Recent content by reddawg

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    Fluids: Energy equation involving head loss

    Yes, I agree with those units. And P is absolute pressure I think.
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    Fluids: Energy equation involving head loss

    ρ = (.5 psi)(144 in2/ft2) / (1716)(609.7) 1716 is R in British units 609.7 is 150 degrees f converted to rankine
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    Fluids: Energy equation involving head loss

    V is equal to the flowrate Q divided by the cross-sectional area of each pipe. From the ideal gas law, density ρ = p/RT = 6.88*10^-5 slugs/ft^3
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    Fluids: Energy equation involving head loss

    Homework Statement See attached image: Homework Equations p/ρg + V^2/2g + z = constant head loss (major) = f * l/D * V^2/2gThe Attempt at a Solution To use the energy equation while incorporating head loss, I need to determine the velocity in each section of pipe. The problem is I don't know...
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    (dp/dz) = ρ0gep/B How do I rearrange that to integrate from z=0 to h?
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    ρ = ρ0ep/B
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    I get: (1/B)*p = ln(ρ/ρ0) when factoring in the initial conditions.
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    So, solving for density using the Bulk Modulus equation: ρ = B*(Δρ/p)
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    That's where I have trouble. I always end up with just [density*gravity*depth]. Is it just (1/g)*(dp/dz) ?
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    That would be just [density*gravity]. So how do I apply that?
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    Integrating Bulk Modulus to Find Pressure at Variable Depth

    Homework Statement See image. Homework Equations pressure = density*gravity*depth The Attempt at a Solution The pressure at 5000 ft according to the book is 322,000 psf. This makes sense because density*gravity*depth = 2*32.2*5000 = 322,000 psf. How do I apply the bulk modulus equation to...
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    Solving Series RLC Circuit: Vrms Across Capacitor & Inductor is Zero

    Ok, I think I get it. Given a phaser representation, tan(φ) = (XL - XC) / R So tan(180) = 0 ⇒ XL = XC which is resonance.
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    Solving Series RLC Circuit: Vrms Across Capacitor & Inductor is Zero

    Homework Statement Given a series RLC circuit: R = 10 Ω L = 0.1 H C = 2 μF Power Source: V = 5sin(500t) If you could change the capacitance what C would you select so Vrms across the capacitor and inductor is zero? Homework Equations ω2 = 1/LC The Attempt at a Solution My instructor's...
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    Calculating Volume of a Double-Lobed Cam Using Polar Coordinates

    Wow, thanks for wasting your time for me! ;-) I guess the cam is pretty small so the volume is more reasonable than I thought.
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    Calculating Volume of a Double-Lobed Cam Using Polar Coordinates

    Ha ha ok. Thanks LCKurtz. When I evaluate the integral (using a calculator) I get 0.79993. This seems awfully low to be a volume of an shape like this. - I checked it twice for errors, I think it's accurate.