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Law Incorporation [45]

3 years ago

14

A heat engine receives 1000kJ/s of heat from a high temperature source at 600◦C and rejects heat to a cold temperature sink at 2

0◦C a. What is the thermal efficiency of this engine? b. What is the power delivered by the engine in watts? c. At what rate is heat rejected to the cold temperature sink?

1 answer:

scoray [572]3 years ago

5 0

Answer:

(A) the thermal efficiency of this engine is 66.44 %

(B) the power delivered by the engine in watts is 664,400 W

(C) The rate at which heat is rejected to the cold temperature sink is 335.6 kJ/s.

Explanation:

Given;

Input Heat received by the engine, $P_{in}$ = 1000 kJ/s

hot temperature, $T_H$ = 600 ⁰C = 600 + 273 = 873 K

cold temperature, $T_C$ = 20 ⁰C = 20 + 273 = 293 K

(A) the thermal efficiency of this engine;

$\eta = 1 - \frac{T_C}{T_H} \\\\\eta = 1 - \frac{293}{873} \\\\\eta = 1 - 0.3356\\\\\eta = 0.6644 = 66.44 \%$

(B) the power delivered by the engine in watts;

$P = \eta P_{in}\\\\P = 0.6644 \times \ 1000kJ/s \ \times \ \frac{1\ kW}{1\ kJ/s} \\\\P = 664.4 \ kW \\\\P = 664,400 \ W$

(C) The rate at which heat is rejected to the cold temperature sink;

$Q_C = Q_H(1-\eta)\\\\Q_C = 1000\ kJ/s(1-0.6644)\\\\Q_C = 335.6 \ kJ/s$

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1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel

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Answer:

1.933 KN-M

Explanation:

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3 years ago

A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and Blie in t

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Answer:

The couples are not all on one axis or plane for that matter but if the A and B connector had to be specified it would go by the yz axis diagonal to the x axis with a magnitude of about 15. The direction of the axis would be pointed up to the second quadrant. Hope this was helpful

Explanation:

8 0

4 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago