Multiple Choice

technician a uses a current output test to check ac generator output. technician b uses a voltage output test to check output. w

expeople1 [14]

Answer:

both

Explanation:

Both the technician are correct, ac generator output can be tested in both ways. The two ways are current output test to check ac generator output. and voltage output test to check output.

7 0

3 years ago

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–

Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

$C_p$ = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = $m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]$

Substitute all parameters in the equation

Δs = $5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

b)

Δs = $m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]$

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = $5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0

3 years ago

A Carnot cycle is to be designed to attain

Anni [7]

Answer:

reservoir is 727°C, then low temperature

Explanation:

Efficiency for any Heat engine = Work done by engine / Heat provided to the engine

Also, for a carnot engine derived formula for Carnot Engine

= 1- T2/T1

Where, T2 = sink reservoir temperature

and T1 = source reservoir temperature

727° C = 1000 K

So, 0.75 = 1-T2 / 1000

So, T2 = 1000*0.25 = 250 K = -23 Kelvin

8 0

3 years ago

Technicians have to determine the flow rate of water in a pipe with the aid of a venturi installation and a mercury au.oaeter. T

Schach [20]

Answer:

Manometric difference x=142.85 mm.

Explanation:

Given :

Pipe diameter $d_1=40 mm$

venturi meter $d_2=20 mm$

We can know that discharge through venturi meter is given as

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}$

$A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}$

$Q=A_1V_1$

$Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s$

$0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}$

h=1.8 m

We know that $h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )$

Where x is the manometric deflection

⇒ $1.8=x\left (\dfrac{13600}{1000}-1\right )$

So x=14.28 mm

Manometric difference x=142.85 mm.

7 0

2 years ago

Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12

Katarina [22]

Answer:

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

$- \dot W_{out} + \dot m \cdot (h_{in} - h_{out}) = 0$

The turbine power output is:

$\dot W_{out} = \dot m\cdot (h_{in}-h_{out})$

The volumetric flow is:

$\dot V = \frac{\pi}{4} \cdot \left( \frac{2}{12}\,ft \right)^{2}\cdot (620\,\frac{ft}{s} )$

$\dot V \approx 13.526\,\frac{ft^{3}}{s}$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,\frac{ft^{3}}{lbm}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu}$

$\dot m = \frac{13.526\,\frac{ft^{3}}{s} }{1.33490\,\frac{ft^{3}}{lbm} }$

$\dot m = 10.133\,\frac{lbm}{s}$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,\frac{BTU}{lbm}$

State 2 (Saturated Vapor)

$h = 1146.1\,\frac{BTU}{lbm}$

The turbine power output is:

$\dot W_{out} = (10.133\,\frac{lbm}{s} )\cdot (1479.1\,\frac{BTU}{lbm}-1146.1\,\frac{BTU}{lbm})$

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

6 0

3 years ago