A 0.5 W, 2000 Ω resistor has a current flow of 0.01 A through it. Is this resistor operating within its power rating?

A solar radiation flux of 700 W/m2 is absorbed in a metal plate that is perfectly insulated on the back surface. The convection

skad [1K]

Answer:

The main solution to this problem is to note that under equilibrium conditions, the rate at which heat is transferred into the metal plate from the sun by radiation must be equal to the rate at which heat is lost from the plate to the surrounding air by convection.

Explanation:

Therefore, qᵃ = qᵇΔT

where

a = Radiation energy source

b = Convection energy source

ΔT = Tp - Ta = Temperature change between the plate and the air under equilibrium condition

Given,

qᵃ = 700 W/m2

qᵇ = 10 W/m2K - 100 W/m2K

Hence, for qᵇ = 10 W/m2K,

ΔT = qᵃ / qᵇ

ΔT = 700/10 = 70K

Similarly, for qᵇ = 100 W/m2K

ΔT = 700/100 = 7K

The temperature change of the plate under equilibrium conditions range from 7K to 70K.

4 0

2 years ago

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 k

In-s [12.5K]

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pm

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3 0

2 years ago

5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energ

GenaCL600 [577]

Answer:

Energy Transfer = 350 kJ

Explanation:

The net work can be determined from an energy balance. That is, with assumption

∆KE + ∆PE + ∆U = Q − W

Where

∆KE = ∆PE = 0 (Since There is no significant change in the kinetic or potential energy of the steam)

The net work is the sum of the work associated with the paddlewheel Wpw

and the work done on the piston Wpiston:

W = Wpw + Wpiston

From the given information, Wpw= −18.5 kJ,

Collecting results:

Wpw + Wpiston = Q − ∆U

Wpiston = Q − ∆U − Wpw= Q − m (u2− u1) − Wpw

Where Q=80kJ, m=5kg, u2 = 2659.6 kJ/kg, u1 = 2709.9 kJ/kg

= 80 kJ − 5 kg (2659.6 − 2709.9)kJ/kg − ( −18. 5 kJ)

= 350 kJ

4 0

2 years ago

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

denis23 [38]

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature $T_\infity = 15$ degree celcius

$T_S = 220 + 273 = 473 K$

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

$Q = hA(T_S - T_{\infty}) \eta$

$= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8$

Q = 125.538 W

5 0

2 years ago

A four-variable logic function that is equal to 1 if any three or all four of its variables are equal to 1 is called a majority

gulaghasi [49]

Answer:

1. Assumption

2.Truth Table

3.K-Map

4.Mapping and Final Expression

5 0

2 years ago