Question

Air flows steadily through a turbine that produces 3.5x 105 ft-lbf/s of work. Using the data below at the inlet and outlet, where the inlet is 10 feet below the outlet, please calculate the heat transferred in units of BTU/hr. You may assume steady flow and ignore viscous work.

Answer 1

The heat transfer rate through the turbine is $\bold{6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}}$.

Explanation:

Express the final form of the overall energy equation.

$\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\iint_{c}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A+\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V+\frac{\delta W_{\mu}}{d t} \ldots \ldots(1)$

Here, the rate of heat addition to the control volume is $\frac{\delta Q}{d t}$, shaft work rate is $\frac{\delta W_{s}}{d t}$, rate of accumulation of energy inside control volume is $\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V$, rate of work accomplished in viscous effects at the surface is $\frac{\delta W_{\mu}}{d t}$, and the net energy efflux from the control volume is

$\iint_{c, s,}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A$

Write the following assumption.

There is no viscous work in the system,  $\frac{\delta W_{\mu}}{d t}=0$

Steady state flow of the system, $\frac{\partial}{\partial t} \iiint_{c, v} e \rho d V=0$

Use the assumption condition in Equation (1).

 $\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\iint_{z}\left(e+\frac{P}{\rho}\right) \rho(\mathbf{v} \cdot \mathbf{n}) d A$

Integrate the above equation.

$\begin{array}{l}\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\rho v A\left[e+\frac{P}{\rho}\right]_{2}-\rho v A\left[e+\frac{P}{\rho}\right] \\=\rho_{1} Q\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right] \\=\rho_{1} A_{1} v_{1}\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right]\end{array}$

Here, internal energy at entry section is $u_{1}$ internal energy at exit section is $u_{2}$  volumetric flow rate is Q, inlet fluid density is $\rho_{1}$, outlet fluid density is $\rho_{2}$, specific energy is e, exit velocity is $v_{2}$, inlet velocity is $v_{1}$, inlet pressure is $P_{1}$, outlet pressure is $P_{2}$, acceleration due to gravity is g and distance between inlet and outlet is z.

Calculate the cross section area at entry section.

$A_{1}=\frac{\pi}{4} d_{1}^{2}$

Here, inlet diameter of pipe is $d_{1}$

Substitute 0.962 ft for $d_{1}$

$\begin{aligned}&A_{1}=\frac{\pi}{4}(0.962 \mathrm{ft})^{2}\\&A_{1}=0.726 \mathrm{ft}^{2}\end{aligned}$

Calculate the heat transfer rate through the turbine.

Substitute  $0.08101 \mathrm{b}_{m} / \mathrm{ft}^{3} \text { for } \rho_{2}, 0.05341 \mathrm{b}_{m} / \mathrm{ft}^{3} \text { for } \rho_{1}, 0.726 \mathrm{ft}^{2} \text { for } A_{1}, 244 \mathrm{ft} / \mathrm{s} \text { for } v_{2}$

$100 \mathrm{ft} / \mathrm{s} \text { for } v_{1}, 3.5 \times 10^{5} \mathrm{Ib}_{m} \mathrm{ft}^{2} / \mathrm{s}^{3} \text { for } \frac{\delta W_{s}}{d t},-16.16 \times 10^{5} \mathrm{ft}^{2} / \mathrm{s}^{2} \text { for } u_{2}-u_{1}, 32.2 \mathrm{ft} / \mathrm{s}^{2} \text { for }$

$g, 400 \mathrm{Ib}_{\mathrm{r}} / \mathrm{in}^{2} \text { for } P_{2}, 150 \mathrm{lb}_{\mathrm{r}} / \mathrm{in}^{2} \text { for } P_{1}, \text { and } 10 \mathrm{ft} \text { for } z_{2}-z_{1} \text { in Equation }(2)$

$\frac{\delta Q}{d t}-\frac{\delta W_{s}}{d t}=\rho_{1} A_{1} v_{1}\left[u_{2}-u_{1}+\frac{v_{2}^{2}-v_{1}^{2}}{2}+\left(P_{2} / \rho_{2}-P_{1} / \rho_{1}\right)+g\left(z_{2}-z_{1}\right)\right]$

$\begin{aligned}&\frac{\delta Q}{d t}=\left(9.98 \times 10^{5}+3.5 \times 10^{5}\right) \mathrm{ftlb}_{\mathrm{f}} / \mathrm{s}\\&=13.48 \times 10^{5} \mathrm{ftb}_{\mathrm{f}} / \mathrm{s}\left[\frac{1 \mathrm{Ba}}{778.17 \mathrm{ftbr}}\right]\left[\frac{3600 \mathrm{s}}{1 \mathrm{hr}}\right]\\&=6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}\end{aligned}$

Hence, the heat transfer rate through the turbine is $6.23 \times 10^{6} \mathrm{Btu} / \mathrm{hr}$.