Recent content by MrMechanic

0.7 = Mass of the Vapor / 0.90 Mass flow rate of the vapor = 0.63 kg/s I got V1 = 0.04704095 m^3/kg and V2 = 0.09196687593 m^3/kg By using interpolation from the steam table of 3MPa i get 353 degrees Celsius. Is there something wrong? And V1 is from the saturated temp of 3MPa which is 233.90...

I see. Thanks for clearing that. From the link that you gave. I found out that I can use X = Mass of vapor / Mass total. I have 0.70 as x but 0.90kg/s is mass of vapor or mass total?

From my understanding the liquid vapor was at 70% quality before it turns into a steam having 0.9kg/s. Either way the problem is really hard to understand. If i can figure out what's initially there before turning into steam.

Homework Statement There are expanded 0.90kg/s of steam at constant pressure from 3MPa and 70% quality to a final state. If the process is nonflow for which W = 121.3kJ/s, find (a) The final temperature, (b) Q, (c) the available part of Q for a sink temperature of to = 27 Celsius Answers : a)...

what are the values of x and its derivatives... same goes for y.. I got confused. But thanks sir.

Homework Statement The projectile A is being tracked by the radar at O. At a given instant, the radar readings are θ = 30degrees, R = 2000m, dR/dt = 200 m/s, and d^2R/dt^2 = 20 m/s^2. Determine the speed of the projectile at that instant. THE ANSWER AT THE BACK IS 299.7m/s [PLEASE SEE...

Oh yeah sorry. It's not kg/m^3 ... It's 9.81 kN/m^3 forgot to change that sorry

The minus sign is not an absolute pressure i think. The density of oil is not given. And also gasoline. In order to get their density you have to multiply their specific gravity to the density of water which is 9.81kg/m^3

When I perform P1 = 0 I get h = 5.235

is this correct? (-40) + (0.82)(9.81)(38-30) + (1.5) (9.81) (h) - (9.81)(h+3) - 20.6 = P1 and i'll perform another on the other side? or should I set P1 = 0?

Specific gravity is the ratio of the density of a substance to the density. that's why it has no unit yes If Patm =14.7 psi = 101.325kPa = 101.325kN/m^2 and P = 101.325 + (0.82)(9.81(38-30) P = 36.9714 kPa What i don't know is how to do I form an equation where i can get the height (h)

Yes the pressure in the airspaces in the tanks are given. I'll start with the left tank P = Patm + (0.82)(9.81)(38-30) is it correct?

Homework Statement Find the elevation at point A. in the figure shown. Homework Equations P = pgh The Attempt at a Solution So i Converted 0.21kg/cm^2 to Kn/m^2 and i got 20.6kN/m^2 or 20.6kPa and -30cm Hg i got 40kN/m^2 or 40kPa

Well, I kinda worked out the formula you gave me. So, here is my solution to (c) ΔP=(0.488)(30/32.2)x(25) = 11.38 lb/in^2 Pgage = 11.38 Pabs = Patm + Pgage Pabs = 14.7 + 11.38 Pabs = 26.08 psia

Pabs = Pair + PH20 + PHg + Patm? So i got this... Pair = 0.02lb/ft^3 /12^3 x (30/32.2) (40) = 1/644 psi PH20 = 62.1 / 12^3 x (30/32.2)(40)= 1.339psi PHg = 0.488 x (30/32) (40) = 18.19psi Is it wrong?