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iVinArrow [24]
3 years ago
5

Special considerations must be given to systems using liquid hydrogen for fuel because of: A. Liquid hydrogen's low density B. L

iquid hydrogen's low storage temperature C. The high flammability of hydrogen vapors D. All of the above
Engineering
1 answer:
vitfil [10]3 years ago
6 0

Answer:

D.All of the above

Explanation:

Properties of hydrogen:

1.It is lighter than air.It has density about 0.089 g/L.

2.Hydrogen  rapidly change from liquid state to gas,so special protection is required to protect it.

3.It is highly flammable gas.

4.Liquid form of hydrogen exits at -432 F .This is very low temperature so special protection requires to keep it in liquid form.          

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True

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7 0
3 years ago
Read 2 more answers
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
4 years ago
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