Explanation:
because it helps in our daily life
Answer:
a) the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air is 1065 kJ/min
Explanation:
Given the data in the question;
[ Outdoor ] ← Q [ W ] Q ← [ House ]
Rate of heat removed from the house; Q = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW
Net-work input; W = 5.25 kW
a) The coefficient of performance of the air conditioner; COP.
COP = Q / W
we substitute
COP = 12.5 kW / 5.25 kW
COP = 2.38
Therefore, the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air.
Q = Q + W
we substitute
Q = 12.5 kW + 5.25 kW
Q = 17.75 kW
Q = ( 17.75 × 60 ) kJ/min
Q = 1065 kJ/min
Therefore, the rate of heat transfer to the outside air is 1065 kJ/min
Answer:
a diameter of D₂ = 0.183 inches would be required
Explanation:
appyling pascal's law
P applied to the hydraulic jack = P required to lift the rock
F₁*A₁ = F₂*A₂
since A₁= π*D₁²/4 , A₂= π*D₂²/4
F₁*π*D₁²/4 = F₂* π*D₂²/4
F₁*D₁²=F₂*D₂²
D₂ = D₁ *√(F₁/F₂)
replacing values
D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches
Answer:
14.27 mm
Explanation:
Force = 80 Newton
Pressure exerted = 5 bar = 5×10⁵ Pa
Pressure exerted = Force/Area
Area = Force/Pressure
Area of the cylinder=πd²/4
Hence diameter of cylinder 14.27 mm