The image above represents semi-log pressure-time slope.

To compute for the semi-log pressure-time slope, four essential parameters are needed and these parameters are **Permeability-Thickness Product (kh), Oil Rate (Q _{o}), Oil Viscosity (μ_{o}) **and

**Oil FVF (B**

_{o}).The formula for calculating the semi-log pressure-time slope:

m = ^{162.6QoBoμo} / _{kh}

Where;

m = Semi-Log Pressure-Time Slope

kh = Permeability-Thickness Product

Q_{o} = Oil Rate

μ_{o} = Oil Viscosity

B_{o} = Oil FVF

Let’s solve an example;

Find the semi-log pressure-time slope, when permeability-thickness product is 12, oil rate is 17, oil viscosity is 21 and oil FVF is 13.7.

This implies that;

kh = Permeability-Thickness Product = 12

Q_{o} = Oil Rate = 17

μ_{o} = Oil Viscosity = 21

B_{o} = Oil FVF = 13.7

m = ^{162.6QoBoμo} / _{kh}

m = ^{162.6 x 17 x 13.7 x 21} / _{12}

m = ^{795260.34} / _{12}

m = 66271.69

Therefore, the **semi-log pressure-time slope **is **66271.69.**

**Calculating the Permeability-Thickness Product when Semi-Log Pressure-Time Slope, Oil Rate, Oil Viscosity and Oil FVF is Given.**

kh = ^{162.6QoBoμo} / _{m}

Where;

kh = Permeability-Thickness Product

m = Semi-Log Pressure-Time Slope

Q_{o} = Oil Rate

μ_{o} = Oil Viscosity

B_{o} = Oil FVF

Let’s solve an example;

Find the permeability-thickness product, when semi-log pressure-time slope is 90, oil rate is 23, oil viscosity is 27 and oil FVF is 14.

This implies that;

m = Semi-Log Pressure-Time Slope = 90

Q_{o} = Oil Rate = 23

μ_{o} = Oil Viscosity = 27

B_{o} = Oil FVF = 14

kh = ^{162.6QoBoμo} / _{m}

kh = ^{162.6 x 23 x 14 x 27} / _{90}

kh = ^{1413644.4} / _{90}

kh = 15707.16

Therefore, the **permeability-thickness product **is **15707.16.**

**Calculating the Oil Rate when Semi-Log Pressure-Time Slope, Permeability-Thickness Product, Oil Viscosity and Oil FVF is Given.**

Q_{o} = ^{m x kh} / _{162.6B}_{o}_{o}

Where;

Q_{o} = Oil Rate

m = Semi-Log Pressure-Time Slope

kh = Permeability-Thickness Product

μ_{o} = Oil Viscosity

B_{o} = Oil FVF

Let’s solve an example;

Find the oil rate when semi-log pressure-time slope is 110, permeability-thickness product is 42, oil viscosity is 31 and oil FVF is 11.

This implies that;

m = Semi-Log Pressure-Time Slope = 110

kh = Permeability-Thickness Product = 42

μ_{o} = Oil Viscosity = 31

B_{o} = Oil FVF = 11

Q_{o} = ^{m x kh} / _{162.6B}_{o}μ_{o}

Q_{o} = ^{110 x 42} / _{162.6 x 11 x 31}

Q_{o} = ^{4620} / _{55446.6}

Q_{o} = 0.0833

Therefore, the **oil rate **is **0.0833.**