Answer:
COP = 0.090
Explanation:
The general formula for COP is:
COP = Desired Output/Required Input
Here,
Desired Output = Heat removed from water while cooling
Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time
Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]
Desired Output = 4.199 W
And the required input can be given as electrical power:
Required Input = Electrical Power = (Current)(Voltage)
Required Input = (2.9 A)(16 V) = 46.4 W
Therefore:
COP = 4.199 W/46.4 W
<u>COP = 0.090</u>
Answer:
coupling is in tension
Force = -244.81 N
Explanation:
Diameter of Hose ( D1 ) = 35 mm
Diameter of nozzle ( D2 ) = 25 mm
water gage pressure in hose = 510 kPa
stream leaving the nozzle is uniform
exit speed and pressure = 32 m/s and atmospheric
<u>Determine the force transmitted by the coupling between the nozzle and hose </u>
attached below is the remaining part of the detailed solution
Inlet velocity ( V1 ) = V2 ( D2/D1 )^2
= 32 ( 25 / 35 )^2
= 16.33 m/s
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Answer:
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Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
