A patient recovering from surgery is being given fluid intravenously. The fluid has a density of 1080 kg/m³, and 6.61 x 10-4 m³ of it flows into the patient every 8 hours. Find the mass flow rate in kg/s.

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**Problem Statement: Mass Flow Rate Calculation**

A patient recovering from surgery is being given fluid intravenously. The fluid has a density of 1080 kg/m³, and 6.61 x 10⁻⁴ m³ of it flows into the patient every 8 hours. Find the mass flow rate in kg/s.

**Solution:**

To find the mass flow rate, we will use the formula:

\[ \text{Mass Flow Rate} = \frac{\text{Density} \times \text{Volume Flow Rate}}{\text{Time}} \]

Given data:
- Density (\( \rho \)): 1080 kg/m³
- Volume (\( V \)): 6.61 x 10⁻⁴ m³
- Time (\( t \)): 8 hours

First, convert the time from hours to seconds:
\[ 8 \text{ hours} = 8 \times 3600 \text{ seconds} = 28,800 \text{ seconds} \]

Now, calculate the volume flow rate (\( \dot{V} \)):
\[ \dot{V} = \frac{V}{t} = \frac{6.61 \times 10⁻⁴ \text{ m³}}{28,800 \text{ s}} \]

Calculate the mass flow rate (\( \dot{m} \)):
\[ \dot{m} = \rho \times \dot{V} = 1080 \text{ kg/m³} \times \left( \frac{6.61 \times 10⁻⁴ \text{ m³}}{28,800 \text{ s}} \right) \]

Perform the calculations:
\[ \dot{V} = \frac{6.61 \times 10⁻⁴}{28,800} = 2.294097 \times 10⁻⁸ \text{ m³/s} \]

\[ \dot{m} = 1080 \text{ kg/m³} \times 2.294097 \times 10⁻⁸ \text{ m³/s} = 2.478445 \times 10⁻⁵ \text{ kg/s} \]

Therefore, the mass flow rate is approximately:
\[ 2.48 \times 10⁻⁵ \text{ kg/s} \]
Transcribed Image Text:**Problem Statement: Mass Flow Rate Calculation** A patient recovering from surgery is being given fluid intravenously. The fluid has a density of 1080 kg/m³, and 6.61 x 10⁻⁴ m³ of it flows into the patient every 8 hours. Find the mass flow rate in kg/s. **Solution:** To find the mass flow rate, we will use the formula: \[ \text{Mass Flow Rate} = \frac{\text{Density} \times \text{Volume Flow Rate}}{\text{Time}} \] Given data: - Density (\( \rho \)): 1080 kg/m³ - Volume (\( V \)): 6.61 x 10⁻⁴ m³ - Time (\( t \)): 8 hours First, convert the time from hours to seconds: \[ 8 \text{ hours} = 8 \times 3600 \text{ seconds} = 28,800 \text{ seconds} \] Now, calculate the volume flow rate (\( \dot{V} \)): \[ \dot{V} = \frac{V}{t} = \frac{6.61 \times 10⁻⁴ \text{ m³}}{28,800 \text{ s}} \] Calculate the mass flow rate (\( \dot{m} \)): \[ \dot{m} = \rho \times \dot{V} = 1080 \text{ kg/m³} \times \left( \frac{6.61 \times 10⁻⁴ \text{ m³}}{28,800 \text{ s}} \right) \] Perform the calculations: \[ \dot{V} = \frac{6.61 \times 10⁻⁴}{28,800} = 2.294097 \times 10⁻⁸ \text{ m³/s} \] \[ \dot{m} = 1080 \text{ kg/m³} \times 2.294097 \times 10⁻⁸ \text{ m³/s} = 2.478445 \times 10⁻⁵ \text{ kg/s} \] Therefore, the mass flow rate is approximately: \[ 2.48 \times 10⁻⁵ \text{ kg/s} \]
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