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8 Secondary Ventilation
Question 8.1 Taking Robertson and Wharton´s (1980) expression for leakage:
L c =
3(Q 1 − Q 2 )(P 1 − P 2 )
2L
P
1.5
1 − P
1.5
2
Demonstrate that this expression is dimensionally equivalent to the general air
leakage law:
Q p = K
√
P
where K is a constant.
Answer
The numerator can be simplified to give: 3ΔQΔP = K 1 ΔQ ΔP.
The denominator, on the other hand, is: K 2 (ΔP
1.5 ), that is:
L c =
3QP
2L((P 1.5 )
= K 3
Q
P
0.5
So, ignoring the Δ notation since this does not affect the dimensions of the
expression and then rearranging:
P
0.5
=
K 3
L c
Q = K 4 Q
Solving for P, if K 4
2
= R, we get:
P = R Q
2
Or solving for Q
Q = K
√
P
Question 8.2 Transform the coefficients θ from Table 8.5, so that the expression 8.36, can be used with SI units. Assume that θ has been determined under
turbulent conditions.
Answer
Taking Eq. 8.36, and solving for θ:
θ =
Q
Lh
γ
i
8 Secondary Ventilation
Question 8.1 Taking Robertson and Wharton´s (1980) expression for leakage:
L c =
3(Q 1 − Q 2 )(P 1 − P 2 )
2L
P
1.5
1 − P
1.5
2
Demonstrate that this expression is dimensionally equivalent to the general air
leakage law:
Q p = K
√
P
where K is a constant.
Answer
The numerator can be simplified to give: 3ΔQΔP = K 1 ΔQ ΔP.
The denominator, on the other hand, is: K 2 (ΔP
1.5 ), that is:
L c =
3QP
2L((P 1.5 )
= K 3
Q
P
0.5
So, ignoring the Δ notation since this does not affect the dimensions of the
expression and then rearranging:
P
0.5
=
K 3
L c
Q = K 4 Q
Solving for P, if K 4
2
= R, we get:
P = R Q
2
Or solving for Q
Q = K
√
P
Question 8.2 Transform the coefficients θ from Table 8.5, so that the expression 8.36, can be used with SI units. Assume that θ has been determined under
turbulent conditions.
Answer
Taking Eq. 8.36, and solving for θ:
θ =
Q
Lh
γ
i