An engine indicator is used to determine the following: (a) speed (d) m.e.p, and IHP (e) BHP (b) temperature (c) volume of cylin
1 answer:
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Answer:
In Btu:
Q=0.001390 Btu.
In Joule:
Q=1.467 J
Part B:
Temperature at midpoint=274.866 C
Explanation:
Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)=
Thermal Conductivity is SI units:
Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft
Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft
T_1=500 C=932 F
T_2=50 C= 122 F
Part A:
In Joules (J)
Heat Q is:
In Btu:
Heat Q is:
PArt B:
At midpoint Length=L/2=0.1 m
On rearranging:
Answer:
change in storage = -310,500 ft^3
intital storage= 3.67 acre ft
Explanation:
Given data:
Rate of inflow = 350 cfs
Rate of outflow = 285 cfs
After 90 min, rate of inflow = 250 cfs
Rate of outflow = 200 cfs
final storage = 10.8 acre-ft
calculating the average inflow and outflow
average inflow
average outlow
total amount of water drain during the period of one hour
= (average outflow - average inflow) *60*90
= (242.5 - 300)*60*90 = -310,500
change in storage is calculate as
= -310,500 ft^3
in cubic meter
= -310500/35.315 = 8792.30 cm^3
in acre-ft
= -310,500/43560 = 7.13 acre ft
initial storage = 10.8 - 7.13 = 3.67 acre ft