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NemiM [27]
3 years ago
6

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

ease TH while keeping To constant, or decrease To while keeping TH constant? Are there any natural limits on the increase in thermal efficiency that might be achieved by such means? Consider how η responds to a change in temperature-if either temperature is changed by some r, which will have a greater effect? Hint: Check the derivative of η with regard to TH and Tc separately
Engineering
1 answer:
alukav5142 [94]3 years ago
7 0

<u></u>\ T_{c} has greater effect.

<u>Explanation</u>:

\eta_{\max }=1-\frac{T_{c}}{T_{A}}

T_{c}\\ = Temperature of cold reservoir

T_{H} = Temperature of hot reservoir

when T_{c} is decreased by 't',

$\eta_{\text {incre }}$ = 1-\frac{\left(\tau_{c}-t\right)}{T_{H}}

=n \ + \frac{t}{T_{n}}      -(i)

when {T_{H}} is increased by 'T'

\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)

\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}

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Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E 5 73 GPa and an ultimate strength of 140
NNADVOKAT [17]

Answer:

81.76 N/mm² ( MPa), 1.71233

Explanation:

Modulus of elasticity = stress / strain

stress = modulus of elastic × strain

strain = ΔL / L = 250.28 mm - 250 mm / 250 mm = 0.00112

Modulus of elasticity E = 73 GPa = 73 × 10³ MPa where 1 MPa = 1 N/mm²

E = 73 × 10³N/mm²

stress =  73 × 10³N/mm²× 0.00112 = 81.76 N/mm² ( MPa)

b) Factor of safety = maximum allowable stress / induced stress = 140 MPa / 81.76 MPa = 1.71233

8 0
4 years ago
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to su
Andrews [41]

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

6 0
4 years ago
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Answer:

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How should the system administrator resolve the issue?
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4 years ago
g 940 The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at o
weqwewe [10]

Answer:

μb = 0.096

μc  = 0.073

Explanation:

member AB:

-800( 4/3 ) + Nb (2) = 0

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Post BC:

summation of force along the y axis=0

Nc + Nb + 150(3/5 ) -50(9.81)=0

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Also (-4/5)(150)(3) + Fb(0.7)= 0

Fb = (4/5)(150)(3)/0.7 = 51.429 N

Likewise alog the x axis,

4/5(150) - Fc -Fb = 0

4/5(150) - Fc -51.429 = 0

Fc = 4/5(150)  -51.429 =68.571 N

μb = Fb/Nb = 51.429/533.3  = 0.096

μc = Fc/Nc = 68.571 / 933.83 = 0.073

7 0
4 years ago
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