At time the valve on an initially empty (perfect vacuum,,) tank is opened and air rushes in. If the tank has a volume of v and the density of air within. the tank increases as, where and are constants, determine the time rate of change of mass within the tank. t = 0VAt time the valve on an initially empty (perfectvacuum,) tank is opened and air rushes in. If thetank has a volume of ✓ and the density of air withinthe tank increases as p= c(1-e), where and areconstants, determine the time rate of change of masswithin the tank.CbdMdt

Given density of air inside the tank as a function of time, ρ=c(1−e−bt)The volume of the tank is…

t = 0 V At time the valve on an initially empty (perfect vacuum,) tank is opened and air rushes in. If the tank has a volume of ✓ and the density of air within the tank increases as p= c(1-e), where and are constants, determine the time rate of change of mass within the tank. C b dM dt

t = 0 V At time the valve on an initially empty (perfect vacuum,) tank is opened and air rushes in. If the tank has a volume of ✓ and the density of air within the tank increases as p= c(1-e), where and are constants, determine the time rate of change of mass within the tank. C b dM dt

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Fluid
A U-tube of uniform cross-section has both arms open to the atmosphere. The U-tube contains three immiscible liquids, with density PA, PB, and pc, as depicted in the diagram below, where po is the atmospheric pressure. We denote h to be the depth of the open surface of liquid A on the left arm, relative to the open surface of liquid Con the right arm. I is the depth of the interface between liquid B and C on the right arm and 2l is the depth of the interface between liquid A and liquid B on the right arm. Po PA (c) Po PC PB е е (a) Explain why the liquid densities need to satisfy the condition PA PB PC. (b) Write down an expression for the pressure at depth z below the surface of liquid A in the left-hand part of the tube. Show that l < h < 2l if PB + PC < PA•
Use an steel tape of length 20 m and a cross-section area of 5 mm 2 and weighing 1.5 kg in , Use the tape by placing it suspended under a tension of 5 kg to measure a distance of 506 m at a temperature of 40 degrees Celsius. the coefficient of linear expansion (a) 0.000018 meters / degree Celsius and it has been placed at a temperature of 20 degrees Celsius under a standard tension of 2.5 kg. Calculate the real distance using the elastic modulus of steel 106 kg / cm?. Q4/
Problem A gas weighs 16 N/m³ at 25°C and at an absolute pressure of 0.25 N/mm². Deter- mine the gas constant and density of the gas.
The initial pressures at A and B are the same at 105 kPa. If the pressure at the liquid at A (SG1 = 1.96) is increased to 132 kPa while keeping the pressure at B constant, determine the new position (Δh, in mm.) of the mercury meniscus inside the tube as shown. Assume no change in the liquid densities. The connecting tube is a 1-cm in diameter, Y1 = 0.58 cm., and the inclination of the tube is 11° with respect to the horizontal. Assume unit weight of air to be negligible. Round off to three decimal places.
Which of the following statements are TRUE, when the cavitation parameter o = 0? 1. The local pressure is reduced to vapor pressure. 2. Cavitation starts 3. Boiling of liquid starts 4. Cavitation stops
answer quickly
A capillary tube of diameter 2 mm is placed in a reservoir of water, and a similar capillary tube is placed in a reservoir of methanol. If both tubes are made of clean glass and the water-glass and methanol-glass contact angles are both approximately equal to 0°. Surface tension of water is 0.073 N/m. Surface tension and density of methanol are 0.0225 N/m and 791 kg/m3, respectively. 8.0 Determine the rise of methanol. a.) 7.4 mm b.) 5.8 mm c.) 14.9 mm d.) 2.9 mm e.) 9.0 Determine the rise of water. a.) 7.4 mm b.) 5.8 mm c.) 14.9 mm d.) 2.9 mm e.).
Give me both right solution according to the question
For fluid undergoing circular motion, the centripetal force is provided by a pressure difference. Consider a fluid element (density p) at radial distance r, which has thickness dr in the radial direction and surface area dA normal to the radial direction. The pressure difference gives rise to a force on the fluid element equal to (p(r + dr) – p(r))dA. (i) Equating the pressure difference to the centripetal force, find an expression for the radial pressure gradient in terms of p, r, and fluid velocity v. (ii) A fluid is undergoing body rotation in the r-0 plane (cylindrical polar co-ordinates). By considering the gravitational pressure head, show that the shape of a free surface z(r) is parabolic. (iii) Find and sketch a similar expression for the free surface of a fluid vortex produced by a rotating cylindrical spindle (i.e. a solid cylindrical rod) of radius Ro.