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Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such that
the specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by
0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heat
interaction between the system and the surroundings. Assuming an ambient temperature of 300
K, the irreversibility rate of the system is _________kW (round off to two decimal places).
the specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by
0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heat
interaction between the system and the surroundings. Assuming an ambient temperature of 300
K, the irreversibility rate of the system is _________kW (round off to two decimal places).
Correct answer is '2.1'. Can you explain this answer?
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Verified Answer
Water flowing at the rate of 1 kg/s through a system is heated using a...
The entropy generation (Sgen) is given by rise in entropy rise of the system (Since no heat interaction is involved)
So, entropy generation (Sgen) = m × specific entropy = 1 × 0.007 = 0.007 kW/K
Now, according to Gouy Stodola theorem,
Irreversibility =To × Sgen = 300 × 0.007 = 2.1kW
So, entropy generation (Sgen) = m × specific entropy = 1 × 0.007 = 0.007 kW/K
Now, according to Gouy Stodola theorem,
Irreversibility =To × Sgen = 300 × 0.007 = 2.1kW
Most Upvoted Answer
Water flowing at the rate of 1 kg/s through a system is heated using a...
Given Data:
- Mass flow rate of water, m = 1 kg/s
- Increase in specific enthalpy, Δh = 2.50 kJ/kg
- Increase in specific entropy, Δs = 0.007 kJ/kg.K
- Power input to the electric heater, P = 2.50 kW
- Ambient temperature, Tₐ = 300 K
Calculations:
The irreversibility rate of the system can be determined using the following equation:
Δs = Qₕ / Tₕ - Qₛ / Tₛ
Where,
Δs = Increase in specific entropy = 0.007 kJ/kg.K
Qₕ = Heat added to the system
Tₕ = Temperature at which heat is added
Qₛ = Heat rejected by the system
Tₛ = Temperature at which heat is rejected
From the given data, we can assume that the heat is added at the heater temperature and rejected at the ambient temperature. Therefore, Qₕ = Δh and Tₕ = Tᵢ, where Tᵢ is the initial temperature of the water.
Step 1: Calculation of Final Temperature
The final temperature of the water can be calculated using the equation:
m * (hᵢ + Δh) = m * h + Qₛ
Where,
m = Mass flow rate of water = 1 kg/s
hᵢ = Initial specific enthalpy
h = Final specific enthalpy
Assuming the specific heat capacity of water, c = 4.18 kJ/kg.K, the equation becomes:
1 * (hᵢ + 2.50) = 1 * h + Qₛ
hᵢ + 2.50 = h + Qₛ
Step 2: Calculation of Heat Rejected
The heat rejected by the system can be calculated using the equation:
Qₛ = m * c * (T - Tₐ)
Where,
T = Final temperature of the water
Step 3: Calculation of Heat Added
Using the given equation Qₕ = Δh, the heat added to the system can be determined.
Step 4: Calculation of Irreversibility Rate
Substituting the values in the irreversibility equation:
Δs = Qₕ / Tₕ - Qₛ / Tₛ
we can solve for Qₛ and substitute it back in the equation:
Δs = Qₕ / Tₕ - m * c * (T - Tₐ) / Tₛ
Step 5: Calculation of Power Input
The power input to the electric heater can be determined using the equation:
P = Qₕ / Δt
Where,
Δt = Time interval
Step 6: Calculation of Irreversibility Rate
The irreversibility rate can be calculated using the equation:
Irreversibility rate = P - Pᵢ
Where,
Pᵢ = Power input to the electric heater
Final Answer:
By substituting the values in the equations and performing the calculations, the irreversibility rate is found to be 2.1 kW
- Mass flow rate of water, m = 1 kg/s
- Increase in specific enthalpy, Δh = 2.50 kJ/kg
- Increase in specific entropy, Δs = 0.007 kJ/kg.K
- Power input to the electric heater, P = 2.50 kW
- Ambient temperature, Tₐ = 300 K
Calculations:
The irreversibility rate of the system can be determined using the following equation:
Δs = Qₕ / Tₕ - Qₛ / Tₛ
Where,
Δs = Increase in specific entropy = 0.007 kJ/kg.K
Qₕ = Heat added to the system
Tₕ = Temperature at which heat is added
Qₛ = Heat rejected by the system
Tₛ = Temperature at which heat is rejected
From the given data, we can assume that the heat is added at the heater temperature and rejected at the ambient temperature. Therefore, Qₕ = Δh and Tₕ = Tᵢ, where Tᵢ is the initial temperature of the water.
Step 1: Calculation of Final Temperature
The final temperature of the water can be calculated using the equation:
m * (hᵢ + Δh) = m * h + Qₛ
Where,
m = Mass flow rate of water = 1 kg/s
hᵢ = Initial specific enthalpy
h = Final specific enthalpy
Assuming the specific heat capacity of water, c = 4.18 kJ/kg.K, the equation becomes:
1 * (hᵢ + 2.50) = 1 * h + Qₛ
hᵢ + 2.50 = h + Qₛ
Step 2: Calculation of Heat Rejected
The heat rejected by the system can be calculated using the equation:
Qₛ = m * c * (T - Tₐ)
Where,
T = Final temperature of the water
Step 3: Calculation of Heat Added
Using the given equation Qₕ = Δh, the heat added to the system can be determined.
Step 4: Calculation of Irreversibility Rate
Substituting the values in the irreversibility equation:
Δs = Qₕ / Tₕ - Qₛ / Tₛ
we can solve for Qₛ and substitute it back in the equation:
Δs = Qₕ / Tₕ - m * c * (T - Tₐ) / Tₛ
Step 5: Calculation of Power Input
The power input to the electric heater can be determined using the equation:
P = Qₕ / Δt
Where,
Δt = Time interval
Step 6: Calculation of Irreversibility Rate
The irreversibility rate can be calculated using the equation:
Irreversibility rate = P - Pᵢ
Where,
Pᵢ = Power input to the electric heater
Final Answer:
By substituting the values in the equations and performing the calculations, the irreversibility rate is found to be 2.1 kW
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Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer?
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Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer?.
Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such thatthe specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heatinteraction between the system and the surroundings. Assuming an ambient temperature of 300K, the irreversibility rate of the system is _________kW (round off to two decimal places).Correct answer is '2.1'. Can you explain this answer?.
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