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inn [45]
4 years ago
13

A dipstick is (a direct,an indirect) measurement device

Engineering
1 answer:
icang [17]4 years ago
3 0
Direct because it’s going right in the spot it needs to be in
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Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi
VladimirAG [237]

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

\eta_{th} = 1 - \frac{T_{L}}{T_{H}}

Where:

T_{L} - Temperature of the cold reservoir, in K.

T_{H} - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}

\eta_{th} = 0.466 \,(46.6\,\%)

Source 2

\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}

\eta_{th} = 0.0798 \,(7.98\,\%)

The power produced by each device is presented below:

Source 1

\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 85.433\,kW

Source 2

\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 146.3\,kW

The source 2 produces the largest amount of power.

8 0
3 years ago
Indicate whether the following statements are true or false for an isothermal process: (A) Q=T(∆S). (B) ∆U=0.(C) The entropy cha
Lena [83]

Answer:

A=False

B=False

C=False

D=False

E=False

F=False

Explanation:

A. In an isothermal process, only the reversibly heat transfer is 0, Q_{rev}=T (\Delta S)

B. Consider the phase change of boiling water. Here, the temperature remains constant but the internal energy of the system increases.

C. This is not true even in reversible process, as can be inferred from the equation in part A.

D. This is only true in reversible processes, but not in all isothermal processes.

E. Consider the phase change of freezing water. Here, the surroundings are increasing their entropy, as they are taking in heat from the system.

F. This is not true if (\Delta U)\neq 0, like in answer B. One case where this is true is in the reversible isothermal expansion (or compression) of an ideal gas.

3 0
3 years ago
A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

thus it is Impossible cycle

(b)P_{out}=2 kW

Q_2=6-2=4 kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

\frac{Q_2}{Q_1}=\frac{T_1}{T_2}

Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

6 0
4 years ago
Plz answer all of these questions!
statuscvo [17]

Answer: all you need to to is go to help me with this question.cm

Explanation:

5 0
3 years ago
Please view the file attached below and help me answer all the questions!!
evablogger [386]

Answe whats the question i can help you

Explanation:

8 0
3 years ago
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