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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Answer:
Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)
Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k
YAll don’t tell my cousin I’m hacking her school resources because I got a 0 in a test and she got a 100
