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2 years ago

If steam enters a turbine at 600K and is exhausted at 400K, calculate the efficiency of the engine.

2 answers:

8 0

The formular for efficiency = change in temprature/ initial temprature
change in temprature = 600 - 400 = 200
efficiency = 200/600 = 0.333 = 33.3%'

ikadub [295]2 years ago

3 0

The correct answer to the question is 33.3 % i.e the efficiency of the engine is 33.3 %.

CALCULATION:

The steam enters a turbine at 600 K.

Hence, the temperature of the source T = 600 K.

The steam is exhausted at a temperature 400 K.

Hence, the temperature of the sink T' = 400 K

We are asked to calculate the efficiency of the engine.

The efficiency of a heat engine whose source is at temperature T K and sink at T' is calculated as -

efficiency $\eta\ =\ 1-\frac{T'}{T}$

$=\ 1-\frac{400\ K}{600\ K}$

$=\ 1-\frac{2}{3}$

$=\ \frac{1}{3}$

$=\ 33.3\%$ [ans]

Hence, the efficiency of the engine is 33.3 %.

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2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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2 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

Greeley [361]

Answer:

I = 215.76 A

Explanation:

The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.

Fm = u_o*I^2*L/2*π*d

where I is the current in each rod, d = 0.0082 m is the distance 27rId

between each, L = 0.85 m is the length of each rod.

Fm = 4π*10^-7*I^2*1.1/2*π*0.0083

The mass of each rod is m = 0.0276 kg

F_m = mg

4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8

I = 215.76 A

note:

mathematical calculation maybe wrong or having little bit error but the method is perfectly fine

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2 years ago

A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. T

AleksandrR [38]

Answer:

$v=8.2158m/s$

Explanation:

(a) Free-body diagram attached.

(b) The stone attached with the string experiences both centripetal (towards the center) and centrifugal (away from the center) forces. The tension of the string counters the centrifugal force until it breaks.

We know that,

Centrifugal force = $\frac{mv^2}{r}$