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3 years ago

Calculate the efficiency of the following appliances:

Physics

1 answer:

lubasha [3.4K]3 years ago

4 0

Answer:

1. The efficiency of the radiator is 90 %

2. The efficiency of the torch is 65 %

3. The efficiency of the car is 40 %

4. The efficiency of the energy saver is 70 %

5. The efficiency of the speaker is 50 %

Explanation:

Efficiency = (Useful energy out ÷ Total energy in) × 100 J

1. Useful energy = 900 J

The total energy in = 1000 J

The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %

2. Useful energy = 65 J

The total energy in = 100 J

The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %

3. Useful energy = 4,000 J

The total energy in = 10,000 J

The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %

4. Useful energy = 700 J

The total energy in = 1,000 J

The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %

5. Useful energy = 50 J

The total energy in = 100 J

The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

Download pdf

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