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nalin [4]
4 years ago
6

A carbon resistor has a resistance of 976 ohms at 0 degrees C. Determine its resistance at 89 degrees C​

Engineering
1 answer:
nignag [31]4 years ago
5 0

Answer:

1028.1184 Ohms

Explanation:

<u>Given the following data;</u>

  • Initial resistance, Ro = 976 Ohms
  • Initial temperature, T1 = 0°C
  • Final temperature, T2 = 89°C

Assuming the temperature coefficient of resistance for carbon at 0°C is equal to 0.0006 per degree Celsius.

To find determine its new resistance, we would use the mathematical expression for linear resistivity;

R_{89} = R_{0} + R_{0}(\alpha T)

Substituting into the equation, we have;

R_{89} = 976 + 976*(0.0006*89)

R_{89} = 976 + 976*(0.0534)

R_{89} = 976 + 52.1184

R_{89} = 1028.1184 \ Ohms

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How do I do this?<br> Blueprints, complete the missing view.
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Which word from the passage best explains what the web in the passage symbolizes
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Blank is common during exercise.
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7 0
2 years ago
An ideal Diesel cycle has a maximum cycle temperature of 2000°C. The state of the air at the beginning of the compression is P1
IrinaVladis [17]

Answer:

Power produced = 90.47 KW

Explanation:

We are given;

R = 0.287 kJ/kg·K

T1 = 15°C = 15 + 273 = 288 K

P1 = 95 KPa

Number of cylinders;N_cyl = 8

Bore;B = 10cm = 0.1m

Stroke;S = 12cm = 0.12m

cp = 1.005 kJ/kg·K

cv = 0.718 kJ/kg·K

k = 1.4

First of all let's find the initial specific volume;

α1 = RT1/P1

α1 = 0.287 * 288/95

α1 = 0.87 m³/kg

Now, let's find the total mass of air from the formula;

m =(V•N_cyl)/α1 =(B²•N_cyl•S•π)/4α1

So, m = (B²•N_cyl•S•π)/4α1

m = (0.1²•8•0.12•π)/(4*0.87)

m = 0.00867 Kg

Now, let's calculate the total mass flow rate;

m' = (m*N_rev)/n'

Where;

N_rev is number of revolutions given as 1500 rpm = 1500/60 rev/s = 25 rev/s

n' is the repetitions per circle = 2.

Thus;

m' = (0.00867*25)/2

m' = 0.108375 kg/s

The temperature at state 2 is gotten from the formula;

T2 = T1*r^(k - 1)

Where r is compression ratio.

We know that formula for compression ratio is;

the ratio of the maximum to minimum volume in the cylinder of an internal combustion engine.

In the question, we are told that minimum volume enclosed in the cylinder is 5 percent of the maximum cylinder volume.

Thus,

r = 100/5 = 20

So, T2 = 288*20^(1.4 - 1)

T2 = 954.563 K

The cut off ratio is gotten from the formula;

r_c = α3/α2 = T3/T2

T3 = 2000°C = 2000 + 273K = 2273K

Thus; r_c = 2273/954.563

r_c = 2.38

The heat input is gotten from the formula;

q_in = cp(T3 - T2)

q_in = 1.005(2273 - 954.563)

q_in = 1325.03 KJ/Kg

The efficiency is gotten from;

η = 1 - [1/(r^(k - 1)]*[((r_c)^(k) - 1)/(k(r_c - 1))]

Thus;

η = 1 - [1/(20^(1.4 - 1)]*[((2.38)^(1.4) - 1)/(1.4(2.38 - 1))]

η = 0.63

Now, the power output is gotten from the equation;

W' = m'•η•q_in

W' = 0.108375*0.63*1325.03

W' = 90.47 KW

7 0
3 years ago
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