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3 years ago

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Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl

ied energy, assuming an environment temperature of 25°C.

1 answer:

ANEK [815]3 years ago

5 0

Answer:

exergy = 33.39 kW

Explanation:

given data

thermal energy reservoir T2 = 1500 K

heat at a rate = 150,000 kJ/h = $\frac{15000}{3600}$ kW = 41.67 kW

environment temperature T1 = 25°C = 298 K

solution

we get here maximum efficiency that is reversible efficiency is express as

reversible efficiency = 1 - $\frac{T1}{T2}$ ...............1

reversible efficiency = 1 - $\frac{298}{1500}$

reversible efficiency = 0.80133

and

the exergy of this supplied energy that is

exergy = efficiency × hat supply ................2

exergy = 0.80133 × 41.67 kW

exergy = 33.39 kW

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A series circuit has 4 identical lamps. The potential difference of the energy source is 60V. The total resistance of the lamps

Alexxx [7]

Answer:

$I=3A$

Explanation:

From the question we are told that:

Number of lamps $N=4$

Potential difference $V=60v$

Total Resistance of the lamp is $R= 20ohms$

Generally the equation for Current I is mathematically given by

$I=\frac{V}{R}$

$I=\frac{60}{20}$

$I=3A$

8 0

3 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is

Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α = $\frac{1}{\pi} (\frac{K}{\sigma Y})^2$ ....................1

put here value we get

α = $\frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2$

α = 15.043 mm

so minimum length of a surface crack is 15.043 mm

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3 years ago

Multiple Choice

12345 [234]

Answer:https://global.oup.com/us/companion.websites/9780199385423/student/ch6/mcq/ just go here

Explanation:

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3 years ago

PLEASE HELP QUICK!!

ivolga24 [154]

R01= 14.1 Ω

R02= 0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω

Read more about resistance here:

brainly.com/question/17563681

#SPJ1

5 0

1 year ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago