Fundamentals Of Engineering Thermodynamics
9th Edition
ISBN: 9781119391388
Author: MORAN, Michael J., SHAPIRO, Howard N., Boettner, Daisie D., Bailey, Margaret B.
Publisher: Wiley,
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Chapter 1, Problem 1.40P
To determine
The pressure at oil water interface and at bottom of the tank.
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3. The reading of an automobile gage is proportional to the gage pressure at the bottom of the
tank as shown. The tank is 32 cm deep. Unit weight of gasoline is 6670 N/m³ and that of air is
11.8 N/m³. Unit weight of water is 9790 N/m³.
Vent
Air
Gasoline
Water
a. Determine the gage reading when the tank is full of gasoline.
b. How many cm of air remains at the top when the gage indicates full and the tank is
contaminated with 3 cm of water?
c. Determine the pressure at the interface of the gasoline and water when the gage indicates full.
Because of a break in a buried oil storage tank, ground water has leaked into the tank to the depth L' = 4 ft. The densities of the water
and oil are, respectively, 62 and 55, each in lb/ft³.
Standpipe open
to atmosphere
Palgage) = i
Determine the pressure at the oil-water interface and at the bottom of the tank, each in lbf/in². (gage). Assume L = 14 ft.
Let g = 32.2 ft/s².
Pb(gage) = i
Patm
Determine the pressure at the oil-water interface, in lbf/in². (gage).
lbf/in².
Oil
Water
lbf/in².
Determine the pressure at the bottom of the tank, in lbf/in². (gage).
Because of a break in a buried oil storage tank, ground water has leaked into the tank to the depth L'= 2 ft. The densities of the water
and oil are, respectively, 62 and 55, each in Ib/ft3.
Standpipe open
to atmosphere
Oil
L'
Water
Determine the pressure at the oil-water interface and at the bottom of the tank, each in Ibf/in². (gage). Assume L = 17 ft.
Let g = 32.2 ft/s?.
Determine the pressure at the oil-water interface, in Ibf/in?. (gage).
Palgage) =
Ibf/in?.
Chapter 1 Solutions
Fundamentals Of Engineering Thermodynamics
Ch. 1 - Prob. 1.2ECh. 1 - Prob. 1.3ECh. 1 - Prob. 1.4ECh. 1 - Prob. 1.5ECh. 1 - Prob. 1.6ECh. 1 - Prob. 1.7ECh. 1 - Prob. 1.8ECh. 1 - Prob. 1.9ECh. 1 - Prob. 1.10ECh. 1 - Prob. 1.11E
Ch. 1 - Prob. 1.12ECh. 1 - Prob. 1.13ECh. 1 - Prob. 1.14ECh. 1 - Prob. 1.1CUCh. 1 - Prob. 1.2CUCh. 1 - Prob. 1.3CUCh. 1 - Prob. 1.4CUCh. 1 - Prob. 1.5CUCh. 1 - Prob. 1.6CUCh. 1 - Prob. 1.7CUCh. 1 - Prob. 1.8CUCh. 1 - Prob. 1.9CUCh. 1 - Prob. 1.10CUCh. 1 - Prob. 1.11CUCh. 1 - Prob. 1.12CUCh. 1 - Prob. 1.13CUCh. 1 - Prob. 1.14CUCh. 1 - Prob. 1.15CUCh. 1 - Prob. 1.16CUCh. 1 - Prob. 1.17CUCh. 1 - Prob. 1.18CUCh. 1 - Prob. 1.19CUCh. 1 - Prob. 1.20CUCh. 1 - Prob. 1.21CUCh. 1 - Prob. 1.22CUCh. 1 - Prob. 1.23CUCh. 1 - Prob. 1.24CUCh. 1 - Prob. 1.25CUCh. 1 - Prob. 1.26CUCh. 1 - Prob. 1.27CUCh. 1 - Prob. 1.28CUCh. 1 - Prob. 1.29CUCh. 1 - Prob. 1.30CUCh. 1 - Prob. 1.31CUCh. 1 - Prob. 1.32CUCh. 1 - Prob. 1.33CUCh. 1 - Prob. 1.34CUCh. 1 - Prob. 1.35CUCh. 1 - Prob. 1.36CUCh. 1 - Prob. 1.37CUCh. 1 - Prob. 1.38CUCh. 1 - Prob. 1.39CUCh. 1 - Prob. 1.40CUCh. 1 - Prob. 1.41CUCh. 1 - Prob. 1.42CUCh. 1 - Prob. 1.43CUCh. 1 - Prob. 1.44CUCh. 1 - Prob. 1.45CUCh. 1 - Prob. 1.46CUCh. 1 - Prob. 1.47CUCh. 1 - Prob. 1.48CUCh. 1 - Prob. 1.49CUCh. 1 - Prob. 1.50CUCh. 1 - Prob. 1.51CUCh. 1 - Prob. 1.52CUCh. 1 - Prob. 1.53CUCh. 1 - Prob. 1.54CUCh. 1 - Prob. 1.55CUCh. 1 - Prob. 1.56CUCh. 1 - Prob. 1.57CUCh. 1 - Prob. 1.58CUCh. 1 - Prob. 1.4PCh. 1 - Prob. 1.5PCh. 1 - Prob. 1.6PCh. 1 - Prob. 1.7PCh. 1 - Prob. 1.8PCh. 1 - Prob. 1.9PCh. 1 - Prob. 1.10PCh. 1 - Prob. 1.11PCh. 1 - Prob. 1.12PCh. 1 - Prob. 1.13PCh. 1 - Prob. 1.14PCh. 1 - Prob. 1.16PCh. 1 - Prob. 1.17PCh. 1 - Prob. 1.18PCh. 1 - Prob. 1.19PCh. 1 - Prob. 1.20PCh. 1 - Prob. 1.21PCh. 1 - Prob. 1.22PCh. 1 - Prob. 1.23PCh. 1 - Prob. 1.24PCh. 1 - Prob. 1.25PCh. 1 - Prob. 1.26PCh. 1 - Prob. 1.27PCh. 1 - Prob. 1.28PCh. 1 - Prob. 1.29PCh. 1 - Prob. 1.30PCh. 1 - Prob. 1.31PCh. 1 - Prob. 1.32PCh. 1 - Prob. 1.33PCh. 1 - Prob. 1.34PCh. 1 - Prob. 1.35PCh. 1 - Prob. 1.36PCh. 1 - Prob. 1.37PCh. 1 - Prob. 1.38PCh. 1 - Prob. 1.39PCh. 1 - Prob. 1.40PCh. 1 - Prob. 1.41PCh. 1 - Prob. 1.42PCh. 1 - Prob. 1.43PCh. 1 - Prob. 1.44PCh. 1 - Prob. 1.45PCh. 1 - Prob. 1.46PCh. 1 - Prob. 1.47PCh. 1 - Prob. 1.48PCh. 1 - Prob. 1.49P
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- A tank is filled with a certain liquid to height of 1.25 m from its bottom. If the pressure gauge at the tank bottom reads 1.46 psi, what is the density of the liquid in lb/ft³. 83.2 Answer is not among the choices a. 51.3 b. 62.4 c. a. 3.42 b. 6.83 d. A hydraulic jack uses two pistons, 0.35 in diameter and 1.0 in diameter, respectively. In order for the 1 inch pistion to raise a load of 100 lb, how much force must be applied at the small piston in pounds? 8.57 12.2 C. d.arrow_forward1. Consider the following diagram. The fluid flowing in the pipe is oil (SG = 0.830). The fluid in the manometer is water (SG = 1.00). Water, pw = 1000 kg/m³ 7 cm Oil, p = 830 kg/m³ a. Determine P₂ - P₁ in kPa. b. Determine the direction of flow. (Left to right, or right to left). Explain why this is the case. c. The height of oil between the manometer and the pipe is unknown. Explain how you can still solve the problem without knowing this height. Would this height ever matter? Hint: consider practical issues of design. Could the height be 0 cm? Could it be 1000 cm?arrow_forwardUsing Boyle's Law Demonstrator, the absolute pressure and the volume of air are respectively 61.15 kPa and 0.5 L. When the oil height is decreased by 2 cm, the new absolute pressure becomes 50.95 kPa. So, the new air volume will be approximately equal to: Select one: a. 0.6 L b. 0.4 L c. It can't be found because there are missed given data. d. 0.2 Larrow_forward
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