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Rufina [12.5K]
2 years ago
6

Q1. Basic calculation of the First law (2’) (a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of hea

t are given off by the spring during this compression. What is the change in internal energy of the spring during the process? (b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?
Engineering
1 answer:
Bogdan [553]2 years ago
4 0

Answer:

(a) ΔU = 125 kJ

(b) ΔU = -110 kJ

Explanation:

<em>(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?</em>

<em />

The work is done to the system so w = 150 kJ.

The heat is released by the system so q = -25 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -25 kJ + 150 kJ = 125 kJ

<em>(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?</em>

<em />

The work is done by the system so w = -100 kJ.

The heat is released by the system so q = -10 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -10 kJ - 100 kJ = -110 kJ

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Explanation:

Step1

In the stress-strain curve of any material, the yield stress is the maximum stress at which material starts yielding.

Step2

Young’s modulus is the constant of proportionality of stress and strain according to hooks law. It is the slope of the slope of the stress-strain curve of the any material under proportional limit.

Step3

Ultimate tensile stress is the maximum stress that induced in the material under application of load.

Step4

Toughness is the strain energy per unit volume up to the fracture point of the stress-strain diagram of any material. This is the area under the curve of stress-strain.

Step5

Point of necking is the point where any material starts necking under application of load in necking region of the stress-strain curve.

Step6

Fracture point is the last point of the stress-strain curve where component fractures under application of load.

All the parameters are shown in below stress-strain curve:

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3 years ago
Discuss three objectives of Tariff and elaborate on three characteristics of it
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Answer:

Three objectives of a tariff are

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2) To protect domestic industries

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Three characteristics of a tariff are;

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3) Fairness

Explanation:

A tariff is an import or export tax placed on goods traded between countries, it serves to control the foreign trade between the two countries and to protect or develop local industry

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3 years ago
Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m
lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

Q = A \times V

where A is Area

A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2

Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s

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3 years ago
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
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Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

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Answer:

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Explanation:

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