Answer:
1028.1184 Ohms
Explanation:
<u>Given the following data;</u>
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;
Substituting into the equation, we have;
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.
Data Given:
x = 27 , 44 , 32 , 47, 23 , 40, 34, 52
y = 30, 19, 24, 13 , 29, 19, 21, 14
It is given that,
∑x = 299
∑y = 167
∑ = 11887
∑ = 3773
We are asked to verify the above values manually in this question.
So,
1. ∑x = 299
Let's verify it:
∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52
∑x = 299
Yes, it is equal to the given value. Hence, verified.
2. ∑y = 167
Let's verify it:
∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14
∑y = 169
No, it is not equal to the given value.
3. ∑ = 11887
Let's verify it:
For this to find, first we need to square all the value of x individually and then add them together to verify.
∑ =
+
+
+
+
+
+
+
∑ = 11,887
Yes, it is equal to the given value. Hence, verified.
4. ∑ = 3773
Let's verify it:
Again, for this we need to find the squares of all the y values and then add them together to verify it.
∑ =
+
+
+
+
+
+
+
∑ = 3,845
No, it is not equal to the given value.
Answer:
A) W' = 15680 KW
B) W' = 17113.87 KW
Explanation:
We are given;
Temperature at state 1; T1 = 290 K
Temperature at state 3; T3 = 1100 K
Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw
Pressure of air into compressor; P_c = 95 kPa
Pressure of air into turbine; P_t = 760 kPa
A) The power assuming constant specific heats at room temperature is gotten from;
W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in
Now, we don't have T4 and T2 but they can be gotten from;
T4 = [T3 × (r_p)^((1 - k)/k)]
T2 = [T1 × (r_p)^((k - 1)/k)]
r_p = P_t/P_c
r_p = 760/95
r_p = 8
Also,k which is specific heat capacity of air has a constant value of 1.4
Thus;
Plugging in the relevant values, we have;
T4 = [(1100 × (8^((1 - 1.4)/1.4)]
T4 = 607.25 K
T2 = [290 × (8^((1.4 - 1)/1.4)]
T2 = 525.32 K
Thus;
W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000
W' = 0.448 × 35000
W' = 15680 KW
B) The power accounting for the variation of specific heats with temperature is given by;
W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in
From the table attached, we have the following;
At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K
At T3 = 1100 K, h3 = 1161.07 KJ/K
At T1 = 290 K, h1 = 290.16 KJ/K
At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K
Thus;
W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000
W' = 17113.87 KW
