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bogdanovich [222]

3 years ago

13

A 200 W vacuum cleaner is powered by an electric motor whose efficiency is 70%. (Note that the electric motor delivers 200 W of

net mechanical power to the fan of the cleaner). Calculate the rate at which this vacuum cleaner supply energy to the room when running. ?

1 answer:

Goshia [24]3 years ago

5 0

Answer:

The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.

Explanation:

power efficiency of electric motor = 70% = 0.70

The power output of the vacuum cleaner = $P_o$ = 200 W

The power output of the vacuum cleaner = $P_i$

$Efficiency=\frac{P_o}{P_i}$

$0.70=\frac{200 W}{P_i}$

$P_i=\frac{200 W}{0.70}=285.71 W$

The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.

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Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at

sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

$h_{f} = 173.358 \\h_{fg} = 2402.522$

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

$h_{2a} = 489.752\\h_{2b} = 313.2$

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

$h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876$

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

$x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119$

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485$

Part b) @ 4 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275$

6 0

3 years ago

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

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Answer:

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