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lisov135 [29]
3 years ago
11

A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A. It is placed in a magnetic field of 0.90 T with

the plane of the coil making an angle of 30° with the magnetic field. What is the torque on the coil?
Physics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

3.08 Nm

Explanation:

N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree

The angle made with the normal of the coil, theta = 90 - 30 = 60 degree

Torque = N I A B Sin Theta

Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60

Torque = 3.08 Nm

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Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s
Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) \frac{E_{d} }{m} = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature (T_{1})of 800 ºR and pressure (P_{1}) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{1}) = 191.81 BTu/lb

Specific entropy (s_{1}) = 0.6956 Btu/(lb.ºR)

At the temperature (T_{2})of 600 ºR and pressure (P_{2}) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{2}) = 143.47 BTu/lb

Specific entropy (s_{2}) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}

Q/m = heat transfer = -2 Btu/lb

V_{1} = 400 ft/s

V_{2} = 100 ft/s

\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}[/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]

The equation above is further simplified to:

\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]

\frac{E_{d} }{m} = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

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