I would simply modify the Bernoulli's equation for an incompressible fluid
$$ p_i + \rho g h_i + \frac{\rho u_i^2}{2} = const$$
to account for pressure loss terms $\Delta p_j$ corresponding to pipe friction and maybe also entrance losses and losses from a couple of 45° elbow bends due to a bent pipe
$$ p_1 + \rho g h_1 + \frac{\rho u_1^2}{2} = p_2 + \rho g h_2 + \frac{\rho u_2^2}{2} + \underbrace{\sum\limits_j \Delta p_j}_\text{Losses}.$$
where in your case $p_1 = p_2 = p_0$ is the atmospheric pressure and $h_1 = h$ and $h_2 = 0$. You can find corresponding empirical K-values for the corresponding losses in the literature. Likely the cross-section of the tank $A_1$ is significantly bigger than the pipe $A_2$ and therefore you might assume that $u_1 \approx 0$, else you might use the continuity equation for an incompressible fluid
$$A_1 u_1 = A_2 u_2$$
to determine the correlation between the speeds in the two cross-sections.
Only considering turbulent flow in a straight pipe with a corresponding pipe friction this pressure drop term is given
$$\Delta p = \underbrace{f \frac{L}{D}}_K \frac{\rho u_2^2}{2}$$
where the friction factor $f$ depends on the Reynolds number inside the pipe with diameter $D$ and length $L$ and a fluid with kinematic viscosity $\nu$ (for water $10^{-6} \frac{m^2}{s}$)
$$Re_D := \frac{u_2 D}{\nu}$$
and the surface roughness and thus the equation
$$ g h + \left( \frac{A_2}{A_1} \right)^2 \frac{u_2^2}{2} = \frac{u_2^2}{2} + f \frac{L}{D} \frac{u_2^2}{2}.$$
has to be solved iteratively for $u_2$. The second term again may be neglected if $A_1 >> A_2$.
Best Answer
Use Bernoulli's equation to derive Torricelli's Law (check any website for this) for the velocity out of the hole; $ v = \sqrt{2 g h(t)} $, where g is gravity and $ h(t) $ is the height of the fluid in the tank at any time.
Write a balance on the mass of the fluid in the tank as:
$$ \text{in - out + gen = accumulation} $$ $$ \rho Q_{in} - \rho Q_{out} = \frac{d(\rho V)}{dt} $$ where the generation term is zero, $\rho$ is the fluid density (constant here) and $Q_{in}$ and $Q_{out}$ is the flow rate in and out of the tank, respectively. $Q_{in}$ is zero so we get: $$ \frac{dV}{dt} = -Q_{out} $$ The flow out is $v A = \sqrt{2 g h(t)} A$, where $A$ is the area of the hole which is calculated by knowing the diameter of the circular hole; given in the problem statement as 1 inch.
The volume of the tank, $V = a_t h(t)$, is the height, $h(t)$ times the area, $a_{t}$.
Putting it all together, we get the separable first order differential equation for the height of the fluid in the tank versus time:
$$ \frac{dh}{dt} = -\frac{A}{a_t}\sqrt{2g}\sqrt{h}$$
Prepare it for Integration $$ \frac{dh}{\sqrt{h}} = -\frac{A}{a_t}\sqrt{2g}{dt}$$
Integrate the equation. The upper bound for $dh$ is $h(t)$. The lower bound is $h(0)=H$ . For $dt$ we integrate from $t$ to $0$: $$ 2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{2g} t$$ Solve for $h(t)$ $$ h(t)=[\sqrt{H} -\frac{A}{2 a_t}\sqrt{2g} t]^2 $$
To find the time when the tank empties, set $h$ equal to zero and solve for $t$:
$$ T= \sqrt{\frac{H}{2g}} \frac{2 a_t}{A}$$
Plug that back in the previous equation to clarify time dependence on t: $$ h(t)=H[1 - t/T]^2 $$
The times to empty the same tank for two different starting heights, $H_1$ and $H_2$ is:
$$ \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} $$
So, finally, if $H_2 = 2 H_1$ as in the problem statement, then the time to empty the tank is not double, but $\sqrt{2}$ times longer.
Make sense?
Paul Safier