1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mariulka [41]
3 years ago
13

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

the high-temperature reservoir is 500 kJ. The overall temperature limits are 1000K and 400K. Both engines produce equal work. What is most nearly the intermediate temperature between the two engines?
Engineering
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

Given:

high temperature reservoir T_{H} =1000k

low temperature reservoir T_{L} =400k

thermal efficiency n_{1}= n_{2}

The engines are said to  operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

n_{1} =1-\frac{T}{T_{H} }

The thermal efficiency of second heat engine can be written as

n_{2} =1-\frac{T_{L} }{T}

The temperature of intermediate reservoir can be defined as  

1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k

You might be interested in
Is a water proof material used around tubs and
Leviafan [203]
I think it’s hard board
4 0
3 years ago
Read 2 more answers
Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a
guapka [62]

Answer:

K_v=12.34

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo =   Discharge / area of the orifice

or

Vo = \frac{0.01}{\frac{\pi}{4}0.045^2}

or

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, V_v =   Discharge / area of the orifice

or

V_v = \frac{0.01}{\frac{\pi}{4}0.0675^2}

or

V_v = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

or

25 m = K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}

where,

K_v is the loss coefficient for the gate valve

on substituting the values, we get

25 m = 10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}

or

K_v\frac{2.79^2}{2\times9.81} = 4.898

or

K_v=12.34

3 0
4 years ago
A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

3 0
3 years ago
It is proposed that a 1,000-MW electric power plant be built with steam as the working fluid with the condenser to be cooled wit
elixir [45]

Answer:

The temperature rise of the river downstream from the power plant is approximately 1.9°C

Explanation:

Pls see the attached files below for the solution as typing it here may be difficult due to the formulas involved in solving the question.

5 0
3 years ago
7. An energy auditor is part of what career field?
Rashid [163]
An energy auditor is part of power operations
6 0
2 years ago
Other questions:
  • Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high
    5·1 answer
  • Determine the phase or phases in a system consisting of H2O at the following conditions and sketch p-v and t-v diagram showing t
    10·1 answer
  • How do i calculate the force acting on the actuator in the suspension system of a fsae car in order to calculate the pressure go
    13·1 answer
  • The part of a circuit that carries the flow of electrons is referred to as the?
    11·1 answer
  • What are the success factors for mechanical engineering?
    8·2 answers
  • How much wood could a wood chuck chuk if a wood chuck could chuck wood
    10·2 answers
  • it creates parts from thin plastic sheets as opposed to plastic pellets. is it 1. Pickling 2. Thermoforming 3. Extrusion​
    13·2 answers
  • Label each of the line types in the drawing below. ( will not mark you brainlest or whatever if you don't at least try to help)
    11·1 answer
  • FOR DARKEND1<br><br>÷<br><br><br>try copy and paste​
    5·2 answers
  • Petroleum engineers work almost exclusively with dry materials.<br><br><br> False<br> True
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!