Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
(d) 2996 kJ
Explanation:
We have given initial volume 
initial pressure 
initial temperature 
we know that during isothermal process the work done is given by

where 
putting all these value in formula of work done

=-2995.8 kJ here negative sign indicates that work is dine on the gas
so wok done =2995.8 kJ
Answer:
Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...
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<h2>
Answer:</h2>
<u>x= 3</u>.
<h2>
Explanation:</h2>
<em>What is presented in this problem is basically an equation in verbal form.</em>
<em />
<h3>1. Write the equation.</h3>

<h3>2. Solve for x.</h3>

<h3>3. Express the result.</h3>
x= 3.