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.
You might be interested in
Answer:
The source temperature is 1248 R.
Explanation:
Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.
Given:
Temperature of the heat sink is 520 R.
Second law efficiency is 60%.
Actual thermal efficiency is 35%.
Calculation:
Step1
Reversible efficiency is calculated as follows:
Step2
Source temperature is calculated as follows:
T = 1248 R.
The heat engine is shown below:
Thus, the source temperature is 1248 R.
Answer:
The theoretical density for Niobium is .
Explanation:
Formula used :
where,
= density of the unit cell
Z = number of atom in unit cell
M = atomic mass
= Avogadro's number
a = edge length of unit cell
We have :
Z = 2 (BCC)
M = 92.91 g/mol ( Niobium)
Atomic radius for niobium = r = 0.143 nm
Edge length of the unit cell = a
r = 0.866 a (BCC unit cell)
On substituting all the given values , we will get the value of 'a'.
The theoretical density for Niobium is .
Answer:
The answer is "conditionally unstable"
Explanation:
The conditional volatility is really a condition of uncertainty, which reflects on whether increasing air is polluted or not. It determines the rate of ambient delay, which has been between humid and dry adiabatic rates. In general, the environment is in an unilaterally unhealthy region.
Classification dependent on ELR:
Larger than 10 m Around 10 and 6
m or less 6
m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9
m, that's why it is "conditionally unstable".