Fundamentals Of Engineering Thermodynamics
9th Edition
ISBN: 9781119391388
Author: MORAN, Michael J., SHAPIRO, Howard N., Boettner, Daisie D., Bailey, Margaret B.
Publisher: Wiley,
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 1, Problem 1.31P
To determine
Pressure of water at a given depth.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
the air, oil, and water system in Figure is at 20 ° C. Knowing that gauge A indicates an absolute pressure of 15 lbf / in2 and that gauge B indicates 1.25 lbf / in2 less than gauge C, calculate (a) the specific weight of the oil in lbf / ft3 and (b) the absolute pressure marked by gauge C in lbf / in2
Consider the manometer in Fig. If the specific weight of fluid B is 20 kN/m3 , what is the absolute pressure, in kPa, indicated by the manometer when the local atmospheric pressure is 720 mmHg?
Consider a column of a planet's atmosphere. The planet's atmosphere is a
compressible ideal gas at rest that obeys the polytropic relation
Po
%3D
3/2
Po
3/2
where pis pressure and pis density. Here, p, and P, are the values of pressure and density,
respectively, at the planet's surface. Take z (altitude) to be positive upward with z=0 at the
surface, take R to be the gas constant for the planet's atmosphere, and take g to be the
downward acceleration due to gravity.
a)
Starting from hydrostatic balance and the polytropic relation above, derive an
expression for the pressure field, p(z), in terms of the given parameters. Leave all
parameters except the polytropic index as algebraic.
b)
Derive an expression for the density field, p(z), in terms of the given parameters.
Leave all parameters except the polytropic index as algebraic.
c)
Derive an expression for the temperature field, T(z), in terms of the given
parameters. Leave all parameters except the polytropic index as algebraic.
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
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- The pressure at 9.85m depth in a fluid is 102,482Pa. Calculate the density of the fluid and present the answer in Pascal with accuracy up to one decimal place. Consider 1 atm = 101325 Pa. Assume the value of gravity to be 9.8 m/s2.arrow_forwardWhat pressure force (in kN) exists on an 80-cm-diameter horizontal area at a depth 100 feet below the surface of a lake? Assume the density of the water is constant at 1000 kg/m³, and atmospheric pressure is 1.00 atm.arrow_forwardA location is located about 4,600 ft above the sea level and is about 80 feet deep. Determine pressure at this depth in the SI units. Assume water to be at rest, the mercury density of 13.6 g/cm³, the water density of 1 g/cm³, local gravitational acceleration of 9.793 m/s?, atmospheric pressure here of 30.2 inHg. Compare with the atmospheric pressure of 12.45 psia at the elevation of 4500 ft.arrow_forward
- T F The specific weight of a fluid is the product of the fluid's density and the acceleration due to gravity. Stronger surface tension leads to higher capillary rise. Absolute pressures are frequently negative. If the pressure of fluid drops below the vapor pressure of that fluid at that temperature, the fluid will cavitate. F F T F F Density can be measured in lb;/ft° in the English system of units. For a hydrostatic incompressible fluid, pressure is independent of depth. A fluid with a high bulk modulus of elasticity is more difficult to compress than one with a low bulk modulus of elasticity. Viscosity is caused, in part, by the surface tension within a fluid. A fluid can resist an applied shear stress by deforming. Pressure increases faster with depth in less dense fluids than in more dense fluids. T F F F F Farrow_forwardA container, having 7 m3 volume, includes 800 kg of granite and air totally. Please determine air mass and overall specific volume. (ρgranite= 2500 kg/m3, ρair= 1.21 kg/m3)arrow_forwardAt a certain point on the wing of the aircraft, the pressure is 37,803 Pa and the density is 45% of the standard sea level values. Determine the temperature (Kelvin) at that point? Use standard metric units and round off answer to the nearest hundredths.arrow_forward
- A vertical composite fluid column whose upper end is open to the atmosphere is composed of 20 inches of mercury (S.G = 13.45), 30 inches of water (density = 62.4 lbm/ft3) and 35 inches of oil (S.G = 0.825). Determine the absolute pressure in psia: a) at the base of the column, b) at the oil-water interface, c) at the water-mercury interface.arrow_forwardDetermine the weight (in Ib;) of an 160-lbm person at sea level (z = 0), in Denver (z = 1610 m) and on top of Mount Everest (z = 8848 m). %3Darrow_forward3. (a) Prove from first principles that for a polytropic atmosphere, the relation between the pressure at two different heights may be modelled using the following relation: P2 - 21 - P1 where p - pressure T- absolute temperature z - altitude above mean sea level R - characteristic gas constant 1 - temperature lapse rate and suffixes 1 and 2 denote the two different altitudes.arrow_forward
- Liquid kerosene flows through a Venturi meter. The pressure of the kerosene in the pipe supports columns of kerosene that differ in height by 12 cm. Determine the difference in pressure between points a and b, in kPa. Does the pressure increase or decrease as the kerosene flows from point a to point b as the pipe diameter decreases? The atmospheric pressure is 101 kPa, Kerosene the specific volume of kerosene is 0.00122 m³/kg, and the acceleration of gravity is g = 9.81 m/s². v=0.00122 kg/m³ Pa 101 kPa g=9.81 m/s² L=12 cmarrow_forwardA tank contains oil (s=0.80), gasoline (s=0.90) and water (s=1). What is the pressure at a depth of 1.6, if the depths of the liquids are respectively, 0.50 m,0.60 m, and 0.80 m? Give your answer in kg/m2.arrow_forwardProve that 69.0 kPa, 7 m of water, 52 cm of mercury (SG = 13.55) are equivalent.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University Press
Mechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSON
Thermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEY
Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Engineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Physics 33 - Fluid Statics (1 of 10) Pressure in a Fluid; Author: Michel van Biezen;https://www.youtube.com/watch?v=mzjlAla3H1Q;License: Standard YouTube License, CC-BY