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Kamila [148]
3 years ago
11

an adiabatic compressor receives 1.5 meter cube per second of air at 30 degrees celsius and 101 kpa. The discharge pressure is 5

05 kpa and the power supplied is 325 kW, what is the discharge temperature

Engineering
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer:

The discharge temperature is 259.82 K

Explanation:

In this question, we are concerned with calculating the discharge temperature

Please check attachment for complete solution

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A gear motor can develop 2 hp when it turns at 450rpm. If the motor turns a solid shaft with a diameter of 1 in., determine the
kramer

Answer:

Maximum shear stress is;

τ_max = 1427.12 psi

Explanation:

We are given;

Power = 2 HP = 2 × 746 Watts = 1492 W

Angular speed;ω = 450 rev/min = 450 × 2π/60 rad/s = 47.124 rad/s

Diameter;d = 1 in

We know that; power = shear stress × angular speed

So,

P = τω

τ = P/ω

τ = 1492/47.124

τ = 31.66 N.m

Converting this to lb.in, we have;

τ = 280.2146 lb.in

Maximum shear stress is given by the formula;

τ_max = (τ•d/2)/J

J is polar moment of inertia given by the formula; J = πd⁴/32

So,

τ_max = (τ•d/2)/(πd⁴/32)

This reduces to;

τ_max = (16τ)/(πd³)

Plugging in values;

τ_max = (16 × 280.2146)/((π×1³)

τ_max = 1427.12 psi

7 0
3 years ago
A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto
vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
3 years ago
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
2 years ago
1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

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

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0
2 years ago
Urgent please help me...
Pepsi [2]
Scuze , am incercat , dat chiar nu stiu. Prefer sa nu raspund si sa scriu asta decat sa raspund gresit.❤️
7 0
2 years ago
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