Answer:
1.444 psi
Explanation:
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in. 2 and a temperature of 60 8 F at the beginning of compression. The maximum temperature in the cycle is 5200 8 R.
Based on this model,
1- Write possible Assumptions no less than three assumptions
2- Draw clear schematic for this problem
3- Determine possible Assumptions no less than three assumptions
4- Draw clear schematic for this problem.
5- calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Based on this model,
1- Write possible Assumptions no less than three assumptions
2- Draw clear schematic for this problem
3- Determine possible Assumptions no less than three assumptions
4- Draw clear schematic for this problem.
5- calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
1 answer:
1) The three possible assumptions are
a) All processes are internally reversible
b) Air, as the working fluid, circulates in a closed-loop
cycle
c) Combustion is represented as a heat-adding process
2) Diagrams are included
5) The net work per cycle is 845.88 kJ/kg
The horsepower produced is approximately 45374 hP
Explanation:
1) The three valid assumptions are
a) All processes are internally reversible
b) Air, the working fluid, moves continuously in a closed-loop
cycle
c) The combustion process is set as a heat addition step
2) Diagrams for illustration are provided
5) The cylinder bore diameter measures 3.7 in., equating to 0.09398 m
The stroke length is 3.4 in., approximately 0.08636 m.
The cylinder volume is calculated as v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume amounts to 16% of the cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ is 9.59 × 10⁻⁵ m³
p₁ is 14.5 lbf/in.² = 99973.981 Pa
T₁ equals 60 F = 288.706 K
For the Otto cycle's T-S diagram,
T₂ calculates to 288.706* = 592.984 K
The peak temperature is T₃ = 5200 R = 2888.89 K
T₄ resolves to 2888.89 / = 1406.5 K
Work performed, W = ×(T₃ - T₂) -
×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power generated in the Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
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Answer:
q' = 5826 W/m
Explanation:
Given:-
- The length of the fin in question, L = 0.15 m
- The fin's surface temperature, Ts = 250°C
- The velocity of free stream air, U = 80 km/h
- The air temperature, Ta = 27°C
- The flow is parallel over both sides of the fin, assuming turbulent flow conditions throughout.
Find:-
What is the heat removal rate per unit width of the fin?
Solution:-
- Steady state conditions are assumed, along with negligible radiation and turbulent flow conditions.
- From Table A-4, we gather air properties (T = 412 K, P = 1 atm ):
Dynamic viscosity, v = 27.85 * 10^-6 m²/s
Thermal conductivity, k = 0.0346 W / m.K
Prandtl number Pr = 0.69
- Compute the Nusselt Number (Nu) corresponding to - turbulent conditions - using the relevant relationship as follows:
Where, Re_L: Average Reynolds number for the entire length of fin:
- The convective heat transfer coefficient (h) can now be derived from:
- The heat loss rate q' per unit width can be established using the convection heat transfer formula and should be multiplied by (x2) since the airflow is present on both sides of the fin:
- Ultimately, the heat loss per unit width from the rectangular fin is q' = 5826 W/m
- The thermal loss per unit width (q') attributed to radiation:
Where, a signifies the Stefan-Boltzmann constant = 5.67*10^-8
- It is observed that radiation losses are not insignificant, accounting for 20% of thermal loss by convection. As the emissivity (e) of the fin is unspecified, this value is dismissed from the calculations as it pertains to the provided information.
Answer:
a) q = 7671 W
T0 = 47.6°C
b) ΔP = 202.3 N/m²
P = 58.2 W
c) hDarray = 2 times hD of an isolated element.
Explanation:
see the image for the solution.
