Carbon tetrachloride (s.g. = 1.6) is pumped at a rate of 2 gpm through a pipe th
ID: 850217 • Letter: C
Question
Carbon tetrachloride (s.g. = 1.6) is pumped at a rate of 2 gpm through a pipe that is inclined upward at an angle a of 30. An inclined tube manometer (with an angle of inclination q of 10) using mercury as the manometer gage fluid (s.g. = 13.6) is connected between two taps on the pipe that are 2 ft apart. The manometer reading L as shown in the figure is 16 in. Note that for a liquid, internal energy can be taken as CvT CpT.
(a) If no heat is lost though the tube wall, what is the temperature rise of the CCl4 over a 100 ft length of the tube?
(b) Why is the manometer made so that it is tilted at an angle instead of being vertical?
Carbon tetrachloride (s.g. = 1.6) is pumped at a rate of 2 gpm through a pipe that is inclined upward at an angle a of 30½. An inclined tube manometer (with an angle of inclination q of 10½) using mercury as the manometer gage fluid (s.g. = 13.6) is connected between two taps on the pipe that are 2 ft apart. The manometer reading L as shown in the figure is 16 in. Note that for a liquid, internal energy can be taken as CvT ½ CpT. (a) If no heat is lost though the tube wall, what is the temperature rise of the CCl4 over a 100 ft length of the tube? (b) Why is the manometer made so that it is tilted at an angle instead of being vertical?Explanation / Answer
Sorry, I answered as a comment to that other guys reply. Yeah. Do the steps I mentioned. Start with bernoullis. Get rid of work and velocity. Come up with an equation for Delta(P). This should be equal to -F which is equal to -Delta(U).
Use circuit approach in the manometer. Use the S.G.s given to find density. Heights can be find using SOHCAHTOA on angles and lengths given.
Once you have Delta(P) you inherently have Delta(U) which you know is equal to CpDelta(T). Look up Cp for CCL4 and solve for Delta(T). Your Delta(T) is going to be very small.