To solve this problem, we need to use the principle of conservation of energy, which is often expressed in the form of the First Law of Thermodynamics. For a steady-flow system such as a steam nozzle, the energy equation simplifies to:

h1 + (V1^2)/2 = h2 + (V2^2)/2

where h1 and h2 are the specific enthalpies at the inlet and outlet, respectively, and V1 and V2 are the velocities at the inlet and outlet, respectively.

Given that the inlet velocity is low, we can assume V1 is approximately 0.

We are given the pressure and temperature at the inlet, so we can look up h1 in the steam tables. At 0.4 MPa and 300 °C, h1 is approximately 3051.2 kJ/kg.

At the outlet, we are given the pressure and the quality (x), so we can find h2 using the equation:

h2 = hf + x*hfg

where hf is the specific enthalpy of saturated liquid and hfg is the specific enthalpy of vaporization, both at the outlet pressure of 0.2 MPa. From the steam tables, hf is approximately 504.7 kJ/kg and hfg is approximately 2201.9 kJ/kg. So, with x = 0.9, we find h2 is approximately 2503.5 kJ/kg.

Substituting these values into the energy equation gives:

3051.2 kJ/kg = 2503.5 kJ/kg + (V2^2)/2

Solving for V2, we find V2 is approximately 956 m/s. So, the correct answer is (b) Exit velocity = 956 m/s.

Question

To solve this problem, we need to use the principle of conservation of energy, which is often expressed in the form of the First Law of Thermodynamics. For a steady-flow system such as a steam nozzle, the energy equation simplifies to:

h1 + (V1^2)/2 = h2 + (V2^2)/2

where h1 and h2 are the specific enthalpies at the inlet and outlet, respectively, and V1 and V2 are the velocities at the inlet and outlet, respectively.

Given that the inlet velocity is low, we can assume V1 is approximately 0.

We are given the pressure and temperature at the inlet, so we can look up h1 in the steam tables. At 0.4 MPa and 300 °C, h1 is approximately 3051.2 kJ/kg.

At the outlet, we are given the pressure and the quality (x), so we can find h2 using the equation:

h2 = hf + x*hfg

where hf is the specific enthalpy of saturated liquid and hfg is the specific enthalpy of vaporization, both at the outlet pressure of 0.2 MPa. From the steam tables, hf is approximately 504.7 kJ/kg and hfg is approximately 2201.9 kJ/kg. So, with x = 0.9, we find h2 is approximately 2503.5 kJ/kg.

Substituting these values into the energy equation gives:

3051.2 kJ/kg = 2503.5 kJ/kg + (V2^2)/2

Solving for V2, we find V2 is approximately 956 m/s. So, the correct answer is (b) Exit velocity = 956 m/s.

Knowee AI · Accepted Answer

To solve this problem, we need to use the principle of conservation of energy, which is often expressed in the form of the First Law of Thermodynamics. For a steady-flow system such as a steam nozzle, the energy equation simplifies to:

h1 + (V1^2)/2 = h2 + (V2^2)/2

where h1 and h2 are the specific enthalpies at the inlet and outlet, respectively, and V1 and V2 are the velocities at the inlet and outlet, respectively.

Given that the inlet velocity is low, we can assume V1 is approximately 0.

We are given the pressure and temperature at the inlet, so we can look up h1 in the steam tables. At 0.4 MPa and 300 °C, h1 is approximately 3051.2 kJ/kg.

At the outlet, we are given the pressure and the quality (x), so we can find h2 using the equation:

h2 = hf + x*hfg

where hf is the specific enthalpy of saturated liquid and hfg is the specific enthalpy of vaporization, both at the outlet pressure of 0.2 MPa. From the steam tables, hf is approximately 504.7 kJ/kg and hfg is approximately 2201.9 kJ/kg. So, with x = 0.9, we find h2 is approximately 2503.5 kJ/kg.

Substituting these values into the energy equation gives:

3051.2 kJ/kg = 2503.5 kJ/kg + (V2^2)/2

Solving for V2, we find V2 is approximately 956 m/s. So, the correct answer is (b) Exit velocity = 956 m/s.

Steam at 0.4 MPa, 300 °C, enters an adiabatic nozzle with a low velocity and leaves at 0.2 MPa with a quality of 90%. Find the exit velocity.*a) Exit velocity = 1200 m/sb) Exit velocity = 956 m/sc) Exit velocity = 1000 m/sd) Exit velocity = 1078 m/s

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