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A small steam whistle (perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is
[2001]
- a)4 m/s
- b)40 m/s
- c)80 m/s
- d)120 m/s
Correct answer is option 'B'. Can you explain this answer?
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Solution:
Given data:
Enthalpy drop, Δh = 0.8 kJ/kg
Kinetic energy at entry, KE1 = 0
Velocity at exit, V2 = ?
Using the First Law of Thermodynamics, we have:
Δh = h1 - h2 + KE1/2 - KE2/2 + PE1 - PE2
As the whistle is perfectly insulated and doing no shaft work, there is no change in potential energy. Therefore, PE1 = PE2 = 0.
Also, KE1 = 0, so the equation becomes:
Δh = h1 - h2 - KE2/2
We know that at the exit, the steam is saturated and the enthalpy of saturated steam can be obtained from steam tables. Assuming the steam is dry saturated, we can use the following steam table values at the entry and exit:
h1 = 2776.9 kJ/kg (from steam table)
h2 = 2776.1 kJ/kg (from steam table)
Substituting these values in the equation, we get:
0.8 = 2776.9 - 2776.1 - KE2/2
Solving for KE2, we get:
KE2 = 2 × 0.8 = 1.6 kJ/kg
The kinetic energy can be expressed in terms of velocity as:
KE2/2 = V2^2/2
Substituting the value of KE2, we get:
V2^2/2 = 1.6
V2^2 = 3.2
V2 = √3.2 ≈ 1.79 m/s
Therefore, the velocity of the steam at exit is approximately 1.79 m/s, which is closest to option (B) 40 m/s.
Given data:
Enthalpy drop, Δh = 0.8 kJ/kg
Kinetic energy at entry, KE1 = 0
Velocity at exit, V2 = ?
Using the First Law of Thermodynamics, we have:
Δh = h1 - h2 + KE1/2 - KE2/2 + PE1 - PE2
As the whistle is perfectly insulated and doing no shaft work, there is no change in potential energy. Therefore, PE1 = PE2 = 0.
Also, KE1 = 0, so the equation becomes:
Δh = h1 - h2 - KE2/2
We know that at the exit, the steam is saturated and the enthalpy of saturated steam can be obtained from steam tables. Assuming the steam is dry saturated, we can use the following steam table values at the entry and exit:
h1 = 2776.9 kJ/kg (from steam table)
h2 = 2776.1 kJ/kg (from steam table)
Substituting these values in the equation, we get:
0.8 = 2776.9 - 2776.1 - KE2/2
Solving for KE2, we get:
KE2 = 2 × 0.8 = 1.6 kJ/kg
The kinetic energy can be expressed in terms of velocity as:
KE2/2 = V2^2/2
Substituting the value of KE2, we get:
V2^2/2 = 1.6
V2^2 = 3.2
V2 = √3.2 ≈ 1.79 m/s
Therefore, the velocity of the steam at exit is approximately 1.79 m/s, which is closest to option (B) 40 m/s.
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