Chapter 11, Problem 11.88P

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To determine

(a)

The type of the turbine.

Answer to Problem 11.88P

Francis turbine

Explanation of Solution

Given Information:

$$

The given table is given for water turbine having the performance data that operates with 49ft head.

Q(m3 /h)	18.7	18.7	18.5	18.3	17.6	16.7	15.1	11.5
RPM	0	500	1000	1500	2000	2500	3000	3500
$$ $\eta$	0	14%	27%	38%	50%	65%	61%	11%

Concept Used:

Brake horse power,

$bhp=\rho gQH\eta$

$\begin{array}{l}Q=flow\\ H=head\\ \eta =efficiency\\ \rho =density\end{array}$

Calculation:

According to the equation,

$bhp=\rho gQH\eta$

$\begin{array}{l}=62.4*0.164*49*0.65\\ =326ft.lbf/s\\ =0.593hp\end{array}$

Specified speed is given as:

$N_{sp}=\frac{rpm*{(bhp)}^{1/2}}{{(H)}^{5/4}}$

$=\frac{2500*{(0.593)}^{1/2}}{{(49)}^{5/4}}\approx 15$

As, the speed with specified power is approximately 15, the type of the turbine is Francis turbine.

Conclusion:

Thus, the type of the turbine is determined.

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To determine

(b)

To compare:

The dimensionless performance plot with the data.

Answer to Problem 11.88P

The given data is different as the speed varies.

Explanation of Solution

Given Information:

$$

The turbine which is similar has the head 150ft and the flow 6.7 ft³ /m.

Concept Used:

Brake horse power,

$bhp=\rho gQH\eta$

$\begin{array}{l}Q=flow\\ H=head\\ \eta =efficiency\\ \rho =density\end{array}$

Calculation:

The given data is different as the speed varies. The other data has the speed which is constant.

Conclusion:

Thus, the dimensionless performance plot with the data is compared.

c,d,e

To determine

The most efficient among the turbine diameter, rotation speed and horsepower.

Answer to Problem 11.88P

Horse power is efficient.

Explanation of Solution

Given Information:

$$

The given table is given for water turbine having the performance data that operates with 49ft head:

Q(m3 /h)	18.7	18.7	18.5	18.3	17.6	16.7	15.1	11.5
RPM	0	500	1000	1500	2000	2500	3000	3500
$$ $\eta$	0	14%	27%	38%	50%	65%	61%	11%

Concept Used:

Flow coefficient,

$C_Q=\frac{Q}{n{D}^3}$

Head coefficient,

$C_H=\frac{gH}{n^2{D}^2}$

Power coefficient:

$C_P=\frac{P}{\rho n^3{D}^5}$

Calculation:

Substituting we have:

Flow coefficient:

$C_Q=\frac{Q}{n{D}^3}$

$=\frac{\left(\frac{16.7}{3600}\right)}{\frac{2500}{60}*{0.0825}^3}\Rightarrow \frac{\left(0.0046\right)}{41.667*{0.0825}^3}=0.198$

Head coefficient:

$C_H=\frac{gH}{n^2{D}^2}$

$=\frac{9.81*49*0.3048}{{(2500/60)}^2{0.0825}^2}=\frac{146.51}{{(41.66)}^2{0.0825}^2}\Rightarrow 12.4$

Power coefficient:

$C_P=\frac{P}{\rho n^3{D}^5}$

$\begin{array}{l}=\frac{326*1.3558W}{998*{(2500/60)}^3{0.0825}^5}\\ =\frac{326*1.3558W}{998*{(41.66)}^3{0.0825}^5}\\ =1.60\end{array}$

With the new data the flow and head coefficients, we have:

$C_Q=\frac{Q}{n{D}^3}\Rightarrow \frac{6.7}{904*{15.7}^3}=0.198$

$C_H=\frac{32.2*150}{{904}^2*{15.7}^2}\Rightarrow 12.4$

$P_2=C_p\rho n_2^3{D}_2^5\Rightarrow 1.60*1.94*{15.1}^3*{1.31}^5$

$=41000ft.lbf/s=74hp$

Using the Moody step-up formula, we have:

$(1-\eta 2)\approx (1.065)/{(4.84)}^{1/4}$

$=0.236$

$\eta 2=0.764$

$=76.4\%$

So, from the new turbine the power is more than the large turbine.

$P2=74hp\left(\frac{76\%}{65\%}\right)$

$=86.5hp$

$\approx 87hp$

Conclusion:

Thus, the most efficient among the turbine diameter, rotation speed and horsepower is determined.