A heating system for a solar house is designed to circulate 120°F hot water from roof-mounted solar collectors through a horizontal heat exchanger embedded in the concrete ﬂoor of the living space as shown in Figure 13.32. Control is provided by a thermostatically activated motorized 3/4 in. globe

Question

A heating system for a solar house is designed to circulate 120°F hot water from roof-mounted solar collectors through a horizontal heat exchanger embedded in the concrete ﬂoor of the living space as shown in Figure 13.32. Control is provided by a thermostatically activated motorized [latex]\frac{3}{4}[/latex] in. globe valve in the bypass line that allows the hot water to return to the solar collectors. The heat exchanger consists of [latex]\frac{3}{4}[/latex] in. Schedule 40 PVC pipe with a total length of 988 ft, and 28 standard 90°elbows, while the bypass line consists of 28 ft of the same pipe. The system is designed for a total ﬂowrate of 80 gal/min. Find the ﬂowrate through the heat exchanger when the globe valve is one-half open.

Accepted Answer

We are asked to determine the ﬂowrate through one branch of a speciﬁed multiple path ﬂow system for a set of deﬁned conditions. Figure 13.32 serves a sketch of the ﬂow system. We will assign node 1 at the inlet of the tee upstream of the heat exchanger and node 2 at the outlet of the tee just downstream of the heat exchanger. Branch 1 contains the branch run of the upstream tee, the heat exchanger, and the branch run of the downstream tee. Branch 2 contains the line run of the upstream tee, the bypass line and globe valve (GV), and the line run of the downstream tee. Applying the ﬁrst of the two principles governing parallel branches, we write Q = Q₁ + Q₂, or

Q₂ = Q - Q₁ (A)

where Q = 80 gal/min [(1 min/60 s)(1 ft³/7.48 gal)] = 1.78 × 10⁻¹ ft³/s. Note that in this case the piping is all of the same diameter, thus we also have [latex]\bar V =\bar V_1 +\bar V_2[/latex]. We now write the total head loss in each branch, accounting for the major and minor losses to obtain

[latex]h_{T1}=2f_1\frac{L_{tee-branch}}{D}\frac{\bar V^2_1}{2}+f_1\frac{L_{1}}{D}\frac{\bar V^2_1}{2}+28f_1\frac{L_{elbow}}{D}\frac{\bar V^2_1}{2}[/latex]

[latex]h_{T2}=2f_2\frac{L_{tee-line}}{D}\frac{\bar V^2_2}{2}+f_2\frac{L_{2}}{D}\frac{\bar V^2_2}{2}+K_{GV}\frac{\bar V^2_2}{2}[/latex]

For convenience, we will write these equations in terms of the ﬂowrates as

[latex]h_{T1}=\left[2f_1\frac{L_{tee-branch}}{D}+f_1\frac{L_{1}}{D}+28f_1\frac{L_{elbow}}{D} ight]\frac{8Q^2_1}{\pi D^2} [/latex]

[latex]h_{T2}=\left[2f_2\frac{L_{tee-line}}{D}+f_2\frac{L_{2}}{D}+f_2\frac{L_{GV}}{D} ight]\frac{8Q^2_2}{\pi D^2} [/latex]

where L_tee-branch/D = 60, L_tee-line/D = 20, L_elbow/D = 30, and L_GV/D = 500 from Table 13.6. Also, from the problem statement we calculate L₁/D = (988 ft/0.06867 ft) = 1.44 × 10⁴ and L₂/D = (28 ft/0.0687 ft) = 408. Thus these two equations become

[latex]h_{T1} = (120 f_1 + 1.44 × 10^4 f_1 + 840 f_1) \frac{8Q^2_1}{\pi D^2} = 1.54 × 10^4 f_1\frac{8Q^2_1}{\pi D^2} [/latex] (B)

[latex]h_{T2}= (40 f_2 + 408 f_2 + 500 f_2)\frac{8Q^2_2}{\pi D^2}= 948 f_2\frac{8Q^2_2}{\pi D^2} [/latex] (C)

Finally, applying the second principle that the total head loss in parallel branches is the same, we write h_T1 = h_T2, and equate (B) and (C) to obtain

[latex]1.54 × 10^4 f_1 \frac{8Q^2_1}{\pi D^2}= 948 f_2 \frac{8Q^2_2}{\pi D^2} [/latex]

Solving for Q₁, after using (A) to eliminate Q₂, we obtain

[latex]Q_1=\frac{Q}{1+\sqrt{\frac{948\ f_2}{1.54 imes 10^4\ f_1}}}[/latex] (D)

We will now use (D) to iterate and solve for Q₁, beginning with a guess for the friction factors. To come up with an initial guess for the friction factors, suppose the ﬂowrate in the branches is the same, and equal to half the total ﬂowrate. For [latex]\frac{3}{4}[/latex] in. Schedule 40 pipe, we ﬁnd D = 0.06867 ft, and A = 3.70 × 10⁻³ ft² (Table 13.4). The average velocity in each branch is

[latex]\bar V=\frac{1}{2}\frac{Q}{A}=\frac{1.78×10^{−1}\ \mathrm{ft} ^3/s }{2(3.70×10^{−3}\ \mathrm{ft} ^2)}=24.0\ \mathrm{ft} /s[/latex]

Using data for water at 120°F, this corresponds to Re = [latex]\bar V D/ u[/latex] = [(24.0 ft/s)× (0.06867 ft)]/(6.067 × 10⁻⁶ ft²/s) = 2.7 × 10⁵, and a friction factor f = 0.014, where we have used the Moody chart and assumed a smooth pipe.

Our ﬁrst iteration now begins by assuming [latex]Q^{(1)}_1 = Q^{(1)}_2 =8.9×10^{−2} \mathrm{ft} ^3/s,\bar V^{(1)}_1 =\bar V^{(1)}_2 =24.0\ \mathrm{ft} ^2/s,Re^{(1)}_1 =Re^{(1)}_2 =2.7×10^5[/latex] and [latex]f^{(1)}_1 =f^{(1)}_2 =0.014[/latex]. Inserting the latter into (D), we ﬁnd

[latex]Q^{(2)}_1=\frac{Q}{1+\sqrt{\frac{948(0.014)}{1.54 imes 10^4(0.014)}}}=0.8Q[/latex]

and from (A) we have [latex]Q^{(2)}_2 =0.2 Q[/latex]. The next iteration now proceeds by using [latex]Q^{(2)}_1 =0.8Q = 0.142 \mathrm{ft} ^3/s, Q^{(2)}_2=0.2Q = 3.56 × 10^{-2} \mathrm{ft} ^3/s,\bar V^{(2)}_1 =38.5 \mathrm{ft} /s, \bar V^{(2)}_2 =9.6 \mathrm{ft} /s, Re^{(2)}_1 =4.4 ×10^5, Re^{(2)}_2 =1.1×10^5[/latex], [latex]f^{(2)}_1 =0.013, ext{and} f^{(2)}_2 =0.018[/latex]. Inserting these values into (D) we ﬁnd

[latex]Q^{(3)}_1=\frac{Q}{1+\sqrt{\frac{948(0.018)}{1.54 imes 10^4(0.013)}}}=0.77Q[/latex]

and from (A) we have [latex]Q^{(3)}_2[/latex] = 0.23Q. These new values turn out to be so close to the earlier values that the friction factors are unchanged. We therefore consider the iteration to have converged, and the ﬁnal values for the ﬂowrates are Q₁ = 0.77Q = 61.6 gal/min through the heat exchanger, and Q₂ = 0.23Q = 18.4 gal/min through the bypass line. The values of various parameters at each iteration are contained in the following table.

Iteration	Q₁	Q₂	[latex]\bar V_1(\mathrm{ft} /s) [/latex]	[latex]\bar V_2(\mathrm{ft} /s) [/latex]	Re₁	Re₂	f₁	f₂
1	0.5Q	0.5Q	24.0	24.0	2.7 × 10⁵	2.7 × 10⁵	0.014	0.014
2	0.8Q	0.2Q	38.5	9.6	4.4 × 10⁵	1.1 × 10⁵	0.013	0.018
3	0.77Q	0.23Q	37.0	11.1	4.2× 10⁵	1.3 × 10⁵	0.013	0.018

TABLE 13.4 Actual Dimensions of Common Nominal Pipe Sizes

Nominal Diameter		Inside Diameter		Flow Area
(in.)	Schedule	ft	cm	ft²	cm²
1/8	40	0.02242	0.683	0.0003947	0.3664
1/8	80	0.01792	0.547	0.0002522	0.2350
1/4	40	0.03033	0.924	0.0007227	0.6706
1/4	80	0.2517	0.768	0.0004974	0.4632
3/8	40	0.04108	1.252	0.001326	1.233
3/8	80	0.03525	1.074	0.0009759	0.9059
1/2	40	0.05183	1.580	0.002110	1.961
	80	0.04550	1.386	0.001626	1.508
	160	0.03867	1.178	0.001174	1.090
3/4	40	0.06867	2.093	0.003703	3.441
	80	0.06183	1.883	0.003003	2.785
	160	0.03867	1.178	0.002043	1.898
1	40	0.08742	2.664	0.006002	5.574
	80	0.07975	2.430	0.004995	5.083
	160	0.06792	2.070	0.003623	3.365
1 1⁄4	40	0.1150	3.504	0.01039	9.643
	80	0.1065	3.246	0.008908	8.275
	160	0.09667	2.946	0.007339	6.816
1 1⁄2	40	0.1342	4.090	0.01313	13.13
	80	0.1250	3.810	0.01227	11.40
	160	0.1115	3.398	0.009764	9.068
2	40	0.1723	5.252	0.02330	21.66
	80	0.1616	4.926	0.02051	19.06
	160	0.1306	4.286	0.01552	13.43
2 1⁄2	40	0.2058	6.271	0.03325	30.89
	80	0.1936	5.901	0.02943	27.35
	160	0.1771	5.397	0.02463	22.88
3	40	0.2557	7.792	0.05134	47.69
	80	0.2417	7.366	0.04587	42.61
	160	0.2187	6.664	0.03755	34.88
3 1⁄2	40	0.2957	9.012	0.06866	63.79
3 1⁄2	80	0.2803	8.544	0.06172	57.33
4	40	0.3355	10.23	0.08841	82.19
	80	0.3198	9.718	0.07984	74.17
	120	0.3020	9.204	0.07163	66.54
	160	0.2865	8.732	0.06447	59.88
5	40	0.4206	12.82	0.1389	129.10
	80	0.4801	13.64	0.1810	168.30
	120	0.3803	11.59	0.1136	105.50
	160	0.3594	10.95	0.1015	94.17
6	40	0.5054	15.41	0.2006	186.50
	80	0.4801	13.64	0.1810	168.30
	120	0.4584	13.98	0.1650	153.50
	160	0.4823	13.18	0.1367	136.40
8	20	0.6771	20.64	0.3601	334.60
	30	0.6726	20.50	0.3553	330.10
	40	0.6651	20.27	0.3474	322.70
	60	0.6511	19.85	0.3329	309.50
	80	0.6354	19.37	0.3171	294.70
	100	0.6198	18.89	0.3017	280.30
	120	0.5989	18.26	0.2817	261.90
	130	0.5834	17.79	0.2673	248.60
	160	0.5678	17.31	0.2532	235.30
10	20	0.8542	26.04	0.5730	332.60
	30	0.8446	25.75	0.5604	520.80
	40	0.8350	25.46	0.5476	509.10
	60	0.8125	24.77	0.5185	481.90
	80	0.7968	24.29	0.4987	463.40
	100	0.7760	23.66	0.4730	439.70
	120	0.7552	23.02	0.4470	416.20
	130	0.7292	22.23	0.4176	388.10
	160	0.7083	21.59	0.3941	366.10
12	20	1.021	31.12	0.8185	760.60
	30	1.008	30.71	0.7972	740.71
	40	0.9948	30.33	0.773	722.50
	60	0.9688	29.53	0.7372	684.90
	80	0.9478	28.89	0.7056	655.50
	100	0.9218	28.10	0.6674	620.20
	120	0.8958	27.31	0.6303	585.80
	130	0.8750	26.67	0.6013	558.60
	160	0.8438	25.72	0.5592	519.60
20	20 (std)	1.604	48.89	2.021	1877.00
20	160	1.339	40.80	1.407	1307.00

TABLE 13.6 Equivalent Lengths for Various Pipe Fittings

Pipe ﬁtting	Description	L_c/D
Ball valve	Fully open	3
Gate valve	Fully open	8
	3/4 open	35
	1/2 open	160
	1/4 open	900
Globe valve	Fully open	340
Globe valve	1/2 open	500
Elbows	45° standard	16
	90° standard	30
	90° long radius	20
180° bend	Close pattern return	50
90° bends	Bend radius = 1 pipe diameter	20
	Bend radius = 2 pipe diameter	12
	Bend radius = 3 pipe diameter	16
	Bend radius = 4 pipe diameter	24
Standard tee	With ﬂow through run	20
Standard tee	With ﬂow through branch	60

Question 13.16: A heating system for a solar house is designed to circulate ...

Related Answered Questions

Verified Answer:

You are to choose the pipe diameter and pump for the system shown in Figure 13.39. The ﬂowrate is 1 ft³/s and the pipe is Schedule 40 commercial steel. The total length including an equivalent length allowance for minor losses is 1800 ft. Choose the pump based on the pump curves shown in ...

Verified Answer:

Gasoline is siphoned from a tank with a smooth hose, i.d. 2 cm, as shown in Figure 13.29. For this grade of gasoline ρ = 719 kg/m³ and µ = 2.92 × 10^−4 (N-s)/m². Find the ﬂowrate. ...

Verified Answer:

Verified Answer:

Find the pressure at the pump inlet shown in Figure 13.22 if the ﬂowrate is 20 gal/min in the 1 in. Schedule 40 galvanized iron inlet pipe. The water is at 60°F. ...

Verified Answer:

Verified Answer:

If the sudden enlargement and contraction in the system described in Example 13.10 are replaced with a conical diffuser and conical nozzle, respectively, and the included angle of each is 10°, by how much is the frictional pressure drop reduced? ...

Verified Answer:

A small water turbine rated at 25 hp has an efficiency of 61% and operates at a ﬂowrate of 10 ft³/s. Find the turbine head and pressure drop across the turbine. ...

Verified Answer:

Verified Answer:

Verified Answer: