True or false. Part of the mission of the NTSB is to determine the probable cause of an accident

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

Aleksandr-060686 [28]

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

5 0

2 years ago

Breh/bro <br><br>what is this, finally I can learn programming

LenKa [72]

Answer:

It's an intoduction to hacking and systematic programming.

Explanation:

Yes, you might be able to grasp a few things from it, but it also may be a way hackers could hack you, by luring you to click it.

8 0

2 years ago

Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid

AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

for this case we can define the following equation for mass flow using the first law of thermodynamics

$m=\frac{Q}{h1-h2}$

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

$Q=\frac{5000}{1123-184.5} =5.328Ibm/hr$

3 0

2 years ago

A cubical picnic chest of length 0.5 m, constructed of sheet styrofoam of thickness 0.025 m, contains ice at 0\[Degree]C. The th

maksim [4K]

Answer:

Rate of heat transfer is 0.56592 kg/hour

Explanation:

Q = kA(T2 - T1)/t

Q is rate of heat transfer in Watts or Joules per second

k is thermal conductivity of the styrofoam = 0.035 W/(mK)

A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2

T1 is initial temperature of ice = 0 °C = 0+273 = 273 K

T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K

t is thickness of styrofoam = 0.025 m

Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s

Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr

6 0

3 years ago

GG(ss) = 1 ss2 + 3 We want to design a feedback control system in a unity feedback structure by adding a controller DD(ss) = ss+

saw5 [17]

Answer:

Explanation:

The explanation is given in the picture that is shown below

3 0

3 years ago