56
2 Environmental Conditions in the Mine
By differentiating and simplifying, Eq. 2.10 is obtained:
14
V e dP + k PdV e = 0
(2.10)
Considering the Clapeyron equation:
PV e = R w T
Differentiating the function (Eq. 2.11):
PdV e = R w dT − V e dP
(2.11)
If in Eq. 2.10, we substitute the corresponding terms for those of Eqs. 2.8 and
2.11, we obtain Eq. 2.12 below:
dH + k(R w dT − dH ) = 0
(2.12)
Working the expression out:
dH + k R w dT − kdH = 0
dH − kdH + k R w dT = 0
(1 − k)dH + k R w dT = 0
Integrating:
(1 − k) ∫ dH + k R w ∫ dT = 0
(1 − k)H + k R w T + C = 0
We obtain:
T =
(k − 1)H
k R w
− C
If we establish initial values of H 0 = 0 and T = T 0 , we find that C = −T 0 . If an
adiabatic process is considered, the equation will be (Eq. 2.13):
T =
(k − 1)H
k R w
+ T 0
(2.13)
As for dry air R w = 29.29 kpm (kg K)
-1 and k = 1.41, we obtain (Eq. 2.14):
T = T 0 + 0.0098 H
(2.14)
14 d(P V k ) = dP V k + P k V k−1 dV = dP V V k−1 + P k V k−1 dV = 0 → V dP + P k dV = 0.
2 Environmental Conditions in the Mine
By differentiating and simplifying, Eq. 2.10 is obtained:
14
V e dP + k PdV e = 0
(2.10)
Considering the Clapeyron equation:
PV e = R w T
Differentiating the function (Eq. 2.11):
PdV e = R w dT − V e dP
(2.11)
If in Eq. 2.10, we substitute the corresponding terms for those of Eqs. 2.8 and
2.11, we obtain Eq. 2.12 below:
dH + k(R w dT − dH ) = 0
(2.12)
Working the expression out:
dH + k R w dT − kdH = 0
dH − kdH + k R w dT = 0
(1 − k)dH + k R w dT = 0
Integrating:
(1 − k) ∫ dH + k R w ∫ dT = 0
(1 − k)H + k R w T + C = 0
We obtain:
T =
(k − 1)H
k R w
− C
If we establish initial values of H 0 = 0 and T = T 0 , we find that C = −T 0 . If an
adiabatic process is considered, the equation will be (Eq. 2.13):
T =
(k − 1)H
k R w
+ T 0
(2.13)
As for dry air R w = 29.29 kpm (kg K)
-1 and k = 1.41, we obtain (Eq. 2.14):
T = T 0 + 0.0098 H
(2.14)
14 d(P V k ) = dP V k + P k V k−1 dV = dP V V k−1 + P k V k−1 dV = 0 → V dP + P k dV = 0.