The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is 3 mm thick and whose interior surface is maintained at a temperature of 36°C. On a calm day the convection heat transfer coefficient at the outer surface is 25 W/m 2 ⋅ K, but with 30 km/h winds it reaches 65 W/m 2 ⋅ K . In both cases the ambient air temperature is − 15 ° C . What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? What will be the skin outer surface temperature for the calm day? For the windy day? What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at − 15 ° C on the windy day?
The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is 3 mm thick and whose interior surface is maintained at a temperature of 36°C. On a calm day the convection heat transfer coefficient at the outer surface is 25 W/m 2 ⋅ K, but with 30 km/h winds it reaches 65 W/m 2 ⋅ K . In both cases the ambient air temperature is − 15 ° C . What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? What will be the skin outer surface temperature for the calm day? For the windy day? What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at − 15 ° C on the windy day?
Solution Summary: The author compares the heat loss per unit area from the skin for the calm day to that of the windy day.
The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is 3 mm thick and whose interior surface is maintained at a temperature of 36°C. On a calm day the convection heat transfer coefficient at the outer surface is
25
W/m
2
⋅
K,
but with 30 km/h winds it reaches
65
W/m
2
⋅
K
.
In both cases the ambient air temperature is
−
15
°
C
.
What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day?
What will be the skin outer surface temperature for the calm day? For the windy day?
What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at
−
15
°
C
on the windy day?
Meh
Battery operated train
Coll CD Af Pair
160,000kg 0.0005 0.15 5m² 1.2kg/m³
19
7et nong
0.98 0.9 0.88
Tesla Prated
Tesla Trated Ywheel ng Jaxle.
270kW
440NM
0.45m 20
2
8.5kgm²
Consider a drive cycle of a 500km trip with 3 stops in
the middle. Other than the acceleration and deceleration
associated with the three stops, the tran maintains.
constant cruise speed velocity of 324 km/hr. The
tran will fast charge at each stop for 15 min at a
rate Peharge = 350 kW
(ผม
τ
(MN
15MIN
Stop
w charging
(350kW
GMIJ
restored during 15
minutes of fast charging at
Calculate the battery energy Pcharge = 350kW
Calculate the net energy gain per stop
t
64
Determice the total battery energy required Ebat
to complete the 500km trip with 3 stops.
etc
DO NOT COPY SOLUTION
The differential equation of a cruise control system is provided by the following equation:
Find the closed loop transfer function with respect to the reference velocity (vr) .
a. Find the poles of the closed loop transfer function for different values of K. How does the poles move as you change K?
b. Find the step response for different values of K and plot in MATLAB. What can you observe?
c. For the given transfer function, find tp, ts, tr, Mp . Plot the resulting step response. G(s) = 40/(s^2 + 4s + 40)
Aswatan gas occupies a space of 0.3 millike cube at a pressure of 2 bar and temperature of 77 degree Celsius it is indicate at constant volume at pressure of 7 parts determine temperature at the end of process mass of a gas changing internal energy change in enthalpy during the process assume CP is equal to 10 1.005 CV is equal to 0.712 is equal to 287
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.