Ok guys i need help with this problem, i think i have an idea of how to do it so

Question

Ok guys i need help with this problem, i think i have an idea of how to do it so the values that i have so far is


Vin=.7m^3 Vout=.7m^3

u1=? h2=?

T1=250 C T2=250 C

Qin=0 Qout=?( what we are tryign to find)

So these are the values i think are what is stated correct me if I am wrong, for the first state
I think it is in saturated mixture because they stated that the volume is half water and half gas.
state 2 only water is coming out so it is saturated liquid. The instructions say make a graph at
10% mass exit increments, so it will be a Q vs Mass Exit graph. So i am assuming we will have to find the Q at each increment of mass exit so at 10%,20%,30%,40% and then graph them. And the final instructions say show full calculations of the 20% mass exit. So what i need help wtih is how to setup the energy equation i think it will be like this for example we are doing it for a mass exit percent of 10%>>>>> -q=(.1)m(h2-u1) the -q is for the q out so it will be loss of energy and .1m is for 10% of total mass is being exited and h2 is because h=u+pv ( v= specific volume) since water is flowing out, there will be flow work. and u1 is not h1 because there is no flow work involved inside the tank. I think you will have to use the steam tables to find some values. So what i want to see from someone is how they went about solving this problem, and what the graph should look like. And please do not provide a answer with alot of holes in it, what i mean is something that is not clear, and i am not looking for an explanation i think i understand what is going on here, but what i need to see is calculation i will provide 5 stars to someone that put an effort to the good calculation. I will not cheat you i promise i just require someone with a good understanding of thermodynamics to help me with this problem.


Thank You

Explanation / Answer

For steam properties, seehttp://www.irc.wisc.edu/properties/ (opens in IE)

At 250 deg C, quality = 0 we get v_f = 799 kg/m3,

At 250 deg C, quality = 1, we get v_g = 20 kg/m3

Hence, m_f = (700/2)*10^-3 * 799 = 279.65 kg

and m_g = (700/2)*10^-3 *20 = 7 kg

m = m_f + m_g = 286.65 kg

Quality x = m_g / m = 0.0244

At T1 = 250 deg C, and x = 0.0244, we get h1 = 1130 kJ/kg

Total enthalpy = m*h1 = 286.65*1130 = 323914.5 kJ

10% of total mass of liquid= 0.1*279.65 = 27.965 kg

Remaining mass of liquid = 279.65 - 27.965 = 251.685 kg

New quality = m_g / m = 7 / (251.685 + 7) = 0.027

At T = 250 deg C and x = 0.027, we get h2 = 1130 kJ/kg

Total enthalpy = (251.685 + 7) *1130 = 292314.05 kJ

Heat transfer = change in enthalpy = 323914.5 - 292314.05 = 31600 kJ

Hence, first data point on graph will be (27.965 kg, 31600 kJ).

Similalrly you can plot other points for 20%, 30%, 40% mass exits.

Due to shortage of time, I am not doing it full but the process is now clear for you.

Please rate 5-stars!! Thanks.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.