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8_murik_8 [283]
4 years ago
8

An inventor proposes an engine that operates between the 27 deg C warm surface layer of the ocean and a 10 deg C layer a few met

ers down. The claim is that this engine can produce 100 kW at a flow of 20 kg/s. Is this possible?
Engineering
1 answer:
SpyIntel [72]4 years ago
8 0

Answer:

Engine not possible

Explanation:

source temperature T1 = 300 K

sink temperature T2= 283 K

therefore, carnot efficiency of the heat engine

η= 1- T_2/T_1

\eta= 1-\frac{T_1}{T_2}

\eta= 1-\frac{283}{300}

= 0.0566

= 5.66%

claims of work produce W = 100 kW,  mass flow rate = 20 kg/s

Q=mc_p(T_1-T_2)

Q=20\times4.18(300-283)

= 1421.2 kW

now \eta= \frac{W}Q}

now \eta= \frac{100}{1421.2}

=7%

clearly, efficiency is greater than carnot efficiency hence the engine is not possible.

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Answer:

c. Both Technicians A and B

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3 years ago
Air is to be compressed so that its volume is reduced to half of its original volume if the initial pressure is 200 kPa and the
Alex_Xolod [135]

Answer:

P_{2} = 527.803\,kPa

Explanation:

The politropic relationship for a isentropic process is:

\frac{P_{2}}{P_{1}} = \left(\frac{V_{1}}{V_{2}}  \right)^{\gamma}

Where \gamma is the ratio of specific heats

The final pressure is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}}\right)^{\gamma}

P_{2} = (200\,kPa)\cdot (2)^{1.4}

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7 0
3 years ago
An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the
Vilka [71]

Answer:

Required Diameter = 302.65 inches

Explanation:

We are given;

Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

Elongation = 0.05 in

Original length = 18 in

We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

Stress = Force/Area

Thus,

Area = Force/stress

So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

(4000 x 4)/π = d²

d² = 5092.96

Required diameter here is;

d = √5092.96

d = 71.36 in

For Strain;

Formula for strain is;

Strain = stress/E

We are given E = 120 ksi

stress = P/A = 24,000/A

strain = elongation/original length = 0.05/18 = 0.00278

Thus;

0.00278 = P/(A•E)

0.00278 = 24000/(120 x A)

Making A the subject to obtain;

A = 24000/(120 x 0.00278)

A_required = 71942 in²

Area = πd²/4

So, 71942 = πd²/4

(71942 x 4)/π = d²

d² = 91599.4

Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

3 0
4 years ago
What does a block tester do?
tatyana61 [14]

Answer:

Block design test. A block design test is a subtest on many IQ test batteries used as part of assessment of human intelligence. It is thought to tap spatial visualization ability and motor skill.

Explanation:

6 0
3 years ago
If the atomic radius of copper is 0.128 nm, calculate the volume of its unit cell in cubic meters.
Alex_Xolod [135]

Answer:

Volume of face centered cubic cell=4.74531*10^{-29} m^3

Explanation:

Consider the face centered cubic cell:

1 atom at each corner of cube.

1 atom at center of each face.

Consider the one face (ABCD) as shown in attachment for calculation:

Length of the all sides of face centered cubic cell is L.

Volume of face centered cubic cell= L^3

Now Consider the figure shown in attachment:

According to Pythagoras theorem on ΔADC.

L^{2}+L^2=(4a)^2     (a is the atomic radius)

L=\frac{4a}{\sqrt{2}} (Put in the formula of Volume)

Volume of face centered cubic cell= L^3

Volume of face centered cubic cell= (\frac{4a}{\sqrt{2}})^3

Volume of face centered cubic cell= (\frac{4(0.128*10^{-9}}{\sqrt{2}})^3

Volume of face centered cubic cell=4.74531*10^{-29} m^3

3 0
4 years ago
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