and vertical dimension of the model is $\frac{1}{10}$ then vertical dimension of prototype.

Solution:

Given:-

Discharge through weir (Prototype)

$Q_p=1.5m^{3}/s$

Horizontal dimension of model=$\frac{1}{50}\times $ Horizontal dimension of prototype

$\frac{Horizontal \ dimension \ of \ prototype}{Horizontal \ dimension \ of \ model}=50$

$(L_r)_{H}=50$

Vertical dimension of model

=$\frac{1}{10}\times $ vertical dimension of prototype

$\frac{vertical \ dimension \ of \ prototype}{vertical \ dimension \ of model}=10$

$(L_r)_{V}$=10

$\frac{Q_p}{Q_m}=(L_r)_{H}\times[(L_r)_V\times ]^{3/2}$

=$50\times 10 ^{3/2}$

=1581.14

$Q_m=\frac{Q_p}{1581.14}$

=$\frac{1.50}{1581.14}$

=$0.000948m^{3}/s$

$Q_{m}=0.948 lit/s$