Question 3.78.E: Steam exiting the turbine of a steam power plant at 100°F is...

Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper pipes (k = 223  Btu / h \cdot ft \cdot{ }^{\circ} F) of inner diameter 0.4 in. and outer diameter 0.6 in. at an average temperature of 70°F. The heat of vaporization of water at 100°F is 1037  Btu / lbm. The heat transfer coefficients are 1500  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F on the steam side and 35  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F on the water side. Determine the length of the tube required to condense steam at a rate of 120  lbm / h.

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Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper tubes. For specified heat transfer coefficients, the length of the tube required to condense steam at a rate of 400 lbm/h is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal properties are constant. 4 Heat transfer coefficients are constant and uniform over the surfaces.

Properties The thermal conductivity of copper tube is given to be k=223  Btu / h \cdot ft \cdot{ }^{\circ} F. The heat of vaporization of water at 100°F is given to be 1037 Btu/lbm.

Analysis The individual resistances are

A_{i}=\pi D_{i} L=\pi(0.4 / 12  ft )(1  ft )=0.105  ft ^{2}

 

A_{o}=\pi D_{o} L=\pi(0.6 / 12  ft )(1  ft )=0.157  ft ^{2}

 

R_{i}=\frac{1}{h_{i} A_{i}}=\frac{1}{\left(35  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F \right)\left(0.105  ft ^{2}\right)}=0.27211  h \cdot ^{\circ} F / Btu

 

R_{\text {pipe }}=\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi k L}=\frac{\ln (3 / 2)}{2 \pi\left(223  Btu / h \cdot ft \cdot{ }^{\circ} F \right)(1  ft )}=0.00029  h \cdot ^{\circ} F / Btu

 

R_{o}=\frac{1}{h_{o} A_{o}}=\frac{1}{\left(1500  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F \right)\left(0.157  ft ^{2}\right)}=0.00425  h \cdot ^{\circ} F / Btu

 

R_{\text {total }}=R_{i}+R_{\text {pipe }}+R_{o}=0.27211+0.00029+0.00425=0.27665  h \cdot ^{\circ} F / Btu

 

The heat transfer rate per ft length of the tube is

\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{\text {total }}}=\frac{(100-70)^{\circ} F }{0.27665^{\circ}  F / Btu }=108.44  Btu / h

 

The total rate of heat transfer required to condense steam at a rate of 400 lbm/h and the length of the tube required is determined to be

\dot{Q}_{\text {total }}=\dot{m} h_{f g}=(120  lbm / h )(1037  Btu / lbm )=124,440  Btu / h

 

\text { Tube length }=\frac{\dot{Q}_{\text {total }}}{\dot{Q}}=\frac{124,440}{108.44}=1148  ft
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