Well as the energy is lost the amplitude decreases starting off at a maximum I guess, so there must be a time where the amount of energy lost means the amplitude falls below 2mm, i mean if I multiply both sides by 3.18 I get that 3.18 x energy lost per cycle = energy stored, at first I thought...
Homework Statement
A plate with a hole in the centre is loaded in tension It has a height of 86mm (W), a thickness of 12 mm, a hole diameter of 33mm and a yield stress of 346MPa. The tensile load is 117kN.
A) Determine the stress on the top edge of the hole
B)Determine the stress on the top...
I'm having to design a screw pump and I'm a bit confused about things, I've never really seen a screw pump before. I've decided upon a twin screw - screw pump and my first question is why do most examples have an outlet at a higher level than the input, surely that just leads to a pressure loss...
Nope completely forgot about that, thanks
Well that leaves me with
[(125*250*2*sin(pi/3) / 291.4 ] *pi/180 = 3.24 which is still off, i can't see where I've made a mistake, I've done problems after this one and been fine, i just can't seem to get this one.
My answers are
Partial b = [((2b-2cCosA)/2*(b^2+c^2-2bcCos(A))^1/2] *δb
= [((250 - 160)/2*((125^2 + 160^2 - 2*160*125*0.5)^0.5))] * 0.5
= (90/291.4) * 0.5 = +-0.154
partial c has the same denominator
[((2c - 2bcos(A)) / 291.4) ]*0.5
= 195/291.4 * 0.5 = +-0.335
Partial A has the same...
Homework Statement
Two Sides of a triangular plate are measured as 125mm and 160mm, each to the nearest millimetre. The included angle is quoted as 60+-1. Calculate the length of the remaining side and the maximum possible error in this result
My first question is to the nearest millimetre...
[SIZE="4"][FONT="System"]I'm current learning MatLab before i start my second year of engineering and i don't really understand a fundamental aspect of it. My problem is the use of the period "." and when it is deemed to be necessary.
I've looked up what this operator does and found...
Honestly my first instincts was to do this. Whats confused me in my teacher solution is that he's said the specific volume of the initial condition(where there is a dryness fraction) will be constant through the whole process as it goes from 8 bar and to 10 bar. But surely the specific volume...
So with linear interpolation :
d is the variable i want, ie pressure.
g is the variable I've been given vg
D1=8 bar
D2=9 bar
G1=0.2403
G2=0.2149
G=0.2367
8+(0.2367-0.2403/0.2149-0.2403)*(9-8) = 8.142
I don't see why the specific volume wouldn't change, surely when it...
Homework Statement
A rigid vessel of volume 2.5 m3 contains steam at a pressure of 6 bar and a dryness fraction of 0.75. If the contents of the vessel are heated, determine:
i) the pressure at which the steam becomes dry saturated,
ii) the temperature of the steam when the pressure reaches...
Thanks, well I'm using "Thermodynamic and Transport Properties of Fluids"
So i can use the entropy of the system i guess.
Am i right in saying for a isentropic process dS=0
Therefore;
@ 400c, 20 bar S=7.126
@ 16KPA Sf= 0.772 Sfg= 7.213
So 7.126 = 0.772 + x7.213
X= 0.88
I think I'm actually okay...
Hm didn't notice the large numbers, its just a past exam question so i guess its Made up. I just don't believes there's enough Information to find X, I am puzzled.
Homework Statement
The power output of an adiabatic steam turbine is 5 MW. If the device can be assumed
to operate as a steady flow device with isentropic expansion, determine the
following:
i) The dryness fraction, x, at the exit from the turbine;
ii) The work output per unit mass of steam...