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kolbaska11 [484]
3 years ago
15

A piston-cylinder assembly contains 3 kg of water, initially at 0.06 bar and 160 C. The water undergoes two processes in series:

D a constant volume process to saturated vapor followed by a cooling process. At the end of the cooling process, the temperature is 20 C and the water is a liquid-vapor mixture with a quality of 25%. Neglect kinetic and potential energy effects. i. (12) Determine the specific volume and specific internal energy of the water at a the initial state, b. the state at the end of the constant volume process, and li. (4) Sketch the T-v and p-v diagrams showing the key states, the ili (6) Find the work and heat transfer for the constant volume process, both in iv. (3) If 5500 KJ of heat is removed from the system during the cooling, c. the state at the end of the cooling process. processes, and relevant isobars or isotherms, kJ, calculate the work and comment whether this work is done on the system or by the system.

Engineering
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

See attachment below

Explanation:

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Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz
butalik [34]

Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

P_1=100 bar,T_=600°C,V_1=35m/s

Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

   h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

4 0
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What phenomenon allows water to reach the top of a building?
Artemon [7]

Answer:

Option C: water pressure.

Explanation:

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