Answer:
Mass flow rate
![\dot{m} = 0.064 (kg)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/c65og9cw36qcpuqdrf7smmg927h0q3766e.png)
Step-by-step explanation:
We know that when fluid is flowing through pipe the value of Reynolds number if 2100 for laminar flow.If Reynolds number is less than 2100 then flow will be laminar flow.
We know that Reynolds number
![Re = (VD)/(\\u )](https://img.qammunity.org/2020/formulas/engineering/college/s7ldmet1rzi5vnupfpqcdtxlrgeqdmhjli.png)
Properties of air at 15°C from standard table
![\\u = 1.5* 10^(-5)](https://img.qammunity.org/2020/formulas/engineering/college/qd2wlmei7i5wx3a8ox47r9u4xh2i6af2vu.png)
Now by putting the values
![2100 = (V* 0.015)/( 1.5* 10^(-5))](https://img.qammunity.org/2020/formulas/engineering/college/sh1qxcyuzxapb3swf7e8jtor48fx3h6otd.png)
So V=2.1 m/s
We know mass flow rate given as
![\dot{m}=\rho AV](https://img.qammunity.org/2020/formulas/engineering/college/4y4h0cfk735tobasq45ciaj4wcssnbejsi.png)
We know that P=ρRT
1500=ρ(0.287)(273+15)
![\rho =18.14 (kg)/(m^3)](https://img.qammunity.org/2020/formulas/engineering/college/v1awryhqqc6j0sfpjkx7n3hvcifn93v6td.png)
![A=(\pi )/(4)* 0.015^2 m^2](https://img.qammunity.org/2020/formulas/engineering/college/3gj2ghplctl7ktbjxgnhemyv50lduqszek.png)
![A=0.00017 m^2](https://img.qammunity.org/2020/formulas/engineering/college/o1xa5wop5t2ajg3l2a87fsh44zjwzy8ai9.png)
![\dot{m}=18.14 * 0.00017 * 2.1](https://img.qammunity.org/2020/formulas/engineering/college/razcqavnpmxpk9fcqoy02ps7ox4fcefoe6.png)
![\dot{m}=0.064(kg)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/7eh3iqicjkavk0ewnskzgfy65edj12m8n1.png)
So mass flow rate
![\dot{m} = 0.064 (kg)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/c65og9cw36qcpuqdrf7smmg927h0q3766e.png)