The U-shaped tube shown on the right contains a nonvolatile liquid of density p. The left...

 

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The U-shaped tube shown on the right contains a nonvolatile liquid of density p. The left arm of the tube is sealed at the top, and a column of air is trapped within it. The right arm of the tube is open. This device can be used as a h Sealed end pressure gauge, as the heights H and h shown in the figure will vary when the open end is exposed to different pressures. To calibrate the device, you expose the open end to atmospheric pressure (P,) and record the height difference Open end between the two liquid interfaces (H,) and the height of the air column (h,). (a) (9 pts) Show that, if the temperature is the same as the calibration temperature, the gauge pressure P at the open end can be calculated from the height h of the air column as P, = [pg(2h-H,)+P.](h, -h)/h. (b) (5 pts) When the open end is exposed to a pressure of 1 bar, the column of air has a height of 10 cm, and the two liquid interfaces are separated by a height of 15 cm. If the pressure at the open end changes to 1.5 bar, what are the new values of both heights (in cm)? Assume that the liquid in the manometer is mercury and both readings are done at ambient temperature (25 °C). (c) (8 pts) Show that, if the gauge pressure at the open end is measured at a temperature T different but not too far from the calibration temperature T,, the gauge pressure calculated using the equation from part (a) should be corrected by adding the term (P, -H,pg)h, (T-T,)/hT,. The U-shaped tube shown on the right contains a nonvolatile liquid of density p. The left arm of the tube is sealed at the top, and a column of air is trapped within it. The right arm of the tube is open. This device can be used as a h Sealed end pressure gauge, as the heights H and h shown in the figure will vary when the open end is exposed to different pressures. To calibrate the device, you expose the open end to atmospheric pressure (P,) and record the height difference Open end between the two liquid interfaces (H,) and the height of the air column (h,). (a) (9 pts) Show that, if the temperature is the same as the calibration temperature, the gauge pressure P at the open end can be calculated from the height h of the air column as P, = [pg(2h-H,)+P.](h, -h)/h. (b) (5 pts) When the open end is exposed to a pressure of 1 bar, the column of air has a height of 10 cm, and the two liquid interfaces are separated by a height of 15 cm. If the pressure at the open end changes to 1.5 bar, what are the new values of both heights (in cm)? Assume that the liquid in the manometer is mercury and both readings are done at ambient temperature (25 °C). (c) (8 pts) Show that, if the gauge pressure at the open end is measured at a temperature T different but not too far from the calibration temperature T,, the gauge pressure calculated using the equation from part (a) should be corrected by adding the term (P, -H,pg)h, (T-T,)/hT,.

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Solution closed onie calirating height f air olumn is ho lamel height f ope y Liquiel 1 lquid column ... View the full answer