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Mariana [72]
3 years ago
11

g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70%

of the carnot effieciency. if 2kj of work are produced, how much heat is rejected to the low temperature reservior
Engineering
1 answer:
Bond [772]3 years ago
6 0

Answer:

<em>Heat rejected to cold body = 3.81 kJ</em>

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

<em>efficiency of the Carnot's engine = 1 - </em>\frac{Tc}{Th}<em> </em>

eff. of the Carnot's engine = 1 - \frac{400}{600}

eff = 1 - 0.25 = 0.75

<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>

work done by heat engine = 2 kJ

<em>eff. of heat engine is gotten as = W/Q</em>

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>

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Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa
DENIUS [597]

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

8 0
3 years ago
The error in the trapezoidal rule is proportional to h^2 True b. False
Lynna [10]

Answer:

b) False

Explanation:

in trapezoidal rule the error is proportional to h^3 and the order of accuracy is proportional to h^3 .

Trapezoidal rule is numerical integration method .Trapezoidal rule is used to find the area of curves.In trapezoidal rule we finds the approximate value of integration.But the real value of integration will not differ to much from the value which finds by using trapezoidal rule.

4 0
3 years ago
A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa.Paddle-wheel work is done on the system until the pressu
AleksandrR [38]

Answer:

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

<u>ΔS = 0.85244 KJ/k</u>

3 0
3 years ago
A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun
Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, \eta = 1*10^{9} psi -sec

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi

Based on Maxwell's equation, the strain is given by:

strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 =  Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

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Answer:

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8 0
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