Answer:
month = input("Input the month (e.g. January, February etc.): ")
day = int(input("Input the day: "))
if month in ('January', 'February', 'March'):
season = 'winter'
elif month in ('April', 'May', 'June'):
season = 'spring'
elif month in ('July', 'August', 'September'):
season = 'summer'
else:
season = 'autumn'
if (month == 'March') and (day > 19):
season = 'spring'
elif (month == 'June') and (day > 20):
season = 'summer'
elif (month == 'September') and (day > 21):
season = 'autumn'
elif (month == 'December') and (day > 20):
season = 'winter'
print("Season is",season)
Explanation:
Answer:
a) 0.684
b) 0.90
Explanation:
Catalyst
EO + W → EG
<u>a) calculate the conversion exiting the first reactor </u>
CAo = 16.1 / 2 mol/dm^3
Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst
Vo = 7.24 dm^3/s
Vm = 800 gal = 3028 dm^3
hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins
next determine the value of conversion exiting the reactor ( Xai ) using the relation below
KIm = 
------ ( 1 )
make Xai subject of the relation
Xai = KIm / 1 + KIm --- ( 2 )
<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>
Xai = 0.684
<u>B) calculate the conversion exiting the second reactor</u>
CA1 = CA0 ( 1 - Xai )
therefore CA1 = 2.5438 mol/dm^3
Vo = 7.24 dm^3/s
To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below
XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )
<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>
XA2 = 0.90
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Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
Answer:
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Answer:
14.52 minutes
<u>OR</u>
14 minutes and 31 seconds
Explanation:
Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.
Specific heat at constant volume at 27°C = 0.718 kJ/kg*K
Initial temperature of room (in kelvin) = 283.15 K
Final temperature (required) of room = 293.15 K
Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg
Heat required at constant volume: 0.718 * (change in temp) * (mass of air)
Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ
Time taken for temperature rise: heat required / (rate of heat change)
Where rate of heat change = 10000 - 5000 = 5000 kJ/hr
Time taken = 1210.26 / 5000 = 0.24205 hours
Converted to minutes = 0.24205 * 60 = 14.52 minutes
