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frozen [14]
3 years ago
10

An insulated 40 ft^3 rigid tank contains air at 50 psia and 580°R. A valve connected to the tank is now opened, and air is allow

ed to escape until the pressure inside drops to 25 psia. The air temperature during this process is kept constant by an electric resistance heater placed in the tank. Determine the electrical work done during this process in Btu assuming variable specific heats.
Engineering
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

The answer is "\bold{W_{in} = 645.434573 \ Btu}".

Explanation:

Its air enthalpy is obtained only at a given temperature by A-17E.  

The solution from the carbon cycle is acquired:

\to \Delta U= W_{in}-m_{out}h_{out}=0\\\\\to W_{in} = (m_1 - m_2)h

           =\frac{Vh}{RT}(P_1- P_2)\\\\= \frac{40 \times 138.66}{0.3704\times 580}(50-25) Btu\\\\= \frac{5,546.4}{214.832}(25) Btu\\\\= 25.8173829(25) Btu\\\\ =645.434573 Btu

W_{in} = 645.434573 \ Btu

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Juli2301 [7.4K]

Answer:

jjj jk to the day of the day of the day of the day of the day

Explanation:

hey hi I am a br of August and I am not able to join the meeting and will

7 0
2 years ago
A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu
suter [353]

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

emf= N\frac{d \phi}{dt}

where;

ΔФ is change in magnetic flux

ΔФ  = BA = B x πr²

ΔФ  = 0.573 x π(0.0395)² = 0.002809 T.m²

emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0
3 years ago
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
Ymorist [56]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
3 years ago
To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01
liq [111]

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

3 0
3 years ago
1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con
QveST [7]

Answer:

Explanation: Here it is: 67 Hope that helps! :)

5 0
3 years ago
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