Respuesta :
The heat transfer rate through the turbine is [tex]\bold{6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}}[/tex].
Explanation:
Express the final form of the overall energy equation.
[tex]\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\iint_{c}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A+\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V+\frac{\delta W_{\mu}}{d t} \ldots \ldots(1)[/tex]
Here, the rate of heat addition to the control volume is [tex]\frac{\delta Q}{d t}[/tex], shaft work rate is [tex]\frac{\delta W_{s}}{d t}[/tex], rate of accumulation of energy inside control volume is [tex]\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V[/tex], rate of work accomplished in viscous effects at the surface is [tex]\frac{\delta W_{\mu}}{d t}[/tex], and the net energy efflux from the control volume is
[tex]\iint_{c, s,}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A[/tex]
Write the following assumption.
There is no viscous work in the system, [tex]\frac{\delta W_{\mu}}{d t}=0\\[/tex]
Steady state flow of the system, [tex]\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V=0[/tex]
Use the assumption condition in Equation (1).
[tex]\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\iint_{z}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A[/tex]
Integrate the above equation.
[tex]\begin{array}{l}\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\rho v A\left[e+\frac{P}{\rho}\right]_{2}-\rho v A\left[e+\frac{P}{\rho}\right] \\=\rho_{1} Q\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right] \\=\rho_{1} A_{1} v_{1}\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right]\end{array}[/tex]
Here, internal energy at entry section is [tex]u_{1}[/tex] internal energy at exit section is [tex]u_{2}[/tex] volumetric flow rate is Q, inlet fluid density is [tex]\rho_{1}[/tex], outlet fluid density is [tex]\rho_{2}[/tex], specific energy is e, exit velocity is [tex]v_{2}[/tex], inlet velocity is [tex]v_{1}[/tex], inlet pressure is [tex]P_{1}[/tex], outlet pressure is [tex]P_{2}[/tex], acceleration due to gravity is g and distance between inlet and outlet is z.
Calculate the cross section area at entry section.
[tex]A_{1}=\frac{\pi}{4} d_{1}^{2}[/tex]
Here, inlet diameter of pipe is [tex]d_{1}[/tex]
Substitute 0.962 ft for [tex]d_{1}[/tex]
[tex]\begin{aligned}&A_{1}=\frac{\pi}{4}(0.962 \mathrm{ft})^{2}\\&A_{1}=0.726 \mathrm{ft}^{2}\end{aligned}[/tex]
Calculate the heat transfer rate through the turbine.
Substitute [tex]0.08101 \mathrm{b}_{m} / \mathrm{ft}^{3} \text { for } \rho_{2}, 0.05341 \mathrm{b}_{m} / \mathrm{ft}^{3} \text { for } \rho_{1}, 0.726 \mathrm{ft}^{2} \text { for } A_{1}, 244 \mathrm{ft} / \mathrm{s} \text { for } v_{2}[/tex]
[tex]100 \mathrm{ft} / \mathrm{s} \text { for } v_{1}, 3.5 \times 10^{5} \mathrm{Ib}_{m} \mathrm{ft}^{2} / \mathrm{s}^{3} \text { for } \frac{\delta W_{s}}{d t},-16.16 \times 10^{5} \mathrm{ft}^{2} / \mathrm{s}^{2} \text { for } u_{2}-u_{1}, 32.2 \mathrm{ft} / \mathrm{s}^{2} \text { for }[/tex]
[tex]g, 400 \mathrm{Ib}_{\mathrm{r}} / \mathrm{in}^{2} \text { for } P_{2}, 150 \mathrm{lb}_{\mathrm{r}} / \mathrm{in}^{2} \text { for } P_{1}, \text { and } 10 \mathrm{ft} \text { for } z_{2}-z_{1} \text { in Equation }(2)[/tex]
[tex]\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\rho_{1} A_{1} v_{1}\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right][/tex]
[tex]\begin{aligned}&\frac{\delta Q}{d t}=\left(9.98 \times 10^{5}+3.5 \times 10^{5}\right) \mathrm{ftlb}_{\mathrm{f}} / \mathrm{s}\\&=13.48 \times 10^{5} \mathrm{ftb}_{\mathrm{f}} / \mathrm{s}\left[\frac{1 \mathrm{Ba}}{778.17 \mathrm{ftbr}}\right]\left[\frac{3600 \mathrm{s}}{1 \mathrm{hr}}\right]\\&=6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}\end{aligned}[/tex]
Hence, the heat transfer rate through the turbine is [tex]6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}[/tex].
