If you touch a downed power line, covered or bare, what's the likely outcome?

A scale on a blue print drawing of a house shows that 666 centimeters represents 333 meters.

Dennis_Churaev [7]

Answer:

545454cm

Explanation:

The blue print drawing of the house shows 666 centimeters, but the real picture of the house is 333 meters. So let the number of cm on the blueprint that represent the distance of 272727 meters be x. Firstly convert the meters to centimeters 666cm = 333m, x=272727m ; then cross multiply, 666cm=33300cm x=27272700cm ; x =(666cm×272727cm)/33300cm =545454cm.

4 0

3 years ago

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

A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1 comma 630 kilogram

Varvara68 [4.7K]

Answer:

attached below

Explanation:

7 0

4 years ago

Solar ovens use heat from the Sun to cook food. They work on a principle similar to that of solar panels used to heat homes. Wha

marishachu [46]

Explanation:

Absorption of the sun's energy (heat) in solar cooking is best achieved when a surface is dark in color, thus the most common solar oven interiors are usually black in color as well as the color of the cookware used for cooking the food. Dark colors absorb the heat, whereas light colors do not absorb heat well.

Metal reflectors should be positioned around the oven to maximize light input to be the most effective.

4 0

3 years ago

Read 2 more answers

Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calcu

diamong [38]

Answer:

125 cm³/min

Explanation:

The material rate of removal is usually given by the formula

Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where

Radial Depth of Cut = 25 mm

Axial depth of cut = 200 mm

Feed rate = 25 mm/min

On multiplying all together, we will then have

MRR = 25 mm * 200 mm * 25 mm/min

MRR = 125000 mm³/min

Or we convert it to cm³/min and have

MRR = 125000 mm³/min ÷ 1000

MRR = 125 cm³/min

4 0

3 years ago