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3 years ago

10

If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas

e

1 answer:

jarptica [38.1K]3 years ago

8 0

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ( $\eta$ ), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$ , then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.

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Nearlyof all serious occupational injuries and illnesses stem from overexertion of repetitive motion.

Ostrovityanka [42]

Answer:

a) 1/2

Explanation:

Overexertion accounted for more than half of all events that resulted in a disabling condition.

Furthermore, 30% of all overexertion cases were reported in the services industry, on the other hand, 25% of injuries resulting from contact with objects and equipment occurred in the manufacturing industry.

The above piece of information is taken from the bureau of labor statistics, Survey of Occupational Injuries and Illnesses

"LOST-WORKTIME INJURIES AND ILLNESSES: CHARACTERISTICS AND RESULTING DAYS AWAY FROM WORK, 2002"

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3 years ago

In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i

Hitman42 [59]

Answer:

All of the above

Explanation:

firstly, a creep can be explained as the gradual deformation of a material over a time period. This occurs at a fixed load with the temperature the same or more than the recrystallization temperature.

Once the material gets loaded, the instantaneous creep would start off and it is close to electric strain. in the primary creep area, the rate of the strain falls as the material hardens. in the secondary area, a balance between the hardening and recrystallization occurs. The material would get to be fractured hen recrstallization happens. As temperature is raised the recrystallization gets to be more.

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2 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

For the following circuit diagram, if A=010 , B= 101.

Fantom [35]

Answer:

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4 0

3 years ago

A concrete mix design calls for 6.5 sacks of cement, a water/cement ratio of 0.45, and an air content of 2.5%. 1. Complete the m

RUDIKE [14]

Answer:

28.6 kg

Explanation:

The final weight can be calculated from the mixing data and formulae which is given as follows:

cement content = $\frac{water content}{water - cement ratio}$

Computing the parameters and checking the tables gives 28.6 kg.

7 0

3 years ago

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