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malfutka [58]
3 years ago
6

The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar rad

iation is incident on the spacecraft at a rate of 1000 W/m2 , determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Engineering
1 answer:
Korolek [52]3 years ago
6 0

Answer:

T_{surf} =3.9°C

Explanation:

solution:

we express the surface temperature from the quality of the emitted and absorbed heat rates:

Q_{abs} =Q_{em} \\\

\alpha Q_{inc}=εσAT^{4} _{surf}

T_{surf} =\sqrt[4]{}\alpha/εσ*\frac{Q_{inc} }{A}........................(1)

where

\alpha=0.2

ε=0.6

A=5.67*10^-8 W/m^2K^4

Q_{inc}= 1000 W/m^2

now putting these value in eq 1

T_{surf} =3.9°C

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Each cout statement has a syntax error. Type the first cout statement, and press Run to observe the error message. Fix the error
saul85 [17]

Answer:

1. cout << "Num: " << songNum << endl;

2. cout << songNum << endl;

3. cout << songNum <<" songs" << endl;

Explanation:

//Full Code

#include <iostream>

using namespace std;

int main ()

{

int songNum;

songNum = 5;

cout << "Num: " << songNum << endl;

cout << songNum << endl;

cout << songNum <<" songs" << endl;

return 0;

}

1. The error in the first cout statement is that variable songnum is not declared.

C++ is a case sensitive programme language; it treats upper case and lower case characters differently.

Variable songNum was declared; not songnum.

2. Cout us used to print a Variable that has already been declared.

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3 years ago
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

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Explanation:

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