Kris and James are working at a construction site that has a significant amount of stagnant water. Which type of hazard are they
2 answers:
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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.
Answer:
a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s
b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2
c) m' = 2.915 kg/s
Explanation:
Given:-
- The inlet pressure, Pi = 1.0 MPa
- The inlet temperature, Ti = 750 K
- Inlet velocity is negligible
- Steady, Isentropic Flow
- The specific heat ratio of air, k = 1.4
- Exit Mach number, Mae = 2
- The throat area, Ath = 20 cm^2
- Gas constant of air, R = 0.287 KJ / kg.K
Find:-
(a) the throat conditions (static pressure, temperature, density, and mach number)
b) the exit plane conditions
c) the mass flow rate
Solution:-
- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).
- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).
- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:
- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:
- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:
- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:
- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:
- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows: