Can someone tell me where to start with this question?
Mesab, How do you determine "m", the initial mass of the gas in tank "A"... You say it's given- but it's not. 'Seems to me you'd need vapor-pressure vs. temperature data for R-12-- and/or the densities of the liquid and gas phases as a function of temperature and pressure... Added later: Thanks for the help from "Left Field" :) I talk about density- you talk about its reciprocal, specific volume. Either way, we need the equivalent of a steam table or Mollier chart for R-12. The question was "where to start"... What's that R-12 Mollier-style chart called?
Calc work-water pumping question...?
Work is force x distance, which is conveniently pounds x feet. Since we are only interested in the height here (as that is the direction we are pumping), we only need to integrate along that path. The differential weight/force being pumped is the volume (LxWxH = 2 x 3 x dH) times the density (80) and the distance is the height that volume is pumped (6 - H). So W = integral( 6 * 80 * dH * (6-H)) from H=0 to 6 W = integral ((2880 - 480H) dH) = 2880H - 240*H^2 = 2880(6) - 240*(6)^2 = 8640 foot-pounds of work. If you are pumping from the spout, replace the 6-H with 11-H, and the answer becomes 23,040 foot-pounds, which makes sense since you have almost tripled the averge height you were lifting. For the 2/3, change the range of the integration to H = 2 - 6, since H = 0 - 2 represents the water at the bottom of the tank that will still be there later. This gives you answers of 3840 and 13,440 foot-pounds respectively, showing that the last couple of feet are the hardest to pump out (because they have the farthest to go...)
If two tanks of area in ratio 1:2 are filled with water to the same height, what will be the pressure at the base of both tank?
The pressure at the bottom of both tanks would be the same.First reaction of most people would be to assume that the tank with larger area is holding more water, meaning more weight and more force, hence more pressure. But this not the case. They have just encountered the ‘Hydrostatic Paradox’.The pressure at a depth inside a fluid, is given by -P = [math]\rho[/math] * g * hGiven same fluid is filled in both, the only parameter which would influence the pressure at the bottom is the depth of the liquid column.In your question you have only mentioned tanks, which we can assume is cylindrical. This result holds good even for unevenly sized tanks.The paradox arises from the fact that people assume that if there is more fluid (meaning more weight), it will result in greater forces, hence greater pressures.Consider this setup, in flask B, there is more fluid, but same base area. So the assumption ( which is wrong) is that pressure at point B would be more.The paradox can be explained by considering how the forces act. he pressure at a point in a static liquid is due entirely to the weight of liquid (plus the atmosphere) directly above it. The liquid exerts a pressure at right angles to the wall. What is important to consider, is that the wall also exerts a force on the liquid, the vertical component of which nullifies the weight.Irrespective of the size/shape of the vessel, the same thing holds true. Only the vertical column over the base contributes to pressure, which is solely dependant on the height of the column.
How do you know your septic system is failing?
Hello - Well, with normal operation here's a simple breakdown. You should have 2 septic tanks and a leach field. The first septic tank is where the solids (poopoo) is caught. The liquids from that should properly run into the second septic tank. Then the second septic tank has a level where the liquids run into the leach field and soak into the ground. If you don't get your septic tank pumped out once every 2 years, the solids could spill into the liquid tank and then spill into the leach field which could possibly clog the tubes that connect the two and leach field itself. Then that's where the back ups begin eg: tubs, showers, toilets then could overflow into your floors. Of course that is the worse case scenario. Now septic tanks do operate at full capacity. The tanks have baffles and transfer tubes located at certain levels (depending on the tank). The solids are what you are concerned with getting pumped out. If you find the 1st tank is filled from top to bottom of the tank with solids, call a pumping company quick. You can tell this by using a 6-7 foot stick or piece of wood and stick in the tank. If it feels like pushing through mud then call a pumping company. If it feels watery, you should be fine. Again - if you haven't pumped it in 2 years, or if you don't know when it was pumped, call a pumping company. I hope this helps out.
I have this thermodynamics questions and i'm stuck! and help?
The initial state of the system is: 1) T1=80C x=0.6 v1=2.043kJ/kg u1=1622kJ/kg where v1 and u1 were obtained from steam tables. The final state is 2) x=1 v2=v1=2.043kJ/kg T2=94.1C u2=2498kJ/kg where T2 and u2 were obtained from steam tables. Using the first law of thermodynamics we have Δu = q - w Δu = 2498 - 1622 = 876 kJ/kg where w=0. This the total heat input is Q = m q = 2*(876) = 1752 kJ
Thermodynamics help. Below I have a question from my homework I do not get?
Two tanks are separated by a partition. Initially tank A contains 2-kg steam at 1 Mpa and 300 C while tnak B contains 3-kg saturated liquid vapor mixture at 150 C with a vapor mass fraction of 5- percent. Now the partition is removed and the two sides are allowed to mix until the mechanical and thermal equilibrium are established. If the pressure at the final state is 300 kPa determine the teperature and quality of the steam at the final state and the amount of heat lost from the tanks
Java code for Volume of vertical cylindrical tank, with variable liquid level.?
import java.text.DecimalFormat; public class MyClass { static final double radius = 18.8; static final DecimalFormat VFMT = new DecimalFormat("0.###"); public static void main(String[] args) { for (int feet=0; feet<=40; ++feet) { for (int inches=0; inches<12; inches+=2) { double depth = ((double) feet) + (1.0 / 12.0) * ((double) inches); double volume = Math.PI * radius * radius * depth; System.out.println("" + feet + "ft, " + inches + "in: vol = " + VFMT.format(volume) + " ft^3"); if (feet == 40) inches = 12; // Skip to end after 40'0" } } } } ===================== Output: 0ft, 0in: vol = 0 ft^3 0ft, 2in: vol = 185.061 ft^3 0ft, 4in: vol = 370.122 ft^3 0ft, 6in: vol = 555.182 ft^3 0ft, 8in: vol = 740.243 ft^3 0ft, 10in: vol = 925.304 ft^3 1ft, 0in: vol = 1110.365 ft^3 1ft, 2in: vol = 1295.425 ft^3 1ft, 4in: vol = 1480.486 ft^3 [...] 39ft, 10in: vol = 44229.52 ft^3 40ft, 0in: vol = 44414.58 ft^3
Help with Thermodynamic (ChemE) question-- Heat capacity?
A steel casting weighing 2 kg has an initial temperature of 500 degrees C; 40 kg of water initially at 25 degrees c is contained in a perfectly insulated steel tank weighing 5 kg. the casting is immersed in the water and the system is allowed to come to equilibrium. What is its final temperature? Ignore effects of expansion or contraction, and assume constant specific heats of 4.18 kJ/(kg K) for water and .50 kJ/(kg K) for steel. I'm stuck because it seems like I need some other measurement--maybe for Q? I think work=0 because it's an insulated system but that doesn't get me very far. And how do I use to specific heats at a time?
Two large tanks connected, determine the mass of salt in each tank?
Two large tanks, each holding 50L of liquid, are interconnected by a pipe with liquid flowing from tank A into tank B at a rate of 5 L/min. The liquid in each tank is kept well stirred. Pure water flows into tank A at a rate of 5L/min. The solution flows out of the system from tank B at 5L/min. If, initially, tank A contains 50kg if salt and tank B contains 100kg, determine the mass of salt in each tank at time t>=0. Here is my complete work: let dQ/dt=the rate of change of salt in the tank at t>=0 and thus dQ/dt=rate in-rate out For Tank A, rate in=(o kg/L)(5L/min) =0 since salt is not entering the tank rate out=(QA(t) kg/50 L) x (5L/min) = QA(t)/10 kg/min dQA(t)/dt=-QA(t)/10kg/min ?????????? For Tank B, rate in=(QA(t)/50L)(5L/min) = QA(t) kg/ 10 min rate out=(QB(t)/50)(5L/min)=QB(t)/10 kg/min So, dQB(t)/dt=QA(t)/10-QB(t)/10 ????????? where QA(0)=50 and QB(0)=100 Your help is appreciated and I hope it's correct!
My septic tank does not drain into the drain field?
Your tank is fine. The system should break down the solids in the first tank faster than you put them in, this is done naturally by bacteria. Then just liquid passes to the second tank and flows out the "leach lines", they are rock channels that clean the gray water before it seeps back into the earth. If you use anti bacterial soaps they kill the natural bacteria in the system, causing the solid to pass into the second tank and allowing the soap scum and other solids to head out into the "leach lines". This plugs the leach lines so they cannot drain into the earth. Your leach lines are not draining so you pump the tank but it only functions until the system fills with water again. The lines will have to be dug up and the rock replaced or a new line dug and filled with drain rock next to the old ones and then tied in. If you keep all anti bacteria soap out of your house and use a bacteria boosting product once every couple of months this will never happen. Unfortunately, once the leach lines are impacted no product will clear them of the soap and solid particles that are plugging them.