Hydrostatic relation

An empirical formula relating pressure and density for seawater w/ temperature held constant is: p/pa = (k+1)(q/qa)^7 - k where p= pressure pa = pressure at the surface k = dimensionless constant q = density qa = density at the surface Using the formula in the hydrostatic relation, determine the pressure as a function of depth. The hydrostatic relation is: p+dgz=constant I am uncertain as to the proper approach for beginning this problem. I know that if I wanted to take the derivative of the empirical formula there must be a z term. I tried to insert P/RT in for q (q=P/RT) and then substitute for T with the relation T=... click for more

Hydrostatic Pressure - where did I go wrong?

Determine the force due to hydrostatic pressure acting on the hinge of the gate shown. The fluid density is d and the width of the elliptical gate (out of the page) is 6H. (I have attached the picture provided) This is the approach I took: F=dg(h*)A d = density g = gravitational constant h* = h bar, the height at which the force is applied A = area I got the answer to be: F=dg(2H)(3H)(H) F=dg6(pi)H^3 The answer is supposed to be: 21(pi)dg(H^3)/8 I know there are forces acting on the stop and hinge but didn't think those were forces due to hydrostatic pressure. I did try to find those but still didn't get the answer provided. Where am I going wrong?

Hydrostatics problem

Consider a square gate of side h that can rotate about a pivot as shown. Determine the depth of the gate's top, H, at which the gate will open if hp=(4/9)h. This is the approach I have taken: F=dgZA where d=density and Z is z bar F=dg(H+h/2)h^2 I found the zcp of center of pressure for the Force to be at Z+I/(ZA) = 2h/3 I then took moments around the top of the gate to find F at stop (Fstop) M=0=(dg(H+h/2)h^2)(h-(2h/3))-Fs(h) Fs=(dg(H+h/2)h)(h-(2h/3)) I then set the moments about the pivot of Fs and F equal to each other which would be when the gate opened. Setting dg(H+h/2)h^2)(h/3)(4h/9)=dg(H+h/2)h^2((2h/3)-(4h/9) I got H=0.5h. The answer is supposed to b... click for more

ENGR 30301 Fluid Mechanics Textbook: Fluid Mechanics, 9th edition

Air at 22C and at pressure of 120 kPa is flowing over a flat plate with a velocity of 4.5 m/s. If the plate is 50 cm wide and the surface in contact with the air is at 82C, calculate the following quantities at x = 42 cm: (a) the local friction coefficient; (b) the average friction coefficient; (c) the drag force; (d) the local convective heat transfer coefficient; (e) the average convective heat transfer coefficient; (f) the rate of heat transfer.

Mass-flow rate

Consider a flow for which the velocity vector is: u=(U/L)(xi-2yj+zk) Compute the mass-flow rate across a plane of area A that is perpendicular to the x axis and lies at x=3L. The fluid density is d. I know that the mass-flow rate, M, is calculated by: M=Double integration [d(u*n)]dA I am having trouble setting up the integral though. I think that dA becomes dy dz but I'm not for sure what the limits will be for those.