A piston cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air at 300K. The air undergoes a const
ant pressure process to a final temp of 400K. During the process heat is transfered to the air by Q=12kJ. Assuming the ideal gas model with k=1.4 and negligible changes in kinetic and potential energy for the air, determine the work done by the paddle on the air and the work done by the air to displace the piston. Gas constant for air is 0.287 kJ/kg*K
note ideal gas model is pV=mRT, and with k=1.4 the total internal energy change of air canbe calculated by U2-U1=(mR/k-1)(delta T)
note ideal gas model is pV=mRT, and with k=1.4 the total internal energy change of air canbe calculated by U2-U1=(mR/k-1)(delta T)
1 answer:
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Answer:
Explanation:
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Answer:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The net work per cycle is 845.88 kJ/kg
The power developed in horsepower ≈ 45374 hP
Explanation:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m
Stroke length = 3.4 in. = 0.08636 m.
The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ = 9.59 × 10⁻⁵ m³
p₁ = 14.5 lbf/in.² = 99973.981 Pa
T₁ = 60 F = 288.706 K
Otto cycle T-S diagram
T₂ = 288.706* = 592.984 K
The maximum temperature = T₃ = 5200 R = 2888.89 K
T₄ = 2888.89 / = 1406.5 K
Work done, W = ×(T₃ - T₂) -
×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power developed in an Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
