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Sedaia [141]
3 years ago
6

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

e waste heat to another reservoir at temperature T. The second engine receives this energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at 300 K. If the thermal efficiencies of both engines are the same, determine the temperature T.
Engineering
1 answer:
Aleksandr-060686 [28]3 years ago
5 0

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

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On the pavement indicate that the adjacent lane is traveling in the same direction and passing is permitted
Mkey [24]

A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.

A broken white line on the pavement show that the adjacent lane is traveling in the same direction and passing is permitted.

<h3>What does pavement markings show?</h3>

Pavement markings are known to be tools that are used to pass infor or messages to roadway users.

Note  that they tell the part of the road that one need to use, give information about conditions ahead, and others

Note that A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.

Learn more about pavement markings from

brainly.com/question/10179521

#SPJ1

6 0
2 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
A 304 stainless steel (yield strength 30 ksi) cylinder has an inner diameter of 4 in and a wall thickness of 0.1 in. If it is su
Natasha_Volkova [10]

Answer:

The value we got is 1.423 ksi which is less than 30 and so the material hasn't failed and yielding hasn't occured.

Explanation:

This is a cylindrical thin walled vessel. Thus;

Hoop stress; σ1 = pr/t

Where ;

p is internal pressure

r_o is radius = 4/2 = 2 in

t is thickness of wall = 0.1 in

Thus;

σ1 = 70 x 2/0.1 = 1400 ksi

Longitudinal stress; σ2 = pr/2t

σ2 = 70 x 2/(0.1 x 2) = 700 ksi

Now, we want to find the normal stress. The inner radius of the circle will be; r_i = r_o - t = 2 - 0.1 = 1.9 in

So, normal stress by axial force is given by;

σ_fx = F/A = F/(π(r_o² - r_i²))

We are given that F = 500 lb

σ_fx = 500/(π(2² - 1.9²))

σ_fx = 408.1 ksi

We can now find the torsion from the formula;

τ = (T•r_o)/J

We are given that T = 70 lb.ft = 70 x 12 lb.in = 840 lb.in

J is the polar moment of inertia and has a formula; ((π/2)(r_o⁴ - r_i⁴))

So,J = ((π/2)(2⁴ - 1.9⁴)) = 4.662 in⁴

Thus,τ_xy = (840 x 2)/4.662 = 360.4 ksi

σ1 is in the y direction and σ2 is in the x direction. Thus;

σ_x = σ1 + σ_fx = 700 + 408.1 = 1108.1 ksi

Also, σ_y = σ1 = 1400 ksi

Now distortion energy can be expressed as;

σ_Y² = σ_x - σ_x•σ_y + (σ_y)² + 3(τ_xy)²

Plugging in the relevant values, we obtain ;

σ_Y² = 1108.1² - (1108.1*1400) + 1400² + 3(360.4)²

So, σ_Y² = 2.02 x 10^(6) psi²

σ_Y = √2.02 x 10^(6)

σ_Y = 1423 psi = 1.423 ksi

The question says the yield strength of the material is 30 ksi.

The value we got is less than 30 and so the material hasn't failed and yielding hasn't occured.

8 0
3 years ago
The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The abso
ser-zykov [4K]

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

8 0
4 years ago
The name of a round stock welding position refers to the
Sedaia [141]
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