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:
d. 2.3 ohms (5.3 amperes)
Explanation:
The calculator's 1/x key makes it convenient to calculate parallel resistance.
Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms
This corresponds to answer choice D.
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<em>Additional comment</em>
This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.
Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that
Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So
We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
