Answer:
COP of the heat pump is 3.013
OP of the cycle is 1.124
Explanation:
W = Q₂ - Q₁
Given
a)
Q₂ = Q₁ + W
= 15 + 7.45
= 22.45 kw
COP = Q₂ / W = 22.45 / 7.45 = 3.013
b)
Q₂ = 15 x 1.055 = 15.825 kw
therefore,
Q₁ = Q₂ - W
Q₁ = 15.825 - 7.45 = 8.375
∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Answer:
Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:
Aug=[A eye(size(A))]
Above command is tested in MATLAB command window and is attached in figure below
Answer:
(i) 169.68 volt
(ii) 16.90 volt
(iii) 16.90 volt
(iv) 108.07 volt
(v) 2.161 A
Explanation:
Turn ratio is given as 10:1
We have given that input voltage 
(i) We know that peak voltage is give by 
(ii) We know that for transformer 
So 

So peak voltage in secondary will be 16.90 volt
(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary
So peak voltage of the rectifier will be 16.90 volt
(iv) Dc voltage of the rectifier is given by 
(v) Now dc current is given by 