Answer:
this might help
Explanation:
https://science.ksc.nasa.gov/mars/msp98/misc/MCO_MIB_Report.pdf
Answer
1056
Explanation:
for example
A soil is to be excavated from a borrow pit which has a density of 1.75g/cc and water content of 12%. The G is 2.7 the soil is compacted to that water content of 18% and dry density of 1.65g/cc. for 1000 m3 of soil used in fill estimate
Quantity of soil to be excavated from pit in m3
Answer:
heat transfer for the process is - 643.3 kJ
Explanation:
given data
mass m = 2 kg
pressure p1 = 500 kPa
temperature t1 = 400°C = 673.15 K
temperature t2 = 40°C = 313.15 K
pressure p2 = 300 kPa
to find out
heat transfer for the process
solution
we know here mass is constant so
m1 = m2
so by energy equation
m ( u2 - u1 ) = Q - W
Q is heat transfer
and in process P = A+ N that is linear spring
so
W = ∫PdV
= 0.5 ( P1+P2) ( V1 - V2)
so for case 1
P1V1 = mRT
put here value
500 V1 = 2 (0.18892) (673.15)
V1 = 0.5087 m³
and
for case 2
P2V2 = nRT
300 V2 = 2 (0.18892) (313.15)
V2 = 0.3944 m³
and
here W will be
W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )
W = -45.72 kJ
and
Q is here for Cv = 0.83 from ideal gas table
Q = mCv ( T2-T1 ) + W
Q = 2 × 0.83 ( 40 - 400 ) - 45.72
Q = - 643.3 kJ
heat transfer for the process is - 643.3 kJ
