressure drop. High MERV filters capture more particles causing them to get congested faster and thereby increasing pressure drop.
Excessive pressure drop can cause overheating and lead to damage of the filter. The pressure drop can be reduced by increasing the surface area of the filter.
The relationship between resistance and the area of the cross section of a wire is inversely proportional . When resistance is increased in a circuit , for example by adding more electrical components , the current decreases as a result.
Answer:
Ponding will occur in 40mins
Explanation:
We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.
Consider checking attachment for the step by step solution.
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