Answer:
COP of heat pump=3.013
COP of cycle=1.124
Explanation
W = Q2 - Q1 ----- equation 1
W = work done
Q2 = final energy
Q1 = initial energy
A) calculate the COP of the heat pump
COP =Q2/W
from equation 1
Q2 = Q1 + W = 15 + 7.45 = 22.45 KW
therefore COP =22.45/7.45 = 3.013
B) COP when cycle is reversed
COP = Q1/W
from equation 1
Q1 + W = Q2 ------ equation 2
Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2
Q1 = 8.375 KW
COP =8.375/7.45 = 1.124
Answer:
W=2 MW
Explanation:
Given that
COP= 2.5
Heat extracted from 85°C
Qa= 5 MW
Lets heat supplied at 150°C = Qr
The power input to heat pump = W
From first law of thermodynamics
Qr= Qa+ W
We know that COP of heat pump given as
W=2 MW
For Carnot heat pump
2.5 T₂ - 895= T₂
T₂=596.66 K
T₂=323.6 °C
Answer:
They communicate ideas very quickly.
Explanation:
Answer:
The following statements are true:
A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction
C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface
E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.
Select ALL statements that are TRUE
B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant
D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed
Explanation:
Answer:
In this era, Sun Ra was among the first of any musicians to make extensive and pioneering use of synthesizers and other various electronic keyboards; he was given a prototype Minimoog by its inventor, Robert Moog.
Explanation:
