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IgorLugansk [536]
3 years ago
7

Saturated steam coming off the turbine of a steam power plant at 40°C condenses on the outside of a 3-cm-outer-diameter, 35-m-lo

ng tube at a rate of 140 kg/h.
Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe.
Engineering
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

93.57 KJ/s

Explanation:

Rate of heat transfer from the steam to the cooling water = mass rate  ×  Heat of vaporization of water at 40 °C

H vaporization of water 40 °C  at saturation pressure = 2406.0 kJ / Kg

rate of heat transfer = 2406.0 kJ / Kg × 140 Kg / ( 60 × 60s) = 93.57 KJ/s

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A_____ transducer is a device that can convert an electronic controller output signal into a standard pneumatic output. A. pneum
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Answer:

The correct answer is

option C. current to pneumatic (V/P)

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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
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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.

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