Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
T₁ = 300 K
= 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + 
(T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
= 1400×
= 1007.6 K
T₅ = T₄ + ( - T₄)/
= 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + (T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
= 1400× = 1007.6 K
T₇ = T₆ + ( - T₆)/ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. 
= cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. 
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Answer:
(a) maximum positive reaction at A = 64.0 k
(b) maximum positive shear at A = 32.0 k
(c) maximum negative moment at C = -540 k·ft
Explanation:
Given;
dead load Gk = 400 lb/ft
live load Qk = 2 k/ft
concentrated live load Pk =8 k
(a) from the influence line for vertical reaction at A, the maximum positive reaction is
= 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k
See attachment for the calculations of (b) & (c) including the influence line
The number of receptacles that are needed for all of these kitchen areas are: C. Four.
<h3>What are
receptacles?</h3>
Receptacles can be defined as types of sockets or series of outlets (openings) that provides a path where current can be taken in a wiring system, so as to run electrical appliances in buildings.
Based on the information provided, the number of receptacles that are needed for all of these kitchen areas are four because one would be used in each area.
Read more on receptacles here: brainly.com/question/23839796
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Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²
x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K
K= 58.66 lb/ft
The damper coefficient for critically damped system
Cc= 12.7 lb.sec/ft
