Ask question

Ask question

All categories

3 years ago

7

An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa

ste heat to a sink at 0 °C. He also claims that this engine produces 3.3 kW while rejecting heat at a rate of 4.4 kW. Is this claim valid?

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)= $\frac{Work produced}{Heat supplied}$

and

Heat Supplied ${\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )$

${Q_s}=3.3+4.4=7.7KW$

η= $\frac{3.3}{7.7}$

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1- $\frac{T_ {sink}}{T_ {source}}$

η=1- $\frac{273}{1023}$

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

You might be interested in

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel

REY [17]

Answer: 10.12m, 20micro meter

Explanation:

8 0

3 years ago

Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are

borishaifa [10]

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

4 0

3 years ago

When you see a street with white markings only, what kind of street is it?

Georgia [21]

Answer:

it's a one way street

3 0

4 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

a)

$-V_{2} =2V_{1}$ ( double)

b)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

b)

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago

In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0

3 years ago