Given: \( \gamma_{h}=1.4 \)
\( V_{1}\approx 0 \), \( T_{1} = T_{0} = 2000^{\circ}\,\text{C} = 2273\,\text{K} \), \( p_{1}=p_{0}=6.8\,\text{MPa} \),
\( p_{e} = p_{\text{amb}} = 10\,\text{kPa} \), \( \text{Thrust} = 10\,\text{MN} \)

To calculate: hydrogen mass flow rate?

The schematic diagram of the problem description is shown in Fig. 1.

Using the molar mass of hydrogen as \( \hat{m}=2.016\,\text{kg/kmol} \) and universal gas constant as \( R_{u}=8314\,\text{J/kmol-K} \), the gas constant of hydrogen can be calculated as, $$ R=\frac{R_{u}}{\hat{m}} = \frac{8314}{2.016} = 4124\,\text{J/kg-K}\ . $$ The heat capacity of hydrogen at constant pressure can be calculated as, $$ c_{p}=\frac{\gamma_{h}R}{\gamma_{h}-1} $$ $$ c_{p} = \frac{1.4\times4124}{1.4-1} = 14434\,\text{J/kg-K}\ . $$

Since the flow through the nozzle can be assumed to be isentropic, $$ \frac{T_{e}}{T_{0}}=\left(\frac{p_{e}}{p_{0}}\right)^{\left(\gamma_{h}-1\right)/\gamma_{h}} $$ $$ T_{e}=T_{0}\left(\frac{p_{e}}{p_{0}}\right)^{\left(\gamma_{h}-1\right)/\gamma_{h}} $$ $$ T_{e}=2273\times\left(\frac{10\times10^{3}}{6.8\times10^{6}}\right)^{\left(1.4-1\right)/1.4}=352.623\,\text{K}\ . $$

Using the conservation of energy equation,

$$ c_{p}T_{0}=c_{p}T_{e}+\frac{V_{e}^{2}}{2} $$ $$ 14434\times2273=14434\times352.623+\frac{V_{e}^{2}}{2} $$

which can be solved for exit velocity as,

$$ V_{e}=\sqrt{2\times\left(14434\times2273-14434\times352.623\right)}=7445.6325\,\text{m/s}\ . $$

Using the conservation of momentum, since the exit plane pressure is same as ambient,

$$ \text{Thrust} = \dot{m}_{e}V_{e} $$

Therefore, the mass flow rate is,

$$ \dot{m}_{e}=\frac{\text{Thrust}}{V_{e}}=\frac{10\times10^{6}}{7445.6325} $$ $$ \boxed{\dot{m}_{e}=1343.07\,\text{kg/s}}\ . $$