Recent content by Reefy

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    Modes of Vibration of 3-DOF Spring Mass System

    Something seems off about that because if I say A = 1, then $$A(1-ω^2) - B = 0 $$ $$→$$ $$B=(1-ω^2)$$ And $$C(1-ω^2) - B = 0$$ $$→$$ $$C = B/(1-ω^2) = (1-ω^2)/(1-ω^2) = 1 $$ But I still have the equation for $$B(2-ω^2) - A - C = B(2-ω^2) - 2 = 0$$ $$→$$ $$B=2/(2-ω^2)$$ which will give me...
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    Modes of Vibration of 3-DOF Spring Mass System

    Ok, if I do that then I get A = C = 1, and then B will equal 1, 0, and -2 for the three different ω
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    Modes of Vibration of 3-DOF Spring Mass System

    Is there another method that you know of? Can I set up a ratio of the amplitudes? My equations after assuming a solution of X1,2,3 = Asinωt, Bsinωt, and Csinωt are $$ A(1-ω^2) - B = 0$$ $$ B(2-ω^2) - A - C = 0 $$ $$ C(1-ω^2) - B = 0 $$
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    Modes of Vibration of 3-DOF Spring Mass System

    1. Homework Statement 2. Homework Equations a11 = a21 = a31 = a12 = a13 a22 = a32 = a23 $$ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} =ω^2m \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} &...
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    Heat Loss Through a Short Cylinder

    1. Homework Statement A short metal cylinder 145 mm in diameter and 145 mm high at 1045 K is suddenly exposed (all sides exposed) to a room air temperature at 300 K with h=25 W/m².K. Assume that for the metal k=40W/m.K, den=7800 kg/m3 and Cp=c=600 J/kg.K. Estimate (a) the time required for...
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    Thermodynamics Gas Mixture - Dew Point Temperature

    Am I supposed to use the humidity ratio to solve this? I thought the 0.1 kg/s was the mass flow rate of dry air ω = 0.622 x (Saturated Pressure)/ (Pressure mixture - Saturated Pressure)
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    Thermodynamics Gas Mixture - Dew Point Temperature

    The saturation pressure at that temperature is 9.14 kPa using linear interpolation with the saturation tables. Ok, so the rate given was for the mixture which is the same for all components. For some reason, I was thinking each component might have a different mass flow rate. Then the part I'm...
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    Thermodynamics Gas Mixture - Dew Point Temperature

    1. Homework Statement Only Number 1, not number 2 2. Homework Equations Dew Point Temperature T = Saturated Temperated at Vapor Pressure Partial Pressure = (mole fraction) x (Mixture Pressure) 3. The Attempt at a Solution The dew point temperature is only dependent on the pressure of...
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    Gas Mixture - Dew Point Temperature

    1. Homework Statement Only Number 1, not number 2 2. Homework Equations Dew Point Temperature T = Saturated Temperated at Vapor Pressure Partial Pressure = (mole fraction) x (Mixture Pressure) 3. The Attempt at a Solution The dew point temperature is only dependent on the pressure of...
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    DC Shunt Motor

    Thanks, I see the mistake I was making! It was pretty simple, I must be tired. Appreciate the help again
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    DC Shunt Motor

    T = PM/ω = EaIa/(2πN)
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    DC Shunt Motor

    Ooohh. Yeah, its the conceptual idea I don't get. I don't get why we have to use mechanical power instead of electrical
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    DC Shunt Motor

    That's what I'm unsure of. I know that Pinput - Poutput = the total losses (Copper, iron, friction). And Copper Loss is I2aRa + I2shRsh
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    DC Shunt Motor

    Ia1 = IL1 - Ish = 25A - 1A = 24A Ia2 = IL2 - Ish = 50A - 1A = 49A
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    DC Shunt Motor

    Pinput(1) = VtIL1 = (250 V)(25 A) = 6250 A at the first speed and it is Pinput(2) = VtIL2 = (250 V)(50 A) = 12500 A at the 2nd speed
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