Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin
e is 630 kJ per kg of steam flowing through the turbine. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, deter- mine (a) the isentropic turbine efficiency, (b) the rate of entropy gen- eration, in kJ/K per kg of steam flowing. Air at 40°F, 1 atm enters a compressor operating at steady state and exits at 620°F, 8.6 atm. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, determine (a) the isentro- pic compressor efficiency and (b) the rate of entropy generation, in Btu/ºr per lb of air flowing. 2.7
1 answer:
Answer:
a) isentropic efficiency = 84.905%
b) rate of entropy generation = .341 kj/(kg.k)
Please kindly see explaination and attachment.
Explanation:
a) isentropic efficiency = 84.905%
b) rate of entropy generation = .341 kj/(kg.k)
The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.
Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.
Please refer to attachment for step by step solution of the question.
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The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
